# Design of Digital Filters

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1 Chapter 8 Desig of Digital Filters Cotets Overview Geeral Cosideratios Desig of FIR Filters Desig of liear-phase FIR filters usig widows Time-delay i desired respose Sidelobes Hilbert trasform (9 phase shift) Desig of FIR filters by frequecy samplig Optimum equiripple liear-phase FIR filters Compariso of FIR methods Desig of IIR Filters from Aalog Filters IIR filter desig by biliear trasformatio Frequecy Trasformatios Desig of Digital Filters Based o Least-Squares Method Summary

4 8.4 c J. Fessler, May 7, 4, 3:8 (studet versio) 8. Desig of FIR Filters A FIR filter of legth M is a LTI system with the followig differece equatio : y[] = M k= b k x[ k]. Note that the book chages the role of M here. Earlier, whe discussig ratioal system fuctios, M was the umber of zeros. Now M is the umber of ozero elemets of h[], which correspods to at most M zeros. (More precisely, we assume b M ad b, but some of the coefficiets i betwee could be zero.) The problem: give δ, δ, c, ad s, we wish to choose M ad {b k } M k= to achieve that goal. We focus o liear-phase FIR filters, because if liear phase is ot eeded, the IIR is probably preferable ayway. We focus o lowpass filters, sice trasformatios ca be made to form highpass, badpass from lowpass, as discussed previously. Impulse respose Clearly for a FIR filter h[] = { b, =,..., M, otherwise. So rather tha writig everythig i terms of b k s, we write it directly i terms of the impulse respose h[]. I fact, for FIR filter desig we usually desig h[] directly, rather tha startig from a pole-zero plot. (A exceptio would be otch filters.) 8.. Symmetric ad atisymmetric FIR filters I focus o the symmetric case. System fuctio: H(z) = M = h[] z. How do we make a filter have liear phase? We previously aswered this i the pole-zero domai. Now we examie it i the time domai. A FIR filter has liear phase if h[] = h[m ], =,,..., M. Example. For M = 5: h[] = {b, b, b, b, b }. Such a FIR filter is called symmetric. Cautio: this is ot eve symmetry though i the sese we discussed previously. This is related, but ot exactly the same as circular symmetry. { } Example. For M = 3 ad h[] = /,, /. Does this filter have liear phase? Is it lowpass or highpass? H() = + e j + e j = e j [ ej + + e j ] = e j ( + cos ), so sice + cos, H() =, which is liear phase. So it works for this particular example, but why does the symmetry coditio esure liear phase i geeral? Cautio. At this poit the book switches from M k= to M k= apparetly. This icosistet with MATLAB, so there are M factors that appear frequetly i the MATLAB calls. I thik that MATLAB is cosistet ad the book makes a udesirable switch of covetio here.

5 c J. Fessler, May 7, 4, 3:8 (studet versio) 8.5 Symmetric real FIR filters, h[] = h[m ], =,..., M, are liear phase. Proof. Suppose M is eve: H(z) = = = = = M = M/ = M/ = M/ = M/ = h[] z h[] z + h[] z + h[] z + M =M/ M/ = M/ = h[] z ] h[] [z + z (M ) M/ = z (M )/ = Thus the frequecy respose is H() = H(z) z=e j M/ = e j(m )/ h[m ] z (M ) h[] z (M ) ] h[] [z (M )/ + z ((M )/ ) = M/ = e j(m )/ = = e j(m )/ H r () M/ [ ( M H r () = h[] cos Phase respose: = (split sum) ( = M ) (symmetry of h[]) (combie) (split phase). ] h[] [e j((m )/ ) + e j((m )/ ) [ ( )] M h[] cos { M H() =, H r () > M + π, H r () <. )]. (Real sice h[] is real.) The case for M odd is similar, ad leads to the same phase respose but with a slightly differet H r ().

