Abstract:
A method of performing space vector modulation for PWM control for creating AC waveforms includes generating and sampling a reference signal to generate reference samples and performing a reference vector approximation to synthesize a reference vector associated with at least one of the reference samples. The reference vector approximation employs active vectors, one or more zero vectors, and one or more pseudo zero vectors in the formation thereof. Another method of performing space vector modulation (SVM) includes generating a reference signal and sampling the reference signal at a sampling frequency to generate a plurality of reference samples. The method also includes performing a reference vector approximation to synthesize a reference vector associated with at least one of the reference samples, wherein the reference vector approximation has a first portion that employs two adjacent active vectors and a remaining portion that employs two non-adjacent active vectors in the formation thereof.

Description:
FIELD 
       [0001]    The present disclosure relates to an apparatus and method for performing space vector modulation for PWM control for creating AC waveforms using pseudo zero vectors or enhanced space vector modulation, or both, for example, in motor control applications. 
       BACKGROUND 
       [0002]    Space vector modulation (SVM) is an algorithm for the control of pulse width modulation (PWM). It is used for the creation of alternating current (AC) waveforms; most commonly to drive three-phase AC powered motors at varying speeds from DC. There are numerous variations of SVM that result in different quality and computational requirements. 
         [0003]    A three phase inverter  100  as shown in  FIG. 1  must be controlled so that at no time are both switches in the same leg turned on or else the DC supply would be shorted. This requirement may be met by the complementary operation of the switches within a leg. That is, if A +  is on, then A −  is off, and vice versa. This leads to eight possible switching vectors for the inverter  100 , V 0  through V 7  with six active switching vectors and two zero vectors, as illustrated in the chart  110  of  FIG. 2 . 
         [0004]    To implement space vector modulation a reference signal V ref  is sampled with a frequency f s  (T s =1/f s ). The reference signal may be generated from three separate phase references using the αβγ transform, for example. The reference vector is then synthesized using a combination of the two adjacent active switching vectors and one or both of the zero vectors. Various strategies of selecting the order of the vectors and which zero vector(s) to use exist. Strategic selection of the vectors will affect the harmonic content and the switching losses. 
       SUMMARY 
       [0005]    The present disclosure is directed to a method of performing space vector modulation for PWM control for creating AC waveforms. The method includes generating and sampling a reference signal to generate reference samples and performing a reference vector approximation to synthesize a reference vector associated with at least one of the reference samples. The reference vector approximation employs active vectors, one or more zero vectors, and one or more pseudo zero vectors in the formation thereof. 
         [0006]    In one embodiment of the method, the pseudo vector comprises a combination of two active vectors that have an angle difference therebetween of 180°. In another embodiment, the two active vectors that in combination form the pseudo vector have a scalar amplitude that is the same. 
         [0007]    In another embodiment of the method, the pseudo vector comprises a combination of three active vectors that have an angle difference therebetween of 120°. In still another embodiment, the three active vectors that in combination form the pseudo vector have a scalar amplitude that is the same. 
         [0008]    In one embodiment of the method, the active vectors in the reference vector approximation comprise adjacent active vectors. In another embodiment, a portion of the reference vector is approximated by two adjacent active vectors, and a remaining portion of the reference vector is approximated by two non-adjacent active vectors. In accordance with one embodiment, the portion of the reference vector driven by active vectors is represented by a variable m, wherein 0≦m≦1, and the remaining portion of the reference vector driven by active vectors is represented by 1−m. When m=1, the entire reference vector driven by active vectors is approximated by two adjacent vectors, and when m=0 the entire reference vector driven by active vectors is approximated by two non-adjacent active vectors. In one embodiment, the non-adjacent active vectors are separated from one another by 120°. 
         [0009]    In accordance with another embodiment of the disclosure, a control system comprises a space vector modulator configured to receive a plurality of reference signal samples and perform a reference vector approximation to synthesize a reference vector associated with at least one of the reference signal samples. The reference vector approximation employs active vectors, one or more zero vectors, and one or more pseudo zero vectors in the formation thereof, wherein the space vector modulator outputs timing signals based on the reference vector approximation. The control system further comprises a pulse width modulation unit configured to receive the timing signals from the space vector modulator, and output pulse width modulation control signals based thereon, and a three phase inverter configured to receive the pulse width modulation control signals and generate alternating current waveforms based thereon. 
         [0010]    In one embodiment of the control system a pseudo vector comprises a combination of two active vectors that have an angle difference therebetween of 180°. In another embodiment, the two active vectors that in combination form the pseudo vector have a scalar amplitude that is the same. 
         [0011]    In another embodiment of the control system a pseudo vector comprises a combination of three active vectors that have an angle difference therebetween of 120°. In still another embodiment, the three active vectors that in combination form the pseudo vector have a scalar amplitude that is the same. 
         [0012]    In one embodiment of the control system the active vectors in the reference vector approximation comprise adjacent active vectors. In another embodiment, a portion of the reference vector is approximated by two adjacent active vectors, and a remaining portion of the reference vector is approximated by two non-adjacent active vectors. In one embodiment, the portion of the reference vector is represented by a variable m, wherein 0≦m≦1, wherein the remaining portion of the reference vector is represented by 1−m. When m=1 the entire reference vector driven by active vectors is approximated by two adjacent active vectors, and when m=0 the entire reference vector driven by active vectors is approximated by two non-adjacent active vectors. In one embodiment, the non-adjacent vectors are separated from one another by 120°. 
         [0013]    In one embodiment of the control system the three phase inverter comprises a first series-connected switch pair connected together at a node that forms a first phase output, a second series-connected switch pair connected together at a node that forms a second phase output, and a third series-connected switch pair connected together at a node that forms a third phase output. The inverter further comprises a shunt resistor connected with a first terminal to a bottom node of each of the first, second and third series-connected switch pairs, and a second terminal coupled to a reference potential, and an amplifier having first and second inputs coupled to the first and second terminals of the shunt resistor, respectively, wherein an output of the amplifier reflects a current level conducting through the shunt resistor. 
     
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
         [0014]    Example embodiments according to the disclosure will be explained in more detail in the following text with reference to the attached figures, in which: 
           [0015]      FIG. 1  is a schematic diagram of a three phase inverter for generating AC waveforms; 
           [0016]      FIG. 2  is a chart illustrating various vectors for switching the switches of the three phase inverter of  FIG. 1 ; 
           [0017]      FIG. 3  is hexagon diagram illustrating the basic voltage space vectors from the chart of  FIG. 2 ; 
           [0018]      FIG. 4  is a hexagon diagram illustrating a reference vector approximation in accordance with conventional techniques; 
           [0019]      FIG. 5  is a graph illustrating normalized time versus reference vector angle for conventional space vector modulation (SVM); 
           [0020]      FIGS. 6A-6E  are hexagon diagrams illustrating a basic pseudo zero vector, wherein  FIGS. 6A-6C  show a combination of two active vectors, and  FIGS. 6D-6E  show a combination of three active vectors; 
           [0021]      FIGS. 7A-7B  show approximations of a reference vector using one pseudo zero vector, such as those illustrated in  FIGS. 6A-6C  according to one embodiment; 
           [0022]      FIG. 8  shows an approximation of a reference vector using two pseudo zero vectors, such as those illustrated in  FIGS. 6A-6C  according to one embodiment; 
           [0023]      FIG. 9  is a graph illustrating a maximum inverter DC link voltage utilization versus A for a reference vector in accordance with the embodiment of  FIGS. 7A-7B ; 
           [0024]      FIG. 10  is a graph illustrating normalized time versus reference vector angle for conventional space vector modulation (SVM) with use of pseudo zero vectors such as that shown in  FIGS. 7A-7B  according to one embodiment; 
           [0025]      FIG. 11  is a graph illustrating a maximum inverter DC link voltage utilization versus A for a reference vector in accordance with the embodiment of  FIG. 8 ; 
           [0026]      FIG. 12  is a graph illustrating normalized time versus reference vector angle for conventional space vector modulation (SVM) with use of pseudo zero vectors such as that shown in  FIG. 8  according to one embodiment; 
           [0027]      FIGS. 13A-13B  are block diagrams illustrating a control system, such as a motor control system, that employs a space vector modulator that employs pseudo zero vectors in approximating reference vectors according to one embodiment of the disclosure; 
           [0028]      FIG. 14  is a space vector hexagon showing a reference vector in polar coordinates and Cartesian coordinates, respectively; 
           [0029]      FIG. 15  is a block diagram illustrating a three phase inverter in a control system such as that illustrated in  FIGS. 13A-13B  that employs single-shunt current sensing according to one embodiment; 
           [0030]      FIGS. 16A-16B  illustrate examples of 5-segment switching sequences for the example of  FIGS. 7A-7B ; 
           [0031]      FIGS. 17A-17B  illustrate examples of 5-segment switching sequences for the example of  FIG. 8 ; 
           [0032]      FIGS. 18A-18B  illustrate examples of 6-segment switching sequences for the example of  FIG. 8 ; 
           [0033]      FIGS. 19A-19B  illustrate examples of 7-segment switching sequences for the example of  FIG. 8 ; 
           [0034]      FIG. 20  illustrates the DC link current corresponding to the switching sequences shown in  FIG. 18A ; 
           [0035]      FIG. 21  is a space vector hexagon illustrating an alternative reference vector approximation with one pseudo zero vector according to one embodiment; 
           [0036]      FIGS. 22A-22B  are space vector hexagons illustrating a reference vector approximation employing enhanced SVM that utilizes three adjacent active vectors according to one embodiment of the disclosure; 
           [0037]      FIGS. 23A-23B  are space vector hexagons illustrating a reference vector approximation employing enhanced SVM under the condition where m=0 such that the reference vector is approximated solely by non-adjacent active vectors in accordance with one embodiment; 
           [0038]      FIG. 24  is a graph illustrating a maximum inverter DC ling voltage utilization versus m for enhanced SVM in accordance with one embodiment of the disclosure; 
           [0039]      FIGS. 25A-25B  are graphs illustrating normalized time versus reference angle plots for enhanced SVM at m=0.8 and m=0.2, respectively, according to one embodiment; 
           [0040]      FIG. 26  is a graph illustrating a normalized time versus reference angle plot for enhanced SVM at m=0 according to one embodiment; 
           [0041]      FIGS. 27A-27B  are block diagrams illustrating a control system, such as a motor control system, that employs enhanced SVM that may include non-adjacent active vectors in approximating reference vectors according to one embodiment of the disclosure; 
           [0042]      FIG. 28  is a space vector hexagon showing a reference vector in enhanced SVM in polar coordinates and Cartesian coordinates, respectively; 
           [0043]      FIG. 29  is a block diagram illustrating a three phase inverter in a control system such as that illustrated in  FIGS. 27A-27B  that employs single-shunt current sensing according to one embodiment; 
           [0044]      FIGS. 30A-30B  illustrate examples of 4-segment switching sequences for enhanced SVM according to one embodiment; 
           [0045]      FIGS. 31A-31B  illustrate examples of 6-segment switching sequences for enhanced SVM according to one embodiment; 
           [0046]      FIGS. 32A-32B  illustrate examples of 3-segment switching sequences for enhanced SVM at m=0 according to one embodiment; 
           [0047]      FIGS. 33A-33B  illustrate examples of 5-segment switching sequences for enhanced SVM at m=0 according to one embodiment; 
           [0048]      FIG. 34  illustrates the DC link current corresponding to the switching sequences shown in  FIG. 31A ; 
           [0049]      FIGS. 35A-35B  are space vector hexagons illustrating special cases of enhanced SVM, wherein  FIG. 35A  illustrates the case of m≧1, and  FIG. 35B  illustrates the case of m≦0, wherein four active vectors are employed to approximate the reference vector, wherein  FIG. 35A  shows adjacent active vectors and  FIG. 35B  shows non-adjacent active vectors. 
       
