Patent Publication Number: US-2010123466-A1

Title: System and Method for Corner Frequency Compensation

Description:
TECHNICAL FIELD 
     The present invention relates generally to a system and method for wireless communications, and more particularly to a system and method for corner frequency compensation in a wireless receiver. 
     BACKGROUND 
     Analog baseband filters (ABF) are widely used in wireless receivers to help eliminate or attenuate unwanted out-of-band interferers/blockers, which may negatively impact the overall performance of the wireless receivers. In general, an ABF may pass desired signals that are in a band of interest while blocking or attenuating undesired signals that are outside the band of interest. 
     An ABF, e.g., a low-pass filter, a band-pass filter, a high-pass filter, and so forth, may be designed so that the band of interest lies within the ABF&#39;s pass band, while the ABF&#39;s stop band encompasses frequencies outside of the band of interest. For example, if the ABF is a low-pass filter, then the ABF&#39;s corner frequency f c ) or cut-off frequency may be set so that it is above the band of interest. Thereby, the ABF will pass signals at frequencies lower than the corner frequency while attenuating signals at frequencies higher than the corner frequency. 
       FIG. 1   a  is a diagram illustrating a magnitude response  100  of an ABF. Magnitude response  100  displays a signal gain (output signal magnitude/input signal magnitude) of the ABF as a function of frequency. As shown in  FIG. 1   a , the ABF is a low-pass filter with a corner frequency at f c . The magnitude response of the ABF at zero hertz (DC) is shown in FIG.  1   a  as G 0  and the magnitude response of the ABF at the corner frequency is shown as G C . By definition, the corner frequency of a filter is a frequency where the magnitude response at the frequency is 
     
       
         
           
             1 
             
               2 
             
           
         
       
     
     of the magnitude response at DC, shown as span  105 . 
     However, during the operation of the wireless receiver, filtering characteristics of the ABF may change. The changes may be due to factors such as variations in operating conditions of the wireless receiver, such as temperature, supply voltage, and so forth. The changes in the operating conditions of the wireless receiver may change the corner frequency of the ABF. Therefore, if the corner frequency is increased, then some out-of-band signals that were previously eliminated (or attenuated) may be allowed to pass, while if the corner frequency is decreased, the some of the desired signals may be eliminated (or attenuated). In either case, the performance of the wireless receiver is degraded. 
       FIG. 1   b  is a diagram illustrating a magnitude response  120  of an ABF wherein the corner frequency of the ABF has increased. The changed corner frequency is shown as f c  and the original corner frequency is shown as f c . Since the corner frequency of the ABF has increased, the magnitude response of the ABF at the original corner frequency is higher (shown as span  125 ). Therefore, the magnitude response of the ABF at the original corner frequency is now greater than 
     
       
         
           
             1 
             
               2 
             
           
         
       
     
     times that of the magnitude response at DC. 
       FIG. 1   c  is a diagram illustrating a magnitude response  140  of an ABF wherein the corner frequency of the ABF has decreased. The changed corner frequency is shown as f″ c  and the original corner frequency is shown as f c . Since the corner frequency of the ABF has decreased, the magnitude response of the ABF at the original corner frequency is lower (shown as span  145 ). Therefore, the magnitude response of the ABF at the original corner frequency is now smaller than 
     
       
         
           
             1 
             
               2 
             
           
         
       
     
     times that of the magnitude response at DC. 
     A prior art technique used to determine a corner frequency of an ABF involves measuring a magnitude response of the ABF at DC and at a desired corner frequency. As discussed previously, if the desired corner frequency is substantially equal to the actual corner frequency of the ABF, then the magnitude response of the ABF at the desired corner frequency (G EC ) is 
     
       
         
           
             1 
             
               2 
             
           
         
       
     
     of the magnitude response at DC. Therefore, if 
     
       
         
           
              
             
               
                 G 
                 EC 
               
               - 
               
                 
                   G 
                   0 
                 
                 
                   2 
                 
               
             
              
           
         
       
     
     is smaller than or equal to a pre-determined error margin, then the actual corner frequency is equal (or about equal) to the desired corner frequency. However, if 
     
       
         
           
             
               
                 G 
                 EC 
               
               &gt; 
               
                 
                   G 
                   0 
                 
                 
                   2 
                 
               
             
             , 
           
         
       
     
     then the actual corner frequency is greater than the desired corner frequency. While, if 
     
       
         
           
             
               
                 G 
                 EC 
               
               &gt; 
               
                 
                   G 
                   0 
                 
                 
                   2 
                 
               
             
             , 
           
         
       
