Abstract:
A system for an eight-phase 45° polyphase filter with amplitude matching, where a full eight-phase 45° split may be achieved by tying together the inputs of two offset four-phase 90° phase splitters. Amplitude matching may be achieved by obtaining those inputs from an additional single four-phase 90° phase splitter. The additional phase splitter can distribute power evenly among the inputs of the two offset phase splitters so as to cancel out the occurrence of any uneven power distribution.

Description:
BACKGROUND OF THE INVENTION 
     1. Field of the Invention 
     This invention relates to polyphase filters and, more specifically, to an eight-phase 45° polyphase filter with amplitude matching. 
     2. Related Art 
     Certain radio receiver architectures require that signals be phase split equally into eight separate phases with constant 45° phase splits between them. For example, direct conversion receivers and subharmonic frequency translators for use in such receivers may have such requirements. In addition, preprocessors for improving the switching characteristics of a local oscillator input to such frequency translators may also have these requirements. 
     One of the most common and most reliable ways of achieving phase splits in general is through the use of passive RC filters, that are commonly called polyphase filters. In the simplest RC series circuit, the voltage across the capacitor will lag the current, and the voltage across the resistor, by 90°. In a differential RC phase splitter, the two inputs are 180° out of phase with each other and four outputs are available, each of which has a phase angle difference of 90° relative to the next output. 
     By selecting the resistor and capacitor values, the phase angle of an output voltage relative to an input voltage can be achieved for a particular frequency. The inputs of two differential RC phase splitters can be connected. The resistor and capacitor values of the two phase splitters can be selected so that there is a 45° phase angle difference between an output of the first phase splitter and a corresponding output of the second phase splitter at a particular frequency. The result is eight outputs with 45° phase splits between them at a particular frequency. 
     The four outputs that are available from a differential RC phase splitter can be amplitude matched at a particular frequency by selection of appropriate resistor and capacitor values. However, the resistor and capacitor values cannot, in general, be selected to achieve both amplitude matching and a particular phase difference relative to the input. 
     SUMMARY 
     A full eight-phase 45° split is achieved by tying together the inputs of two offset four-phase 90° phase splitters. The resistor and capacitor values are selected to obtain the 45° phase angle difference between corresponding outputs of the two offset phase splitters. Amplitude matching is achieved by obtaining the inputs for that stage of the polyphase filter from an additional single four-phase 90° phase splitter, whose resistor and capacitor values are selected to achieve amplitude matching. At a particular frequency, the additional phase splitter can distribute power evenly among the four inputs of the two offset phase splitters, so as to cancel out an uneven power distribution which otherwise would occur between the outputs of the two offset phase splitters. 
     Other systems, methods, features and advantages of the invention will be or will become apparent to one with skill in the art upon examination of the following figures and detailed description. It is intended that all such additional systems, methods, features and advantages be included within this description, be within the scope of the invention, and be protected by the accompanying claims. 
    
    
     BRIEF DESCRIPTION OF THE FIGURES 
     The invention can be better understood with reference to the following figures. The components in the figures are not necessarily to scale, emphasis instead being placed upon illustrating the principles of the invention. 
     FIG. 1 is a circuit diagram of an RC series circuit with a sinusoidal voltage source. 
     FIG. 2 is a phasor diagram of voltages in FIG.  1 . 
     FIG. 3 is a circuit diagram of a two-phase 90° polyphase filter. 
     FIG. 4 is a circuit diagram of a four-phase 90° polyphase filter with two inputs. 
     FIG. 5 is a circuit diagram of an eight-phase 45° polyphase filter with two inputs. 
     FIG. 6 is a circuit diagram of an eight-phase 45° polyphase filter with four inputs. 
     FIG. 7 is a block diagram illustrating an example of phase-splitting to achieve an eight-phase 45° phase split with amplitude matching. 
     FIG. 8 is a circuit diagram of an eight-phase 45° polyphase filter with two inputs. 
     FIG. 9 a  is a graph representing phase angle (between an output and an input) versus frequency. 
     FIG. 9 b  is a graph representing the difference between the phase angles of two outputs versus frequency. 
    