    
    
     DETAILED DESCRIPTION 
       [0050]    In some cases, the same reference symbols are used in the following text for objects and functional units that have the same or similar functional characteristics. Furthermore, optional features of the various example embodiments can be combined with one another or replaced by one another. 
         [0051]    For industry and automotive motor control applications, SVM is ubiquitously used in the sinusoidal commutation control (such as V/f, FOC (field oriented control), and DTC (direct torque control)) of PMSM (permanent magnet synchronous motor) and ACIM (alternating current induction motor) to generate sine waveforms from three-phase inverters. Sinusoidal commutation motor control with a single-shunt current sensing resistor inserted in the inverter DC link is a desirable solution when compared to dual-shunt and triple-shunt current sensing techniques, owing to its important advantages such as low cost, simplicity and etc. However, two current samples within one pulse width modulation (PWM) cycle are needed for correct motor phase current reconstruction with single-shunt current sensing. However, with conventional SVM techniques, the accurate current construction is difficult in the following situations: (1) The reference voltage space vector is crossing a sector border as only one current sample can be measured (this occurs in many instances) and (2) When the modulation index is low and the sampling intervals are too short, none of the current samples can be taken (this normally happens in ultra-low speed motor control). 
         [0052]    The present disclosure proposes a new concept of a pseudo zero vector (alternatively this may be called a quasi zero vector, or a synthetic zero vector) to SVM to address the above-mentioned problems. With the new SVM using pseudo zero vectors, one is able to provide low-cost, high-quality, more reliable, and unique motor control solutions (e.g., sensorless FOC with single-shunt current sensing) to customers. The new SVM with pseudo zero vectors may also be used in three-phase power inverter control for uninterruptible power supplies, renewable energy, and etc. 
         [0053]    The space vector diagram (a regular hexagon)  120  and reference vector approximation  130  of the existing conventional SVM are shown in  FIGS. 3 and 4 , respectively. {right arrow over (V)} 1  to {right arrow over (V)} 6  are active vectors. {right arrow over (V)} 0  and {right arrow over (V)} 7  do not generate any voltage difference in the inverter outputs and are the only two zero vectors (or passive vectors) in existing SVM. The revolving reference vector {right arrow over (V)} ref =|V ref |·e jθ  is approximated by two adjacent active vectors (e.g., {right arrow over (V)} 1 , {right arrow over (V)} 2  in sector A) and one or both of the existing zero vectors (e.g., {right arrow over (V)} 0  only). The plane of the space vector hexagon is dissected in six sectors from A to F and the angle θ of {right arrow over (V)} ref  is transformed into the relative angle θ rel  in each sector. Using reference vector in sector A as an example, the following portion shows the calculations of existing or conventional SVM. 
         [0054]    Using volt-second balancing: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       V 
                       → 
                     
                     ref 
                   
                   = 
                   
                     
                       
                         
                           T 
                           0 
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         0 
                       
                     
                     + 
                     
                       
                         
                           T 
                           1 
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         1 
                       
                     
                     + 
                     
                       
                         
                           T 
                           2 
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         2 
                       
                     
                   
                 
               
               
                 
                   ( 
                   1 
                   ) 
                 
               
             
             
               
                 
                   
                     T 
                     S 
                   
                   = 
                   
                     
                       T 
                       0 
                     
                     + 
                     
                       T 
                       1 
                     
                     + 
                     
                       T 
                       2 
                     
                   
                 
               
               
                 
                   ( 
                   2 
                   ) 
                 
               
             
           
         
       
     
         [0000]    Solving the Equations (1) and (2), one obtains: 
         [0000]        T   1   =K  sin(60°−θ rel )· T   S   (3)
 
         [0000]        T   2   =K  sin(θ rel )· T   S   (4)
 
         [0000]    Adding the Equations (3) and (4), one obtains: 
         [0000]        T   1   +T   2   =K  sin(60°+θ rel )· T   S   (5)
 
         [0000]    So the zero vector time is: 
         [0000]        T   0   =T   S −( T   1   +T   2 )=[1− K  sin(60°+θ rel )]· T   S   (6)
 
         [0000]    where: 
         [0055]    T 0 —Time of the zero vector(s) is/are applied. The zero vector(s) can be {right arrow over (V)} 0  [000], or {right arrow over (V)} 7  [111], or both 
         [0056]    T 1 —Time of the 1 st  active vector (e.g., {right arrow over (V)} 1  in sector A) is applied within one sampling period, 
         [0057]    T 2 —Time of the 2 nd  active vector (e.g., {right arrow over (V)} 2  in sector A) is applied within one sampling period, 
         [0058]    K— 
         [0000]    
       
         
           
             K 
             = 
             
               
                 3 
               
               · 
               
                 
                    
                   
                     V 
                     ref 
                   
                    
                 
                 
                   V 
                   
                     D 
                      
                     
                         
                     
                      
                     C 
                   
                 
               
               · 
               
                  
                 
                   V 
                   ref 
                 
                  
               
             
           
         
       
     
         [0000]    is the amplitude of {right arrow over (V)} ref , and V DC  is the inverter DC link voltage, 
         [0059]    T S —Sampling period, e.g., T S =50 μs 
         [0000]    As T 0 ≧0 (or T 1 +T 2 ≦T S ) all the time, so K≦1. We have inverter DC link voltage utilization without over-modulation: 
         [0000]    
       
         
           
             
               
                 
                   η 
                   = 
                   
                     
                       
                          
                         
                           V 
                           ref 
                         
                          
                       
                       
                         V 
                         
                           D 
                            
                           
                               
                           
                            
                           C 
                         
                       
                     
                     ≤ 
                     
                       1 
                       
                         3 
                       
                     
                   
                 
               
               
                 
                   ( 
                   7 
                   ) 
                 
               
             
           
         
       
     
         [0060]    Plots of normalized time T 1    140  and T 2    150  of Equations (3) and (4) are shown in  FIG. 5 . It is obvious that either T 1  or T 2  is close or equal to zero at the border of each sector (e.g., near θ=0° or 60° in sector A), which is the root cause of the above-mentioned problem (1) for motor control with SVM and single-shunt current sensing. There have been numerous prior attempts to solve this problem, but each solution has its disadvantages. 
         [0061]    For example, motor phase current construction from DC link current via a single-shunt resistor is done by limiting T 1  and T 2  with a time T min  (which is PWM dead time+driver delay+ADC sampling time) for ADC to sample the correct current values. However, limiting T 1  and T 2  will generate a distorted voltage vector, and therefore result in high torque ripple, high vibration &amp; acoustic noise, and even unstable motor control with high dynamic loads. A very fast ADC is required to optimize the system performance in such instances. 
         [0062]    Another conventional solution is modifying the SVM switching pattern to a minimum measurement time window in order to allow two current samples to be taken. This pattern modification could generate some current ripple; moreover, more CPU resources are needed to implement the algorithm due to modification of patterns and correction of the same modifications. 
         [0063]    Another conventional solution employs use of asymmetrical PWM pulses (with two PWM pulses shifted to obtain enough time for current sampling while duty cycles for all the PWM pulses preserved) only partly solves the problem. It can be found that both T 1  and T 2  are close or equal to zero if K is very small or K=0, which causes the problem (2) mentioned above. Using the asymmetrical PWM pulses just mentioned can only partly solve this problem. 
         [0064]    The present disclosure proposes a new concept of a pseudo zero vector to SVM, with which the approximation of the reference vector with more than two active vectors becomes relatively easy and straightforward. (In contrast, existing SVM uses only two adjacent active vectors to approximate the reference vector). The pseudo zero vectors shown in  FIGS. 6A-6E  complement the existing two zero vectors and expand the SVM theory, giving more choice of zero vectors in the approximation of the reference vector. 
         [0065]    A basic pseudo zero vector is either a combination of two active vectors  160 ,  170 , or  180  as shown in  FIGS. 6A ,  6 B and  6 C, or a combination of three active vectors  190  or  200  as shown in  FIGS. 6D and 6E . The pseudo zero vector time T z  shown in  FIGS. 6A-6E  can be time-variant, θ-angle-dependent, or constant, depending on the different application requirements. These pseudo zero vectors have similar effects of existing zero vectors (i.e., they do not generate any voltage difference in the inverter outputs). Any combinations of the pseudo zero vectors also give the same effects. The pseudo zero vectors can be used in similar ways as the existing zero vectors are used (i.e., one, two, or a combination of zero vectors being used in the approximation of a reference vector). 
         [0066]    The advantages of the new concept of a pseudo zero vector according to the present disclosure are that there are many more choices of “zero vectors” in the SVM approximation. The approximation of the reference vector with more than two active vectors becomes very easy and convenient. Additionally, the corresponding calculations are similar to those of the existing SVM, and thus are relatively simple and fast. 
         [0067]    Two SVM examples according to the present disclosure are provided below with selective utilization of the pseudo zero vectors. Both new SVM examples have non-zero time intervals for two single-shunt current samplings within any PWM cycle, and hence can easily solve the above-mentioned problem (1). Further, and advantageously, new SVM Example #2 can completely solve problem (2) mentioned above. 
         [0068]    In the new SVM examples, engineers can tune the pseudo zero vector time T z  to obtain different current sampling intervals based on different system requirements and hardware designs, so as to achieve the best performance of motor control with single-shunt current sensing. Particularly, as the long ADC sampling intervals can be obtained easily by selecting longer pseudo zero vector time T z , it is also possible to use a low-speed, common and low-cost operational amplifier (if any) for DC link current signal amplification to further lower the system cost. 
         [0069]    Table 1 summarizes the basic voltage space vectors for the SVM with the new concept of a pseudo zero vector. 
         [0000]    
       
         
               
             
               
               
               
             
               
               
               
               
             
           
               
                 TABLE 1 
               
             
             
               
                   
               
               
                 Basic voltage  
               
               
                 space vectors of SVM with new concept of pseudo zero vector 
               
             
          
           
               
                   
                 Basic Voltage Space Vector 
                 Remark 
               
               
                   
                   
               
             
          
           
               
                   
                 Active 
                 {right arrow over (V)} 1  [100] 
                 These are all the 
               
               
                   
                 Vectors 
                 {right arrow over (V)} 2  [110] 
                 eight basic voltage 
               
               
                   
                   
                 {right arrow over (V)} 3  [010] 
                 space vectors of 
               
               
                   
                   
                 {right arrow over (V)} 4  [011] 
                 existing SVM. 
               
               
                   
                   
                 {right arrow over (V)} 5  [001] 
                   
               
               
                   
                   
                 {right arrow over (V)} 6  [101] 
                   
               
               
                   
                 Existing 
                 {right arrow over (V)} 0  [000] 
                   
               
               
                   
                 Zero 
                 {right arrow over (V)} 7  [111] 
                   
               
               
                   
                 Vectors 
                   
                   
               
               
                   
                   
               
               
                   
                 Pseudo Zero Vectors 
                 
                   
                     
                       
                         
                           
                             
                               
                                 
                                   
                                     
                                       T 
                                       z 
                                     
                                     
                                       T 
                                       S 
                                     
                                   
                                    
                                   
                                     
                                       V 
                                       -&gt; 
                                     
                                     1 
                                   
                                 
                                 + 
                                 
                                   
                                     
                                       T 
                                       z 
                                     
                                     
                                       T 
                                       S 
                                     
                                   
                                    
                                   
                                     
                                       V 
                                       -&gt; 
                                     
                                     4 
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 
                                   
                                     
                                       T 
                                       z 
                                     
                                     
                                       T 
                                       S 
                                     
                                   
                                    
                                   
                                     
                                       V 
                                       -&gt; 
                                     
                                     2 
                                   
                                 
                                 + 
                                 
                                   
                                     
                                       T 
                                       z 
                                     
                                     
                                       T 
                                       S 
                                     
                                   
                                    
                                   
                                     
                                       V 
                                       -&gt; 
                                     
                                     5 
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 
                                   
                                     
                                       T 
                                       z 
                                     
                                     
                                       T 
                                       S 
                                     
                                   
                                    
                                   
                                     
                                       V 
                                       -&gt; 
                                     
                                     3 
                                   
                                 
                                 + 
                                 
                                   
                                     
                                       T 
                                       z 
                                     
                                     
                                       T 
                                       S 
                                     
                                   
                                    
                                   
                                     
                                       V 
                                       -&gt; 
                                     
                                     6 
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 
                                   
                                     
                                       T 
                                       z 
                                     
                                     
                                       T 
                                       S 
                                     
                                   
                                    
                                   
                                     
                                       V 
                                       -&gt; 
                                     
                                     1 
                                   
                                 
                                 + 
                                 
                                   
                                     
                                       T 
                                       z 
                                     
                                     
                                       T 
                                       S 
                                     
                                   
                                    
                                   
                                     
                                       V 
                                       -&gt; 
                                     
                                     3 
                                   
                                 
                                 + 
                                 
                                   
                                     
                                       T 
                                       z 
                                     
                                     
                                       T 
                                       S 
                                     
                                   
                                    
                                   
                                     
                                       V 
                                       -&gt; 
                                     
                                     5 
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 
                                   
                                     
                                       T 
                                       z 
                                     
                                     
                                       T 
                                       S 
                                     
                                   
                                    
                                   
                                     
                                       V 
                                       -&gt; 
                                     
                                     2 
                                   
                                 
                                 + 
                                 
                                   
                                     
                                       T 
                                       z 
                                     
                                     
                                       T 
                                       S 
                                     
                                   
                                    
                                   
                                     
                                       V 
                                       -&gt; 
                                     
                                     4 
                                   
                                 
                                 + 
                                 
                                   
                                     
                                       T 
                                       z 
                                     
                                     
                                       T 
                                       S 
                                     
                                   
                                    
                                   
                                     
                                       V 
                                       -&gt; 
                                     
                                     6 
                                   
                                 
                               
                             
                           
                         
                           
                       
                     
                   
                 
                 New concept for SVM. T z  can be time-variant, θ- angle-dependent, or constant. 
               
