Abstract:
A method and apparatus for digitizing a signal. The method comprises the steps of receiving an input analog signal, splitting the received input analog signal into a plurality of signals and frequency converting at least one of the signals in accordance with a predetermined periodic function having a predetermined frequency. The signals are then digitized and combined mathematically to form a single output stream that is a substantially correct representation of the original input analog signal.

Description:
CROSS-REFERENCE TO RELATED APPLICATIONS 
     This application is a continuation of U.S. patent application Ser. No. 10/693,188 filed Oct. 24, 2003 now U.S. Pat. No. 7,058,548, currently pending, which in turn claims the benefit of U.S. Provisional Patent Application 60/420,937 filed Oct. 24, 2002. The entire contents of each of these prior applications are incorporated herein by reference. 
    
    
     FIELD OF THE INVENTION 
     The present invention relates to a high bandwidth real-time digital storage oscilloscope (DSO) incorporating heterodyning to increase the bandwidth of a typical oscilloscope design. 
     BACKGROUND OF THE INVENTION 
     A digital storage oscilloscope (DSO) is the primary tool utilized by engineers to view electrical signals. As signals get ever faster, it is very beneficial to have DSOs capable of digitizing, displaying and analyzing these faster signals. The capability of a DSO to digitize fast signals is determined by its bandwidth and sample rate. 
     Because one of the uses of the DSO is to design and analyze new electronic devices, DSOs employed in the design of high speed electronics must operate at a bandwidth sufficient to capture and display signals from these devices. Providing ever-faster sampling circuitry is one way to increase bandwidth, but is subject to practical limitations in hardware development. There are a number of other methods that have been traditionally employed in an attempt to provide additional bandwidth. One of the more common methods is called equivalent time sampling and is based upon triggering repeatedly on a periodic event. Data from multiple trigger events can be combined to provide a more complete and accurate view of the waveform. This technique is employed in a conventional sampling oscilloscope. A sampling oscilloscope repeatedly triggers on an event and acquires only a few points of the periodically repeating waveform (sometimes only one point of the waveform) in response to each trigger event. After repeated triggers, the acquired data points are combined to display a high effective sampling rate representation of the waveform. However, such a sampling oscilloscope requires a repetitive input signal so that the representation of the waveform can be generated over many triggers. This method also makes certain desirable types of analyses of the waveform, such as cycle-to-cycle jitter, impossible. 
     A common problem in complex signal analysis is that a non-repetitive signal must be analyzed. In fact, it is very often the case that a non-repetitive event is the cause of some failure in an electronic system. It is the function of the test equipment to help the user identify the cause of the failure. Therefore, a piece of test equipment that requires repetitive signals may be of limited usefulness. Situations when the user wants to capture or look at infrequent or non-repetitive events require a DSO capable of operating with high bandwidth and sample rate in response to a single trigger. Such a DSO that allows for acquisition of a signal based upon only a single trigger event is called a real-time oscilloscope. Acquisitions taken utilizing only a single trigger event are called single-shot acquisitions. 
     While architectural techniques, such as channel combination, interleaving, and the like are generally available for designing high sample rate real time oscilloscopes, designing real time oscilloscopes that simultaneously achieve high bandwidth and high sample rate is another issue. Increasing bandwidth in such a real time oscilloscope is typically dealt with through direct application of ever faster, very good high-speed electronics. Increasing the sample rate is generally achieved by the application of various interleaving techniques. As noted above, a real-time DSO is more useful in certain situations than a sampling oscilloscope, even though a sampling oscilloscope provides high bandwidth and effective sampling rate. This is because unlike a sampling oscilloscope, a real time DSO does not require the input signal to be repetitive. However, in order to allow for acquisition of enough sampling points in real time to accurately reproduce the acquired signal, the bandwidth of the real-time scope may be limited. 
     Even with current design techniques it is difficult to achieve simultaneously very high bandwidths and sampling rates. 
     SUMMARY OF THE INVENTION 
     This invention pertains generally to systems that digitize waveforms; and more specifically systems that convert an analog input signal to a digital output signal whereby the digital signal consists of an array of numbers that represent the amplitude of the analog waveform at known times. This invention also pertains to systems with limited bandwidth where there is a need for higher bandwidth. This invention therefore addresses systems incapable of accurately digitizing very rapidly changing signals. 
     The most specific application of this invention is to real-time Digital Storage Oscilloscopes (DSOs) where high demands are placed on the bandwidth of signals digitized in a single-shot acquisition. 
     The invention accordingly comprises, in a preferred embodiment, a method and apparatus for digitizing a signal. The method comprises the steps of receiving an input analog signal and splitting the received input analog signal into a plurality of band limited signals. The frequency of at least one of the band limited signals is frequency converted in accordance with a predetermined periodic function having a predetermined frequency. After frequency conversion, the band limited signals are digitized and subsequently combined mathematically to form a single output digital datastream that is a substantially correct representation of the original input signal. 
     It is therefore an object of the invention to overcome the drawbacks of the prior art. 
     Still other objects and advantages of the invention will in part be obvious and will in part be apparent from the specification and the drawings. 
     The invention accordingly comprises the several steps and the relation of one or more of such steps with respect to each of the others, and the apparatus embodying features of construction, combinations of elements and arrangement of parts that are adapted to effect such steps, all as exemplified in the following detailed disclosure, and the scope of the invention will be indicated in the claims. 
    
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
       For a more complete understanding of the invention, reference is made to the following description and accompanying drawings, in which: 
         FIG. 1  is a block diagram depicting a digitizing system constructed in accordance with the invention; 
         FIG. 2  is a block diagram showing one possible method of the extension of this technique to 4 channels using a mixing frequency that is at the low side of the frequency band of interest (low side conversion); 
         FIG. 3  is a block diagram showing another possible method of the extension of this technique to 4 channels using a mixing frequency that is at the high side of the frequency band of interest (high side conversion); 
         FIGS. 4–57  are useful in explaining the advantages obtained by the present invention. 
     