     
     then the actual corner frequency is smaller than the desired corner frequency. In either case, the actual corner frequency may be adjusted by tuning resistive (R), capacitive (C), inductive (L), and/or other parameters of the ABF and the measurement repeated until the actual corner frequency of the ABF has been tuned or compensated to the desired corner frequency. 
     SUMMARY OF THE INVENTION 
     These and other problems are generally solved or circumvented, and technical advantages are generally achieved, by embodiments of a system and a method for corner frequency compensation in a wireless receiver. 
     In accordance with an embodiment, a method for computing a corner frequency of a filter is provided. The method includes injecting a signal into an input of the filter, measuring a phase response of the filter at the frequency, and computing a corner frequency of the filter using the phase response. The signal has value substantially only at a frequency outside of a passband of the filter. 
     In accordance with another embodiment, a method for adjusting a corner frequency of a filter is provided. The method includes computing the corner frequency of the filter, wherein the computing uses a measured phase response of the filter at a frequency outside of a passband of the filter, determining if the computed corner frequency is different from a desired corner frequency by more or less than a threshold. The method also includes adjusting parameters of the filter to alter the corner frequency, and repeating the computing and the determining if the computed corner frequency differs from the desired corner frequency by more than the threshold. The method further includes leaving the parameters of the filter unchanged if the computed corner frequency differs from the desired corner frequency by less than the threshold. 
     In accordance with another embodiment, a receiver is provided. The receiver includes a signal path coupled to a signal input and to a baseband unit, the signal path includes a mixer that demodulates a signal provided by the signal input and changes the signal to a baseband signal, a filter coupled to the mixer, an analog to digital converter, and a decimation filter coupled to the analog to digital converter. The filter attenuates undesired signals outside of a frequency band of interest, the analog to digital converter digitizes the filtered baseband signal, and the decimation filter reduces a number of samples produced by the analog to digital converter. The receiver also includes a signal generator coupled to the filter, a demodulator coupled to the decimation filter and to the signal generator, a value estimator coupled to the demodulator, and a processor coupled to the value estimator. The signal generator generates an out-of-band signal having substantially value only at a frequency outside of a passband of the filter, the demodulator demodulates the reduced digital sample stream produced by the decimation filter with the out-of-band signal, the value estimator extracts a direct current (DC) component from the demodulated digital sample stream and selects a value of the DC component. The processor computes a phase response of the filter from the value provided by the value estimator. 
     An advantage of an embodiment is that out-of-band signals are used in the compensation of the corner frequency. Since the out-of-band signals will be subsequently eliminated in the wireless receiver, the out-of-band signals do not negatively impact the performance of the wireless receiver by introducing any distortion to desired in-band signals. 
     A further advantage of an embodiment is that the phase response is used in the compensation of the corner frequency. The use of phase response may make the compensation less sensitive to noise, numerical error, and so forth, which may be present in the process of estimating the corner frequency. Therefore, the compensation of the corner frequency may be performed with greater accuracy. 
     The foregoing has outlined rather broadly the features and technical advantages of the present invention in order that the detailed description of the embodiments that follow may be better understood. Additional features and advantages of the embodiments will be described hereinafter which form the subject of the claims of the invention. It should be appreciated by those skilled in the art that the conception and specific embodiments disclosed may be readily utilized as a basis for modifying or designing other structures or processes for carrying out the same purposes of the present invention. It should also be realized by those skilled in the art that such equivalent constructions do not depart from the spirit and scope of the invention as set forth in the appended claims. 
    
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
       For a more complete understanding of the embodiments, and the advantages thereof, reference is now made to the following descriptions taken in conjunction with the accompanying drawings, in which: 
         FIG. 1   a  is a plot illustrating a magnitude response of a filter; 
         FIG. 1   b  is a plot illustrating a magnitude response of a filter wherein the corner frequency of the filter has increased; 
         FIG. 1   c  is a plot illustrating a magnitude response of a filter wherein the corner frequency of the filter has decreased; 
         FIG. 2   a  is a diagram of a high-level view of a portion of a wireless transceiver; 
         FIG. 2   b  is a diagram of a portion of a receiver; 
         FIG. 2   c  is a diagram of a demodulation unit and a direct current (DC) estimation unit; 
         FIG. 3   a  is a flow diagram of a computing of a corner frequency of a filter using out-of-band signals; 
         FIG. 3   b  is a flow diagram of a computing of a corner frequency of a filter using out-of-band signals, wherein a receiver containing the filter has components with non-zero phase shifts; and 
         FIG. 4  is a flow diagram of a sequence of events in the compensating of a corner frequency of a filter. 
     
    
    