               
                   
                   
               
             
          
         
       
     
         [0070]    An element of the disclosure is the introduction of the new concept of a pseudo zero vector to SVM. With the new pseudo zero vectors, there are many more choices of zero vectors compared to the only two in existing or conventional SVM, which gives more flexibility for the approximation of the reference vector as well as the design of the SVM switching sequences. The approximation of a reference vector with more than two active vectors becomes relatively easy and convenient. In addition, the corresponding calculations are similar to that of existing SVM, and hence relatively simple and fast. 
         [0071]    This section shows two examples of selective utilization of pseudo zero vectors in SVM to straightforwardly solve the single-shunt motor control problems that the existing SVM has difficulties in dealing with effectively. 
       New SVM Example #1 
       [0072]    Using sector A as an example as shown in  FIGS. 7A and 7B , the reference vector {right arrow over (V)} ref    210 ,  220  is approximated by two adjacent active vectors, one pseudo zero vector, and one or both of the existing zero vectors. Combining the time of the same active vectors (e.g: 
         [0000]    
       
         
           
             
               
                 T 
                 2 
                 ′ 
               
               
                 T 
                 S 
               
             
              
             
               
                 V 
                 → 
               
               2 
             
              
             
                 
             
              
             and 
              
             
                 
             
              
             
               
                 T 
                 z 
               
               
                 T 
                 S 
               
             
              
             
               
                 V 
                 → 
               
               2 
             
           
         
       
     
         [0000]    in  FIG. 7A ), {right arrow over (V)} ref  is actually approximated by three active vectors (e.g., {right arrow over (V)} 1 , {right arrow over (V)} 2  and {right arrow over (V)} 4  in  FIG. 7B  and {right arrow over (V)} 1 , {right arrow over (V)} 2  and {right arrow over (V)} 5  in  FIG. 7A ). This new SVM Example #1 is elaborated in greater detail below. 
       New SVM Example #2 
       [0073]    Using sector A as an example as shown in  FIG. 8 , the reference vector {right arrow over (V)} ref    230  is approximated by two adjacent active vectors, two pseudo zero vectors, and one or both of the existing zero vectors. Combining the time of the same active vectors (i.e.: 
         [0000]    
       
         
           
             
               
                 
                   T 
                   1 
                   ′ 
                 
                 
                   T 
                   S 
                 
               
                
               
                 
                   V 
                   → 
                 
                 1 
               
                
               
                   
               
                
               and 
                
               
                   
               
                
               
                 
                   T 
                   z 
                 
                 
                   T 
                   S 
                 
               
                
               
                 
                   V 
                   → 
                 
                 1 
               
             
             , 
             
               
                 
                   T 
                   2 
                   ′ 
                 
                 
                   T 
                   S 
                 
               
                
               
                 
                   V 
                   → 
                 
                 2 
               
                
               
                   
               
                
               and 
                
               
                   
               
                
               
                 
                   T 
                   z 
                 
                 
                   T 
                   S 
                 
               
                
               
                 
                   V 
                   → 
                 
                 2 
               
             
           
         
       
     
         [0000]    ), {right arrow over (V)} ref  is actually approximated by four active vectors. The new SVM Example #2 is elaborated in greater detail below as well. 
         [0074]    Table 2 compares and summarizes the existing SVM and the new SVM examples with pseudo zero vectors. 
         [0000]    
       
         
               
             
               
               
               
               
               
             
               
               
               
               
               
               
             
               
               
               
               
               
               
               
               
             
               
               
               
               
               
               
               
               
               
               
             
           
               
                 TABLE 2 
               
             
             
               
                   
               
               
                 Comparison of existing SVM and new SVM examples using pseudo zero vectors 
               
             
          
           
               
                   
                   
                   
                 Active  
                   
               
               
                   
                   
                 {right arrow over (V)} ref  Angle 
                 Vectors For {right arrow over (V)} ref    
                   
               
             
          
           
               
                   
                   
                 θ 
                   
                 Approximation 
                   
               
             
          
           
               
                   
                   
                 (limit 0° ≦ θ &lt; 360°) 
                   
                   
                   
                 
                   Note 
                   1 
                 
                   
               
             
          
           
               
                   
                 Sector 
                 Start 
                 End 
                 θ rel   
                 1 st   
                 2 nd   
                 3 rd   
                 4 th   
                 Formulas 
               
               
                   
               
               
                 Existing SVM 
                 A B C D E F 
                  0°  60° 120° 180° 240° 300° 
                  60° 120° 180° 240° 300° 360° 
                 θ-0° θ-60° θ-120° θ-180° θ-240° θ-300° 
                 {right arrow over (V)} 1  {right arrow over (V)} 2  {right arrow over (V)} 3  {right arrow over (V)} 4  {right arrow over (V)} 5  {right arrow over (V)} 6   
                 {right arrow over (V)} 2  {right arrow over (V)} 3  {right arrow over (V)} 4  {right arrow over (V)} 5  {right arrow over (V)} 6  {right arrow over (V)} 1   
                 — — — — — — 
                 — — — — — — 
                 T 1  = Ksin(60° − θ rel ) · T S  T 2  = Ksin(θ rel ) · T S  T 0  = T s  − (T 1  + T 2 ) 
                 where                 K     =         3     ·            V   ref            V   DC                       and                          V   ref            V   DC       ≤     1     3                       
 
               
               
                   
               
               
                 New SVM Example #1 
                 A1   A2 B1 B2 C1 C2 D1 D2 E1 E2 F1 F2 
                  0°   Θ Tr   60°  60° + Θ Tr  120° 120° + Θ Tr  180° 180° + Θ Tr  240° 240° + Θ Tr  300° 300° + Θ Tr   
                 Θ Tr   Note   2   60°  60° + Θ Tr  120° 120° + Θ Tr  180° 180° + Θ Tr  240° 240° + Θ Tr  300° 300° + Θ Tr  360° 
                 θ-0°   θ-0° θ-60° θ-60° θ-120° θ-120° θ-180° θ-180° θ-240° θ-240° θ-300° θ-300° 
                 {right arrow over (V)} 1    {right arrow over (V)} 1  {right arrow over (V)} 2  {right arrow over (V)} 2  {right arrow over (V)} 3  {right arrow over (V)} 3  {right arrow over (V)} 4  {right arrow over (V)} 4  {right arrow over (V)} 5  {right arrow over (V)} 5  {right arrow over (V)} 6  {right arrow over (V)} 6   
                 {right arrow over (V)} 2    {right arrow over (V)} 2  {right arrow over (V)} 3  {right arrow over (V)} 3  {right arrow over (V)} 4  {right arrow over (V)} 4  {right arrow over (V)} 5  {right arrow over (V)} 5  {right arrow over (V)} 6  {right arrow over (V)} 6  {right arrow over (V)} 1  {right arrow over (V)} 1   
                 {right arrow over (V)} 5    {right arrow over (V)} 4  {right arrow over (V)} 6  {right arrow over (V)} 5  {right arrow over (V)} 1  {right arrow over (V)} 6  {right arrow over (V)} 2  {right arrow over (V)} 1  {right arrow over (V)} 3  {right arrow over (V)} 2  {right arrow over (V)} 4  {right arrow over (V)} 3   
                 —   — — — — — 
                 1). For 0° ≦ θ rel  &lt; Θ Tr :  T 1  = Ksin(60° − θ rel ) · T S  T 2  = [Ksin(θ rel ) + λ] · T S  2). For Θ Tr  ≦ θ rel  &lt; 60°:  T 1  = [Ksin(60° − θ rel ) + λ] · T S  T 2  = Ksin(θ rel ) · T S  For both conditions 1) &amp; 2): T 3  = T z  = λT S    Note   3  T 0  = T s  − (T 1  + T 2  + T 3 ) 
                   where                 K     =       3     ·            V   ref            V   DC           ,     λ   =       T   z       T   S         ,   and                        V   ref            V   DC       ≤       1   -     2      λ         3                       
   
               
               
                   
               
               
                 New SVM Example #2 
                 A B C D E F 
                  0°  60° 120° 180° 240° 300° 
                  60° 120° 180° 240° 300° 360° 
                 θ-0° θ-60° θ-120° θ-180° θ-240° θ-300° 
                 {right arrow over (V)} 1  {right arrow over (V)} 2  {right arrow over (V)} 3  {right arrow over (V)} 4  {right arrow over (V)} 5  {right arrow over (V)} 6   
                 {right arrow over (V)} 2  {right arrow over (V)} 3  {right arrow over (V)} 4  {right arrow over (V)} 5  {right arrow over (V)} 6  {right arrow over (V)} 1   
                 {right arrow over (V)} 4  {right arrow over (V)} 5  {right arrow over (V)} 6  {right arrow over (V)} 1  {right arrow over (V)} 2  {right arrow over (V)} 3   
                 {right arrow over (V)} 5  {right arrow over (V)} 6  {right arrow over (V)} 1  {right arrow over (V)} 2  {right arrow over (V)} 3  {right arrow over (V)} 4   
                 T 1  = [Ksin(60° − θ rel ) + λ] · T S  T 2  = [Ksin(θ rel ) + λ] · T S  T 3  = T z  = λT S  T 4  = T z  = λT S  T 0  = T s  − (T 1  + T 2  + T 3  + T 4 ) 
                   where                 K     =       3     ·            V   ref            V   DC           ,     λ   =       T   z       T   S         ,   and                        V   ref            V   DC       ≤       1   -     4      λ         3                       
 
               
               
                   
               
               
                   Note   1 : 
               
               
                 The 3 rd  and 4 th  active vectors (if any) come from the pseudo zero vectors being used. 
               
               
                   Note   2 : 
               
               
                 Θ Tr  is a transition angle for new sectors (e.g.: A1, A2, B1, B2, and etc), and 0° &lt; Θ Tr  &lt; 60°. 
               
               
                   Note   3 : 
               
               
                 Pseudo zero vector time T z  = λT S  ≧ T min , where T min  is PWM dead time + driver delay + ADC sampling time. 
               