    
    
     DETAILED DESCRIPTION OF THE PREFERRED EMBODIMENTS 
       FIG. 1  is a block diagram showing a high bandwidth digital storage oscilloscope (DSO) architecture according to a first embodiment of the present invention.  FIG. 1  shows two channels of a DSO combined to digitize waveforms in a manner that effectively doubles the system bandwidth. It should be understood that the bandwidth can be tripled, quadrupled etc. by utilizing three, four or more channels in combination. There is no inherent limitation on the number of channels that may be utilized. 
     As is shown in  FIG. 1  an input signal  100  is provided at the input. Viewed from the frequency-domain perspective, the input signal might have a frequency content shown at  103 . In a standard configuration input signal  100  directly enters a first channel, CH 1  at  104 . This signal passes through an analog front end  105 , and on to an ADC  106  which digitizes the waveform. The channel has a finite bandwidth, as shown by  108  which results in a digitized waveform of finite bandwidth  107 . For the purpose of future explanation, the cutoff frequency at which the bandwidth is limited is designated as F. 
     The first embodiment of the present invention involves the addition of analog circuitry  112  and  102  between the input  100  and the two channels  104  and  109  and downstream processing of the digital data to work with this additional hardware. In accordance with the present invention, the signal at input  100  with example frequency content  103  enters a power splitter  112 , or power amplifier, diplexer, or the like, as appropriate. The splitter  112  provides a termination to the input signal and provides at its two equally attenuated outputs the same signal attenuated by a known amount. One output of splitter  112  directly connects to CH 1  channel ( 104 ). The signal proceeds through front-end  105  and is digitized in a conventional manner by ADC  106  as described above. Because the combination of front-end  105  and ADC  106  is bandwidth limited, as shown in  108 , this results in a bandwidth limited acquisition with frequency content shown by  107 . CH 1  channel ( 104 ) is designated as containing the low frequency (LF) portion of the signal. Note, however, that the high and low frequency signals could be switched, therefore processing the low frequency signal and passing the high frequency signal through without the processing to be described below. The path through the other output of splitter  112  will be described below and is designated as the high frequency (HF) path. 
     The invention only adds the power splitter or diplexer directly in the signal path of CH 1  channel ( 104 ). In other words, the low frequency signal path with the splitter removed looks identical to the signal path of the DSO not utilizing this invention. Very high quality splitters and diplexers with very high bandwidth are readily commercially available and as such, do not serve to degrade the signal path. 
     The HF signal path will now be explained in greater detail. It is well known that frequencies can be translated through the use of a process called heterodyning. Heterodyning is achieved through the time-domain multiplication of a signal with a sinusoidal signal, resulting in a frequency converted signal. It is well known that if a sinusoid with a frequency f 0  is mixed with another sinusoid with a frequency f 1 , the result is two sinusoids at sum and difference frequencies (i.e. sinusoids at frequencies f 0 +f 1  and f 0 −f 1  which will be referred to as images) with each sinusoid being half the amplitude of the product of the amplitudes of f 0  and f 1 : 
                 [       A   0     ·     cos   ⁡     (     2   ·   π   ·     f   0       )         ]     ·     [       A   1     ·     cos   ⁡     (     2   ·   π   ·     f   1       )         ]       →           A   0     ·     A   1       2     ·     [       cos   ⁡     (     2   ·   π   ·     (       f   0     +     f   1       )       )       +     cos   ⁡     (     2   ·   π   ·     (       f   0     -     f   1       )       )         ]             
In  FIG. 1 , the signal from the second output of splitter  112  (signal  101 ) enters a high pass filter  113 . High-pass filter  113  is designed to reject to the greatest extent possible all frequencies below frequency F as shown in  114 . The result of high pass filtering the input signal with frequency content as shown in  103  is shown in  115 . The output of  113  is heterodyned with a sinusoid at frequency F utilizing mixer  116 . The result of mixing the signal with frequency content shown by  115  with the sinusoidal signal having a frequency F shown in  117  is the frequency content shown in  118 .  118  shows that two images of the content shown in  115  are produced at the sum and difference frequencies, as noted above. In cases where the cutoff of the high-pass filter is inadequate, thereby allowing overlap between the defined low and high frequency bands, the heterodyning frequency may be chosen slightly higher such that a dead-band is utilized to prevent the low frequency edges of the high-pass filter output from folding back into the pass-band.
 