     DETAILED DESCRIPTION OF ILLUSTRATIVE EMBODIMENTS 
     The making and using of the embodiments are discussed in detail below. It should be appreciated, however, that the present invention provides many applicable inventive concepts that can be embodied in a wide variety of specific contexts. The specific embodiments discussed are merely illustrative of specific ways to make and use the invention, and do not limit the scope of the invention. 
     The embodiments will be described in a specific context, namely a wireless Cartesian receiver utilizing an analog baseband filter to help reduce or eliminate unwanted out-of-band interferers. The invention may also be applied, however, to other forms of wireless receivers, such as non-Cartesian wireless receivers. Furthermore, the invention may also be applied to other communications devices desiring to use analog filters to help reduce or eliminate interferers that may be present outside of a desired frequency band. The invention is especially important for a full duplex communication system such as the third-generation cellular communication system based on WCDMA. In this kind of system, the receiver continuously receives signals from the air. An advantage of the invention is that out-of-band signals are used in the compensation of the corner frequency. Since the out-of-band signals will be subsequently eliminated in the wireless receiver, the out-of-band signals do not negatively impact the performance of the wireless receiver by introducing any distortion to desired in-band signals. 
       FIG. 2   a  shows a diagram that illustrates a high-level view of a portion of a wireless transceiver  200 . Wireless transceiver  200  includes a transmitter  205 , a receiver  210 , a front-end module (FEM)  215 , and an antenna  220 . FEM  215  may include a power amplifier, antenna switches, duplexer, diplexer, SAW filters, and so forth. Transmitter  205  may be used to provide signal processing necessary to transmit information from a baseband unit over the air using antenna  220 , while the receiver may be used to provide signal processing to provide information received over the air via antenna  220  to the baseband unit. 
     Generally, in a transmitter (TX), a digital signal from a digital baseband unit may be processed (for example, filtering, digital-to-analog conversion, etc.) and then modulated onto an RF carrier signal. The RF signal containing the modulated digital signal may then be amplified and radiated through an antenna. This modulation (or up-conversion) in a transmitter may require the use of a local oscillator (LO) or an RF frequency synthesizer (for example, a phase-locked loop). Phase modulation may also be performed at the LO when a polar architecture is adopted for the transmitter. Generally, in a receiver (RX), a received RF signal may be amplified by a low-noise amplifier (LNA) and then down-converted by mixers to an analog baseband signal. There may also be filters between the LNA and the mixers. The analog baseband signal may then be filtered by analog baseband filters and may be further amplified. The baseband signal may then be digitized by an ADC. The down-conversion in the mixers generally requires the use of a local oscillator (LO). The transmitter and the receiver may share a common LO or have separate transmit (TX) LO and receive (RX) LO. 
       FIG. 2   b  is a diagram illustrating a portion of receiver  210 . The portion of receiver  210  shown in  FIG. 2   b  may be used to determine a corner frequency of an ABF used in receiver  210 , as well as to provide compensation for the corner frequency of the ABF if it is determined that the corner frequency has changed. Collectively, when receiver  210  determines the corner frequency and provides compensation of the corner frequency may be referred to as corner frequency compensation. Receiver  210  may periodically perform corner frequency compensation. Alternatively, receiver  210  may perform corner frequency compensation when a specified event occurs. Examples of events may include prior to receiving a transmission, when performance of wireless transceiver  200  reaches a threshold, when measured performance metrics reach a threshold, when a request to perform corner frequency compensation is received, and so forth. 
     As shown in  FIG. 2   b , receiver  210  is a Cartesian receiver and has two signal paths, a first signal path for an I-phase and a second signal path for a Q-phase. The discussion of receiver  210  will focus on the first signal path (I-phase). However, the two signal paths are substantially identical and the discussion of the first signal path will also apply to the second signal path. 
     A received signal, such as one received by antenna  220  and processed by FEM  215 , may be provided to receiver  210 . Once at receiver  210 , the received signal may be amplified by amplifiers, such as a low-noise amplifier and/or a transconductance amplifier. After amplification, the amplified received signal may be provided to the both the first signal path and the second signal path. 
     In the first signal path, the amplified received signal may be down-converted to an analog baseband signal by mixer  240 . Mixer  240  may multiply the amplified received signal with an I-phase output of a local oscillator (LO). In the second signal path, the amplified received signal may similarly be down-converted by another mixer that multiplies it with a Q-phase output of the LO. After conversion to the analog baseband signal by mixer  240 , an ABF  242  may be used to filter the analog baseband signal to eliminate or attenuate out-of-band signals. ABF  242  may be a first-order, a second-order, a third-order, or higher low-pass filter, a high-pass filter, a band-pass filter, and so on. The order and type of ABF  242  may be dependent on factors such as the nature of the in-band signal, the nature of the out-of-band signals, the closeness of the out-of-band signals to the in-band signal, and so forth. An ADC  244  may then digitize the output of ABF  242 . A decimation filter (RCF)  246  may then be used to reduce the number of digitized samples of the output of ADC  244 . The digitized and decimated baseband signal may then be provided to a baseband unit for further processing. 
     Receiver  210  also includes a first signal generator (SIG GEN I)  248  that may be used to generate an out-of-band signal for use in performing corner frequency compensation. In general, a single signal generator may be used to generate the various out-of-band signals to be used in corner frequency compensation. However, logically, a different signal generator may be used to generate each of the various out-of-band signals. Therefore, since the out-of-band signal used in the first signal path may be different than an out-of-band signal used in the second signal path, a second signal generator (SIG GEN Q)  249  may be used to generate the out-of-band signal for the second signal path. 
     The out-of-band signal generated by SIG GEN I  248  may be a cosine wave at a desired frequency and the out-of-band signal generated by SIG GEN Q  249  may be a sine wave at the same desired frequency. SIG GEN I  248  (and SIG GEN Q  249 ) may be implemented as a memory containing a look-up table containing entries descriptive of the out-of-band signal to be generated, for example. Alternatively, an oscillator along with a delay may be used to generate the out-of-band signals needed for the first signal path and the second signal path. Although the discussion focuses on sine and cosine out-of-band signals, other signal forms may be used as the out-of-band signals. Therefore, the discussion of sine and cosine out-of-band signals should not be construed as being limiting to either the scope or the spirit of the embodiments. 