             
          
         
       
     
       New SVM Example #1 
     SVM with One Pseudo Zero Vector 
       [0075]    As shown in  FIGS. 7A and 7B , the reference vector in each existing sector (i.e., A, B, C, D, E, and F) can be approximated by two different sets of active vectors. A transition angle Θ Tr  (0°&lt;Θ Tr &lt;60°) is introduced to get new sectors A1, A2, B1, B2, and so on as shown in Table 2. Therefore in  FIG. 7A  the reference vector  210  is in sector A1, and in  FIG. 7B  the reference vector  220  is in sector A2. At the transition angles the reference vector approximation transits from one set of active vectors to another set of active vectors for the new SVMs. Θ Tr  can be different in different existing sectors. For simplicity, we can choose the same value, e.g., Θ Tr =30°, for all the sectors. 
         [0076]    Calculations when 0°≦θ rel &lt;Θ Tr    
         [0077]    Using the reference vector  210  in sector A as shown in  FIG. 7A  as an example, the following shows the calculations when 0°≦θ rel &lt;Θ Tr . Using volt-second balancing: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       V 
                       → 
                     
                     ref 
                   
                   = 
                   
                     
                       
                         
                           T 
                           0 
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         0 
                       
                     
                     + 
                     
                       
                         
                           T 
                           1 
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         1 
                       
                     
                     + 
                     
                       
                         
                           
                             
                               T 
                               2 
                               ′ 
                             
                             + 
                             
                               T 
                               z 
                             
                           
                           
                              
                             
                               T 
                               2 
                             
                           
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         2 
                       
                     
                     + 
                     
                       
                         
                           
                             T 
                             z 
                           
                           
                              
                             
                               T 
                               3 
                             
                           
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         5 
                       
                     
                   
                 
               
               
                 
                   ( 
                   8 
                   ) 
                 
               
             
             
               
                 
                   
                     T 
                     S 
                   
                   = 
                   
                     
                       T 
                       0 
                     
                     + 
                     
                       T 
                       1 
                     
                     + 
                     
                       
                         ( 
                         
                           
                             T 
                             2 
                             ′ 
                           
                           + 
                           
                             T 
                             z 
                           
                         
                         ) 
                       
                       
                          
                         
                           T 
                           2 
                         
                       
                     
                     + 
                     
                       
                         T 
                         z 
                       
                       
                          
                         
                           T 
                           3 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   9 
                   ) 
                 
               
             
           
         
       
     
         [0000]    Choose T z  more than or equal to T min  (which is PWM dead time+driver delay+ADC sampling time). For simplicity, select 
         [0000]        T   3   =T   z   =λT   S   ≧T   min   (10)
 
         [0000]    where: λ—A constant and 
         [0000]    
       
         
           
             
               λ 
               = 
               
                 
                   
                     T 
                     z 
                   
                   
                     T 
                     S 
                   
                 
                  
                 
                   ( 
                   
                     0 
                     ≤ 
                     λ 
                     ≤ 
                     1 
                   
                   ) 
                 
               
             
             , 
           
         
       
     
         [0000]    e.g., if T S =50 μs and T min =2 μs, we can select 
         [0000]    
       
         
           
             λ 
             = 
             
               
                 1 
                 25 
               
               . 
             
           
         
       
     
       Solve Equations (8) to (10) to get 
       [0078]        T   1   =K  sin(60°−θ rel )· T   S  
 
         [0000]        T   2   =T   2   ′+T   z   =[K  sin(θ rel )+λ]·T s   (12)
 
         [0000]        T   1   +T   2   +T   3   =[K  sin(60°+θ rel )+2 λ]·T   S   (13)
 
         [0000]    So the zero vector time is 
         [0000]        T   0   =T   S −( T   1   +T   2   +T   3 )=[(1−2λ)− K  sin(60°+θ rel )]· T   S   (14)
 
         [0000]    where: 
         [0079]    T 0 —Time of existing zero vector(s) is applied. The zero vector(s) can be {right arrow over (V)} 0  [000], or {right arrow over (V)} 7  [111], or both 
         [0080]    T z —Time of pseudo zero vector is applied 
         [0081]    T 1 —Time of the 1 st  active vector is applied within one sampling period 
         [0082]    T 2 —Time of the 2 nd  active vector is applied within one sampling period 
         [0083]    T 3 —Time of the 3 rd  active vector is applied within one sampling period, which is part of the pseudo zero vector being used 
         [0000]    
       
         
           
             
               K 
               - 
               K 
             
             = 
             
               
                 3 
               
               · 
               
                 
                   
                      
                     
                       V 
                       ref 
                     
                      
                   
                   
                     V 
                     DC 
                   
                 
                 . 
               
             
           
         
       
     
         [0000]    |V ref | is the amplitude of {right arrow over (V)} ref , and V DC  is the inverter DC link voltage 
         [0084]    T S —Sampling period 
         [0000]    As T 0 ≧0 all the time, from Equation (14) we find K≦1-2λ, so the inverter DC link voltage utilization without over-modulation is 
         [0000]    
       
         
           
             
               
                 
                   η 
                   = 
                   
                     
                       
                          
                         
                           V 
                           ref 
                         
                          
                       
                       
                         V 
                         DC 
                       
                     
                     ≤ 
                     
                       
                         1 
                         - 
                         
                           2 
                            
                           λ 
                         
                       
                       
                         3 
                       
                     
                   
                 
               
               
                 
                   ( 
                   15 
                   ) 
                 
               
             
           
         
       
     
         [0085]    Calculations when Θ Tr ≦θ rel &lt;60° 
         [0086]    Using the reference vector  220  in sector A as shown in  FIG. 7B  as an example, the following shows the calculations when Θ Tr ≦θ rel &lt;60°. Using volt-second balancing: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       V 
                       → 
                     
                     ref 
                   
                   = 
                   
                     
                       
                         
                           T 
                           0 
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         0 
                       
                     
                     + 
                     
                       
                         
                           
                             
                               T 
                               1 
                               ′ 
                             
                             + 
                             
                               T 
                               z 
                             
                           
                           
                              
                             
                               T 
                               1 
                             
                           
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         1 
                       
                     
                     + 
                     
                       
                         
                           T 
                           2 
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         2 
                       
                     
                     + 
                     
                       
                         
                           
                             T 
                             z 
                           
                           
                              
                             
                               T 
                               3 
                             
                           
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         4 
                       
                     
                   
                 
               
               
                 
                   ( 
                   16 
                   ) 
                 
               
             
             
               
                 
                   
                     T 
                     S 
                   
                   = 
                   
                     
                       T 
                       0 
                     
                     + 
                     
                       
                         ( 
                         
                           
                             T 
                             1 
                             ′ 
                           
                           + 
                           
                             T 
                             z 
                           
                         
                         ) 
                       
                       
                          
                         
                           T 
                           1 
                         
                       
                     
                     + 
                     
                       T 
                       2 
                     
                     + 
                     
                       
                         T 
                         z 
                       
                       
                          
                         
                           T 
                           3 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   17 
                   ) 
                 
               
             
           
         
       
     
         [0000]    Similarly, select T z =λT S , i.e., 
         [0000]        T   3   =λT   s   ≧T   min   (18)
 
       Solve Equations (16) and (17) to get 
       [0087]        T   1   =T   1   ′+T   z   =[K  sin(60°−θ rel )+λ]·T S   (19)
 
         [0000]        T   2   =K  sin(θ rel )· T   S   (20)
 
         [0000]        T   1   +T   2   +T   3   =[K  sin(60°+θ rel )+2 λ]·T   S   (21)
 
         [0000]    So the zero vector time is 
         [0000]        T   0   =T   S −( T   1   +T   2   +T   3 )=[(1−2λ)− K  sin(60°+θ rel )]· T   S   (22)
 
         [0000]    Likewise, it can be found that the inverter DC link voltage utilization without over-modulation is the same as Equation (15). So the new SVM Example #1 has one maximum inverter DC link voltage utilization without over-modulation, which is 
         [0000]    
       
         
           
             
               
                 
                   
                     η 
                     max 
                   
                   = 
                   
                     
                       1 
                       - 
                       
                         2 
                          
                         λ 
                       
                     
                     
                       3 
                     
                   
                 
               
               
                 
                   ( 
                   23 
                   ) 
                 
               
             
           
         
       
     
         [0000]    A plot of Equation (23) is shown in  FIG. 9 , which is the maximum inverter DC link voltage  240  utilization versus λ for Example 1. When λ=0, the new SVM becomes the existing SVM and 
         [0000]    
       
         
           
             η 
             ≤ 
             
               
                 1 
                 
                   3 
                 
               
               . 
             
           
         
       
     
         [0088]    Plots of normalized time T 1   250 , T 2    260 , and T 3    270  for the new SVM Example #1 are shown in  FIG. 10  with Θ Tr =30° for all the sectors. It can be found that T 1 , T 2  and T 3  are all non-zero (except when K is very small or K=0). So the new SVM Example #1 can easily solve the problem (1) mentioned previously. Using single-shunt current sensing, at any time it is a good choice to measure inverter DC link current during two time intervals T 1  and T 2 . 
       New SVM Example #2 
     SVM with Two Pseudo Zero Vectors 
       [0089]    Using the reference vector  230  in sector A as shown in  FIG. 8  as an example, the following shows the calculations. Using volt-second balancing: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       V 
                       → 
                     
                     ref 
                   
                   = 
                   
                     
                       
                         
                           T 
                           0 
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         0 
                       
                     
                     + 
                     
                       
                         
                           
                             
                               T 
                               1 
                               ′ 
                             
                             + 
                             
                               T 
                               z 
                             
                           
                           
                              
                             
                               T 
                               1 
                             
                           
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         1 
                       
                     
                     + 
                     
                       
                         
                           
                             
                               T 
                               2 
                               ′ 
                             
                             + 
                             
                               T 
                               z 
                             
                           
                           
                              
                             
                               T 
                               2 
                             
                           
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         2 
                       
                     
                     + 
                     
                       
                         
                           
                             T 
                             z 
                           
                           
                              
                             
                               T 
                               3 
                             
                           
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         4 
                       
                     
                     + 
                     
                       
                         
                           
                             T 
                             z 
                           
                           
                              
                             
                               T 
                               4 
                             
                           
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           5 
                         
                         → 
                       
                     
                   
                 
               
               
                 
                   ( 
                   24 
                   ) 
                 
               
             
             
               
                 
                   
                     T 
                     S 
                   
                   = 
                   
                     
                       T 
                       0 
                     
                     + 
                     
                       
                         ( 
                         
                           
                             T 
                             1 
                             ′ 
                           
                           + 
                           
                             T 
                             z 
                           
                         
                         ) 
                       
                       
                          
                         
                           T 
                           1 
                         
                       
                     
                     + 
                     
                       
                         ( 
                         
                           
                             T 
                             2 
                             ′ 
                           
                           + 
                           
                             T 
                             z 
                           
                         
                         ) 
                       
                       
                          
                         
                           T 
                           2 
                         
                       
                     
                     + 
                     
                       
                         2 
                          
                         
                           T 
                           z 
                         
                       
                       
                          
                         
                           
                             T 
                             3 
                           
                           + 
                           
                             T 
                             4 
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   25 
                   ) 
                 
               
             
           
         
       
     
         [0000]    Similarly, select T z =λT S , i.e., 
         [0000]        T   3   =T   4   =λT   S   ≧T   min   (26)
 
       Solve Equations (24) to (26) to get 
       [0090]        T   1   =T   1   ′+T   z   =[K  sin(60°−θ rel )+λ]·T S   (27)
 
         [0000]        T   2   =T   2   ′+T   z   =[K  sin(θ rel )+λ]·T S   (28)
 
         [0000]        T   1   +T   2   +T   3   +T   4   =[K  sin(60°+θ rel )+4 λ]·T   S   (29)
 
         [0000]    So the zero vector time is 
         [0000]        T   0   =T   S −( T   1   +T   2   +T   3   +T   4 )=[(1−4λ)− K  sin(60°+θ rel )]· T   S   (30)
 
         [0000]    where: 
         [0091]    T 0 —Time of existing zero vector(s) is applied. The zero vector(s) can be {right arrow over (V)} 0  [000], or {right arrow over (V)} 7  [111], or both 
         [0092]    T z —Time of pseudo zero vectors are applied 
         [0093]    T 1 —Time of the 1 st  active vector is applied within one sampling period 
         [0094]    T 2 —Time of the 2 nd  active vector is applied within one sampling period 
         [0095]    T 3 , T 4 —Time of the 3 rd  and 4 th  active vectors are applied within one sampling period, which are part of the pseudo zero vectors being used 
         [0096]    K— 
         [0000]    
       
         
           
             K 
             = 
             
               
                 3 
               
               · 
               
                 
                    
                   
                     V 
                     ref 
                   
                    
                 
                 
                   V 
                   DC 
                 
               
               · 
               
                  
                 
                   V 
                   ref 
                 
                  
               
             
           
         
       
     
         [0000]    is the amplitude of {right arrow over (V)} ref , and V DC  is the inverter DC link voltage 
         [0097]    T S —Sampling period 
         [0000]    As T 0 ≧0 all the time, from Equation (30) we find K≦1-4λ, so the inverter DC link voltage utilization without over-modulation is 
         [0000]    
       
         
           
             
               
                 
                   η 
                   = 
                   
                     
                       
                          
                         
                           V 
                           ref 
                         
                          
                       
                       
                         V 
                         DC 
                       
                     
                     ≤ 
                     
                       
                         1 
                         - 
                         
                           4 
                            
                           λ 
                         
                       
                       
                         3 
                       
                     
                   
                 
               
               
                 
                   ( 
                   31 
                   ) 
                 
               
             
           
         
       
     
         [0000]    So the maximum inverter DC link voltage utilization without over-modulation is 
         [0000]    
       
         
           
             
               
                 
                   
                     η 
                     max 
                     ′ 
                   
                   = 
                   
                     
                       1 
                       - 
                       
                         4 
                          
                         λ 
                       
                     
                     
                       3 
                     
                   
                 
               
               
                 
                   ( 
                   32 
                   ) 
                 
               
             
           
         
       
     
         [0000]    Plot of Equation (32) is shown in  FIG. 11  at  280 , which shows the maximum inverter DC link voltage utilization versus A for new Example 2. When λ=0, the new SVM becomes the existing SVM and 
         [0000]    
       
         
           
             η 
             ≤ 
             
               
                 1 
                 
                   3 
                 
               
               . 
             