     The output of mixer  116  connects to CH 2  channel ( 109 )—the high frequency channel. The signal passes through a front-end  110  and is digitized by an ADC  111 . Since like CH 1  channel ( 104 ), the combination of front-end  110  and ADC  111  is bandwidth limited, as shown in  119  this results in a bandwidth limited acquisition with frequency content shown by  120 . The digitizing through ADCs  106  and  111  occur substantially simultaneously in parallel, so channels CH 1  and CH 2  are substantially simultaneously acquired. If the digitizing does not take place in such a substantially simultaneous manner, the resulting digital data may be stored and processed to compensate for any transmission time differences between the two channels. 
     To summarize, the signals seen at the inputs to each of the channels are as follows. The LF CH 1  channel ( 104 ) sees the input waveform directly. The HF CH 2  channel ( 109 ) sees only the frequency content in the frequency band extending upward from F. Through the use of heterodyning, the HF CH 2  channel sees the frequency content of an input signal at frequency F+ΔF at frequency ΔF. Thus, the input signal has been frequency converted to a lower frequency band. Although there were two images (seen in  118 ), the second image is rejected due to the finite channel bandwidth  119 . It should also be noted that part of the first image may also be at least partially rejected depending on the choice of frequencies. Said differently, the LF CH 1  channel acquires the low frequency content of the input signal from 0 to F, while the HF CH 2  channel acquires the frequency content of the input signal from F to 2·F. This HF CH 2  signal is frequency converted to a lower frequency band, namely from frequency band F to 2·F to the frequency band of 0 to F so it fits into the bandwidth of the front end. Thus, the band from DC to 2F (2F frequency content) input signal has been made to fit into two channels each of bandwidth F. 
     Both the LF and HF signals are then digitized. It is assumed that both channels are sampling at a rate consistent with the requirements of a channel having a bandwidth F. Such a sufficiently sampled channel (satisfying the Nyquist Criterion) allows for the complete reconstruction of the signal. Each channel may be upsampled utilizing upsamplers  121  and  122  so that during processing, undesirable portions of the image may be sufficiently separated from the desired image, and therefore rejected. This upsampling is performed utilizing a method such as SinX/X interpolation (or any other well known method for this process) to interpolate between acquired sample points. The method and validity of this and other methods of interpolation are well known to those skilled in the art. 
     The data from the HF (CH 2 ) channel is then digitally heterodyned at a digital heterodyning mixer  123 , preferably numerically by combination with a digitally synthesized periodic signal, such as a sinusoid for example, having substantially the same frequency F ( 124 ) as, and having a substantially fixed phase relationship with, a local oscillator of analog mixer  116  in the analog HF signal path. The result of digital heterodyning mixer  123  is two images of the HF signal shown by  125 . Each image appears respectively at frequencies from 0 to F and from F to 2·F. The first, low frequency image is mirrored about F and is not used. The second image is a replica of the actual high frequency content of the input signal. The output of digital mixer  123  passes through an image reject filter plus equalization  126 , which has a frequency response shown by  127 , for rejection of the lower frequency image. The result is the frequency content shown in  128 . 
     The result at this point is two digital waveforms, one representing the low frequency portion of the input signal  107  and the other representing the high frequency portion of the input signal  128 . 
     Because both signals passed through imperfect channels, they may be equalized separately to compensate for non-ideal magnitude and phase characteristics, including transmission delay of the front-end and digitizing systems. The equalizer for the CH 1  channel ( 129 ) is shown in this example with its response  130  being an ideal low-pass filter. This results in no change between the frequency content shown in  108  and the equalized content shown in  131  with the understanding that this would not necessarily be the case if there were imperfections in the signal  131 . The equalizer for the CH 2  channel is shown integrated with the image reject filter  126 . Both equalizers also have the difficult job of preparing the signals to accommodate the cross-over from LF to HF. 
     Finally, the waveform resulting from the LF channel equalizer  129  is combined with the waveform resulting from the HF channel equalizer  126  by a combiner  132 . The result of this combination is shown by the combination of the LF frequency content in  131  and the HF content in  128 , shown graphically in  133 . Thus, the output  135  is a high-bandwidth, high-sample rate acquisition with the frequency content as shown in  134 . A detailed numerical example stepping through this process is provided below. 
     To summarize the effect, the input signal with frequency content  103  normally would be digitized by one channel to form an acquisition with frequency content shown in  108 . Instead, as a result of this embodiment of the invention, two channels are utilized and the resulting acquisition has twice the bandwidth as demonstrated by the signal frequency content shown in  134 . Note that the two other benefits of traditional time interleaving—that of doubling the sample rate and allocation of two channels of memory to one signal—are still achieved by this invention. 
     Multiple channels could be combined in similar arrangements whereby the system bandwidth is increased by a factor equal to the number of combined channels. An example showing how this technique may be used to extend the bandwidth four times using four channels is shown in  FIG. 2 .  FIG. 2  depicts a low side conversion. In this case, frequency bands B, C and D in  201  are frequency translated down by heterodyning, and then is digitized by channels  2 ,  3  and  4  respectively. Frequency band A in  201  is digitized directly by channel  1 . The filter blocks shown ( 202 ) are optional. The goal is to sufficiently isolate each desired frequency band. This can be done, in this example, using a highpass filter, a bandpass filter or no filter at all. In the case of no filter, there will be images of the adjacent frequency band that will be digitized by the channel. As long as they do not overlap, these can be removed using DSP techniques in the recombination DSP block ( 203 ), as will be described below. This recombination technique is shown in a first example for a low side conversion (using three channels) in  FIGS. 4–26 . 
     Another example of how this technique can be applied is shown in  FIG. 3 .  FIG. 3  depicts a high side conversion. Careful inspection of  FIG. 3  will show that the major difference between  FIG. 2  and  FIG. 3  is the frequency used in heterodyning the frequency band of interest into the low frequency band of the corresponding acquisition channel.  FIG. 2  uses a frequency on the low side of the band of interest (F 1 ) to heterodyne frequency band “B” to the frequency band of the acquisition system whereas  FIG. 3  uses a frequency on the high side of the band of interest (F 2 ) to heterodyne frequency band “B” to the frequency band of the acquisition system. Note that the translated frequency band is “reversed” if high side conversion is used. That is, the highest frequency in frequency band “B” (F 2 ) becomes the lowest frequency in the translated band. This will be corrected in the reconstruction by using F 2  as the up-conversion frequency in the DSP reconstruction of the signal. This recombination technique is shown in a second example for a high side conversion (using two channels) in  FIGS. 27–57 . 
     When recombining the signals to generate the final output signal, the phase and or associated delay of the translation frequency should be known in order to reconstruct the original signal. This knowledge of the phase of the translation frequency can be passed to the process for the recombination by, for example, summing a pilot tone having a substantially predefined phase relationship to a local oscillator of the mixer used for the heterodyning process into the signal channel so that the phase of this pilot tone may be determined upon recombination, and used to compensate for any phase changes of any of the signals to be recombined. Alternatively, the heterodyning mixer phase may be locked to the sample clock, thus providing output signals to be recombined having the same phase. 
     EXAMPLES 
     The first example shows how a step can be digitized using two 5 GHz bands and low side downconversion. 
     
       
         
           
             
               
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 f ( t ):=if [ t&lt;TD, 0,[−1−ω0·( t−TD )]· e   [−ω0·(t−TD)] +1]
 
     To simulate the behavior of the analog components, it is modeled digitally with an extremely high sample rate. 
     
       
         
           
             
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       FIG. 4  shows a picture of the simulated 45 ps step. 
     
       
         
           
             
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               ⁢ 
               
                   
               
               ⁢ 
               FFT 
             
           
         
       
       
         
           
             
               
                 N 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 H 
               
               := 
               
                 
                   K 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   H 
                 
                 2 
               
             
             ⁢ 
             
                 
             
           
         
       
       
         
           
             
                 
             
             ⁢ 
             
               
                 n 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 h 
               
               := 
               
                 0 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 … 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 NH 
               
             
           
         
       
       
         
           
             
               fh 
               nh 
             
             := 
             
               
                 
                   n 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   h 
                 
                 
                   N 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   H 
                 
               
               · 
               
                 
                   F 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   
                     S 
                     hi` 
                   
                 
                 2 
               
             
           
         
       
     
       FIG. 5  shows the frequency content of the simulated step. 
     As we know, the scope does not have the bandwidth to digitize this signal. Therefore, we apply the method of this invention. First, we will utilize a system bandwidth of 5 GHz. then, we develop bandpass filters that select 5 GHz bands of the signal. Note that because the system is bandlimited, it is not actually necessary to utilize bandpass filters—only high pass filters need be utilized, but bandpass filters are used to simplify the discussion. Furthermore, the first band does not even need a filter—the scopes limited bandwidth will do this for us. (inside the scope, a digital low pass filter would be utilized to provide the hard bandwidth limiting).
 