     The out-of-band signal generated by SIG GEN I  248  may be converted into an analog signal by a DAC  250  and then combined (added, for example) with the analog baseband signal produced by mixer  240 . The out-of-band signal in combination with the analog baseband signal may then be filtered by ABF  242 , digitized by ADC  244 , and decimated by RCF  246 . In addition to being provided to the baseband unit, the output of RCF  246  may also be provided to a demodulation unit  252 , which may demodulate the output of RCF  246  with the out-of-band signals, as generated by SIG GEN I  248  and SIG GEN Q  249 . Additionally, a DC estimation unit  254  may be used to detect a maximum and a minimum value of the output of demodulation unit  252 . Collectively, demodulation unit  252  and DC estimation unit  254  may be used to measure a phase response of ABF  242 . The maximum and the minimum values of the output of demodulation unit  252  may be provided to a processor, such as a script processor, that may make use of the maximum and the minimum values to compute the corner frequency of ABF  242  and make any adjustments to ABF  242  as needed. 
     Like demodulation unit  252  and DC estimation unit  254 , a demodulation unit  258  and a DC estimation unit  260  may be used to measure a phase response of an ABF in the second signal path. Maximum and minimum values of the output of demodulation unit  258  may also be provided to the processor, where the corner frequency of the ABF in the second signal path may be computed and any adjustments to the ABF may be performed if needed. 
     Although shown in  FIG. 2   b  as having a separate demodulation unit and DC estimation unit for each signal path, it may be possible to utilize a single demodulation unit and DC estimation unit for both signal paths. The two signal paths may share the demodulation unit and the DC estimation unit. This may lead to a reduction in hardware requirements, thereby potentially reducing complexity and cost. 
       FIG. 2   c  is a diagram illustrating a detailed view of demodulation unit  252  and DC estimation unit  254 . Although  FIG. 2   c  illustrates a detailed view of demodulation unit  252  and DC estimation unit  254 , demodulation unit  258  and DC estimation unit  260  may be substantially similar. Therefore, the discussion of demodulation unit  252  and DC estimation unit 254 also applies to demodulation unit  258  and DC estimation unit  260 . 
     Demodulation unit  252  comprises a first multiplier  270  and a second multiplier  271 . First multiplier  270  may be used to demodulate the output of RCF  246  with the out-of-band signal produced by SIG GEN I  248 , while second multiplier  271  may be used to demodulate the output of RCF  246  with the out-of-band signal produced by SIG GEN Q  249 . 
     DC estimation unit  254  includes a first filter  274  and a second filter  275 , preferably low-pass filters, to remove unwanted high-frequency components from outputs of first multiplier  270  and second multiplier  271 , respectively. First filter  274  and second filter  275  may be used to produce DC signals from outputs of first multiplier  270  and second multiplier  271 . Filter characteristics of first filter  274  and second filter  275  may be similar and should be set so that as much of non-DC signals present in the outputs of first multiplier  270  and second multiplier  271  as possible are eliminated (or attenuated). 
     After filtering, a first min/max detector  278  and a second min/max filter  279  may be used to select values (minimum and maximum output values) from the outputs of first filter  274  and second filter  275 . First min/max detector  278  and second min/max detector  279  may be implemented comparators and memories, with the comparators used to compare a current output of a filter (either first filter  274  or second filter  275 ) with minimum and maximum values stored in the memories. Outputs from first min/max detector  278  and second detector  279  may be provided to a processor that may be used to compute the corner frequency of ABF  242  and make necessary adjustments if needed. 
       FIG. 3   a  is a flow diagram illustrating a flow chart  300  in the computing of a corner frequency of a filter using out-of-band signals. Flow chart  300  may be descriptive of operations taking place in a wireless receiver performing corner frequency compensation. The computing of the corner frequency of a filter, such as an ABF, may be performed periodically. Alternatively, the computing of the corner frequency may occur when a specified event occurs. For example, the computing of the corner frequency may occur prior to receiving a transmission, when performance of the wireless transceiver reaches a threshold, when measured performance metrics reach a threshold, when a request to perform corner frequency compensation is received, a detected change in operating conditions, a detected change in supply voltage, and so forth. 
     The computing of the corner frequency may begin with an injection of out-of-band signals (block  305 ). Referencing  FIG. 2   b , the out-of-band signals may be injected by SIG GEN I  248  and SIG GEN Q  249 . Examples of out-of-band signals may include cosine waves, sine waves, and so forth. Without loss of generality, let SIG GEN I  248  inject a cosine wave defined as d 0  cos(2πf IF nT) and SIG GEN Q  249  inject a sine wave defined as d 0  sin(2πf IF nT). 
     After the out-of-band signals have been injected, a phase response of the ABFs of wireless receiver  200 , such as ABF  242  and a corresponding ABF of the second signal path, may be measured (block  310 ). In order to simplify measurement of the phase response, the following assumptions are made: 1) a phase shift introduced by ABF  242  is φ I ; 2) a phase shift introduced by ADC  244  is zero (typically true if f IF  is not significantly larger than the corner frequency of ABF  242 ); 3) a phase shift of RCF  246  is known and is zero (also typically true if f IF  is not significantly larger than the corner frequency of ABF  242 ); 4) a phase shift of DAC  250  is zero. Similar assumptions are also made for the second signal path. 
     If the phase shift due to ADC  244 , RCF  246 , and/or DAC  250  is non-zero, then the phase shift may be measured using a technique similar to what is described below for measuring the phase shift for use in computing the corner frequency, but with adjustments made to ABF  242  so that the corner frequency of ABF  242  is at its maximum value. By maximizing the corner frequency of ABF  242 , the phase shift introduced by ABF  242  may be negligible. The measured phase shift due to ADC  244 , RCF  246 , and/or DAC  250  may then be removed (for example, subtracted) from a measured phase shift due to ABF  242  and ADC  244 , RCF  246 , and/or DAC  250 . This may then yield the phase shift due to ABF  242  alone.  FIG. 3   b  is a flow diagram illustrating a flow chart  350  in the computing of a corner frequency of a filter using out-of-band signals, wherein the phase shift due to ADC  244 , RCF  246 , and/or DAC  250  is non-zero. 
     Denote the output of RCF  246  as d 1  cos(2πf IF nT+θ I ). From θ I , θ I  may be computed as being equal to either θ I  or θ I  minus the phase shift due to DAC  250 , RCF  246 , and/or ADC  224  (as discussed above). 
     Referencing  FIG. 2   c , an output of multiplier  270  is expressible as: 
     
