Fourier Transforms in Radar and Signal Processing_6
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Fourier Transforms in Radar and Signal Processing_6 Interpolation for Delayed Waveform Time Series 119The FIR filter coefficients from the sampled impulse response are given by 2 22 r T 2) h r = h (rT ) = exp (−4 (5.49)where T = 1/F is the sampling interval. If we take coefficients to the −40-dBlevel, then we have 8 2 2r m T 2 = 4 ln (10), or 2 √ ln (10)/2 F F rm = = 0.342 (5.50)where ±r m are the indexes of the first and last coefficients. We can now estimate the amount of computation required to producethe simulated clutter directly. With F = 104 Hz and = 10 Hz, we seethat r m = 342, so there are 685 taps, and this is the number of complexmultiplications needed for each output sample (in addition to generatingthe inputs from a normal distribution).5.4.2 Efficient Clutter Waveform Generation Using InterpolationIn this case we generate Gaussian clutter with the required bandwidth butat a much lower sampling rate f s , and then interpolate to obtain the samplesat the required rate F (Figure 5.20). Thus we will need F /f s times as manyinterpolations as samples. From Section 5.2 above we know that with moder-ate oversampling rates, we can achieve good interpolation with very few taps.Figure 5.20 Gaussian waveform generation with interpolation.120 Fourier Transforms in Radar and Signal ProcessingLet the number of taps in the interpolation filter be m and the number inthe Gaussian FIR filter is, from (5.50), 0.684f s / ( +1, which we neglect),so that the average number of complex multiplications per output sampleis = m + (0.684 f s / ) / (F /f s ) = m + 0.684 f s2 / F (5.51) In Figure 5.12 we see that with an oversampling factor of 3, we needonly four taps, weighted above the −40-dB level, to interpolate up to themaximum time shift of half the sampling interval. Using these figures, wehave m = 4 and f s = 24 (as the effective bandwidth of the waveform istaken to be 8 in Section 5.3.1 above), and from (5.51) we obtain = 4.4,a factor of over 150 lower than in the direct sampling case. There will haveto be F /2f s sets of four weights (or 21 sets in this example) to interpolatefrom −1/2f s to +1/2f s .5.5 ResamplingAn application of interpolation is to obtain a resampled time series. In thiscase, data has been obtained by sampling some waveform at one frequencyF 1 , but the series that would have been obtained by sampling this waveformat a different frequency F 2 is now required. We consider first the case whereF 1 /F 2 is rational and so can be expressed in the form n 1 /n 2 , with n 1 andn 2 mutually prime (with no common factor). To illustrate the method, wetake n 1 = 4 and n 2 = 7, as shown in Figure 5.21. Over a time interval T= n 1 T 1 = n 2 T 2 , the pattern repeats, where T 1 = 1/F 1 and T 2 = 1/F 2 , andif the output sequence is timed so that some samples are at zero shift relativeto the input, then there will be further time shifts of ± T, ±2 T, . . . , upto ± (n 2 − 1)/2 for n 2 odd, or −n 2 /2 + 1 and +n 2 /2 for n 2 even, where T = T /n 1 n 2 . In Figure 5.21 the required time shifts for the differentFigure 5.21 Resampling. Interpolation for Delayed Waveform Time Series 121pulses are shown in units of T, and we see that the values required arefrom −3 T to +3 T. Over a period of four input pulse intervals, there areseven output pulses, as required, with seven different delays, one being zero.We also see that if the frequency ratio were inverted in this figure, so thatthe input samples are shown by the dashed lines and the outputs by thecontinuous lines, then time shifts of −1, +2, +1, and 0 only, relative to thenearest input sample, are required. If the input sequence is oversampled, we can use the results of Section5.3.2 above to reduce the size of the sampling FIR filters and so achievequite economical resampling, requiring only a few multiplications for eachoutput sample. Only n 2 − 1 time shifts are needed, and the number ofdistinct vectors defining the FIR filter coefficients is only (n 2 − 1)/2 (n 2odd) or n 2 /2 (n 2 even) (as the set of coefficients is the same for positiveand negative shifts, applied in reverse order, with a shift of the input sequence)and these can be precomputed and stored. If the output sequence is at arather higher frequency than the input, as in Figure 5.21, the maximumtime shifts, up to half an output sample interval, will be rather less thanhalf an input interval, and this can also be used to reduce the length of theFIR interpolation filters, as shown in the figu ...