6 8.6 c J. Fessler, May 7, 4, 3:8 (studet versio) I fact, the odd M case is eve easier by otig that h[ + (M )/] is a eve fuctio, so its DTFT is real, so the DTFT of h[] is e j(m )/ times a real fuctio. This proof does ot work for M eve sice the (M )/ is ot a iteger so we caot use the shift property. What about pole-zero plot? (We wat to be able to recogize FIR liear-phase filters from pole-zero plot.) From above, H(z) = z (M )/ M/ = h[] [ z (M )/ + z ((M )/ )] so H ( z ) = z (M )/ M/ = h[] [ z (M )/ + z ((M )/ )] = z M H(z). Thus H(z) = z (M ) H ( z ). So if q is a zero of H(z), the /q is also a zero of H(z). Furthermore, i the usual case where h[] is real, if q is a zero of H(z), the so is q. Example. Here are two pole-zero plots of such liear-phase filters. Im(z) Im(z) r r Re(z) 4 Re(z) What is differece betwee this ad all-pass filter? It was poles ad zeros i reciprocal relatioships for all-pass filter. Now we kow coditios for FIR filters to be liear phase. How do we desig oe? Delays I cotiuous time, the delay property of the Laplace trasform is x a (t τ) L e sτ X a (s). How do we build a circuit that delays a sigal? Sice e sτ is ot a ratioal fuctio, so it caot be implemeted exactly usig RLC circuits. So eve though a time delay system is a LTI system, we caot build it usig RLC compoets! We ca make a approximatio, e.g., which is ratioal i s, so we ca desig such a RLC circuit. e sτ sτ + s τ Or, we ca use more mechaical approaches to delay like a tape loop. Picture. write, read, erase head. delay / tape velocity. Aother approach is to rely o sigal propagatio time dow a log wire, ad tap ito the wire at various places for various delays. What about i discrete time? We just eed a digital latch or buffer (flip flops) to hold the bits represetig a digital sigal value util the ext time poit.

7 c J. Fessler, May 7, 4, 3:8 (studet versio) Desig of liear-phase FIR filters usig widows Perhaps the simplest approach to FIR filter desig is to take the ideal impulse respose h d [] ad trucate it, which meas multiplyig it by a rectagular widow, or more geerally, to multiply h d [] by some other widow fuctio, where h d [] = π H d () e j d. π π Typically h d [] will be ocausal or at least o-fir. {, c, Example. As show previously, if H d () =, otherwise, We ca create a FIR filter by widowig the ideal respose: ( = rect ) c the h d [] = c π sic ( c π ). h[] = w[] h d [] = where the widow fuctio w[] is ozero oly for =,..., M. What is the effect o the frequecy respose? W() = { hd [] w[], =,..., M, otherwise, M = w[] e j ad by time-domai multiplicatio property of DTFT, aka the widowig theorem: where π deotes π-periodic covolutio. H() = W() π H d () = π π π H d (λ) W( λ) dλ, (8-) I words, the ideal frequecy respose H d () is smeared out by the frequecy respose W() of the widow fuctio. What would the frequecy respose of the ideal widow be? W() = π δ() = impulse. Such a widow would cause o smearig of the ideal frequecy respose. However, the correspodig widow fuctio would be w[] =, which is ocausal ad o-fir. So i practice we must make tradeoffs. Example. rectagular widow. w[] = {, =,...,, otherwise, So by the shift property of the DTFT: h d [] = sic ( ( ) ) ad h[] = So the resultig frequecy respose is H() = e j [ + 4 π cos()]. Picture This is oly a 3-tap desig. Let us geeralize ext. k= with H d () = e j rect ( ) π, i.e., c = π/. { sic ( ( ),, sic π δ( πk), a Dirac ) } where sic ( ) = /π.64.