           
         
       
     
         [0098]    Plots of normalized time T 1   290 , T 2    300 , T 3  and T 4    310  for new SVM Example #2 are shown in  FIG. 12 . It can be found that both T 1  and T 2  are longer than T min  in all conditions (as long as λT S ≧T min ), even when K=0 (i.e.: |V ref |=0). So the New SVM Example #2 not only can solve the problem (1), but also can completely solve the problem (2) mentioned above. Using single-shunt current sensing, at any time it is a good choice to measure inverter DC link current during two time intervals T 1  and T 2 . 
         [0099]    Usage of new SVM in motor control: The connections of SVM in a motor control are shown in  FIGS. 13A and 13B . The input to the new SVM can be the polar coordinates (i.e., radial coordinate |V ref | and angular coordinate θ) of the reference vector {right arrow over (V)} ref  as shown in  FIG. 13A , which has been discussed above. The inputs to the SVM can also be the Cartesian coordinates (V α , V β ) of the reference vector {right arrow over (V)} ref  in the α-β Cartesian coordinate system, as shown in  FIG. 13B . The coordinate systems in the SVM space vector hexagon are shown in  FIG. 14 . The Polar-to-Cartesian Transform is: 
         [0000]        V   α   =|V   ref |cos(θ)  (33)
 
         [0000]        V   β   =|V   ref |sin(θ)  (34)
 
         [0000]    With Equations (33) and (34), all the formulas listed in Table 2 can be transferred to format with inputs of V α  and V β . For example, the time calculations of New SVM Example #1 in sector A1 become 
         [0000]    
       
         
           
             
               
                 
                   
                     T 
                     1 
                   
                   = 
                   
                     
                       
                         
                           3 
                         
                          
                         
                           T 
                           S 
                         
                       
                       
                         2 
                          
                         
                           V 
                           DC 
                         
                       
                     
                     · 
                     
                       ( 
                       
                         
                           
                             3 
                           
                            
                           
                             V 
                             α 
                           
                         
                         - 
                         
                           V 
                           β 
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   35 
                   ) 
                 
               
             
             
               
                 
                   
                     T 
                     2 
                   
                   = 
                   
                     
                       
                         
                           
                             3 
                           
                            
                           
                             T 
                             S 
                           
                         
                         
                           V 
                           DC 
                         
                       
                       · 
                       
                         V 
                         β 
                       
                     
                     + 
                     
                       λ 
                        
                       
                           
                       
                        
                       
                         T 
                         S 
                       
                     
                   
                 
               
               
                 
                   ( 
                   36 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       
                         
                           T 
                           3 
                         
                         = 
                           
                          
                         
                           T 
                           z 
                         
                       
                     
                   
                   
                     
                       
                         = 
                           
                          
                         
                           λ 
                            
                           
                               
                           
                            
                           
                             T 
                             S 
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   37 
                   ) 
                 
               
             
             
               
                 
                   
                     T 
                     0 
                   
                   = 
                   
                     
                       T 
                       S 
                     
                     - 
                     
                       ( 
                       
                         
                           T 
                           1 
                         
                         + 
                         
                           T 
                           2 
                         
                         + 
                         
                           T 
                           3 
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   38 
                   ) 
                 
               
             
           
         
       
     
         [0100]      FIG. 13A  shows a control system  320 , for example, a motor control system. The control system  320  includes a space vector modulator (SVM)  330  according to the present disclosure that utilizes pseudo zero vectors. The SVM  330  receives a reference signal or reference samples and synthesizes one or more reference vectors based thereon, wherein at least one of the reference vectors employ one or more pseudo zero vectors as described herein. Based on the synthesized reference vector(s), the SVM modulator  330  outputs timing signals  340  to a PWM unit  350  that receives the timing signals  340  and generates PWM control signals  360 . The PWM control signals  360  are provided to a three-phase inverter circuit  370  which generates output signals u, v, and w to drive a load  380  such as a three-phase motor.  FIG. 13B , as highlighted above, is similar to  FIG. 13A , but illustrates receiving an input reference signal in Cartesian coordinates rather than polar coordinates, but generally operates in the same manner as that described above. 
         [0101]    Inverter with single-shunt current sensing: The connection of a three-phase two-level voltage source inverter and a motor are shown in  FIG. 15  at  400 . The six switching devices of the inverter  400 , which could be MOSFET, IGBT or similar parts, are controlled by microcontroller PWM signals. As illustrated, the inverter comprises a first pair of series-connected switches  410 , a second pair of series-connected switches  420 , and a third pair of series-connected switches  430 . Each of the series-connected pairs  410 ,  420 , and  430  are connected at a node that forms an output u, v, w that connects to a respective phase of the load  440 . Each of the series-connected pairs  410 ,  420 ,  430  also couple together at a terminal  450  that couples to a first terminal of a shunt resistor  460  that has a second terminal connected to a reference potential  470 . An amplifier  480  has input terminals coupled to the first and second terminals of the shunt resistor  460 , respectively, wherein an output of the amplifier  480  reflects a current conducting through the shunt resistor  460 . 
         [0102]    The motor windings can be wired in a star (as shown) or a delta configuration. SVM is used to control the PWM to create three-phase sinusoidal waveforms to the motor windings. The shunt resistor R shunt    460  is inserted into the inverter DC link to sense the DC link current. If needed, the amplifier  480  is used to amplify the resistor voltage drop which is proportional to the DC link current. Note that a Hall sensor, a current transformer, or other current sensors can replace the shunt resistor to sense the DC link current. 
         [0103]    Compared to dual-shunt and triple-shunt current sensing, single-shunt current sensing has the following important advantages: 
         [0000]    1) Cost reduction as a result of only one current sensor, one amplifier (if any), and one ADC channel are used. In contrast, dual-shunt current sensing and triple-shunt current sensing need multiple current sensors, amplifiers (if any) and ADC channels.
 
2) No need to calibrate amplifier gains and offsets (which may due to component tolerance, fluctuating temperature, aging, and etc) since the same current sensing circuit and ADC channel are used for all the current measurements of motor phases.
 
3) Simpler and easier electronic schematics and PCB design.
 
Switching sequence design: There are a lot of switching sequence combinations for the new SVMs, depending on different sequencings of active/zero vectors, splitting of the duty cycles of the vectors, and choice of existing zero vectors (i.e., to choose zero vector {right arrow over (V)} 0  [000], or {right arrow over (V)} 7 [111], or both). It is difficult to list all the switching sequences here. This section only gives some examples of the switching sequences, which can be easily implemented using Infineon microcontrollers or other type microcontrollers.  FIGS. 16A and 16B  show examples of 5-segment switching sequences  500 ,  510  for new SVM Example #1.  FIGS. 17A and 17B  show examples of 5-segment switching sequences  520 ,  530  for new SVM Example #2,  FIGS. 18A and 18B  show examples of 6-segment switching sequences  540 ,  550 , and  FIGS. 19A and 19B  show examples of 7-segment switching sequences  560 ,  570 .
 
         [0104]    Current reconstruction: Two/three motor phase currents can be reconstructed by using single-shunt current sensing. In each PWM cycle, the inverter DC link current is measured at least twice during two different active vector segments to get two motor phase currents. The ADC samplings are normally triggered near the center of the active vector segments to avoid current transitions. Table 3 shows the inverter DC link current of different PWM segments. As an example,  FIG. 20  shows the DC link current I DClink    580  corresponding to the switching sequences  540  shown earlier in  FIG. 18A . Two phase currents can be measured at two PWM active vector segments which are more than or equal to T min , e.g.: I DClink =−I W  during T 2  and I DClink =I U  during T 1  in  FIG. 20  since T 2 ≧T min  and T 1 ≧T min . With two phase currents, the third motor phase current can be calculated easily because I U +I V +I W =0. 
         [0000]    
       
         
               
             
               
               
               
               
             
               
               
               
               
               
             
           
               
                 TABLE 3 
               
             
             
               
                   
               
               
                 Inverter DC link current of different PWM segments 
               
             
          
           
               
                   
                   
                 Inverter DC 
                   
               
               
                   
                   
                 link current 
               
               
                   
                 PWM Segment 
                 I DClink   Note 1   
                 Remark 
               
               
                   
                   
               
             
          
           
               
                   
                 Active 
                 {right arrow over (V)} 1  [100] 
                 I U   
                 Take current 
               
               
                   
                 Vectors 
                 {right arrow over (V)} 2  [110] 
                 −I W   
                 sample for 
               
               
                   
                   
                 {right arrow over (V)} 3  [010] 
                 I V   
                 motor phase 
               
               
                   
                   
                 {right arrow over (V)} 4  [011] 
                 −I U   
                 current 
               
               
                   
                   
                 {right arrow over (V)} 5  [001] 
                 I W   
               
               
                   
                   
                 {right arrow over (V)} 6  [101] 
                 −I V   
               
               
                   
                 Zero 
                 {right arrow over (V)} 0  [000] 
                 0 
                 If needed, can 
               
               
                   
                 Vectors 
                 {right arrow over (V)} 7  [111] 
                 0 
                 be used for 
               
               
                   
                   
                   
                   
                 channel offset 
               
               
                   
                   
                   
                   
                 calibration 
               
               
                   
                   
               
               
                   
                   Note 1 I U , I V  and I W  are the currents of motor phases U, V and W, respectively. 
               
             
          
         
       
     
         [0105]    Using a reference vector {right arrow over (V)} ref  in sector A as an example,  FIG. 21  shows an alternative approximation of the reference vector  590  with one pseudo zero vector in accordance with another embodiment of the disclosure. Clearly, calculations for T 1  and T 2  are the same as those for the existing SVM, and the reference vector  590  is actually approximated by four active vectors. It is another example of selective utilization of a pseudo zero vector to easily solve the problems mentioned above. 
         [0106]    As highlighted above, SVM is regularly employed for industry and automotive motor control applications, and used in the sinusoidal commutation control (such as V/f, FOC, and DTC) of PMSM and ACIM to generate sine waveforms from three-phase inverters. Sinusoidal commutation motor control with a single-shunt current sensing resistor inserted in the inverter DC link is a desired solution when compared to dual-shunt and triple-shunt current sensing techniques, owing to its important advantages such as low cost, simplicity and etc. However, two current samples within one PWM cycle are needed for correct motor phase current reconstruction with single-shunt current sensing. But with existing SVM, the accurate current construction is difficult when the reference voltage space vector is crossing a sector border, as in that situation only one current sample can be measured. It is a problem for most of the normal speed motor control with SVM and single-shunt current sensing. 
         [0107]    The enhanced SVM proposed in accordance with one embodiment of the disclosure can solve the above-mentioned problem. With the new SVM technology, one can offer low-cost, high-quality, more reliable, and unique motor control solutions (e.g., sensorless FOC with single-shunt current sensing) to customers. The enhanced SVM may also be used in three-phase power inverter control for uninterruptible power supplies, renewable energy, and etc. 
         [0108]    To solve the problems highlighted above associated with conventional SVM, the present disclosure according to the present embodiment introduces innovative approximations of the reference vector with different active vectors, instead of two adjacent active vectors in existing SVM. New approximations of the enhanced SVM are shown in  FIGS. 22A-22B . One portion (i.e., m{right arrow over (V)} ref  with 0≦m≦1) of the reference vectors  600 ,  610  is approximated by two adjacent active vectors just like the existing SVM, and the remaining portion (1−m){right arrow over (V)} ref  is approximated by two non-adjacent and 120°-separated active vectors. The enhanced SVM is elaborated below. One special case of the enhanced SVM is when m=1, the entire reference vector {right arrow over (V)} ref  is approximated by two adjacent active vectors in each sector, and the enhanced SVM becomes the existing SVM as shown in  FIG. 4 . Another special case of the enhanced SVM is when m=0, the entire reference vector {right arrow over (V)} ref  is approximated by two non-adjacent and 120°-separated active vectors in each sector. So the approximations  600 ,  610  in  FIGS. 22A and 22B  become approximations  620 ,  630  in  FIGS. 23A and 23B , respectively. The enhanced SVM becomes another new SVM (which is called “Enhanced SVM at m=0” in this disclosure) and which is described in detail below. 
         [0109]    Table 4 compares and summarizes the existing SVM and proposed new SVMs. 
         [0000]    
       