    
     DETAILED DESCRIPTION 
     FIG. 1 is a circuit diagram, for discussion purposes, of a prior art RC series circuit with a sinusoidal voltage source V i  (bolding indicates a phasor) driving a current I through resistor R and capacitor C. FIG. 2 is a phasor diagram of voltages in FIG.  1 . The voltage across resistor R is IR, and the voltage across capacitor C is I(1/jωC). V i =IR+I(1/jωC). The voltage across R, IR, always leads V i  by a phase angle φ, and the voltage across C, I(1/jωC), always lags IR by 90°. The phase angle φ=arctan(1/ωRC). The voltages across R and across C have equal amplitudes when R=1/ωC or ωRC=1. When that occurs, φ=45°, and the voltage across R leads V i  by 45° and the voltage across C lags V i  by 45°. 
     FIG. 3 is a circuit diagram, for discussion purposes, of a two-phase 90° polyphase filter. (“Polyphase filter” and “phase splitter” are used interchangeably.) In FIG.  3 : 
     
       
         
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     V 1  and V 2  are 90° out of phase with respect to each other, regardless of the frequency, the value of R, or the value of C. However, the amplitudes of V 1  and V 2  are equal only when ωRC=1, i.e., when ω=1/RC. When that occurs, the phase angles of V 1  and V 2  with respect to V i  are −45° and +45°, respectively. Each side of FIG. 3 is effectively an RC series circuit as in FIG.  1 . 
     FIG. 4 is a circuit diagram of a differential four-phase 90° polyphase filter. Inputs V i1  and V i2  are of equal amplitude and 180° out of phase with each other. As V i1 =−V i2 , as each of the resistors has the same value, and as each of the capacitors has the same value, the symmetry renders points A and B virtual grounds. The circuit of FIG. 4 is a combination of two circuits like the circuit of FIG.  3 . Consequently, V o1  lags V o2  by 90°, and they straddle V i1  with respect to phase angle. Similarly, V o3  lags V o4  by 90°, and they straddle V i2  with respect to phase angle. Due to the symmetry again, V o1  and V o2  are 180° out of phase with V o3  and V o4 , respectively. Therefore, the phase angle of each of the four outputs is 90° different than the next output. As discussed above, amplitude matching occurs only when ω=1/RC. When that occurs, V o1 , V o2 , V o3  and V o4  have phase angles of −45°, 45°, 135° and 225°, respectively, with respect to V i1 . 
     FIG. 5 is a circuit diagram of a differential eight-phase 45° polyphase filter. It is a combination of two circuits like the circuit of FIG. 4, with the respective inputs of the two circuits connected and the respective virtual grounds of the two circuits connected. Inputs V i1  and V i2  are of equal amplitude and 180° out of phase with each other. Each of outputs V o11 , V o12 , V o13  and V o14  is 90° out of phase with the next output and, similarly, each of outputs V o21 , V o22 , V o23  and V o24  is 90° out of phase with the next output. 
     As illustrated in FIG. 2, the phase angle of an output voltage relative to an input voltage can be achieved for a particular frequency, by selection of the resistor and capacitor values. In the circuit of FIG. 5, the values of R 1 C 1  and R 2 C 2  can be chosen so that V o21 , V o22 , V o23  and V o24  lead V o11 , V o12 , V o13  and V o14 , respectively, by 45° for a particular frequency (for an unloaded filter). For example, V o11 , leads V i1  by 22.5° when 1/ωR 1 C 1 =tan 22.50=0.41421, and V o21  leads V i1  by 67.5° when 1/ωR 2 C 2 =tan 67.5°=2.41421 (for an unloaded filter). In that case, the phase angles of V 011 , V o21 , V o12 , V o22 , V o13 , V o23 , V o14  and V o24  are 22.5°, 67.5°, 112.5°, 157.5°, 202.5°, 247.5°, 292.5° and 337.5°, respectively, with respect to V i1  for an unloaded filter. In that case:              R   1          C   1           R   2          C   2         =         1     ω                   R   2          C   2           1     ω                 R1C1         =       2.41421   0.41421     ≅   5.7                   (     1     ω                   R   1          C   1         )          (     1     ω                   R   2          C   2         )       =         (   2.41421   )          (   0.41421   )       =   1                            
     (which necessarily follows because 22.5° and 67.5° are complementary angles, and their tangents are reciprocals);          1       ω   2          R   1          C   1          R   2          C   2         =     1                 and                 R   1          C   1          R   2          C   2       =     1     ω   2                              
     This example results in eight outputs with 45° phase splits between them at a particular frequency. However, the two relationships:              R   1          C   1           R   2          C   2         ≅   5.7                          
     and              R   1          C   1          R   2          C   2       =     1     ω   2         ,                          
     require that          1       R   1          C   1         &lt;   ω   &lt;       1       R   2          C   2         .                            
