Nov'04CS3291: Section 41 University of Manchester Department of Computer Science CS3291 Digital Signal Processing '04-'05 Section 4: ‘A design technique for FIR digital filters’
Nov'04CS3291: Section Introduction: FIR digital filter of order M implemented by programming the signal flow graph shown below. Its difference equation is: y[n] = a 0 x[n] + a 1 x[n-1] + a 2 x[n-2] a M x[n-M]
Nov'04CS3291: Section 43 Its impulse-response is {..., 0,..., a 0, a 1, a 2,..., a M, 0,...} Its frequency-response is: M H(e j ) = a n e - j n n=0 Consider problem of choosing a 0, a 1,..., a M such that H( e j ) is close to some target frequency response. Use the inverse DTFT:
Nov'04CS3291: Section 44 Assume we require low-pass filter whose gain-response approximates ideal 'brick-wall' gain-response:. If we take phase-response ( )=0 for all , the required frequency-response is:
Nov'04CS3291: Section 45 By the inverse DTFT, = (1/3)sinc(n/3) for all n.
Nov'04CS3291: Section 46 x sinc(x)
Nov'04CS3291: Section 47 A digital filter with this impulse-response would have exactly the required low-pass frequency-response. But {h[n]} has non-zero samples extending from n = - to , Not a finite impulse-response. Also not causal.
Nov'04CS3291: Section 48 To produce a realisable impulse-response: (1) Truncate {h[n]} to a FIR by setting h[n] to zero for all values of n outside the range -M/2 n M/2 ( assume M is the order of the required FIR filter and is even ). (2) Delay resulting sequence by M/2 samples to ensure that the first non-zero sample occurs at n = 0.
Nov'04CS3291: Section 49 n M=10 h[n] = (1/3)sinc(n/3) Starting with ideal impulse response:
Nov'04CS3291: Section 410 n n h[n] M=10 Truncate to M/2 Delay by M/2 samples
Nov'04CS3291: Section 411 Resulting causal impulse response realised by setting a n = h[n] for n=0,1,2,...,M. Taking M=4, for example, the finite impulse response obtained for the /3 cut-off low-pass specification is : {..,0,..,0, 0.14, 0.28, 0.33, 0.28, 0.14, 0,..,0,..}
Nov'04CS3291: Section 412 Gain & phase responses given in Figure 4.4. z-1 z-1 z-1 z-1 x[n] y[n] Resulting FIR filter is as shown in Figure 4.1 with a 0 =0.14, a 1 =0.28, a 2 =0.33, a 3 =0.28, a 4 =0.14. ( Note:4th order FIR filter has 4 delays & 5 multipliers ).
Nov'04CS3291: Section 413 G( dB /3 -6 dB Fig 4.4
Nov'04CS3291: Section 414 Truncation of {h[n]} to M/2 & the M/2 samples delay produces gain & phase responses different from those originally specified. Gain-response: cut-off rate not sharp, two "ripples" appear in stop-band, peak of the first ripple at about -21dB. Phase-response: not zero for all as orig specified, ( ) = - ( M/2 ) for | | /3; linear phase in pass-band slope arctan(M/2) with M = 4. phase-delay M/2 = 2 samples.
Nov'04CS3291: Section 415 Question.Why does delaying {h[n]} by M/2 produce this effect? Answer: If DTFT of {h[n]} is H(e j ) = G( )e j ( ), the DTFT of the delayed impulse-response: Delaying {h[n]} by M/2 samples multiplies H(e j ) by e -M j/2 which increases ( ) by -M / 2. Increases phase-delay - ( ) / by M/2 sampling intervals.
Nov'04CS3291: Section 416 Can we improve low-pass filter by increasing order to ten? Taking 11 terms of { (1 / n ) sin (n / 3) } we get, after delaying by 5 samples: {...0,-0.055,-.069, 0,.138,.276,.333,.276,.138,0,-.069,-.055,0,...}. If we analyse, by computer, a tenth order FIR filter with this impulse-response, we obtain the gain response shown in fig 4.5. The graph was produced by the following MATLAB statements:
Nov'04CS3291: Section 417 for n=1:11; a(n)= *sinc((n-6)/3); end; H=freqz(a,1,1000); whitebg; plot( [0:999]/1000,20*log10(abs(H) ); axis( 0,1,-60,10] ); grid on; xlabel( ‘rel_freq / pi ); ylabel( ‘Gain (dB)’ ) Alternative graph may be produced by the single statement: freqz( [ , , 0, 0.138, 0.276, 0.333, 0.276, 0.138, 0, , ] );
Nov'04CS3291: Section 418 A note on the phase-response graphs produced by ‘freqz’ Plots ( ) against rather than phase lag - ( ). Segments at phase intervals of 2 or 360 o -()-() ()() // 22 44 22 1 freqz
Nov'04CS3291: Section 419 Gain response of 10th order lowpass FIR filter with C = /3
Nov'04CS3291: Section 420 Gain response of 20th order lowpass FIR filter with C = /3
Nov'04CS3291: Section 421 Cut-off rate for 10 th order FIR filter sharper than for 4th order. More stop-band ripples. Gain at peak of first ripple after cut-off remains at -21 dB. This effect is due to Gibb's phenomenon. Linear phase in passband with phase delay of 5 samples. Going to 20th order produces even faster cut-off rates, & more stop-band ripples, but main stop-band ripple remains at about -20dB. To improve matters we need to discuss " windowing ".