         
               
             
               
               
               
               
               
             
               
               
               
               
               
               
             
               
               
               
               
               
               
               
               
               
             
           
               
                 TABLE 4 
               
             
             
               
                   
               
               
                 Comparison of existing SVM and proposed enhanced SVMs 
               
             
          
           
               
                   
                   
                 {right arrow over (V)} ref  Angle 
                 Active  
                   
               
             
          
           
               
                   
                   
                 θ 
                   
                 Vectors For {right arrow over (V)} ref    
                   
               
               
                   
                   
                 (limit 0° ≦ θ &lt; 360°) 
                   
                 Approximation 
                   
               
             
          
           
               
                   
                 Sector 
                 Start 
                 End 
                 θ rel   
                 1 st   
                 2 nd   
                 3 rd   
                 Formulas 
               
               
                   
               
               
                 Existing SVM (A special case of Enhanced SVM at m = 1) 
                 A B C D E F 
                  0°  60° 120° 180° 240° 300° 
                  60° 120° 180° 240° 300° 360° 
                 θ-0° θ-60° θ-120° θ-180° θ-240° θ-300° 
                 {right arrow over (V)} 1  {right arrow over (V)} 2  {right arrow over (V)} 3  {right arrow over (V)} 4  {right arrow over (V)} 5  {right arrow over (V)} 6   
                 {right arrow over (V)} 2  {right arrow over (V)} 3  {right arrow over (V)} 4  {right arrow over (V)} 5  {right arrow over (V)} 6  {right arrow over (V)} 1   
                 — — — — — — 
                 T 1  = Ksin(60° − θ rel ) · T S  T 2  = Ksin(θ rel ) · T S  T 0  = T s  − (T 1  + T 2 ) 
         where                 K     =           3     ·            V   ref            V   DC                       and                          V   ref            V   DC         ≤       1     3                   
 
               
               
                   
               
               
                 Enhanced SVM (0 ≦ m ≦ 1) 
                 AB   BC CD DE EF FA 
                 Θ Tr   Note   1   60° + Θ Tr  120° + Θ Tr  180° + Θ Tr  240° + Θ Tr  300° + Θ Tr   
                  60° + Θ Tr    120° + Θ Tr  180° + Θ Tr  240° + Θ Tr  300° + Θ Tr  Θ Tr   
                 θ-0°   θ-60° θ-120° θ-180° θ-240° θ-300°  
                 {right arrow over (V)} 1    {right arrow over (V)} 2  {right arrow over (V)} 3  {right arrow over (V)} 4  {right arrow over (V)} 5  {right arrow over (V)} 6   
                 {right arrow over (V)} 2    {right arrow over (V)} 3  {right arrow over (V)} 4  {right arrow over (V)} 5  {right arrow over (V)} 6  {right arrow over (V)} 1   
                 {right arrow over (V)} 3    {right arrow over (V)} 4  {right arrow over (V)} 5  {right arrow over (V)} 6  {right arrow over (V)} 1  {right arrow over (V)} 2   
                 1). For Θ Tr  ≦ θ rel  &lt; 60°:  T 1  = {square root over (1 − m(1 − m))} · K · sin(Θ 1  − θ rel ) · T S    Note   2  T 2  = mKsin(θ rel ) · T S  T 3  = (1 − m)Ksin(θ rel ) · T S  2). For 60° ≦ θ rel  &lt; 60° + Θ Tr :  T 1  = (1 − m)Ksin(120° − θ rel ) · T S  T 2  = mKsin(120° − θ rel ) · T S  T 3  = {square root over (1 − m(1 − m))} · K · sin(θ rel  − Θ 3 ) · T S    Note   3  For both conditions 1) and 2): T 0  = T s  − (T 1  + T 2  + T 3 ) 
           where                 K     =       3     ·            V   ref            V   DC           ,       and                          V   ref            V   DC         ≤       1       9   -     3        m        (     3   -   m     )                           
 
               
               
                   
               
               
                 Enhanced SVM at m = 0 (A special case of Enhanced SVM) 
                 AB BC CD DE EF FA 
                 Θ Tr   60° + Θ Tr  120° + Θ Tr  180° + Θ Tr  240° + Θ Tr  300° + Θ Tr   
                  60° + Θ Tr  120° + Θ Tr  180° + Θ Tr  240° + Θ Tr  300° + Θ Tr  Θ Tr   
                 θ-0° θ-60° θ-120° θ-180° θ-240° θ-300° 
                 {right arrow over (V)} 1  {right arrow over (V)} 2  {right arrow over (V)} 3  {right arrow over (V)} 4  {right arrow over (V)} 5  {right arrow over (V)} 6   
                 — — — — — — 
                 {right arrow over (V)} 3  {right arrow over (V)} 4  {right arrow over (V)} 5  {right arrow over (V)} 6  {right arrow over (V)} 1  {right arrow over (V)} 2   
                 T 1  = Ksin(120° − θ rel ) · T S  T 3  = Ksin(θ rel ) · T S  T 0  = T s  − (T 1  + T 3 ) 
         where                 K     =           3     ·            V   ref            V   DC                       and                          V   ref            V   DC         ≤     1   3           
 
               
               
                   
               
               
                   Note   1 : 
               
               
                 Θ Tr  is a transition angle for new sectors (i.e., AB, BC CD, DE, EF, and FA), and 0° &lt; Θ Tr  &lt; 60°. 
               
               
                   Note   2 : 
               
               
                           Θ   1     =         arctan                   (       3         2      m     -   1       )       +     k                 π                 where                 k                 is                 an                   integer   .              Select                   k                 so                 that                 60      °       ≤     Θ   1     ≤     120      °                 for                 0     ≤   m   ≤   1.           Note   3 : 
               
               
                 
                   
                     
                       
                         
                           Θ 
                           3 
                         
                         = 
                         
                           
                             
                               arctan 
                                
                               
                                   
                               
                                
                               
                                 ( 
                                 
                                   
                                     
                                       3 
                                     
                                      
                                     m 
                                   
                                   
                                     2 
                                     - 
                                     m 
                                   
                                 
                                 ) 
                               
                             
                             + 
                             
                               k 
                                
                               
                                   
                               
                                
                               π 
                                
                               
                                   
                               
                                
                               where 
                                
                               
                                   
                               
                                
                               k 
                                
                               
                                   
                               
                                
                               is 
                                
                               
                                   
                               
                                
                               an 
                                
                               
                                   
                               
                                
                               
                                 integer 
                                 . 
                                 
                                     
                                 
                                  
                                 Select 
                               
                                
                               
                                   
                               
                                
                               k 
                                
                               
                                   
                               
                                
                               so 
                                
                               
                                   
                               
                                
                               that 
                                
                               
                                   
                               
                                
                               0 
                                
                               ° 
                             
                           
                           ≤ 
                           
                             Θ 
                             3 
                           
                           ≤ 
                           
                             60 
                              
                             ° 
                              
                             
                                 
                             
                              
                             for 
                              
                             
                                 
                             
                              
                             0 
                           
                           ≤ 
                           m 
                           ≤ 
                           
                             1. 
                               
                           
                         
                       
                     
                   
                 
               
             
          
         
       
     
         [0110]    As highlighted previously, enhanced SVM has advantages over conventional SVM techniques. For example, enhanced SVM is well suited for three-phase motor control with single-shunt current sensing, so it can fully use the advantages of a single-shunt current sensing technique as elaborated in greater detail below. Another advantage of the enhanced SVM is that customers can tune the factor m to obtain different current sampling intervals based on different system requirements and hardware designs, so as to achieve the best performance of motor control with single-shunt current sensing. As three adjacent active vectors are used for the approximation in the enhanced SVM, if necessary, it is possible to take three ADC samplings of the inverter DC link current within one PWM cycle to get three motor phase currents directly (which will be discussed in greater detail below). It will be useful for the application cases that the sum of three motor winding currents is not zero but still use single-shunt current sensing. It is also possible to take only two ADC samplings of the interested motor phase current (e.g., I U  and I V  only) directly within any PWM cycle. 
         [0111]    The enhanced SVM at m=0 is also well suited for motor control with single-shunt current sensing. It has long current sampling time intervals within every PWM cycle, so it is possible to use a low-speed, common and low-cost operational amplifier for current signal amplification to further lower the system cost. The enhanced SVM at m=0 has a lower DC link bus voltage utilization. It is not a problem for applications with high DC link voltage, e.g., low-cost PMSM ceiling fan drive with very high DC link voltage (up to 400V DC due to the use of power factor correction). 
         [0112]    Enhanced SVM: Using the reference vector in sector A as an example as shown in  FIGS. 22A and 22B , the reference vector in each existing sector (i.e., A, B, C, D, E, or F) can be approximated by two different sets of active vectors. A transition angle Θ Tr  (0°&lt;Θ Tr &lt;60°) is introduced to get new combined sectors AB, BC CD, DE, EF, and FA as shown in Table 1. At the transition angles the reference vector approximation transits from one set of active vectors to another set of active vectors for the new SVMs. Θ Tr  can be different in different existing sectors. For simplicity, we can choose the same value, e.g., Θ Tr =30°, for all the sectors. 
         [0113]    Calculations when Θ Tr ≦θ rel &lt;60°: Using reference vector in sector A as shown in  FIG. 22A  as an example, the following shows the calculations when Θ Tr ≦θ rel &lt;60°. Using volt-second balancing: 
         [0000]    
       
         
           
             
               
                 
                   
                     m 
                      
                     
                       
                         V 
                         → 
                       
                       ref 
                     
                   
                   = 
                   
                     
                       
                         
                           T 
                           0 
                           ′ 
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         0 
                       
                     
                     + 
                     
                       
                         
                           T 
                           1 
                           ′ 
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         1 
                       
                     
                     + 
                     
                       
                         
                           T 
                           2 
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         2 
                       
                     
                   
                 
               
               
                 
                   ( 
                   39 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       ( 
                       
                         1 
                         - 
                         m 
                       
                       ) 
                     
                      
                     
                       
                         V 
                         → 
                       
                       ref 
                     
                   
                   = 
                   
                     
                       
                         
                           T 
                           0 
                           ″ 
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         0 
                       
                     
                     + 
                     
                       
                         
                           T 
                           1 
                           ″ 
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         1 
                       
                     
                     + 
                     
                       
                         
                           T 
                           3 
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         3 
                       
                     
                   
                 
               
               
                 
                   ( 
                   40 
                   ) 
                 
               
             
           
         
       
     
         [0000]    Add both sides of Equations (39) and (40), 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       V 
                       → 
                     
                     ref 
                   
                   = 
                   
                     
                       
                         
                           
                             
                               T 
                               0 
                               ′ 
                             
                             + 
                             
                               T 
                               0 
                               ″ 
                             
                           
                           
                              
                             
                               T 
                               0 
                             
                           
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         0 
                       
                     
                     + 
                     
                       
                         
                           
                             
                               T 
                               1 
                               ′ 
                             
                             + 
                             
                               T 
                               1 
                               ″ 
                             
                           
                           
                              
                             
                               T 
                               1 
                             
                           
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         1 
                       
                     
                     + 
                     
                       
                         
                           T 
                           2 
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         2 
                       
                     
                     + 
                     
                       
                         
                           T 
                           3 
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
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                         3 
                       
                     
                   
                 
               
               
                 
                   ( 
                   41 
                   ) 
                 
               
             
             
               
                 
                   
                     T 
                     S 
                   
                   = 
                   
                     
                       
                         ( 
                         
                           
                             T 
                             0 
                             ′ 
                           
                           + 
                           
                             T 
                             0 
                             ″ 
                           
                         
                         ) 
                       
                       
                          
                         
                           T 
                           0 
                         
                       
                     
                     + 
                     
                       
                         ( 
                         
                           
                             T 
                             1 
                             ′ 
                           
                           + 
                           
                             T 
                             1 
                             ″ 
                           
                         
                         ) 
                       
                       
                          
                         
                           T 
                           1 
                         
                       
                     
                     + 
                     
                       T 
                       2 
                     
                     + 
                     
                       T 
                       3 
                     
                   
                 
               
               