     That is, with this example neither part of the circuit in FIG. 5 can be at its amplitude matched point. Because inputs V i1  and V i2  are of equal amplitude and 180° out of phase with each other, amplitude matched outputs could have been achieved if ω=         ω   =     1   RC       ,                          
     but then the desired 45° phase splits between the outputs would not occur. 
     FIG. 6 is a circuit diagram similar to FIG. 5, but there are four inputs V i1 , V i2 , V i3  and V i4 . The intermediate points which were virtual grounds in FIG. 5 are additional voltage inputs in FIG.  6 . In the circuit of FIG. 6, for an unloaded filter            V   o11     -     V   i2       =         I   1          R   1       =           (       V   i1     -     V   i2       )          R   1           R   1     +     1     j                 ω                   C   1             =       (       V   i1     -     V   i2       )                       j                 ω                   R   1          C   1         1   +     j                 ω                   R   1          C   1                                 V   o11     =         V   i1            j                 ω                   R   1          C   1         1   +     j                 ω                   R   1          C   1             +       V   i2          (     1   -       j                 ω                   R   1          C   1         1   +     j                 ω                   R   1          C   1             )                     =         V   i1            j                 ω                   R   1          C   1         1   +     j                 ω                   R   1          C   1             +       V   i2          1     1   +     j                 ω                   R   1          C   1                                            
     Similarly, the other seven outputs in FIG. 6 can be expressed in terms of the inputs as follows:          V   012     =         V   i2            j                 ω                   R   1          C   1         1   +     j                 ω                   R   1          C   1             +       V   i3          1     1   +     j                 ω                   R   1          C   1                         V   o13     =         V   i3            j                 ω                   R   1          C   1         1   +     j                 ω                   R   1          C   1             +       V   i4          1     1   +     j                 ω                   R   1          C   1                         V   014     =         V   i4            j                 ω                   R   1          C   1         1   +     j                 ω                   R   1          C   1             +       V   i1          1     1   +     j                 ω                   R   1          C   1                         V   021     =         V   i1            j                 ω                   R   2          C   2         1   +     j                 ω                   R   2          C   2             +       V   i2          1     1   +     j                 ω                   R   2          C   2                         V   o22     =         V   i2            j                 ω                   R   2          C   2         1   +     j                 ω                   R   2          C   2             +       V   i3          1     1   +     j                 ω                   R   2          C   2                         V   023     =         V   i3            j                 ω                   R   2          C   2         1   +     j                 ω                   R   2          C   2             +       V   i4          1     1   +     j                 ω                   R   2          C   2                         V   o24     =         V   i4            j                 ω                   R   2          C   2         1   +     j                 ω                   R   2          C   2             +       V   i1          1     1   +     j                 ω                   R   2          C   2                                      
     Each output has a contribution from two inputs. The transfer functions applied to the two inputs have the same denominator, and the numerator of one transfer function is entirely “real” and the numerator of the other transfer function is entirely “imaginary.” 
     If V i2  leads V i1  by 90°, then each of those inputs will add in-phase for V o11  and each of those inputs will add in-phase for V o21 . Similarly, if each input leads the preceding input by 90°, then the two inputs to each output will add in-phase. 
     Continuing with the circuit of FIG. 6, if, in addition to leading the preceding input by 90°, each of the inputs V i1 , V i2 , V i3  and V i4  has the same amplitude, then 
     
       
           V   i2   =jV   i1 , 
       
     
     
       
           V   i3   =jV   i2   =−V   i1 , and 
       
     
     
       
           V   i4   =jV   i3   =−Jv   i1 . 
       