Nov'04CS3291: Section Windowing: To design FIR filters we multiplied {h[n]} as calculated by inverse DTFT by a rectangular window sequence {r[n]} where
Nov'04CS3291: Section 423 Causes sudden transition to zero at window edges. It is these transitions that produce the stop-band ripples. Levels of ripples reduced if {r[n]} replaced by non-rectangular window sequence { w[n] }. Produces a more gradual transition at the window edges. Simple non-rectangular window sequence is Hann window It is a "raised cosine "
Nov'04CS3291: Section 424 Other types of window exist ( e.g. Hamming, Kaiser ). Multiplying {h[n]} by {w[n]} instead of {r[n]} gradually tapers impulse-response towards zero at window edges. Consider again low-pass filter example with rel cut-off /3. Ideal impulse-response was found to be: {h[n]} = { …....., 0.14, 0.28, 0.33, 0.28, 0.14, ………} When M = 4, the Hann window {w[n]} = {..,0,..,0, 0.25, 0.75, 1, 0.75, 0.25, 0,..,0,..} Multiplying term by term & delaying by M/2 = 2 samples we get: {..,0,..,0, 0.04, 0.21, 0.33, 0.21, 0.04, 0,..,0,..}
Nov'04CS3291: Section 425 Resulting "Hann-windowed" FIR filter of order 4 is as shown below with a 0 =0.04, a 1 =0.21, a 2 =0.33, a 3 = 0.21, a 4 = Its gain-response is approximately as shown in Figure 4.9. z-1 z-1 z-1 z-1 x[n] y[n] a0 a1 a3 a4 a2a2
Nov'04CS3291: Section 426 G( dB dB 2 Fig. 4.9
Nov'04CS3291: Section 427 Figure 4.9: 4 th order FIR filter (Hann)
Nov'04CS3291: Section Effect of windowing on freq-response of FIR digital filter Effect is to gradually reduce amplitude of ideal impulse-response towards zero at edges of window rather than to abruptly truncate. Effect on gain-response of FIR filter obtained is: i) to greatly reduce stop-band ripples ( good ). ii) to reduce the cut-off rate ( bad ). Phase-response is not affected in the pass-band. We can improve the cut-off rate by going to higher orders. Graphs below are for 10 th & 20 th order ( Hann windowed ):
Nov'04CS3291: Section 429 Tenth order FIR filter with C = /3 ( Hann window )
Nov'04CS3291: Section 430 MATLAB program to design & graph 10 th order FIR lowpass filter with Hann window w=hanning(11); for n=1:11; a(n) = *sinc((n-6)/3)*w(n); end; h = freqz(a,1,1000); plot([0:999]/1000,20*log10(abs(h)),'k'); axis([0,1,-50,0]); grid on; xlabel('Rel_freq / pi'); ylabel('Gain(dB)');
Nov'04CS3291: Section th order FIR filter with C = /3 (Hann window)
Nov'04CS3291: Section Highpass, band-pass & band-stop linear phase FIR filters Can be designed almost as easily as low-pass. Remember to define required gain-response G I ( ) from - to + Make G I (- ) = G I ( ). Band-pass filter with pass-band from f S /8 to f S /4 has following gain response ideally:-
Nov'04CS3291: Section 433 Applying the inverse DTFT (not forgetting negative ): Taking ( ) = 0 for all initially as before, we obtain: Can evaluate this, but there is a nicer way
Nov'04CS3291: Section 434 H(e j ) = H 1 (e j ) H 2 (e j ) -- /4 /2-- - /2- /4 11 = (1/2)sinc(n/2) - (1/4)sinc(n/4) for - < n <
Nov'04CS3291: Section 435 Exercise: By a similar method, or otherwise, show that the impulse-response for an ideal zero phase 'brick-wall' high-pass filter with cut-off frequency /6 radians per sample (i.e. one twelfth of the sampling frequency) is: h[n] = sinc(n) - (1/6)sinc(n/6) for - < n < Exercise: Derive the impulse-response for an ideal zero phase 'brick-wall' band-stop digital filter with cut-off frequencies L and U radians per sample.