8 8.8 c J. Fessler, May 7, 4, 3:8 (studet versio) Example. rectagular widow. W() = M = w[] = {, =,..., M, otherwise, e j = = e j(m )/ W r (), where W r () = si(m/) si(/), M, =. I this case the ideal frequecy respose is smeared out by a sic-like fuctio, because W r () M sic ( M π ). The fuctio si(m/) M si(/) is available i MATLAB as the diric fuctio: the Dirichlet or periodic sic fuctio. How much is the ideal frequecy respose smeared out? The width of the mai lobe of W() is 4π/M, because the first zeros of W r () are at = ±π/m. As M icreases, width of mai lobe W() decreases, so arrower trasitio bad. Example. Here is the case c = π/4 for various values of M. Rectagular Widow Widow Respose Magitude Respose.8 w[] W r () / M.6.4. H() π π. π π Rectagular Widow Widow Respose Magitude Respose.8 w[] W r () / M.6.4. H() π π. π π How did I make these figures? Usig H = freqz(h,, om) sice b = h for causal FIR filters.

9 c J. Fessler, May 7, 4, 3:8 (studet versio) 8.9 Time-delay i desired respose The term e j(m )/ i W() above comes from the fact that the rectagular widow is ot cetered aroud =, but rather is time-shifted to be cetered aroud = (M )/. This phase term will cause additioal distortio of H d (), uless H d () is also phase-shifted to compesate. For a lowpass filter with cutoff c, widowed by a legth-m widow fuctio, the appropriate desired respose is: H d () = { e j(m )/, c,, otherwise, = H d () = {, c,, otherwise, so that This is illustrated below. Note that from (8-), H() = π π π H d (λ) W( λ) dλ = π h d [] = ( [ c π sic c M ]). π c e jλ(m )/ j( λ)(m )/ si(( λ) M/) e dλ c si(( λ) /) = e j(m )/ c si(( λ) M/) dλ. π c si(( λ) /) Thus there is a overall delay of (M )/ samples from such a legth-m causal FIR filter. Without the phase term, the covolutio itegral is severely affected Audio applicatio, F s = 44kHz ad say M = 45. The delay i samples is (M )/ =. The time delay is M T = /44kHz =.5msec. The speed of soud i air is about 33meters / secod, so 3.3 meters away takes about msec. Thus a.5msec delay is well withi the tolerable rage for audio. Ideal Frequecy Respose.3.5 Ideal Impulse Respose H d () h d [] π π. Magitude Respose.3 Causal FIR Filter H() π π h[] Magitude Respose.3 Causal FIR Filter Shifted by (M )/ H().6 h[] π π.

10 8. c J. Fessler, May 7, 4, 3:8 (studet versio) Sidelobes The rectagular widow has high sidelobes i W(), ad the sidelobe amplitude is relatively uaffected by M. Sidelobes cause large passbad ripple, related to Gibbs pheomea. Sidelobes caused by abrupt discotiuity at edge of widow. Solutio: use some other widow fuctio. Examples: Bartlett, Blackma, Haig, Hammig, Kaiser, Laczos, Tukey. MATLAB s widow fuctio has 6 choices! All have lower sidelobes tha rectagular, hece less passbad ripple. Tradeoff? Wider trasitio bad for the same M compared to rectagular widow. Example. Haig widow: w[] = ( ) π cos. M What is the frequecy respose W()? Somewhat messy. How do we plot W()? W = freqz(w,, om) which i tur uses the DFT/FFT with N M via zero paddig. Cautio: this is the ha fuctio i MATLAB, ot the haig fuctio! Rectagular Widow 3 Norm. Widow Respose w[] M=9 W() / W() (db) π π Haig Widow 3 Norm. Widow Respose w[] M=9 W() / W() (db) π π Hammig Widow 3 Norm. Widow Respose w[] M=7 W() / W() (db) π π As M icreases, mai lobe width decreases. As M icreases, some of the sidelobe amplitudes decrease, but peak sidelobe amplitude remais approximately costat. Agai, the price is eed larger M tha for rectagular widow for same trasitio width. The various widows tradeoff mai lobe width with peak sidelobe amplitude. Haig mai lobe width is 8π/M, but peak sidelobe is -3dB compared with -3dB for rectagular.