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Fourier Transforms in Radar and Signal Processing_6 Interpolation for Delayed Waveform Time Series 119The FIR filter coefficients from the sampled impulse response are given by 2 22 r T 2) h r = h (rT ) = exp (−4 (5.49)where T = 1/F is the sampling interval. If we take coefficients to the −40-dBlevel, then we have 8 2 2r m T 2 = 4 ln (10), or 2 √ ln (10)/2 F F rm = = 0.342 (5.50)where ±r m are the indexes of the first and last coefficients. We can now estimate the amount of computation required to producethe simulated clutter directly. With F = 104 Hz and = 10 Hz, we seethat r m = 342, so there are 685 taps, and this is the number of complexmultiplications needed for each output sample (in addition to generatingthe inputs from a normal distribution).5.4.2 Efficient Clutter Waveform Generation Using InterpolationIn this case we generate Gaussian clutter with the required bandwidth butat a much lower sampling rate f s , and then interpolate to obtain the samplesat the required rate F (Figure 5.20). Thus we will need F /f s times as manyinterpolations as samples. From Section 5.2 above we know that with moder-ate oversampling rates, we can achieve good interpolation with very few taps.Figure 5.20 Gaussian waveform generation with interpolation.120 Fourier Transforms in Radar and Signal ProcessingLet the number of taps in the interpolation filter be m and the number inthe Gaussian FIR filter is, from (5.50), 0.684f s / ( +1, which we neglect),so that the average number of complex multiplications per output sampleis = m + (0.684 f s / ) / (F /f s ) = m + 0.684 f s2 / F (5.51) In Figure 5.12 we see that with an oversampling factor of 3, we needonly four taps, weighted above the −40-dB level, to interpolate up to themaximum time shift of half the sampling interval. Using these figures, wehave m = 4 and f s = 24 (as the effective bandwidth of the waveform istaken to be 8 in Section 5.3.1 above), and from (5.51) we obtain = 4.4,a factor of over 150 lower than in the direct sampling case. There will haveto be F /2f s sets of four weights (or 21 sets in this example) to interpolatefrom −1/2f s to +1/2f s .5.5 ResamplingAn application of interpolation is to obtain a resampled time series. In thiscase, data has been obtained by sampling some waveform at one frequencyF 1 , but the series that would have been obtained by sampling this waveformat a different frequency F 2 is now required. We consider first the case whereF 1 /F 2 is rational and so can be expressed in the form n 1 /n 2 , with n 1 andn 2 mutually prime (with no common factor). To illustrate the method, wetake n 1 = 4 and n 2 = 7, as shown in Figure 5.21. Over a time interval T= n 1 T 1 = n 2 T 2 , the pattern repeats, where T 1 = 1/F 1 and T 2 = 1/F 2 , andif the output sequence is timed so that some samples are at zero shift relativeto the input, then there will be further time shifts of ± T, ±2 T, . . . , upto ± (n 2 − 1)/2 for n 2 odd, or −n 2 /2 + 1 and +n 2 /2 for n 2 even, where T = T /n 1 n 2 . In Figure 5.21 the required time shifts for the differentFigure 5.21 Resampling. Interpolation for Delayed Waveform Time Series 121pulses are shown in units of T, and we see that the values required arefrom −3 T to +3 T. Over a period of four input pulse intervals, there areseven output pulses, as required, with seven different delays, one being zero.We also see that if the frequency ratio were inverted in this figure, so thatthe input samples are shown by the dashed lines and the outputs by thecontinuous lines, then time shifts of −1, +2, +1, and 0 only, relative to thenearest input sample, are required. If the input sequence is oversampled, we can use the results of Section5.3.2 above to reduce the size of the sampling FIR filters and so achievequite economical resampling, requiring only a few multiplications for eachoutput sample. Only n 2 − 1 time shifts are needed, and the number ofdistinct vectors defining the FIR filter coefficients is only (n 2 − 1)/2 (n 2odd) or n 2 /2 (n 2 even) (as the set of coefficients is the same for positiveand negative shifts, applied in reverse order, with a shift of the input sequence)and these can be precomputed and stored. If the output sequence is at arather higher frequency than the input, as in Figure 5.21, the maximumtime shifts, up to half an output sample interval, will be rather less thanhalf an input interval, and this can also be used to reduce the length of theFIR interpolation filters, as shown in the figu ...
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