BW:=5
 
system bandwidth utilized for each band (GHz)
 
     Make low pass and bandpass filters for each band.
 
 nn:= 1 . . .  NH− 1
 
 Mfl   nh :=if( fh   nh   ≦BW,  1,0)
 
 Mfh   nh :=if( BW&lt;fh   nh ≦2· BW, 1,0)
 
 Mfl   NH+nn   :=Mfl   NH−nn  
 
 Mfh   NH+nn   :=Mfh   NH−nn  
 
       FIG. 6  shows the response of the low frequency band filter. 
       FIG. 7  shows the response of the high frequency band filter. 
     Apply these filters to the input waveform. 
     
       
                 
         
             
             
         
      
     
       FIG. 8  shows the frequency domain result of applying the low frequency band filter to the simulated step. 
       FIG. 9  shows the frequency domain result of applying the high frequency band filter to the simulated step. 
     Calculate the inverse FFT of these filtered bands.
 
 xfl:=ICFFT ( Xfl )
 
 xfh:=ICFFT ( Xfh )
 
 xf:=ICFFT ( Xf )
 
       FIG. 10  shows the input waveform. 
       FIG. 11  shows the input waveform bandlimited to 10 GHz. 
       FIG. 12  shows the time domain result of applying the low frequency band filter to the simulated step. 
       FIG. 13  shows the time domain result of applying the high frequency band filter to the simulated step. 
     It is useful to add these two signals together and compare them to the input waveform.  FIG. 14  shows this. You will note the sum is not identical to the input because the system has limited the bandwidth at 10 GHz. The 10 GHz bandwidth limited signal is the best that we will be able to provide. 
     It is also useful to compare the low frequency and actual input waveforms directly.  FIG. 15  shows this. The point of  FIG. 15  is to demonstrate the problem that this invention is designed to solve. The limited bandwidth slows the edge of the step. This simulates the analog waveform that gets sampled by a digitizer with a front-end bandwidth of 5 GHz. Our goal is to digitize the actual waveform with a much higher bandwidth. 
     First, the high frequency band is applied to the mixer.
 
 F   mixer0 :=1 ·BW  
 
Φ mixer0   :=rnd (2·π)
 
     The frequency of the high frequency mixer is at the cutoff frequency of the first band. apply the mixer
 
 xfhm   kh   :=xfh   kh ·2·cos(2·π· F   mixer0     th     kh +Φ mixer0 )
 
     Look at the frequency content.
 
 Xfhm:=CFFT ( xfhm )
 
     Low pass filter the mixer outputs. 
     
       
                 
         
             
             
         
      
     
     Note again that the typical manner of low pass filtering the mixer outputs would be to use the scope front-end. This filtering is being shown here as actual low pass filters applied. 
       FIG. 16  shows the effect at the high frequency band mixer and image reject filter output in the frequency domain. 
     take the inverse FFF to generate the analog mixer output signals—the analog signals input to the channel digitizers.
 
 xfhml:=ICFFT ( Xfhml )
 
       FIGS. 17 and 18  show the low frequency and high frequency band signal due to the processing of the simulated step input signal. 
     It is interesting to see the sum of these two waveforms their sum does not produce a waveform that is useful in representing the originally input waveform. This is shown in  FIG. 19 . This is shown to exemplify that further processing of the digitized waveforms is required to achieve the objective. 
     At this point, the waveforms are digitized. The waveforms must be sampled at a rate sufficient to satisfy the Nyquist Criterion. For this example, this means that they must be sampled at at least 2 times BW, or 10 GS/s. After the waveforms have been digitized, they are immediately upsampled using SinX/x interpolation. This is possible because all digitized waveforms are bandlimited. It is useful to upsample the waveforms to a sample rate capable of meeting the Nyquist Criterion for the system bandwidth—I have chosen 40 GS/s. The upsampling is trivial and for the purpose of this example, I simply use a 40 GS/s digitizer with the understanding that the exact same waveform would result from sampling the waveform at 10 GS/s and upsampling by a factor of 4. 
     
       
         
           
             
               F 
               ⁢ 
               
                   
               
               ⁢ 
               S 
             
             := 
             
               40 
               ⁢ 
               
                   
               
               ⁢ 
               upsampled 
               ⁢ 
               
                   
               
               ⁢ 
               digitizer 
               ⁢ 
               
                 
                     
                 
                 ⁢ 
                 
                     
                 
               
               ⁢ 
               sample 
               ⁢ 
               
                   
               
               ⁢ 
               rate 
             
           
         
       
       
         
           
             D 
             := 
             
               
                 
                   
                     F 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       S 
                       hi 
                     
                   
                   
                     
                       F 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       S 
                     
                     ⁢ 
                     
                         
                     
                   
                 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 D 
               
               = 
               
                 25 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 upsampling 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 factor 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 for 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 analog 
                 ⁢ 
                 
                   
                       
                   
                   ⁢ 
                   
                       
                   
                 
                 ⁢ 
                 waveform 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 model 
               
             
           
         
       
       
         
           
             K 
             := 
             
               
                 
                   
                     K 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     H 
                   
                   D 
                 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 k 
               
               := 
               
                 
                   0 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   … 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   K 
                 
                 - 
                 1 
               
             
           
         
       
     
     Sample the waveforms. 
     
       
         
           
             
               t 
               k 
             
             := 
             
               
                 
                   k 
                   
                     F 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     S 
                   
                 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 
                   x 
                   
                     l 
                     k 
                   
                 
               
               := 
               
                 
                   
                     xfl 
                     
                       k 
                       · 
                       D 
                     
                   
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   
                     x 
                     
                       h 
                       k 
                     
                   
                 
                 := 
                 
                   
                     
                       xfhml 
                       
                         k 
                         · 
                         D 
                       
                     
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       x 
                       k 
                     
                   
                   := 
                   
                     
                       
                         xh 
                         
                           k 
                           · 
                           D 
                         
                       
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         w 
                         k 
                       
                     
                     := 
                     
                       wh 
                       
                         k 
                         · 
                         D 
                       
                     
                   
                 
               
             
           
         
       
     
     Generally, at this point, we would apply the sharp cutoff filter. If a sharp cutoff analog filter was not used, we&#39;d have to satisfy the Nyquist Criterion such that any extra frequency content would not fold back into the 5 GHz band. I&#39;ve already applied a sharp cutoff filter to the analog signal, so this is not necessary. 
     Also, at this point, some magnitude and phase compensation would probably be necessary to account for non-ideal channel frequency response characteristics. This example shows the signal digitized with ideal digitizers with ideal frequency response characteristics. 
     Next, the high and very high frequency waveforms are mixed up to there appropriate frequency location and digitally bandpass filtered. 
     Note that these digital mixers know the phase of the analog mixers—some mechanism must be provided for determining this—either through a pilot tone or locking of the mixer phase to the sample clock. 
     Apply digital mixers
 
 x   hm     k     :=x   h     k   ·(2·cos(2·π F   mixer0     t     k +Φ mixer0))  
 
     Bandpass filter the mixer outputs. 
     