11 c J. Fessler, May 7, 4, 3:8 (studet versio) 8. Example. Suppose we wat a FIR filter desig of a lowpass filter with H d () = widow with M = 5. The the desired impulse respose is: h d [] = 4 sic ( 4 ( (M )/) ). {, < π/4, otherwise, desiged usig the Haig I MATLAB, here is how we compute ad display the impulse respose ad the frequecy respose. % fig_wi_example.m M = 5; = [:(M-)]; om = lispace(-pi, pi, ); % for displayig frequecy respose oc = pi/4; % cutoff frequecy % desired impulse respose: hd = ilie( oc/pi * sic(oc/pi*(-(m-)/)),, oc, M ); Hd = ilie( *(abs(om) < oc), om, oc ); h = hd(, oc, M).* ha(m) ; % Haig widow applied to ideal impulse resp. clf, subplot(3) stem(, h, filled ), stem_fix axis([ M- -.,.3]), xlabel, ylabel h[] title(spritf( Haig Lowpass, M=%d, M)) subplot(3) H = freqz(h,, om); plot(om, *log(abs(h)), -, om, *log(max(hd(om,oc),eps)), -- ) xlabel \omega, ylabel H(\omega) (db) title Magitude Respose axisy(-8, ), xaxis_pi -p p/4 p % savefig fig_wi_example.3 Haig Lowpass, M=5 Magitude Respose.5. h[].5..5 H() (db) π π/4 π If we are usatisfied with the width of the trasitio bad, or the sidelobe amplitude, or the passbad ripple, the what could we do? Icrease M, try other widow fuctios, ad/or try other filter desig methods. Filter desig by widowig i MATLAB The fir commad i MATLAB is its tool for widow-based FIR filter desig. I the precedig example, we could have simply typed h = fir(4,.5, ha(5), oscale ) to get exactly the same desig.

12 8. c J. Fessler, May 7, 4, 3:8 (studet versio) Example. Digital phaser or flage. Time varyig pole locatios i cascade of st ad d-order allpass filters, Sum output of allpass cascade with origial sigal, creatig time-varyig otches FIR differetiator H d () = j skim 8..6 Hilbert trasform (9 phase shift) H d () = j sg() skip 8..3 Desig of FIR filters by frequecy samplig For the frequecy samplig method of FIR filter desig, to desig a M-poit FIR filter we specify the desired frequecy respose at a set of equally-spaced frequecy locatios: H d (), k =,..., M. = π M k I other words, we provide equally spaced samples over [, π). Picture Recall from the DTFT formula that if h d [] is ozero oly for =,..., M, the Thus, at the give frequecy locatios, we have H d () = M = h d [] e j. ( ) M π H d M k = h d [] e j π M k, k =,..., M. = This is the formula for the M-poit DFT discussed i Ch. 5 (ad i EECS 6). So we ca determie h d [] from { ( H π d M k)} M by usig the iverse DFT formula (or h = ifft(h) i MATLAB): k= h d [] = M M k= ( ) π H d M k e j π M k. This will be the impulse respose of the FIR filter as desiged by the frequecy samplig method. If we wat( h d [] to be real, the H d () must be Hermitia symmetric, i.e., Hd () = H d( ) = H d (π ). So if we specify H π d M k) to be some value, we kow that Hd( π M k) ( = H d π π M k) ( = H π d M (M k)) (, so H π d M (M k)) is also specified. Thus, software such as MATLAB s fir commad oly requires ( H d () oly o the iterval [, π]. If h d [] is to be real, the it also follows that H d (π) = H π ) M d M must be real valued whe M is eve. If you choose H d () to be liear phase, the the desiged h[] will be liear phase. But you are ot required to choose H d () to be liear phase! The book discusses may further details, but the above big picture is sufficiet for this class.