       
         
           
             
               
                 
                   
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     and an output of multiplier  271  is expressible as: 
     
       
         
           
             
               
                 
                   
                     M 
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                       . 
                     
                   
                 
               
             
           
         
       
     
     After first filter  274  and second filter  275  remove high frequency terms from the above, they become DC signals expressible as: 
     
       
         
           
             
               L 
               I 
             
             = 
             
               
                 1 
                 2 
               
                
               
                 d 
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     L I  and L Q  may then be provided to first min/max detector  278  and second min/max detector  279  and the phase of the output of RCF  246  may be solved, 
     
       
         
           
             
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     As such, the phase response of ABF  242  φ I  may be measured at frequency f IF . 
     Similarly, the measuring of the phase response of the ABF in the second signal path, φ Q , at frequency f IF  with the output of SIG GEN Q  249  as d 0  sin(2πf IF nT) may also be performed. Let an output of an RCF in the second signal path be expressible as d 2  sin(2πf IF nT+θ Q ). Then, an output of a multiplier in an I signal path, similar to multiplier  270 , is expressible as: 
     
       
         
           
             
               
                 
                   
                     
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                            
                           π 
                            
                           
                               
                           
                            
                           
                             f 
                             IF 
                           
                            
                           nT 
                         
                         ) 
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                   
                     
                       1 
                       2 
                     
                      
                     
                       d 
                       0 
                     
                      
                     
                       
                         d 
                         2 
                       
                        
                       
                         [ 
                         
                           
                             sin 
                              
                             
                               ( 
                               
                                 θ 
                                 Q 
                               
                               ) 
                             
                           
                           + 
                           
                             sin 
                              
                             
                               ( 
                               
                                 
                                   4 
                                    
                                   π 
                                    
                                   
                                       
                                   
                                    
                                   
                                     f 
                                     IF 
                                   
                                    
                                   nT 
                                 
                                 + 
                                 
                                   θ 
                                   Q 
                                 
                               
                               ) 
                             
                           
                         
                         ] 
                       
                     
                   
                 
               
             
           
         
       
     
     and an output of a multiplier in a Q signal path, similar to multiplier  271 , is expressible as: 
     
       
         
           
             
               
                 
                   
                     
                       M 
                       ~ 
                     
                     Q 
                   
                   = 
                   
                     
                       d 
                       2 
                     
                      
                     
                       sin 
                        
                       
                         ( 
                         
                           
                             2 
                              
                             π 
                              
                             
                                 
                             
                              
                             
                               f 
                               IF 
                             
                              
                             nT 
                           
                           + 
                           
                             θ 
                             Q 
                           
                         
                         ) 
                       
                     
                      
                     
                       d 
                       0 
                     
                      
                     
                       sin 
                        
                       
                         ( 
                         
                           2 
                            
                           π 
                            
                           
                               
                           
                            
                           
                             f 
                             IF 
                           
                            
                           nT 
                         
                         ) 
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                   
                     
                       1 
                       2 
                     
                      
                     
                       d 
                       0 
                     
                      
                     
                       
                         
                           d 
                           2 
                         
                          
                         
                           [ 
                           
                             
                               cos 
                                
                               
                                 ( 
                                 
                                   θ 
                                   Q 
                                 
                                 ) 
                               
                             
                             + 
                             
                               cos 
                                
                               
                                 ( 
                                 
                                   
                                     4 
                                      
                                     π 
                                      
                                     
                                         
                                     
                                      
                                     
                                       f 
                                       IF 
                                     
                                      
                                     nT 
                                   
                                   + 
                                   
                                     θ 
                                     Q 
                                   
                                 
                                 ) 
                               
                             
                           
                           ] 
                         
                       
                       . 
                     
                   
                 
               
             
           
         
       
     
     After filters remove high frequency terms, the remaining DC signals expressible as: 
     
       
         
           
             
               
                 L 
                 ~ 
               
               I 
             
             = 
             
               
                 1 
                 2 
               
                
               
                 d 
                 0 
               
                
               
                 d 
                 2 
               
                
               
                 sin 
                  
                 
                   ( 
                   
                     θ 
                     Q 
                   
                   ) 
                 
               
             
           
         
       
       
         
           
             
               
                 L 
                 ~ 
               
               Q 
             
             = 
             
               
                 1 
                 2 
               
                
               
                 d 
                 0 
               
                
               
                 d 
                 2 
               
                
               
                 
                   cos 
                    
                   
                     ( 
                     
                       θ 
                       Q 
                     
                     ) 
                   
                 
                 . 
               
             
           
         
       
     
     {tilde over (L)} I  and {tilde over (L)} Q  may then be measured by min/max detectors and the phase of the output of the RCF in the second signal path, θ Q , may be solved, 
     
       
         
           
             
               θ 
               Q 
             
             = 
             
               
                 
                   tan 
                   
                     - 
                     1 
                   
                 
                  
                 
                   ( 
                   
                     
                       
                         L 
                         ~ 
                       
                       I 
                     
                     
                       
                         L 
                         ~ 
                       
                       Q 
                     
                   
                   ) 
                 
               
               . 
             
           
         
       
     
     As such, the phase response of the ABF in the second signal path, φ Q , may be measured at frequency f IF . 
     Referencing back to  FIG. 3   a , after measuring the phase response of ABF  242  and a corresponding ABF in the second signal path (block  310 ), the corner frequency of ABF  242  and the corresponding ABF in the second signal path may be computed using the measured phase responses (block  315 ). 
     The computing of the corner frequency may be dependent on the order of the ABF. For example, if the ABF is a first order low-pass filter with a transfer function expressible as: 
     
       
         
           
             
               
                 H 
                  
                 
                   ( 
                   s 
                   ) 
                 
               
               = 
               
                 
                   G 
                   0 
                 
                 
                   1 
                   + 
                   
                     j 
                      
                     
                       f 
                       
                         f 
                         c 
                       
                     
                   
                 
               
             
             , 
           
         
       
     
     where f c  is the corner frequency of the ABF and G 0  is the magnitude response at DC, then the phase response of the ABF is expressible as: 
     
       
         
           
             ϕ 
             = 
             
               - 
               
                 
                   
                     tan 
                     
                       - 
                       1 
                     
                   
                    
                   
                     ( 
                     
                       f 
                       
                         f 
                         c 
                       
                     
                     ) 
                   
                 
                 . 
               