     
     In that case,          V   o11     =     j                   V   i1            1   +     ω                   R   1          C   1           1   +     j                 ω                   R   1          C   1                       V   o12     =       -                V   i1              1   +     ω                   R   1          C   1           1   +     j                 ω                   R   1          C   1                       V   013     =       -   j                     V   i1            1   +     ω                   R   1          C   1           1   +     j                 ω                   R   1          C   1                       V   014     =                  V   i1            1   +     ω                   R   1          C   1           1   +     j                 ω                   R   1          C   1                       V   o21     =     j                   V   i1            1   +     ω                   R   2          C   2           1   +     j                 ω                   R   2          C   2                       V   o22     =       -                V   i1              1   +     ω                   R   2          C   2           1   +     j                 ω                   R   2          C   2                       V   023     =       -   j                     V   i1            1   +     ω                   R   2          C   2           1   +     j                 ω                   R   2          C   2                       V   024     =       V   i1            1   +     ω                   R   2          C   2           1   +     j                 ω                   R   2          C   2                                    
     Each of outputs V o11 , V o12 , V o13  and V o14  will have the same amplitude, and each of outputs V o21 , V o22 , V o23  and V o24  will have the same amplitude. The ratio of V o11 to V o21  is              (     1   +     ω                   R   1          C   1         )          (     1   +     j                 ω                   R   2          C   2         )           (     1   +     jω                   R   1          C   1         )          (     1   +                ω                   R   2          C   2         )         =         (     1   +     ω                   R   1          C   1         )     +     j        (         ω   2          R   1          C   1          R   2          C   2       +     ω                   R   2          C   2         )             (     1   +     ω                   R   2          C   2         )     +     j        (         ω   2          R   1          C   1          R   2          C   2       +     ω                   R   1          C   1         )                                  
     If R 1 C 1 R 2 C 2 =1/ω 2 , then the ratio of V o11 , to V o21  is            (     1   +     ω                   R   1          C   1         )     +     j        (     1   +     ω                   R   2          C   2         )             (     1   +     ω                   R   2          C   2         )     +     j        (     1   +     ω                   R   1          C   1         )                                
     The absolute value of both numerator and the denominator is the same, V o11  and V o21  have the same amplitude, and all eight of the outputs will have the same amplitude. 
     Therefore, for an unloaded filter, the eight outputs in the circuit of FIG. 6 will be amplitude matched, each of outputs V o11 , V o12 , V o13  and V o14  will lead the previous output by 90°, and each of outputs V o21 , V o22 , V o23  and V o24  will lead the previous output by 90°, if R 1 C 1 R 2 C 2 =           R   1          C   1          R   2          C   2       =     1     ω   2                              
     and each of the inputs V i1 , V i2 , V i3  and V i4  has the same amplitude and leads the preceding input by 90°. In addition, if                R   1          C   1           R   2          C   2         ≅   5.7     ,                          
     then the phase angles of V o11 , V o21 , V o12 , V o22 , V o13 , V o23 , V o14  and V o24  will be 22.5°, 67.5°, 112.5°, 157.5°, 202.5°, 247.5°, 292.5° and 337.5°, respectively, with respect to (V i1 −V i2 ). This would be the eight-phase 45° phase splitting with amplitude matching which is sought. Since (V i1 −V i2 )=(1−j)V i1 , (V i1 −V i2 ) lags V i1  by 45°. 
     FIG. 7 is a block diagram illustrating an example of phase-splitting to achieve an eight-phase 45° phase split with amplitude matching. Two signals V i1  and V i2 , of equal amplitude and 180° out of phase with each other, can be split (12) to four intermediate signals of equal amplitude, each one of which is 90° out of phase with the next one of the four signals. The second input, V i2 , can already be available in some embodiments and, in other embodiments, can readily be obtained by inverting V i1  as is well known in the art. In one embodiment, the amplitude matching of the resulting four intermediate signals can be achieved for a particular frequency by appropriate selection of component values in a four-phase 90° polyphase filter. The amplitude-matched point can result in the four intermediate signals having phase angles of 45°, 135°, 225°, and 315° with respect to V i1  in FIG.  7 . This was explained regarding the circuit illustrated in FIG.  4 . 
