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Comparison of Transient Respon of FFT PFT and Polypha DFT Filter Banks
By John Lillington
CTO, RF Engines Limited
This brief paper examines the important differences between the respons of FFT and Polypha DFT filter banks. In general, whilst the Polypha DFT (or, indeed, any filter bank, such as the PFT) has much better adjacent channel rejection in a steady-state condition, the transient respon can actually be wor. This is a conquence of the impul respon of the filter. The differences are illustrated by a simple ca study of typical 1024 bin filter banks.
1 Introduction.
The subject of filter bank frequency respon is a complex one. Most texts on the subject deal with the steady-state respon but in the real world, radar and other burst-mode signals are very transient in nature. It is, therefore, important to understand the transient performance of filter banks. In general, the better the frequency discrimination properties, the longer the time taken to ttle. This is becau sharper filters need more taps and have longer impul respons. Conquently, during
the ttling period, the filter taps are not filled and the frequency respon to adjacent channels can actually be wor than that of the unweighted FFT filter bank.
The following discussion makes comparisons between FFT and Polypha DFT filter banks and looks at both the steady state and the transient conditions. The subject of oversampling and so-called “minimum pha” filters are also briefly covered.
2 Filter Bank Steady State Frequency
Respons.
2.1 Comparison of Frequency Respons.
The effective filter respon of a pipelined FFT is the well-known Sinx/x (Sinc) function which, for many applications, provides inadequate adjacent channel rejection. The u of simple window functions such as Hanning, Kair, Blackman-Harris etc. improves the filter sidelobes but at the expen of the width of the main lobe. This is illustrated in Figs. 1 and 2
Figure 1. Comparison of FFT and Filter Bank
Frequency Respons.
Input Bandwidth < +Channel Spacing = F
-Fs/2+Fs/2
Frequency
F s/ 2
山东文化>如何开户买港股Input S/R = Fs (Complex)
s
/N
Per-Channel Output S/R = F s/N (Complex)
FFT (N=32) with Kair Weighting
This may be improved by using properly designed true filter banks such as the PFT (Ref.[1]) or Polypha DFT (Ref.[4]) as shown in Figs. 1 and 3.
干预0-Fs/2+Fs/2
Frequency Channel Spacing = F s /N
Per-Channel Output S/R = F s /N (Complex)
Superior Cut-Off &
Stop-Band Performance 2.2 Effective Noi Bandwidth. Another uful measure of filter performance is that of Effective Noi Bandwidth (ENB). Table 1 below gives a comparison between
various filter banks. Table 1 Window Type ENB ENB
(dB)
Uniform 1.0 0.00
Hanning 1.503 1.77
Kair 1.656 2.19
Blackman-Harris 2.006 3.02
3 Tap Poly, 0.1 dB Ripple, 100dB Stopband
透明的英文
1.756
2.44 4 Tap Poly, 0.1 dB Ripple, 100dB Stopband
1.527 1.84 5 Tap Poly, 0.1 dB Ripple, 100dB Stopband 1.452 1.62
8 Tap Poly, 0.1 dB Ripple, 100dB Stopband
1.276 1.06
This shows, for example, that a Blackman-Harris
weighted FFT, whilst giving good sidelobes, will lo over 3 dB in signal-to-noi (S/N) whereas an 8 tap Polypha DFT (i.e. a window 8x the transform length, as explained below) gives good sidelobes and low ENB, losing only 1 dB S/N. 3 General Time Domain Considerations.
3.1 Windowing of Time Domain Samples.
For simple windowing, the number of samples in the window will equal that of the transform itlf. To achieve better filters requires the number of samples in the window to exceed that of the transform. This window can, in theory, be of any length. In the Polypha DFT ca, it is normally an integer number (2x, 3x etc.) times the transform length but in the WOLA implementation (Ref.[5]) it can be any length. Thus, for example, a 1024 bin filter bank might require 4096 input samples (4x frame length) for the first full output frame. For the PFT filter bank (Ref.[1]), the number of samples required is equivalent to the impul respon of the cascaded filters but the net effect is similar in that the number of samples required to achieve the first full frame of output data far exceeds the frame length.
3.2 Oversampling.
Another factor affecting the required processing throughput rate is the degree of oversampling required at the output filters. Even to achieve critical sampling (i.e. sample rate just equal to Nyquist)
requires overlapped processing. Using the above example, 4096 samples are needed to form the first full output frame of 1024 bins. To achieve critical sampling means that the next frame of 1024 bins must be contiguous in time with the previous frame
and that the output sample rate must be equal to the
input sample rate. This can be achieved only if the
圆葱怎么做好吃start of the next t of 4096 samples is delayed by 1024 samples which requires 4 overlapped process. To achieve over-sampling, incread overlap is required. Using the above example again,
2x oversampling would require the 4096 samples to
be delayed by only 512 samples which requires 8 overlapped process. A factor of 2x oversampling is standard in the PFT and higher factors can easily be achieved by simply not decimating in some of the stages, as shown in Fig. 4 below.
Stages 1 or 2)
For the Polypha DFT, there are two elegant architectures that can be ud. The WOLA structure is shown in Fig. 5 below.
In this illustration, the window is around 4x the transform length, the windowed samples being sliced into 4 ctions, overlapped in time and added before applying the FFT.
The degree of oversampling is determined simply by the number of new samples, M, introduced for each new transform. The lower the value of M, the higher the oversampling factor.
For integer oversampling, the polypha structure of Fig. 6 below may be ud. The input sample stream is decimated into K phas, filtered by an N-tap filter (K=1024 and N=4 in the above example) before applying the K-point DFT. Oversampling is achieved by up-sampling (zero stuffing) by a factor I. Thus, for 2x oversampling, I=2 etc. Note that this architecture can only be ud with integer oversampling and gives exactly the same results as the WOLA structure with integer oversampling.