             
           
         
       
     
     Therefore, if the measured phase response at f I  is equal to φ 1 , then the corner frequency may be computed as: 
     
       
         
           
             
               f 
               c 
             
             = 
             
               
                 
                   f 
                   1 
                 
                 
                   tan 
                    
                   
                     ( 
                     
                       ϕ 
                       1 
                     
                     ) 
                   
                 
               
               . 
             
           
         
       
     
     The estimated term in the phase response measurement is tan(φ 1 ), not φ 1 ). This may imply that in calculating f c , mathematical functions, such as tan and tan −1 , do not have to be used. Since the mathematical functions, such as tan and tan −1 , tend to be sensitive to noise, numerical errors, and so forth, the estimation of tan(φ 1 ) and the computing of the corner frequency may be less affected by noise, numerical errors, and so on. 
     If the ABF is a second-order low-pass filter with a transfer function expressible as: 
     
       
         
           
             
               H 
                
               
                 ( 
                 s 
                 ) 
               
             
             = 
             
               
                 
                   G 
                   0 
                 
                 
                   
                     
                       ( 
                       
                         s 
                         
                           ω 
                           n 
                         
                       
                       ) 
                     
                     2 
                   
                   + 
                   
                     
                       1 
                       Q 
                     
                      
                     
                       ( 
                       
                         s 
                         
                           ω 
                           n 
                         
                       
                       ) 
                     
                   
                   + 
                   1 
                 
               
               . 
             
           
         
       
     
     Then the phase response of the ABF is expressible as: 
     
       
         
           
             
               ϕ 
               = 
               
                 - 
                 
                   
                     tan 
                     
                       - 
                       1 
                     
                   
                   ( 
                   
                     
                       
                         1 
                         Q 
                       
                        
                       
                         ( 
                         
                           ω 
                           
                             ω 
                             n 
                           
                         
                         ) 
                       
                     
                     
                       1 
                       - 
                       
                         
                           ( 
                           
                             ω 
                             
                               ω 
                               n 
                             
                           
                           ) 
                         
                         2 
                       
                     
                   
                   ) 
                 
               
             
             , 
           
         
       
     
     where ω n  is the natural frequency. 
     Assuming that the phase response at two out-of-band frequencies have been measured (φ 1  and φ 2  at f 1  and f 2 , respectively), then: 
     
       
         
           
             
               ϕ 
               1 
             
             = 
             
               - 
               
                 
                   tan 
                   
                     - 
                     1 
                   
                 
                  
                 
                   ( 
                   
                     
                       
                         1 
                         Q 
                       
                        
                       
                         ( 
                         
                           
                             ω 
                             1 
                           
                           
                             ω 
                             n 
                           
                         
                         ) 
                       
                     
                     
                       1 
                       - 
                       
                         
                           ( 
                           
                             
                               ω 
                               1 
                             
                             
                               ω 
                               n 
                             
                           
                           ) 
                         
                         2 
                       
                     
                   
                   ) 
                 
               
             
           
         
       
       
         
           
             
               
                 ϕ 
                 2 
               
               = 
               
                 - 
                 
                   
                     tan 
                     
                       - 
                       1 
                     
                   
                    
                   
                     ( 
                     
                       
                         
                           1 
                           Q 
                         
                          
                         
                           ( 
                           
                             
                               ω 
                               2 
                             
                             
                               ω 
                               n 
                             
                           
                           ) 
                         
                       
                       
                         1 
                         - 
                         
                           
                             ( 
                             
                               
                                 ω 
                                 2 
                               
                               
                                 ω 
                                 n 
                               
                             
                             ) 
                           
                           2 
                         
                       
                     
                     ) 
                   
                 
               
             
             , 
           
         
       
     
     where ω 1 =2πf 1  and ω 2 =2πf 2 . From the above two equations, two unknowns Q and ω n  (filter parameters) need to be solved. Once solved, the corner frequency may then be computed from Q and ω n . 
     To solve for the two unknowns Q and ω n , rewrite the equations for φ 1  and φ 2  above as: 
     
       
         
           
             
               
                 - 
                 
                   tan 
                    
                   
                     ( 
                     
                       ϕ 
                       1 
                     
                     ) 
                   
                 
               
               = 
               
                 
                   
                     
                       
                         1 
                         Q 
                       
                        
                       
                         ( 
                         
                           
                             ω 
                             1 
                           
                           
                             ω 
                             n 
                           
                         
                         ) 
                       
                     
                     
                       1 
                       - 
                       
                         
                           ( 
                           
                             
                               ω 
                               1 
                             
                             
                               ω 
                               n 
                             
                           
                           ) 
                         
                         2 
                       
                     
                   
                    
                   
                     
 
                   
                   - 
                   
                     tan 
                      
                     
                       ( 
                       
                         ϕ 
                         2 
                       
                       ) 
                     
                   
                 
                 = 
                 
                   
                     
                       
                         1 
                         Q 
                       
                        
                       
                         ( 
                         
                           
                             ω 
                             2 
                           
                           
                             ω 
                             n 
                           
                         
                         ) 
                       
                     
                     
                       1 
                       - 
                       
                         
                           ( 
                           
                             
                               ω 
                               2 
                             
                             
                               ω 
                               n 
                             
                           
                           ) 
                         
                         2 
                       
                     
                   
                   . 
                   