       
         
           
             N 
             := 
             
               
                 
                   K 
                   2 
                 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 n 
               
               := 
               
                 0 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 … 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 N 
               
             
           
         
       
       
         
           
             
               f 
               n 
             
             := 
             
               
                 n 
                 N 
               
               · 
               
                 
                   F 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   S 
                 
                 2 
               
             
           
         
       
       
         
           
             
               X 
               ⁢ 
               
                   
               
               ⁢ 
               hm 
             
             := 
             
               
                 
                   CFFT 
                   ⁡ 
                   
                     ( 
                     
                       x 
                       hm 
                     
                     ) 
                   
                 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 X 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 lm 
               
               := 
               
                 CFFT 
                 ⁡ 
                 
                   ( 
                   
                     x 
                     l 
                   
                   ) 
                 
               
             
           
         
       
       
         
           
             
               Xfhm 
               n 
             
             := 
             
               
                 
                   if 
                   ⁡ 
                   
                     ( 
                     
                       
                         
                           f 
                           n 
                         
                         &lt; 
                         
                           
                             1 
                             · 
                             B 
                           
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           W 
                         
                       
                       , 
                       0 
                       , 
                       
                         X 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         h 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           m 
                           n 
                         
                       
                     
                     ) 
                   
                 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 X 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 f 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 h 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 
                   m 
                   n 
                 
               
               := 
               
                 if 
                 ⁡ 
                 
                   ( 
                   
                     
                       
                         f 
                         n 
                       
                       &gt; 
                       
                         
                           2 
                           · 
                           B 
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         W 
                       
                     
                     , 
                     0 
                     , 
                     
                       X 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         fhm 
                         n 
                       
                     
                   
                   ) 
                 
               
             
           
         
       
       
         
           
             nn 
             := 
             
               
                 1 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 … 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 N 
               
               - 
               1 
             
           
         
       
       
         
           
             
               Xfhm 
               
                 N 
                 + 
                 nn 
               
             
             := 
             
               
                 Xfhm 
                 
                   N 
                   - 
                   
                     n 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     n 
                   
                 
               
               _ 
             
           
         
       
       
         
           
             
               
                 X 
                 h 
               
               := 
               
                 CFFT 
                 ⁡ 
                 
                   ( 
                   
                     x 
                     h 
                   
                   ) 
                 
               
             
             ⁢ 
             
                 
             
           
         
       
       
         
           
             
               X 
               l 
             
             := 
             
               CFFT 
               ⁡ 
               
                 ( 
                 
                   x 
                   l 
                 
                 ) 
               
             
           
         
       
     
       FIG. 20  shows the low frequency band waveform frequency content. 
       FIG. 21  shows the high frequency band waveform frequency content before and after digital mixing and filtering. 
       FIG. 22  shows the results of summing the output waveforms. We have acquired the waveform with a 10 GHz bandwidth utilizing two 5 GHz bandwidth channels. 
     Now lets see how the time domain waveforms compare.
 
 xfhm:=Re ( ICFFT ( Xfhm ))
 
       FIG. 23  shows a comparison of the analog low frequency portion of the input waveform to the digitized and processed low frequency waveform result.  FIG. 23  shows that these are identical. 
       FIG. 24  shows a comparison of the analog high frequency portion of the input waveform to the mixed, digitized and digitally remixed and processed high frequency waveform result.  FIG. 24  shows that these are identical. 
       FIGS. 25 and 26  show how successive addition of the processed band outputs causes the resulting waveform to more and more closely approximate the input waveform.  FIG. 26  shows that the result is identical to the input waveform band limited to 10 GHz and shows that the 10 GHz bandwidth limited step is fully recreated. 
     The second example shows how a step can be digitized using three 5 GHz bands and high side downconversion. 
     
       
         
           
             rt 
             = 
             
               .035 
               ⁢ 
               
                   
               
               ⁢ 
               risetime 
               ⁢ 
               
                   
               
               ⁢ 
               of 
               ⁢ 
               
                   
               
               ⁢ 
               edge 
               ⁢ 
               
                 
                     
                 
                 ⁢ 
                 
                     
                 
               
               ⁢ 
               specified 
               ⁢ 
               
                   
               
               ⁢ 
               
                 ( 
                 
                   n 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   s 
                 
                 ) 
               
             
           
         
       
       
         
           
             
               f 
               bw 
             
             = 
             
               
                 
                   0.344 
                   rt 
                 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 
                   f 
                   bw 
                 
               
               = 
               9.829 
             
           
         
       
       
         
           
             
               ω 
               ⁢ 
               
                   
               
               ⁢ 
               0 
             
             = 
             
               
                 
                   1.5542 
                   · 
                   π 
                   · 
                   
                     f 
                     bw 
                   
                 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 Calculate 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 the 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 center 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 frequency 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 for 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 the 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 
                   system 
                   . 
                   
                     
 
                   
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   
                     ω0 
                     
                       2 
                       · 
                       π 
                     
                   
                 
               
               = 
               
                 15.274 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 center 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 frequency 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 
                   ( 
                   GHz 
                   ) 
                 
               
             
           
         
       
       
         
           
             
               T 
               ⁢ 
               
                   
               
               ⁢ 
               D 
             
             = 
             
               5 
               ⁢ 
               
                   
               
               ⁢ 
               time 
               ⁢ 
               
                   
               
               ⁢ 
               delay 
               ⁢ 
               
                   
               
               ⁢ 
               for 
               ⁢ 
               
                 
                     
                 
                 ⁢ 
                 
                     
                 
               
               ⁢ 
               step 
               ⁢ 
               
                 
                     
                 
                 ⁢ 
                 
                     
                 
               
               ⁢ 
               edge 
               ⁢ 
               
                   
               
               ⁢ 
               
                 ( 
                 
                   n 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   s 
                 
                 ) 
               