                 
                   ( 
                   42 
                   ) 
                 
               
             
           
         
       
     
       Solve Equations (39) and (40) to get 
       [0114]        T   1   ′=mK  sin(60°−θ rel )· T   S   (43)
 
         [0000]        T   1 ″=(1 −m ) K  sin(120°−θ rel )· T   S   (44)
 
         [0000]        T   2   =mK  sin(θ rel )· T   S   (45)
 
         [0000]        T   3 =(1 −m ) K  sin(θ rel )· T   S   (46)
 
         [0000]    where: 
         [0115]    T 0 —Time of zero vector(s) is applied. The zero vector(s) can be {right arrow over (V)} 0  [000], or {right arrow over (V)} 7 [111], or both 
         [0116]    T 1 —Time of the 1 st  active vector is applied within one sampling period 
         [0117]    T 2 —Time of the 2 nd  active vector is applied within one sampling period 
         [0118]    T 3 —Time of the 3 rd  active vector is applied within one sampling period 
         [0119]    K— 
         [0000]    
       
         
           
             K 
             = 
             
               
                 3 
               
               · 
               
                 
                    
                   
                     V 
                     ref 
                   
                    
                 
                 
                   V 
                   DC 
                 
               
               · 
               
                  
                 
                   V 
                   ref 
                 
                  
               
             
           
         
       
     
         [0000]    is the amplitude of {right arrow over (V)} ref , and V DC  is the inverter DC link voltage 
         [0120]    T S —Sampling period 
         [0000]    Add both sides of Equations (43) and (44), it can be found that 
         [0000]        T   1   =T   1   ′+T   1 ″=√{square root over (1 −m (1 −m ))}· K  sin(Θ 1 −θ rel )· T   S   (47)
 
         [0000]    where Θ 1  is an angle depending on the factor m only, and 
         [0000]    
       
         
           
             
               
                 
                   
                     Θ 
                     1 
                   
                   = 
                   
                     
                       arctan 
                       ( 
                       
                         
                           3 
                         
                         
                           
                             2 
                              
                             m 
                           
                           - 
                           1 
                         
                       
                       ) 
                     
                     + 
                     
                       k 
                        
                       
                           
                       
                        
                       π 
                     
                   
                 
               
               
                 
                   ( 
                   48 
                   ) 
                 
               
             
           
         
       
     
         [0000]    where k is an integer and k=0, ±1, ±2, ±3, or . . . . Select k so that 60°≦Θ 1 ≦120° for 0≦m≦1, e.g.: Θ 1 =120° when m=0.
 
Add both sides of Equations (45), (46) and (47) to get
 
         [0000]        T   1   +T   2   +T   3 =√{square root over (3 −m (3 −m ))}· K  sin(Θ 2 +θ rel )· T   S   (49)
 
         [0000]    where: Θ 2 —An angle depending on m only, and 
         [0000]    
       
         
           
             
               
                 Θ 
                 2 
               
               = 
               
                 arctan 
                 ( 
                 
                   
                     3 
                   
                   
                     3 
                     - 
                     
                       2 
                        
                       m 
                     
                   
                 
                 ) 
               
             
             , 
           
         
       
     
         [0000]    e.g.: Θ 2 =30° when m=0. The zero vector time is 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       
                         
                           T 
                           0 
                         
                         = 
                           
                          
                         
                           
                             
                               T 
                               0 
                               ′ 
                             
                             + 
                             
                               T 
                               0 
                               ″ 
                             
                           
                           = 
                           
                             
                               T 
                               S 
                             
                             - 
                             
                               ( 
                               
                                 
                                   T 
                                   1 
                                 
                                 + 
                                 
                                   T 
                                   2 
                                 
                                 + 
                                 
                                   T 
                                   3 
                                 
                               
                               ) 
                             
                           
                         
                       
                     
                   
                   
                     
                       
                         = 
                           
                          
                         
                           
                             [ 
                             
                               1 
                               - 
                               
                                 
                                   
                                     
                                       3 
                                       - 
                                       
                                         ( 
                                         
                                           3 
                                           - 
                                           m 
                                         
                                         ) 
                                       
                                     
                                   
                                   · 
                                   K 
                                 
                                  
                                 
                                     
                                 
                                  
                                 
                                   sin 
                                    
                                   
                                     ( 
                                     
                                       
                                         Θ 
                                         2 
                                       
                                       + 
                                       
                                         θ 
                                         rel 
                                       
                                     
                                     ) 
                                   
                                 
                               
                             
                             ] 
                           
                           · 
                           
                             T 
                             S 
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   50 
                   ) 
                 
               
             
           
         
       
     
         [0000]    As T 0 ≧0 (or T 1 +T 2 +T 3 ≧T S ) all the time, so √{square root over (3−m(3−m))}·K≦1, so the inverter DC link voltage utilization without over-modulation is 
         [0000]    
       
         
           
             
               
                 
                   η 
                   = 
                   
                     
                       
                          
                         
                           V 
                           ref 
                         
                          
                       
                       
                         V 
                         
                           D 
                            
                           
                               
                           
                            
                           C 
                         
                       
                     
                     ≤ 
                     
                       1 
                       
                         
                           9 
                           - 
                           
                             3 
                              
                             
                               m 
                                
                               
                                 ( 
                                 
                                   3 
                                   - 
                                   m 
                                 
                                 ) 
                               
                             
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   51 
                   ) 
                 
               
             
           
         
       
     
         [0121]    Calculations when 60°≦θ rel &lt;60°±Θ Tr : Using reference vector in sector A as an example as shown in  FIG. 22B , the following show the slightly different calculations when 60°≦θ rel &lt;60°+Θ Tr . Similarly we have 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       V 
                       → 
                     
                     ref 
                   
                   = 
                   
                     
                       
                         
                           T 
                           0 
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         0 
                       
                     
                     + 
                     
                       
                         
                           T 
                           1 
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         6 
                       
                     
                     + 
                     
                       
                         
                           T 
                           2 
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         1 
                       
                     
                     + 
                     
                       
                         
                           
                             
                               T 
                               3 
                               ′ 
                             
                             + 
                             
                               T 
                               3 
                               ″ 
                             
                           
                           
                              
                             
                               T 
                               
                                 3 
                                  
                                 
                                     
                                 
                               
                             
                           
                         
                         
                           T 
                           S 
                         
                       
                        
                       
                         
                           V 
                           → 
                         
                         2 
                       
                     
                   
                 
               
               
                 
                   ( 
                   52 
                   ) 
                 
               
             
             
               
                 
                   
                     T 
                     S 
                   
                   = 
                   
                     
                       T 
                       0 
                     
                     + 
                     
                       T 
                       1 
                     
                     + 
                     
                       T 
                       2 
                     
                     + 
                     
                       
                         ( 
                         
                           
                             T 
                             3 
                             ′ 
                           
                           + 
                           
                             T 
                             3 
                             ″ 
                           
                         
                         ) 
                       
                       
                          
                         
                           T 
                           3 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   53 
                   ) 
                 
               
             
           
         
       
     
         [0000]    Solve Equations (52) and (53), we have 
         [0000]        T   1 =(1 −m ) K  sin(120°−θ rel )· T   S   (54)
 
         [0000]        T   2   =mK  sin(120°−θ rel )· T   S   (55)
 
         [0000]        T   3   =T   3   ′+T   3 ″=√{square root over (1 −m (1 −m ))}· K  sin(θ rel −Θ 3 )· T   S   (56)
 
         [0000]    where: Θ 3 —An angle depending on the factor m only, and 
         [0000]    
       
         
           
             
               Θ 
               3 
             
             = 
             
               
                 arctan 
                 ( 
                 
                   
                     
                       3 
                     
                      
                     m 
                   
                   
                     2 
                     - 
                     m 
                   
                 
                 ) 
               
               + 
               
                 k 
                  
                 
                     
                 
                  
                 
                   π 
                   . 
                 
               
             
           
         
       
     
         [0000]    k is an integer and k=0, ±1, ±2, ±3, or . . . . Select k so that 0°≦Θ 3 ≦60° for 0≦m≦1, e.g.: Θ 3 =0° when m=0. The zero vector time is 
         [0000]    
       
         
           
             
               
                 
                   
                     T 
                     0 
                   
                   = 
                   
                     
                       
                         T 
                         S 
                       
                       - 
                       
                         T 
                         1 
                       
                       - 
                       
                         T 
                         2 
                       
                       - 
                       
                         T 
                         3 
                       
                     
                     = 
                     
                       
                         [ 
                         
                           1 
                           - 
                           
                             
                               
                                 
                                   3 
                                   - 
                                   
                                     m 
                                      
                                     
                                       ( 
                                       
                                         3 
                                         - 
                                         m 
                                       
                                       ) 
                                     
                                   
                                 
                               
                               · 
                               K 
                             
                              
                             
                                 
                             
                              
                             
                               sin 
                                
                               
                                 ( 
                                 
                                   
                                     Θ 
                                     4 
                                   
                                   + 
                                   
                                     θ 
                                     rel 
                                   
                                 
                                 ) 
                               
                             
                           
                         
                         ] 
                       
                       · 
                       
                         T 
                         S 
                       
                     
                   
                 
               
               
                 
                   ( 
                   57 
                   ) 
                 
               
             
           
         
       
     
         [0000]    where: Θ 4 —An angle depending on m only, and 
         [0000]    
       
         
           
             
               
                 Θ 
                 4 
               
               = 
               
                 arctan 
                 [ 
                 
                   
                     
                       3 
                     
                      
                     
                       ( 
                       
                         1 
                         - 
                         m 
                       
                       ) 
                     
                   
                   
                     3 
                     - 
                     m 
                   
                 
                 ] 
               
             
             , 
           
         
       
     
         [0000]    e.g.: Θ 4 =30° when m=0. Likewise, it can be found that the inverter DC link voltage utilization without over-modulation is the same as Equation (51). So the enhanced SVM has one maximum inverter DC link voltage utilization without over-modulation, which is 
         [0000]    
       
         
           
             
               
                 
                   
                     η 
                     
                       m 
                        
                       
                           
                       
                        
                       ax 
                     
                   
                   = 
                   
                     1 
                     
                       
                         9 
                         - 
                         
                           3 
                            
                           
                             m 
                              
                             
                               ( 
                               
                                 3 
                                 - 
                                 m 
                               
                               ) 
                             
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   58 
                   ) 
                 
               
             
           
         
       
     
         [0122]    Plot of Equation (58) is shown in  FIG. 24  at  640 . When m=1, the enhanced SVM becomes the existing SVM and 
         [0000]    
       
         
           
             
               η 
               ≤ 
               
                 1 
                 
                   3 
                 
               
             
             , 
           
         
       
     
         [0000]    which has been stated in earlier Equation (7); when m=0, it becomes the special case of the enhanced SVM at m=0 and 
         [0000]    
       
         
           
             η 
             ≤ 
             
               1 
               3 
             
           
         
       
     
         [0000]    (to be discussed in greater detail below). 
         [0123]    Plots of normalized time T 1   650 , T 2   660 , and T 3    670  for the enhanced SVM are shown in  FIGS. 25A and 25B  with Θ Tr =30° for all the sectors, with m=0.8 and 0.2 as examples, respectively. It is obvious that T 1 , T 2  and T 3  are all non-zero. Using single-shunt current sensing, at any time it is a good choice to measure inverter DC link current during two time intervals which is longer than the third time (e.g., T 2  and T 3  when 0°≦θ&lt;30° as shown in  FIG. 25A , T 1  and T 3  for all θ as shown in  FIG. 25B ). At Θ Tr =30°, it can be found that the minimum current sampling time, which is limited by T 1 , T 2  or T 3 , is 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       T 
                       minimum 
                     
                     = 
                     
                       
                         
                           
                             3 
                           
                            
                           
                             ( 
                             
                               1 
                               - 
                               m 
                             
                             ) 
                           
                         
                         2 
                       
                        
                       
                         
                           KT 
                           S 
                         
                          
                         
                           ( 
                           
                             0.5 
                             ≤ 
                             m 
                             ≤ 
                             1 
                           
                           ) 
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   or 
                 
               
               
                 
                   ( 
                   59 
                   ) 
                 
               
             
             
               
                 
                   
                     T 
                     minimum 
                   
                   = 
                   
                     
                       
                         1 
                         - 
                         m 
                       
                       2 
                     
                      
                     
                       
                         KT 
                         S 
                       
                        
                       