                     
 
                   
                    
                   Then 
                 
               
             
             , 
             
               
 
             
              
             
               
                 
                   tan 
                    
                   
                     ( 
                     
                       ϕ 
                       1 
                     
                     ) 
                   
                 
                 
                   tan 
                    
                   
                     ( 
                     
                       ϕ 
                       2 
                     
                     ) 
                   
                 
               
               = 
               
                 
                   
                     1 
                     - 
                     
                       
                         ( 
                         
                           
                             ω 
                             2 
                           
                           
                             ω 
                             n 
                           
                         
                         ) 
                       
                       2 
                     
                   
                   
                     1 
                     - 
                     
                       
                         ( 
                         
                           
                             ω 
                             1 
                           
                           
                             ω 
                             n 
                           
                         
                         ) 
                       
                       2 
                     
                   
                 
                  
                 
                   
                     ω 
                     1 
                   
                   
                     ω 
                     2 
                   
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     tan 
                      
                     
                       ( 
                       
                         ϕ 
                         1 
                       
                       ) 
                     
                   
                   
                     tan 
                      
                     
                       ( 
                       
                         ϕ 
                         2 
                       
                       ) 
                     
                   
                 
                  
                 
                   
                     ω 
                     2 
                   
                   
                     ω 
                     1 
                   
                 
               
               = 
               
                 
                   
                     
                       
                         
                           ω 
                           n 
                           2 
                         
                         - 
                         
                           ω 
                           2 
                           2 
                         
                       
                       
                         
                           ω 
                           n 
                           2 
                         
                         - 
                         
                           ω 
                           1 
                           2 
                         
                       
                     
                     . 
                     
                       
 
                     
                      
                     Let 
                   
                    
                   
                       
                   
                    
                   k 
                 
                 = 
                 
                   
                     
                       tan 
                        
                       
                         ( 
                         
                           ϕ 
                           1 
                         
                         ) 
                       
                     
                     
                       tan 
                        
                       
                         ( 
                         
                           ϕ 
                           2 
                         
                         ) 
                       
                     
                   
                    
                   
                     
                       ω 
                       2 
                     
                     
                       ω 
                       1 
                     
                   
                 
               
             
             , 
             then 
           
         
       
       
         
           
             
               
                 ω 
                 n 
                 2 
               
               - 
               
                 ω 
                 2 
                 2 
               
             
             = 
             
               k 
                
               
                 ( 
                 
                   
                     ω 
                     n 
                     2 
                   
                   - 
                   
                     ω 
                     1 
                     2 
                   
                 
                 ) 
               
             
           
         
       
       
         
           
             
               ω 
               n 
               2 
             
             = 
             
               
                 
                   k 
                    
                   
                       
                   
                    
                   
                     ω 
                     1 
                     2 
                   
                 
                 - 
                 
                   ω 
                   2 
                   2 
                 
               
               
                 k 
                 - 
                 1 
               
             
           
         
       
       
         
           
             
               ω 
               n 
             
             = 
             
               
                 
                   
                     
                       k 
                        
                       
                           
                       
                        
                       
                         ω 
                         1 
                         2 
                       
                     
                     - 
                     
                       ω 
                       2 
                       2 
                     
                   
                   
                     k 
                     - 
                     1 
                   
                 
               
               . 
             
           
         
       
     
     It then follows that 
     
       
         
           
             Q 
             = 
             
               
                 1 
                 
                   
                     tan 
                      
                     
                       ( 
                       
                         ϕ 
                         1 
                       
                       ) 
                     
                   
                    
                   
                     ( 
                     
                       
                         
                           ω 
                           1 
                           2 
                         
                         
                           ω 
                           n 
                           2 
                         
                       
                       - 
                       1 
                     
                     ) 
                   
                    
                   
                     
                       ω 
                       n 
                     
                     
                       ω 
                       1 
                     
                   
                 
               
               . 
             
           
         
       
     
     At the corner frequency of the second-order ABF, 
     
       
         
           
             
               
                  
                 
                   
                     j 
                      
                     
                       
                         ω 
                         c 
                       
                       
                         ω 
                         n 
                       
                     
                      
                     
                       1 
                       Q 
                     
                   
                   + 
                   1 
                   - 
                   
                     
                       ω 
                       c 
                       2 
                     
                     
                       ω 
                       n 
                       2 
                     
                   
                 
                  
               
               = 
               
                 2 
               
             
             , 
           
         
       
     
     where ω c =2πf c . This implies that: 
     
       
         
           
             
               
                 
                   ( 
                   
                     1 
                     - 
                     
                       
                         ( 
                         
                           
                             ω 
                             c 
                           
                           
                             ω 
                             n 
                           
                         
                         ) 
                       
                       2 
                     
                   
                   ) 
                 
                 2 
               
               + 
               
                 
                   
                     ( 
                     
                       
                         ω 
                         c 
                       
                       
                         ω 
                         n 
                       
                     
                     ) 
                   
                   2 
                 
                  
                 
                   1 
                   
                     Q 
                     2 
                   
                 
               
             
             = 
             2. 
           
         
       
     
     Let  
     
       
         
           
             
               x 
               = 
               
                 
                   ( 
                   
                     
                       ω 
                       c 
                     
                     
                       ω 
                       n 
                     
                   
                   ) 
                 
                 2 
               
             
             , 
           
         
       
     
     then 
     
       
         
           
             
               
                 
                   ( 
                   
                     1 
                     - 
                     x 
                   
                   ) 
                 
                 2 
               
               + 
               
                 x 
                 
                   Q 
                   2 
                 
               
             
             = 
             2 
           
         
       
       
         
           
             
               
                 x 
                 2 
               
               + 
               
                 
                   ( 
                   
                     
                       1 
                       
                         Q 
                         2 
                       
                     
                     - 
                     2 
                   
                   ) 
                 
                  
                 x 
               
               - 
               1 
             
             = 
             0. 
           