             
           
         
       
       
         
           
             
               H 
               ⁡ 
               
                 ( 
                 s 
                 ) 
               
             
             = 
             
               
                 
                   
                     ω 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       0 
                       2 
                     
                   
                   
                     
                       ( 
                       
                         
                           s 
                           2 
                         
                         + 
                         
                           
                             
                               ω 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               0 
                             
                             Q 
                           
                           · 
                           s 
                         
                         + 
                         
                           ω 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             0 
                             2 
                           
                         
                       
                       ) 
                     
                     · 
                     s 
                   
                 
                 · 
                 
                   ⅇ 
                   
                     
                       
                         - 
                         s 
                       
                       · 
                       TD 
                     
                     ⁢ 
                     
                         
                     
                   
                 
               
               ⁢ 
               Laplace 
               ⁢ 
               
                   
               
               ⁢ 
               transform 
               ⁢ 
               
                   
               
               ⁢ 
               of 
               ⁢ 
               
                 
                     
                 
                 ⁢ 
                 
                     
                 
               
               ⁢ 
               the 
               ⁢ 
               
                 
                     
                 
                 ⁢ 
                 
                     
                 
               
               ⁢ 
               step 
               ⁢ 
               
                 
                     
                 
                 ⁢ 
                 
                     
                 
               
               ⁢ 
               specified 
             
           
         
       
     
     The inverse Laplace transform provides the time-domain step waveform.
 
 f ( t )=if[ t&lt;TD, 0,[−1−ω0·( t−TD )]· e   [−ω0·(t−TD)] +1]
 
     To simulate the behavior of the analog components, it is modeled digitally with an extremely high sample rate. 
     
       
         
           
             
               F 
               ⁢ 
               
                   
               
               ⁢ 
               
                 S 
                 hi 
               
             
             = 
             
               1000 
               ⁢ 
               
                   
               
               ⁢ 
               sample 
               ⁢ 
               
                 
                     
                 
                 ⁢ 
                 
                     
                 
               
               ⁢ 
               rate 
               ⁢ 
               
                   
               
               ⁢ 
               for 
               ⁢ 
               
                   
               
               ⁢ 
               simulating 
               ⁢ 
               
                 
                     
                 
                 ⁢ 
                 
                     
                 
               
               ⁢ 
               analog 
               ⁢ 
               
                   
               
               ⁢ 
               system 
               ⁢ 
               
                   
               
               ⁢ 
               
                 ( 
                 GHz 
                 ) 
               
             
           
         
       
       
         
           
             
               K 
               ⁢ 
               
                   
               
               ⁢ 
               H 
             
             = 
             
               
                 10000 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 k 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 h 
               
               = 
               
                 
                   0 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   … 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   K 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   H 
                 
                 - 
                 1 
               
             
           
         
       
       
         
           
             
               t 
               ⁢ 
               
                   
               
               ⁢ 
               
                 h 
                 kh 
               
             
             = 
             
               
                 
                   k 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   h 
                 
                 
                   F 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   
                     S 
                     hi 
                   
                 
               
               ⁢ 
               
                   
               
               ⁢ 
               time 
               ⁢ 
               
                   
               
               ⁢ 
               of 
               ⁢ 
               
                   
               
               ⁢ 
               each 
               ⁢ 
               
                   
               
               ⁢ 
               point 
               ⁢ 
               
                   
               
               ⁢ 
               
                 ( 
                 
                   ns 
                 
                 ) 
               
             
           
         
       
     
     Utilize a raised cosine window to minimize effects of the FFT. 
     
       
         
           
             
               
                 
                   
                     wh 
                     kh 
                   
                   = 
                   
                     
                       1 
                       2 
                     
                     - 
                     
                       
                         
                           1 
                           2 
                         
                         · 
                         cos 
                       
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         ( 
                         
                           2 
                           · 
                           π 
                           · 
                           
                             kh 
                             
                               KH 
                               - 
                               1 
                             
                           
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     xh 
                     kh 
                   
                   = 
                   
                     f 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       
                         ( 
                         
                           th 
                           kh 
                         
                         ) 
                       
                       · 
                       
                         wh 
                         kh 
                       
                     
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     Calculate 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     the 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     windowed 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       step 
                       . 
                     
                   
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     t 
                     10 
                   
                   = 
                   
                     
                       
                         .53181160838961202015 
                         ω0 
                       
                       + 
                       
                         TD 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           t 
                           10 
                         
                       
                     
                     = 
                     5.006 
                   
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     t 
                     90 
                   
                   = 
                   
                     
                       
                         3.8897201698674290579 
                         ω0 
                       
                       + 
                       
                         TD 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           t 
                           90 
                         
                       
                     
                     = 
                     5.041 
                   
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       t 
                       90 
                     
                     - 
                     
                       t 
                       10 
                     
                   
                   = 
                   
                     0.035 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     Verify 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     that 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     risetime 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     is 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       correct 
                       . 
                     
                   
                 
               
               
                 
                     
                 
               
             
           
         
       
     
       FIG. 27  shows a picture of the simulated 35 ps step. 
     
       
         
           
             
               
                 
                   Xh 
                   = 
                   
                     CFFT 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       ( 
                       xh 
                       ) 
                     
                   
                 
               
               
                 
                   Calculate 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   the 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   FFT 
                 
               
             
             
               
                 
                   NH 
                   = 
                   
                     KH 
                     2 
                   
                 
               
               
                 
                   nh 
                   = 
                   
                     
                       0 
                       ⁢ 
                       … 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       NH 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         fh 
                         nh 
                       
                     
                     = 
                     
                       
                         nh 
                         NH 
                       
                       · 
                       
                         
                           FS 
                           hi 
                         
                         2 
                       
                     
                   
                 
               
             
           
         
       
     
       FIG. 28  shows the frequency content of the simulated step. 
     As we know, the scope does not have the bandwidth to digitize this signal. Therefore, we apply the method of this invention. First, we will utilize a system bandwidth of 5 GHz. then, we develop bandpass filters that select 5 GHz bands of the signal. Note that because the system is bandlimited, it is not actually necessary to utilize bandpass filters—only high pass filters need be utilized, but bandpass filters are used to simplify the discussion. Furthermore, the first band does not even need a filter—the scopes limited bandwidth will do this for us. (inside the scope, a digital low pass filter would be utilized to provide the hard bandwidth limiting).
 