                         ( 
                         
                           0 
                           ≤ 
                           m 
                           &lt; 
                           0.5 
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                   ( 
                   60 
                   ) 
                 
               
             
           
         
       
     
         [0000]    A user can choose the value of m based on the required maximum DC link voltage utilization, and the minimum time needed for the current sampling (i.e., make sure T minimum ≧T min , where T min  is PWM dead time+driver delay+ADC sampling time). 
         [0124]    Enhanced SVM at m=0: Similarly, the calculations of the enhanced SVM at m=0 are shown below: 
         [0000]        T   1   =K  sin(120°−θ rel )· T   S   (61)
 
         [0000]        T   3   =K  sin(θ rel )· T   S   (62)
 
         [0000]        T   1   +T   3 =√{square root over (3)}· K  sin(30°+θ rel )· T   S   (63)
 
         [0000]        T   0   =T   S −( T   1   +T   3 )=[1−√{square root over (3)} ·K  sin(30°+θ rel )]· T   S   (64)
 
         [0000]    Note that Equations (61) to (64) work for both  FIGS. 23A and 23B . The inverter DC link voltage utilization without over-modulation becomes 
         [0000]    
       
         
           
             
               
                 
                   η 
                   = 
                   
                     
                       
                          
                         
                           V 
                           ref 
                         
                          
                       
                       
                         V 
                         
                           D 
                            
                           
                               
                           
                            
                           C 
                         
                       
                     
                     ≤ 
                     
                       1 
                       3 
                     
                   
                 
               
               
                 
                   ( 
                   65 
                   ) 
                 
               
             
           
         
       
     
         [0000]    Plot of normalized time T 1    680  and T 3    690  with transition angle Θ Tr =30° for all the sectors is shown in  FIG. 26 . It is obvious that both T 1  and T 3  are non-zero. With Θ Tr =30° the shortest time for the current ADC sampling is 
         [0000]    
       
         
           
             
               
                 
                   
                     T 
                     minimum 
                     ′ 
                   
                   = 
                   
                     
                       1 
                       2 
                     
                     · 
                     
                       KT 
                       S 
                     
                   
                 
               
               
                 
                   ( 
                   66 
                   ) 
                 
               
             
           
         
       
     
         [0125]    One advantageous element of the disclosure according to this embodiment is that instead of two adjacent active vectors being employed to synthesize a reference vector as in existing SVM, the enhanced SVM uses three adjacent active vectors or two non-adjacent active vectors for the approximation of reference vector {right arrow over (V)} ref . In the new enhanced SVMs, non-zero intervals can be achieved for multiple current samplings within one PWM cycle, and the reconstruction of phase currents with single-shunt current sensing can be easily done without introducing distortion to the resulted motor phase current. 
         [0126]    Usage of new SVM in motor control: The connections of SVM in a motor control system  700  are shown in  FIGS. 27A and 27B .  FIG. 27A  shows a portion of a control system  700 , for example, a motor control system. The control system  700  includes a space vector modulator (SVM)  710  according to the present disclosure that utilizes the enhanced SVM using three adjacent active vectors or two non-adjacent active vectors for synthesizing a reference vector. The enhanced SVM  710  receives a reference signal or reference samples and synthesizes one or more reference vectors based thereon, wherein at least one of the reference vectors employ either three adjacent active vectors or two non-adjacent active vectors as described herein. Based on the synthesized reference vector(s), the SVM modulator  710  outputs timing signals  720  to a PWM unit  730  that receives the timing signals  720  and generates PWM control signals  740 . The PWM control signals  740  are provided to a three-phase inverter circuit  750  which generates output signals u, v, and w to drive a load  760  such as a three-phase motor. 
         [0127]    The input to the new enhanced SVM can be the polar coordinates (i.e., radial coordinate |V ref | and angular coordinate θ) of the reference vector {right arrow over (V)} ref  as shown in  FIG. 27A , which has been discussed above. The inputs to the SVM can also be the Cartesian coordinates (V α , V β ) of the reference vector {right arrow over (V)} ref  in the α-β Cartesian coordinate system, as shown in  FIG. 27B . 
         [0128]    The coordinate systems in the SVM space vector hexagon are shown in  FIG. 28  at  770 . The Polar-to-Cartesian Transform is: 
         [0000]        V   α   =|V   ref |cos(θ)  (67)
 
         [0000]        V   β   =|V   ref |sin(θ)  (68)
 
         [0000]    With Equations (67) and (68), all the formulas listed in Table 4 can be transferred to format with inputs of V α  and V β . For example, the time calculations of the enhanced SVM at m=0 in sector AB become 
         [0000]    
       
         
           
             
               
                 
                   
                     T 
                     1 
                   
                   = 
                   
                     
                       
                         
                           3 
                         
                          
                         
                           T 
                           S 
                         
                       
                       
                         2 
                          
                         
                           V 
                           
                             D 
                              
                             
                                 
                             
                              
                             C 
                           
                         
                       
                     
                     · 
                     
                       ( 
                       
                         
                           
                             3 
                           
                            
                           
                             V 
                             α 
                           
                         
                         + 
                         
                           V 
                           β 
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   69 
                   ) 
                 
               
             
             
               
                 
                   
                     T 
                     3 
                   
                   = 
                   
                     
                       
                         
                           3 
                         
                          
                         
                           T 
                           S 
                         
                       
                       
                         V 
                         
                           D 
                            
                           
                               
                           
                            
                           C 
                         
                       
                     
                     · 
                     
                       V 
                       β 
                     
                   
                 
               
               
                 
                   ( 
                   70 
                   ) 
                 
               
             
             
               
                 
                   
                     T 
                     0 
                   
                   = 
                   
                     
                       T 
                       S 
                     
                     - 
                     
                       ( 
                       
                         
                           T 
                           1 
                         
                         + 
                         
                           T 
                           3 
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   71 
                   ) 
                 
               
             
           
         
       
     
         [0129]    Inverter with single-shunt current sensing: The connection  780  of a three-phase two-level voltage source inverter  790  and a motor  800  are shown in  FIG. 29 . As illustrated, the inverter  790  comprises a first pair of series-connected switches  810 , a second pair of series-connected switches  820 , and a third pair of series-connected switches  830 . Each of the series-connected pairs  810 ,  820 , and  830  are connected at a node that forms an output u, v, w that connects to a respective phase of the load  800 . Each of the series-connected pairs  810 ,  820 ,  830  also couple together at a terminal  850  that couples to a first terminal of a shunt resistor  860  that has a second terminal connected to a reference potential  870 . An amplifier  880  has input terminals coupled to the first and second terminals of the shunt resistor  860 , respectively, wherein an output of the amplifier  880  reflects a current conducting through the shunt resistor  860 . 
         [0130]    The six switching devices of the inverter, which could be MOSFET, IGBT or similar parts, are controlled by microcontroller PWM signals. The motor windings can be wired in star (as shown) or delta. The enhanced SVM discussed above using either three adjacent active vectors or two non-adjacent active vectors to synthesize the reference vector is used to control the PWM to create three-phase sinusoidal waveforms to the motor windings. The shunt resistor R shunt    860  is inserted into the inverter DC link to sense the DC link current. If needed, the amplifier  880  is used to amplify the resistor voltage drop which is proportional to the DC link current. Note that a Hall sensor, a current transformer, or other current sensors can replace the shunt resistor to sense the DC link current. 
         [0131]    There are a lot of switching sequence combinations for the new enhanced SVMs, depending on different sequencings of active/zero vectors, splitting of the duty cycles of the vectors, and choice of existing zero vectors (i.e., to choose zero vector {right arrow over (V)} 0  [000], or {right arrow over (V)} 7  [111], or both). It is difficult to list all the switching sequences here. This section only gives some examples of the switching sequences, which can be easily implemented using Infineon microcontrollers, or other microcontrollers. 
         [0132]    Switching sequence design for enhanced SVM:  FIGS. 30A and 30B  show examples of 4-segment switching sequences for the enhanced SVM at  890 ,  900 , and  FIGS. 31A and 31B  show examples of 6-segment switching sequences at  910 ,  920 .  FIGS. 32A and 32B  show examples of 3-segment switching sequences for the enhanced SVM at m=0 at  930 ,  940 , and  FIGS. 33A and 33B  show examples of 5-segment switching sequences at m=0 at  950 ,  960 . 
         [0133]    Current reconstruction: two/three motor phase currents can be reconstructed by using single-shunt current sensing. In each PWM cycle, inverter DC link current is measured at least twice during two different active vector segments to get two motor phase currents. The ADC samplings are normally triggered near the center of the active vector segments to avoid current transitions. 
         [0134]    Table 5 shows inverter DC link current of different PWM segments. As an example,  FIG. 34  shows the DC link current I DClink    970  corresponding to the switching sequences  910  shown earlier in  FIG. 31A  for the enhanced SVM. Two phase currents can be measured at two PWM active vector segments which are more than or equal to T min , e.g.: I DClink =−I W  during T 2 /2 and I DClink =I V  during T 3  in  FIG. 34  if T 2 /2≧T min  and T 3 ≧T min . With two phase currents, the third motor phase current can be calculated easily because I U +I V +I W =0. 
         [0135]    It can be found from  FIG. 34  that for the enhanced SVM, it is possible to take three ADC samplings of inverter DC link current within one PWM cycle to get three motor phase currents directly, if all the three active vector segments are longer than T min . It will be useful for the application cases with I U +I V +I W ≠0. It is also possible to take only two ADC samplings of the interested motor phase current (e.g.: I U  and I V  only) directly within any PWM cycle. 
         [0000]    
       
         
               
             
               
               
               
               
             
               
               
               
               
               
             
           
               
                 TABLE 5 
               
             
             
               
                   
               
               
                 Inverter DC link current of different PWM segments 
               
             
          
           
               
                   
                   
                 Inverter DC 
                   
               
               
                   
                   
                 link current 
               
               
                   
                 PWM Segment 
                 I DClink   Note 1   
                 Remark 
               
               
                   
                   
               
             
          
           
               
                   
                 Active 
                 {right arrow over (V)} 1  [100] 
                 I U   
                 Take current 
               
               
                   
                 Vectors 
                 {right arrow over (V)} 2  [110] 
                 −I W   
                 sample for 
               
               
                   
                   
                 {right arrow over (V)} 3  [010] 
                 I V   
                 motor phase 
               
               
                   
                   
                 {right arrow over (V)} 4  [011] 
                 −I U   
                 current 
               
               
                   
                   
                 {right arrow over (V)} 5  [001] 
                 I W   
               
               
                   
                   
                 {right arrow over (V)} 6  [101] 
                 −I V   
               
               
                   
                 Zero 
                 {right arrow over (V)} 0  [000] 
                 0 
                 If needed, can 
               
               
                   
                 Vectors 
                 {right arrow over (V)} 7  [111] 
                 0 
                 be used for 
               
               
                   
                   
                   
                   
                 channel offset 
               
               
                   
                   
                   
                   
                 calibration 
               
               
                   
                   
               
               
                   
                   Note 1 I U , I V  and I W  are the currents of motor phases U, V and W, respectively. 
               
             
          
         
       
     
         [0136]    To solve the problem as mentioned above, it is possible to approximate the reference vector {right arrow over (V)} ref  in alternative ways as shown in  FIGS. 35A and 35B , where n≧0. Using reference vector in sector A as examples, in  FIG. 35A  at  980 , (1+n){right arrow over (V)} ref  is approximated by two adjacent active vectors just like the existing SVM, and a reverse portion n{right arrow over (V)} ref  is approximated by two non-adjacent and 120°-separated active vectors; in  FIG. 35B  at  990 , (1+n){right arrow over (V)} ref  is approximated by two non-adjacent and 120°-separated active vectors, and a reverse portion n{right arrow over (V)} ref  is approximated by two adjacent active vectors. The alternative solutions shown in  FIGS. 35A and 35B  use four active vectors to approximate the reference vector, while the solutions mentioned earlier use three or two. The approximation shown in  FIG. 35A  can be considered a special case of the enhanced SVM at m≧1, and the one shown in  FIG. 35B  can be considered a special case of the enhanced SVM at m≦0. 
         [0137]    The exemplary embodiments described above represent only an illustration of the principles of the present invention. It is self-evident that modifications and variations of the arrangements and details described herein may be of interest to other specialists. The aim is therefore that the invention should be restricted only by the scope of protection of the following patent claims and not by the specific details, which have been presented herein on the basis of the description and the explanation of the exemplary embodiments.