Nov'04CS3291: Section 436 Technique not restricted to “conventional” gain-responses. It is not difficult to design a linear phase filter whose gain- response approximates that below:
Nov'04CS3291: Section Summary of design technique To design an FIR digital filter of even order M, with gain response G I ( ) and linear phase by the windowing method, 1) Set H(e j ) = G I ( ) the required gain-response. This assumes ( ) = 0. 2) IDTFT to produce the ideal impulse-response {h[n]}. 3) Window to M/2 using chosen window. 4) Delay windowed impulse-response by M/2 samples. 5) Realise by setting multipliers of FIR filter.
Nov'04CS3291: Section 438 Instead of obtaining H( e j ) = G I ( ), we get e -j M/2 G( ) G( ) is distorted version of G I ( ) due to windowing. Phase-response is ( ) = - M/2 which is linear phase. - ( ) / = M/2 for all . Filter coeffs symmetric about M/2. e.g. {…2, -3, 5, 7, 5, -3, 2, …} M =6 (even) {…, 1, 3, 5, 5, 3, 1, …} M=5 (odd) This is because the filters are linear phase.
Nov'04CS3291: Section 439 = e -5j /2 (e 5j /2 +3e 3j /2 +5e j /2 +5e -j /2 +3e -3j /2 +e -5j /2 ) = e -5j /2 (2cos(2.5 ) + 6cos (1.5 ) + 10cos( /2) ) = G( )e j ( ) with ( ) = -5 /2. Hence ( ) / = -5/2 = constant, so H(e j ) is linear phase. FIR digital filters whose impulse-responses are symmetric are linear phase. Let {h[n]} = {…, 1, 3, 5, 5, 3, 1, …}
Nov'04CS3291: Section 440 Windowing or "Fourier series approxn technique ". It is possible to design FIR filters which are not linear phase
Nov'04CS3291: Section Further applications of windowing design technique Technique even more powerful than has been yet indicated Not restricted to linear phase filters. Consider some further examples of its use Fractional sampling interval delay filter: Differentiator: Hilbert transformer: See notes for details
Nov'04CS3291: Section Fractional sampling interval delay filter : Gain-response is one for all , but whose phase-response is ( ) = . Therefore: e -j / 2 : 0 H(e j ) = e j / 2 : - 0
Nov'04CS3291: Section Differentiator: Would produce cos( t) from sin ( t). FIR filter can act as a differentiator if it outputs {( /T) cos( n)} when the input is {sin( n)} for any in the range - to . Required gain-response is therefore G I ( ) = /T. Instead of specifying a phase response of zero, specify a required phase lead of /2. Required frequency response is: ( /T) e j / 2 : 0 j /T : 0 H(e j ) = = ( /T e - j / 2 : - 0 -j /T : - 0
Nov'04CS3291: Section Hilbert transformer: G( ) = 1 and ( ) = - /2 for 0 < < . “Quadrature phase” rather than linear phase. A linear phase component is added to the 90o phase shift when we delay the windowed impulse response for causality. The required frequency response is: e - j / 2 : 0 -j : 0 H(e j ) = 0 : = 0 = 0 : = 0 e j / 2 : - 0 j : - 0
Nov'04CS3291: Section 445 Designing & implementing FIR digital filters using SP Toolbox Design of linear phase FIR digital filters by windowing technique carried out by command 'FIR1'. Applied to segment of sampled sound by command 'filter'. Design & implement 128 th order FIR band-pass digital filter with passband 300 Hz to 3.4 k Hz Apply it to wav file of mono music sampled at kHz. Notes: (1) FIR cut-off frequencies specified relative to f S /2. (2) By default FIR1 uses Hamming window; (other windows such as Hann can be specified). (3) Filter scaled so centre of pass-band has magnitude one.