BW=5
 
system bandwidth utilized for each band (GHz)
 
     Make low pass and bandpass filters for each band.
 
 nn= 1 . . .  NH− 1
 
 Mfl   nh =if( fh   nh   ≦BW, 1,0)  Mfl   NH+nn   =Mfl   NH−nn  
 
       FIG. 29  shows the response of the low frequency band filter.
   Mfh   nh =if( BW&lt;fh   nh ≦2· BW, 1,0)  Mfh   NH+nn   =Mfh   NH−nn    
       FIG. 30  shows the response of the high frequency band filter.
   Mfhh   NH+nn   =Mfhh   NH−nn      Mfhh   nh =if(2· BW&lt;fh   nh ≦3· BW, 1,0) 
       FIG. 31  shows the response of the very high frequency band filter. 
     Apply these filters to the input waveform. 
     
       
                 
         
             
             
         
      
     
       FIG. 32  shows the frequency domain result of applying the low frequency band filter to the simulated step. 
       FIG. 33  shows the frequency domain result of applying the high frequency band filter to the simulated step 
       FIG. 34  shows the frequency domain result of applying the very high frequency band filter to the simulated step. 
     Calculate the inverse FFT of these filtered bands.
 
 xfl=ICFFT ( Xfl )
 
 xfh=ICFFT )( Xfh )
 
 xfhh=ICFFT ( Xfhh )
 
 xf=ICFFT ( Xf )
 
       FIG. 35  shows the band limited input waveform. 
       FIG. 36  shows the time domain result of applying the low frequency band filter to the simulated step. 
       FIG. 37  shows the time domain result of applying the high frequency band filter to the simulated step. 
       FIG. 38  shows the time domain result of applying the very high frequency band filter to the simulated step. 
     It is useful to add these three signals together and compare them to the input waveform.  FIG. 39  shows this. You will note the sum is not identical to the input because the system has limited the bandwidth at 15 GHz. The 15 GHz bandwidth limited signal is the best that we will be able to provide. 
     It is also useful to compare the low frequency and actual input waveforms directly.  FIG. 40  shows this. The point of  FIG. 40  is to demonstrate the problem that this invention is designed to solve. The limited bandwidth slows the edge of the step. This simulates the analog waveform that gets sampled by a digitizer with a front-end bandwidth of 5 GHz. Our goal is to digitize the actual waveform with a much higher bandwidth. First, the high frequency and very high frequency bands are applied to the mixers.
 
 F   mixer0   =BWΦ   mixer0   =rnd (2·π)
 
     The frequency of the high frequency mixer is at the twice the cutoff frequency of the first band. The frequency of the very high frequency mixer is three times the cutoff frequency of the first band.
 
 F   mixer1 =2· BW  
 
Φ mixer1   =rnd (2·π)
 
     Apply the mixers.
 
 xfhm   kh   =xfh   kh ·2·cos(2·π· F   mixer0     th     kh +Φ mixer0 )  xfhhm   kh   =xfhh   kh ·2·cos(2 ·π·F   mixer1     th     kh +Φ mixer1 )
 
     Look at the frequency content.
 
 Xfhm=CFFT ( xfhm )
 
 Xfhhm=CFFT ( xfhhm )
 
     Low pass filter the mixer outputs. 
     
       
                 
         
             
             
         
      
     
     Note again that the typical manner of low pass filtering the mixer outputs would be to use the scope front-end. This filtering is being shown here as actual low pass filters applied. 
       FIG. 41  shows the effect at the high frequency band mixer and image reject filter output in the frequency domain. 
       FIG. 42  shows the effect at the very high frequency band mixer and image reject filter output in the frequency domain. 
     Take the inverse FFT to generate the analog mixer output signals—the analog signals input to the channel digitizers.
 
 xfhml=ICFFT ( Xfhml )
 
 xfhhml=ICFFT ( Xfhhml )
 
       FIGS. 43 ,  44  and  45  show the low frequency, high frequency, and very high frequency band digitizer input signal due to the processing of the simulated step input signal. 
     It is interesting to see the sum of these three waveforms their sums do not produce a waveform that is useful in representing the originally input waveform. This is shown in  FIG. 46 . This is shown to exemplify that further processing of the digitized waveforms is required to achieve the objective. 
     At this point, the waveforms are digitized. The waverforms must be sampled at a rate sufficient to satisfy the Nyquist Criterion. For this example, this means that they must be sampled at least 2 times BW, or 10 GS/s. After the waveforms have been digitized, they are immediately unsampled using SinX/x interpolation. This is possible because all digitized waveforms are bandlimited. It is useful to unsample the waveforms to a sample rate capable of meeting the Nyquist Criterion for the system bandwidth—I have chosen 40 GS/s. The upsampling is trivial and for the purpose of this example, I simply use a 40 GS/s digitizer with the understanding that the exact same waveform would result from sampling the waveform at 10 GS/s and upsampling by a factor of 4. 
     
       
         
           
             
               
                 
                   FS 
                   = 
                   40 
                 
               
               
                 
                   upsampled 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   digitizer 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   sample 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   rate 
                 
               
             
             
               
                 
                   D 
                   = 
                   
                     
                       FS 
                       hi 
                     
                     FS 
                   
                 
               
               
                 
                   D 
                   = 
                   
                     25 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     upsampling 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     factor 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     for 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     analog 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     waveform 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     model 
                   
                 
               
             
             
               
                 
                   K 
                   = 
                   
                     KH 
                     D 
                   
                 
               
               
                 
                   k 
                   = 
                   
                     
                       0 
                       ⁢ 
                       … 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       K 
                     
                     - 
                     1 
                   
                 
               
             
           
         
       
     
     Sample the waveforms
 
 t   k   =k/FS  
 
 x   l     k     =xfl   k·D  
 
 x   h     k     =xfhml   k·D  
 
 x   k   =xh   k·D  
 
 w   k   =wh   k·D  
 
 x   hh     k     =xfhhml   k·D  
 
     Generally, at this point, we would apply the sharp cutoff filter. If a sharp cutoff analog filter was not used, we&#39;d have to satisfy the Nyquist Criterion such that any extra frequency content would not fold back into the 5 GHz band. I&#39;ve already applied a sharp cutoff filter to the analog signal, so this is not necessary. 
     Also, at this point, some magnitude and phase compensation would probably be necessary to account for non-ideal channel frequency response characteristics. This example shows the signal digitized with ideal digitizers with ideal frequency response characteristics. Next, the high and very high frequency waveforms are mixed up to there appropriate frequency location and digitally bandpass filtered. 
     Note that these digital mixers know the phase of the analog mixers—some mechanism must be provided for determining this—either through a pilot tone or locking of the mixer phase to the sample clock. 
     Apply digital mixers.
 
 x   hm     k     =x   h     k   ·(2·cos(2·π· F   mixer0     t     k +Φ mixer0 ))
 
 x   hhm     k     =x   hh     k   ·(2·cos(2·π· F   mixer1     t     k +Φ mixer1 ))
 
     Bandpass filter the mixer outputs. 
     