Figure 6. Polypha DFT Filter Bank
Implementation.
4 Transient Analysis.
4.1 General Parameters.
Although the Polypha, WOLA or PFT filter banks give much better adjacent channel rejection in the steady-state condition (i.e. for a steady signal condition and all filter taps filled), the transient respon is quite another matter. Some early work on this subject is covered in Ref.[2] and is recommended reading. In the following discussion, the general parameters ud are:-
• No. of Bins = 1024
• Input Sample Rate = 102.4 Ms/s (complex) • Output Sample Rate = 204.8 Ms/s (2x oversampled)
• No. of Polypha Taps = 5 (Equiv. To 5120 point window)
• Filter Stopband = -85 dB
• Filter Passband Ripple = 0.2 dB (pk-pk) • Filter Overlap = 75%
The last parameter is a measure of the filter cut-off rate. The filters are designed to be flat to the edge of the bin and then reach the stopband level by 75% of the width of the adjacent bin.
The transient results from a step function input of a sinewave input at a frequency offt from the centre by one half bin. This prents the worst spectral leakage ca for the FFT.
The graphs prented below are for normalid power spectra (10*Log10{I 2+Q 2}) and reprent the pipelined output at the 2x oversampled rate (204.8 Ms/s in this ca). An individual frame of 1024 samples contains the interleaved frequency bins from –Fs/2 to +Fs/2, where Fs is the input complex sample rate.
4.2 FFT Transient Respon.
Fig. 7 below gives the respon for an unweighted FFT. This does not show the “dead” time caud by the need to collect 1024 samples before the first output can be procesd nor any other latency due to the actual hardware processing. Becau it is 2x oversampled, the first 1024 point frame has not reached steady state. This can be explained by the fact that the first frame is produced in 5uS (1024 points at 204.8 Ms/s sample rate) whereas the input
Figure 7. Twice Oversampled, Unweighted, 1024
Bin Pipelined FFT Transient Respon. sample rate of 102.4 Ms/s means that only 512 samples are available (the remainder being zeros).The first frame is shown in more detail in Fig. 8 below.
Figure 8. Frame 1 from Pipelined FFT
Figure 9. Frame 2 from Pipelined FFT
Fig.9 shows the cond frame which has now reached the steady state condition. This exhibits the expected Sinx/x sidelobe respon for a signal which is one half-bin offt in frequency. This can be i
河北教育考试院网mproved, as discusd above, by windowing the input data but at the expen of broadening the central lobe.
4.3 Polypha DFT Transient Respon.
Fig. 10 below gives the respon for the 5 tap Polypha DFT.
Polypha DFT Transient Respon. Clearly the transient respon is long compared with the FFT and has not fully ttled to steady state until frame 10. Figs. 11, 12 and 13 show frames 1, 5 and
10 respectively.
Figure 11. 2x Oversampled Polypha DFT
Frame 1
Figure 12. 2x Oversampled Polypha DFT
Frame 5
Figure 13. 2x Oversampled Polypha DFT
Frame 10
Even by frame 5, the effective filter frequency respon is no better than an unweighted FFT. This is becau the filter taps are still only half-full. It takes 10 frames (5 taps x oversampling factor) to reach the full steady state.
4.4 Effect of Oversampling.
It might be thought that the transients can be reduced through oversampling the output of the filter bank but this is not true . The transients are a function of the filter impul respon and the only effect of oversampling is to allow the transient to be en in more detail. This may be illustrated by the following example:-
• No. of Bins = 1024
•
Input Sample Rate = 6.4 Ms/s (complex)
• Output Sample Rate = 204.8 Ms/s (32x
oversampled)
• No. of Polypha Taps = 5 (Same effective filter as ud above)
Fig. 14 below gives the transient respon for this
32 times oversampled ca. To keep the output sample rate at 204.8 Ms/s (limited by the maximum device output rate), the input rate must be reduced
to 6.4 Ms/s
Polypha DFT Transient Respon. Figures 16 and 17 below show the respons at frames 80 and 160 for the 32x oversampled ca which correspond exactly to tho of frames 5 and 10 for the 2x-oversampled ca (Figs.12 & 13). This demonstrates clearly that there is no time advantage to be gained by oversampling. Frame 1 for the 32x ca (Fig.15) is even less well developed than Frame 1 for the 2x ca (Fig. 11) since there are now only 32 of the 5120 samples available. All this provides is more detail on the transient respon as may be clearly en by comparing Figs. 10 and 14 (note the different time axis scales due to the change of input sample rate from 102.4 Ms/s to 6.4
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Ms/s).
Figure 15. 32x Oversampled Polypha DFT
Frame 1
Figure 16. 32x Oversampled Polypha DFT
Frame 80
Figure 17. 32x Oversampled Polypha DFT
Frame 160
Allowing for the lower input sample rate, it may be en that the only effect of oversampling is to obrve greater detail in the transient respon.
4.5 Effect of Minimum Pha FIR Filters.
This is discusd in some detail in Ref.[2]. A more accurate name would be “minimum group delay filters” since, in a similar manner to IIR filters, the group delay at centre-band may be reduced at the expen of a non-linear pha respon. A typical example is shown in Fig. 18 below.
Figure 18. 2x Oversampled 5 Tap, Minimum
Pha, 1024 Bin, Polypha DFT Transient
Respon.
The parameters are the same for tho of Fig. 10 except that the filter tap coefficients are different. It may be en that the amplitude respon of the filter has a much faster “attack”. This does not nece
ssarily mean that the filter actually ttles faster