Nov'04CS3291: Section 446 clear all; [x, fs, nbits] = wavread('caprice.wav'); %To design: wlow = 2 * pi * 300 / fs ; % radians/sample rel to fs wup = 2 * pi * 3400 / fs ; % radians/sample rel to fs a = fir1(128, [wlow wup] / pi ) ; freqz ( a, 1) ; % plot gain & phase % To implement: L=length(x); y = filter(a, 1, x ); % Remember the delay of 64 samples. wavwrite(x,fs,nbits,'capnew.wav');
Nov'04CS3291: Section 447 Remez Exchange Algorithm method: Better than windowing technique, but more complicated. Available in MATLAB. Design 40 th order FIR lowpass filter whose gain is unity (0 dB) in range 0 to 0.3 radians/sample & zero in range 0.4 to . The 41 coefficients will be found in array ‘a’. Produces 'equi-ripple' gain-responses where peaks of stop-band ripples are equal rather than decreasing with increasing frequency. Highest peak in stop-band lower than for FIR filter of same order designed by windowing technique to have same cut-off rate. There are 'equi-ripple' pass-band ripples.
Nov'04CS3291: Section 448 a = remez (40, [0, 0.3, 0.4,1],[1, 1, 0, 0] ); h = freqz (a,1,1000); plot([0:999]/1000,20*log10(abs(h)),'k'); axis([0,1,-50,0]); grid on; xlabel('Rel_freq / pi'); ylabel('Gain(dB)');
Nov'04CS3291: Section 449 Gain of 40 th order FIR lowpass filter designed by “ Remez ”
Nov'04CS3291: Section Some implementation issues Fixed point implementation of FIR digital filters FIR digital filters often implemented in mobile equipment. Low power fixed point DSP processors are norm. Typically with a basic 16-bit word-length. Must be programmed using only integer arithmetic. Take 4 th order FIR filter with impulse response: {… , 0.21, 0.33, 0.21, 0.04, …...}. Rounding each coeff to nearest integer clearly a mistake. Multiply all coeffs by a large constant then round:. A 0 = 4, A 1 = 21, A 2 = 33, A 3 = 21, A 4 = 4. We must divide the output by same constant, in this case 100. Instead of 100, we choose a power of two for the constant. Dividing by a power of two (e.g. 1024) is very simple.
Nov'04CS3291: Section 451 The larger the constant, the more accurate the coefficients. Careful not to choose too large a constant If integers produced get too large, we risk overflow Difficult balancing act between inaccuracy & overflow. FIR filters easy to program in fixed point arithmetic. Never become unstable as there is no feedback. In some cases, overflows can be allowed to occur Can risk overflow more readily with FIR digital filters than with IIR digital filters, and thus have greater coefficient accuracy. Scaling by 1024, is adopting a 'Q-format' of ten. Programmer assumes a binary point to exist ten bit positions from the right within the 16-bit word.
Nov'04CS3291: Section 452 MATLAB implementation of 4 th order low-pass filter using integer arithmetic only: A = [ ] ; x = [ ] ; while 1 x(1) = input( 'X = '); Y = A(1)*x(1); for k = 5 : -1: 2 Y = Y + A(k)*x(k); x(k) = x(k-1); end; Y = round( Y / 100) ; disp([' Y = ' num2str(Y)]); end;
Nov'04CS3291: Section Alternative signal flow graphs Signal flow-graph below has same impulse-response as fig 4.1. Same input x[n] applied to all multipliers. Could have advantages & other alternatives exist. z AMAM A0A0 A M-2 A M-1 x[n] y[n]
Nov'04CS3291: Section 454 PROBLEMS: 1 Design a 10th order FIR low-pass digital filter with cut-off at f S /4 with and without a Hann window. Use MATLAB to compare the gain responses obtained. 2 Design a 10th order FIR bandpass digital filter with lower and upper cut-off frequencies at /4 and /2 respectively. 3 Write a MATLAB program for one of these filters using integer arithmetic only. 4 Design a 4th order FIR high-pass filter with cut-off at /3. 5 Do all FIR filters have exactly linear phase responses? 6 Show that if { h[n] } is real, H( e -j ) = H*( e j ) and hence that: 7 Show that if H( e j ) is linear phase with a phase delay N samples, with N an integer, its impulse-response is symmetric about n=N.
Nov'04CS3291: Section Design 6th order FIR filter which is linear phase & delays input by half a sampling interval (plus usual 3 for causality). 9. Design 6th order linear phase FIR filter whose gain-response approximates that shown in fig Show that if input to ideal Hilbert transformer is {cos( n)} output will be {sin( n)} for all . 11.Design 8th order FIR approx to all-pass Hilbert transformer. Sketch a graph of its phase-response. 12. Show that ( ) = -k corresponds to a delay of k samples. 13 Give H(z), {h[n]} and a program for Fig 4.15 & mention any other advantages it may have fig Rearrange figs 4.1 and 4.15 for greater efficiency when M is even & the filter is linear phase. 15 Explain why IIR digital filters cannot have exactly linear phase responses, whereas FIR filters can.