       
         
           
             
               
                 
                   
                       
                   
                   ⁢ 
                   
                     
                       
                         
                           N 
                           = 
                           
                             K 
                             2 
                           
                         
                       
                       
                         
                             
                         
                       
                       
                         
                             
                         
                       
                     
                     
                       
                         
                             
                         
                       
                       
                         
                           n 
                           = 
                           
                             0 
                             ⁢ 
                             … 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             N 
                           
                         
                       
                       
                         
                             
                         
                       
                     
                     
                       
                         
                           
                             f 
                             n 
                           
                           = 
                           
                             
                               n 
                               N 
                             
                             · 
                             
                               FS 
                               2 
                             
                           
                         
                       
                       
                         
                             
                         
                       
                       
                         
                             
                         
                       
                     
                   
                 
               
             
             
               
                 
                   
                     
                       
                         
                           Xhm 
                           = 
                           
                             CFFT 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               ( 
                               
                                 x 
                                 hm 
                               
                               ) 
                             
                           
                         
                         ⁢ 
                         
                             
                         
                       
                     
                     
                       
                         Xhhm 
                         = 
                         
                           CFFT 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             ( 
                             
                               x 
                               hhm 
                             
                             ) 
                           
                         
                       
                     
                     
                       
                         Xlm 
                         = 
                         
                           CFFT 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             ( 
                             
                               x 
                               l 
                             
                             ) 
                           
                         
                       
                     
                   
                 
               
             
             
               
                 
                   
                     
                       
                         
                           Xfhm 
                           n 
                         
                         = 
                         
                           if 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             ( 
                             
                               
                                 
                                   f 
                                   n 
                                 
                                 &gt; 
                                 BW 
                               
                               , 
                               
                                 Xhm 
                                 n 
                               
                               , 
                               0 
                             
                             ) 
                           
                         
                       
                     
                     
                       
                         
                           Xfhhm 
                           n 
                         
                         = 
                         
                           if 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             ( 
                             
                               
                                 
                                   f 
                                   n 
                                 
                                 &gt; 
                                 
                                   2 
                                   · 
                                   BW 
                                 
                               
                               , 
                               
                                 Xhhm 
                                 n 
                               
                               , 
                               0 
                             
                             ) 
                           
                         
                       
                     
                   
                 
               
             
             
               
                 
                   nn 
                   = 
                   
                     
                       1 
                       ⁢ 
                       … 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       N 
                     
                     - 
                     1 
                   
                 
               
             
             
               
                 
                   
                     Xfhm 
                     
                       N 
                       + 
                       nn 
                     
                   
                   = 
                   
                     
                       
                         
                           Xfhm 
                           
                             N 
                             - 
                             nn 
                           
                         
                         _ 
                       
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         Xfhhm 
                         
                           N 
                           + 
                           nn 
                         
                       
                     
                     = 
                     
                       
                         Xfhhm 
                         
                           N 
                           - 
                           nn 
                         
                       
                       _ 
                     
                   
                 
               
             
             
               
                 
                   
                     X 
                     h 
                   
                   = 
                   
                     
                       CFFT 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         ( 
                         
                           x 
                           h 
                         
                         ) 
                       
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         X 
                         l 
                       
                     
                     = 
                     
                       
                         CFFT 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           ( 
                           
                             x 
                             l 
                           
                           ) 
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           X 
                           hh 
                         
                       
                       = 
                       
                         CFFT 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           ( 
                           
                             x 
                             hh 
                           
                           ) 
                         
                       
                     
                   
                 
               
             
           
         
       
     
       FIG. 47  shows the low frequency band waveform frequency content. 
       FIG. 48  shows the high frequency band waveform frequency content before and after digital mixing and filtering. 
       FIG. 49  shows the very high frequency band waveform frequency content before and after digital mixing and filtering. 
       FIG. 50  shows the results of summing the output waveforms. We have acquired the waveform with a 15 GHz bandwidth utilizing three 5 GHz channels. 
     Now let&#39;s see how the time domain waveforms compare.
 
 xfhm=Re ( ICFFT ( Xfhm ))
 
 xfhhm=Re ( ICFFT ( Xfhhm ))
 
       FIG. 51  shows a comparison of the analog low frequency portion of the input waveform to the digitized and processed low frequency waveform result.  FIG. 51  shows that these are identical. 
       FIG. 52  shows a comparison of the analog high frequency portion of the input waveform to the mixed, digitized and digitally remixed and processed high frequency waveform result.  FIG. 52  shows that these are identical. 
       FIG. 53  shows a comparison of the analog very high frequency portion of the input waveform to the mixed, digitized and digitally remixed and processed very high frequency waveform result.  FIG. 53  shows that these are identical. 
       FIGS. 54 ,  55  and  56  show how successive addition of the processed band outputs causes the resulting waveform to more and more closely approximate the input waveform. 
       FIG. 57  shows that the result is identical to the input waveform band limited to 15 GHz and shows that the 15 GHz bandwidth limited step is fully recreated. 
     It should be obvious to one skilled in the art that there are many combinations of translation frequencies and filter choices that will accomplish the objective of this invention. Each has different tradeoffs and implementation considerations depending on the specific application. 
     It will be understood that the above description of the present invention is susceptible to various modifications, changes and adaptations, and the same are intended to be comprehended within the meaning and range of equivalents of the appended claims. The most obvious modification, for example, is the use of more than two channels. 
     It will thus be seen that the objects set forth above, among those made apparent from the preceding description, are efficiently attained and, since certain changes may be made in carrying out the above method and in the constructions set forth without departing from the spirit and scope of the invention, it is intended that all matter contained in the above description shall be interpreted as illustrative and not in a limiting sense. 
     It is also to be understood that the following claims are intended to cover all of the generic and specific features of the invention herein described and all statements of the scope of the invention which, as a matter of language, might be said to fall therebetween.