         
       
     
     The above equation yields a solution 
     
       
         
           
             x 
             = 
             
               
                 
                   
                     ( 
                     
                       2 
                       - 
                       
                         1 
                         
                           Q 
                           2 
                         
                       
                     
                     ) 
                   
                   + 
                   
                     
                       
                         
                           ( 
                           
                             2 
                             - 
                             
                               1 
                               
                                 Q 
                                 2 
                               
                             
                           
                           ) 
                         
                         2 
                       
                       + 
                       4 
                     
                   
                 
                 2 
               
               . 
             
           
         
       
     
     Therefore,  
     
       
         
           
             
               ω 
               c 
             
             = 
             
               
                 
                   x 
                 
                  
                 
                   ω 
                   n 
                 
                  
                 
                     
                 
                  
                 and 
                  
                 
                     
                 
                  
                 
                   f 
                   c 
                 
               
               = 
               
                 
                   
                     
                       x 
                     
                      
                     
                       ω 
                       n 
                     
                   
                   
                     2 
                      
                     
                         
                     
                      
                     π 
                   
                 
                 . 
               
             
           
         
       
     
     Furthermore, as discussed above, the avoidance of using math functions, such as tan and tan −1 , may make the proposed approach less sensitive to the noise and numerical errors in estimating the corner frequency. 
     In general, for an n-th order ABF, measurements of the phase response of the ABF may be made at n different out-of-band frequencies. The n phase responses may then be used to solve for n different unknowns, which in turn, may be used to compute the corner frequency. In order to help ensure that the assumptions made above remain valid, the n different out-of-band frequencies should be relatively close to the corner frequency, however, they should be far enough apart so that they are not substantially a single frequency. 
       FIG. 4  is a flow diagram illustrating a sequence of events  400  in the compensating of a corner frequency of a filter using out-of-band signals. Sequence of events  400  may be descriptive of operations taking place in a wireless receiver performing corner frequency compensation. The compensation of the corner frequency of a filter, such as an ABF, may be performed periodically. Alternatively, the computing of the corner frequency may occur when a specified event occurs. For example, the computing of the corner frequency may occur prior to receiving a transmission, when performance of the wireless transceiver reaches a threshold, when measured performance metrics reach a threshold, when a request to perform corner frequency compensation is received, a detected change in operating conditions, a detected change in supply voltage, and so forth. 
     The corner frequency compensation may begin with an injection of out-of-band signals (block  405 ). In general, the number of out-of-band signals injected may be dependent on the ABF&#39;s order. For an n-th order ABF used in a Cartesian wireless receiver, n pairs of out-of-band signals may be injected. Each pair of out-of-band signals comprises an out-of-band signal for an I-phase signal path and a Q-phase signal path. For example, a pair of out-of-band signals may include a cosine wave and a sine wave, with the cosine wave and the sine wave having the same frequency. 
     After injecting a pair of out-of-band signal at a frequency, measurements of the ABF&#39;s phase response at the frequency of the out-of-band signals may be made (block  410 ). Generally, if multiple pairs of out-of-band signals are to be used, a pair of out-of-band signals may be injected one at a time and the measurement of the ABF&#39;s phase response made before another pair is injected. 
     If multiple pairs of out-of-band signals are to be injected, the injecting (block  405 ) and the measuring (block  410 ) may be repeated for each pair of out-of-band signals. Once all of the pairs of out-of-band signals have been injected and the phase response of the ABF has been measured, then the corner frequency of the ABF may be computed using the measured phase response (block  415 ). 
     The computed corner frequency of the ABF may then be compared with a desired corner frequency of the ABF (block  420 ). If the computed corner frequency and the desired corner frequency are equal or differ by less than a threshold, then the corner frequency does not need compensation and may terminate. However, if the computed corner frequency and the desired corner frequency differ by more than the threshold, then corner frequency compensation is needed. The corner frequency compensation may be performed by adjusting capacitor, resistor, or inductor values in the ABF (block  425 ). The capacitor, resistor, or inductor values may be adjusted by switching in and/or out different capacitors/resistors/inductors in the ABF, digitally adjusting capacitor/resistor/inductor values based on a control word, and so forth. 
     Since in a Cartesian wireless receiver there are two signal paths (the I-phase and Q-phase signal paths), any adjustments should be made to both signal paths at substantially the same time. By making the adjustments to both signal paths at the same time, potential mismatches in the two signal paths may be minimized. Additionally, the injection of the out-of-band signals should also occur at substantially the same time, again to minimize potential mismatches in the two signal paths. 
     After adjusting the ABF (block  425 ), the corner frequency compensation may be repeated to determine if the adjustments are sufficient. The corner frequency compensation may be repeated until the computed corner frequency differs from the desired corner frequency by less than the threshold. 
     In addition to using the computed corner frequency to compare and then adjust the ABF, other relevant parameters of the transfer function of the ABF (such as Q and ω n ) may also be used to compare with their desired values to guide and assist in the tuning of the ABF. 
     Although the embodiments and their advantages have been described in detail, it should be understood that various changes, substitutions and alterations can be made herein without departing from the spirit and scope of the invention as defined by the appended claims. Moreover, the scope of the present application is not intended to be limited to the particular embodiments of the process, machine, manufacture, composition of matter, means, methods and steps described in the specification. As one of ordinary skill in the art will readily appreciate from the disclosure of the present invention, processes, machines, manufacture, compositions of matter, means, methods, or steps, presently existing or later to be developed, that perform substantially the same function or achieve substantially the same result as the corresponding embodiments described herein may be utilized according to the present invention. Accordingly, the appended claims are intended to include within their scope such processes, machines, manufacture, compositions of matter, means, methods, or steps.