     The four intermediate signals can be converted ( 14 ) to a first group of four output signals, each one of which is 90° out of phase with the next one. The four intermediate signals also can be converted (16) to a second group of four output signals, each one of which is 90° out of phase with the next one. The first and second groups of output signals in FIG. 7 can be offset from each other, respectively, by 45°. This was explained regarding the circuit illustrated in FIG.  5 . For an unloaded filter, in one embodiment, this can be achieved for a particular frequency by appropriate selection of component values in two four-phase 90° polyphase filters with the same input signals. This can result in the eight output signals having phase angles of 22.5°, 67.5°, 112.5°, 157.5°, 202.5°, 247.5°, 292.5°, and 337.5° with respect to V i1 . Amplitude matching of the eight output signals in FIG. 7 can be achieved because of the amplitude matching of the four intermediate signals. This was explained regarding the circuit illustrated in FIG.  6 . 
     FIG. 8 is a circuit diagram of a differential eight-phase 45° polyphase filter, implementing the block diagram of FIG.  7 . Inputs V i1  and V i2  are of equal amplitude and 180° out of phase with each other. The first stage is similar to the circuit of FIG.  4 . When          ω   =     1       R   3          C   3           ,                          
     the four outputs of the first stage (i.e., the intermediate signals which are inputs to the second stage) will be 90° phase-split and amplitude matched. These are the input conditions discussed above for one example of the circuit of FIG.  6 . The first stage outputs (second stage inputs) will have phase angles of 45°, 135°, 225° and 315° with respect to V i1 . 
     The second stage of the circuit of FIG. 8 is similar to the circuit of FIG.  6 . For an unloaded filter, when          ω   =     1       R   3          C   3           ,         R   1          C   1          R   2          C   2       =     1     ω   2         ,       and                       R   1          C   1           R   2          C   2           ≅   5.7     ,                          
     the second stage outputs will be 45° phase-split and amplitude matched. They will have phase angles of 22.5°, 67.5°, 112.5°, 157.5°, 202.5°, 247.5°, 292.5° and 337.5° with respect to V i1 . 
     FIG. 9 a  is an illustration representing a plot of phase angle φ (between an output and an input in the circuit of FIG. 8) versus log ω. As illustrated in FIG. 2, a phase angle φ=arctan(1/ωRC). The phase angle φ approaches 90° asymptotically as 1/ωRC increases and is effectively a constant 90° when ω is less than about an order of magnitude less than 1/RC. The phase angle φ approaches 0° asymptotically as 1/ωRC decreases and is effectively a constant 0° when ω is more than about an order of magnitude more than 1/RC. The change in φ for a given change in ω is greatest when ω=1/RC, and φ=45°. It is also relatively constant around φ=45°, and then diminishes toward zero as φ approaches 0° or 90°. 
     FIG. 9 a  is an illustration of two plots, φ 1  and φ 2 , representing the phase angle for two different values of RC, R 1 C 1  and R 2 C 2 , respectively. If R 1 C 1 =(K)(R 2 C 2 ) where K is a constant, then φ 1  for ω=ω a  has the same value as φ 2  for ω=(1/K)ω a , since φ=arctan(1/ωRC). Since log((1/K)ω)=log ω+log (1/K), and since log (1/K) is a constant, the plot representing φ 1  is just a translation along the log ω axis of the plot representing φ 2 . 
     FIG. 9 b  is an illustration representing a plot of the difference in phase angle Δφ between φ 2  and φ 1  in FIG. 9 a  versus log ω. If                R   1          C   1           R   2          C   2         ≅   5.7     ,                          
     then φ 1  will be 22.5° when φ 2  is 67.5°. This was explained regarding the circuit illustrated in FIG.  5 . In that case, Δφ would peak at that point and would equal 45°. As ω changes in either direction from that point, one of the two phase angles (φ 1  or φ 2 ) will move closer to 45° and the other will move further away. For a given change in ω, the phase angle getting closer to 45° will increase (or decrease) more than the other phase angle, and Δφ will decrease as represented in FIG. 9 b.    
     For the 22.5°/67.5° phase-split at the operating frequency, Δφ remains relatively flat over a wide band (approximately the desired operating frequency ±50%). However, a circuit such as in FIG. 5 would have a severe amplitude mismatch at its output as a consequence of the offset nature of the two RC constants. The addition of the first stage, as in the example of FIG. 8 permits the eight-phase 45° split with amplitude matching. The first stage distributes power evenly among the four inputs to the second stage, cancelling out the uneven distribution which would occur otherwise in the second stage. 
     The circuits discussed above were analyzed in a voltage mode, but the same principles apply to current mode signals. Using current inputs and outputs would result in the same phase splitting and amplitude matching. 
     While various embodiments of the application have been described, it will be apparent to those of ordinary skill in the art that many more embodiments and implementations are possible that are within the scope of the invention. Accordingly, the invention is not to be restricted except in light of the attached claims and their equivalents.