13 c J. Fessler, May 7, 4, 3:8 (studet versio) 8.3 Example. The treble boost revisited. Double the amplitude of all frequecy compoets above c = π/. % fig_freq_sample.m Hd = ilie( exp(-i*om*(m-)/).* ( + (abs(om) > pi/)), om, M ); M = 9; ok = [:(M-)]/M * *pi; Hk = Hd(mod(ok+pi,*pi)-pi, M); % trick: [-pi,pi] specificatio of H(\omega) h = ifft(hk); h = reale(h, war ); % h = fir(m-, [.5.5 ], [ ], boxcar(m) ); om = lispace(-pi,pi,); clf, pl = 3; subplot(pl+), plot(om, abs(hd(om,m))) hold o, stem(ok(ok >= ), abs(hk(ok >= )), filled ), stem_fix, hold off xlabel \omega, ylabel H_d(\omega) axisy([.5]), xaxis_pi -p p/ p subplot(pl+), stem(:(m-), h, filled ), stem_fix, title(spritf( M=%d,M)) xlabel, ylabel h[], axis([ M ]) H = freqz(h,, om); subplot(pl+3), plot(om, abs(h), -, om, abs(hd(om,m)), --, ok, abs(hk), o ) xlabel \omega, ylabel H(\omega), axisy([.5]), xaxis_pi -p p/ p % savefig fig_freq_sample.5.5 M=9.5 H d ().5 h[].5 H() π π/ π π π/ π I MATLAB, use h = fir(m-, [.5.5 ], [ ], boxcar(m) ) to desig (almost) the above filter.

14 8.4 c J. Fessler, May 7, 4, 3:8 (studet versio) 8..4 Optimum equiripple liear-phase FIR filters The widow method has a mior disadvatage, that it is difficult to precisely specify p ad s, sice these two result from the smearig. All we really specify is c, the cutoff. A ideal liear-phase desig procedure would be as follows. Specify p, s, δ, δ, ad ru a algorithm that returs the miimum M that achieves that desig goal, as well as the impulse respose h[], =,..., M, where, for liear phase, h[] = h[m ]. To my kowledge, there is o such procedure that is guarateed to do this perfectly. However, we ca come close usig the followig iterative procedure. Choose M, ad fid the liear-phase h[] whose frequecy respose is as close to H d () as possible. If it is ot close eough, the icrease M ad repeat. How ca we measure closeess of two frequecy respose fuctios? Pictures of H d () ad H(). Possible optios iclude the followig. π π π H d() H() d, average absolute error π π π H d() H() d, average squared error π π π H d() H() W() d, average weighted squared error [ ] π /p, π H d() H() p W() d weighted Lp error, p π max H d () H(), maximum error (p = ) max W(H d () H()), maximum weighted error (p = ) I this sectio, we focus o the last choice, the maximum weighted error betwee the desired respose ad the actual frequecy respose of a FIR filter. We wat to fid the FIR filter that miimizes this error. How to fid this? Not by brute force search or trial ad error, but by aalysis! If W() the Cosider the case of a lowpass filter. Note that H d () = E() = W(H d () H()) = W() H d () H(). { e j(m )/, c,, > c = e j(m )/ H dr () where H dr () = Also recall that for M eve ad h[] liear phase ad symmetric, So H r () is the sum of M/ cosies. The error ca be simplified as follows: H() = e j(m )/ H r (), where H r () = M/ = {, c,, > c. ( ( )) M h[] cos. E() = W() H d () H() = W() e j(m )/ H dr () e j(m )/ H r () = W() H dr () H r (). Thus we oly eed to cosider the real parts of the desired vs actual frequecy respose.

16 8.6 c J. Fessler, May 7, 4, 3:8 (studet versio) Example. digital crossover for audio F s = 44kHz F pass = 4.4kHz p = πf pass /F s =.π F stop = 6.6kHz s = πf stop /F s =.3π 3 bads: pass, trasitio, stop M = 6 for ow; revisit shortly. % fig_remez_example.m M = 6; f = [..3 ]; Hdr = [ ]; W = [ ]; h = remez(m-, f, Hdr, W); stem(:m-, h, filled ) freqz(h,, ) Equiripple (zoom).5. M=6 h[].5..5 H() (db) H() π π.4. π.4 π Haig widow (zoom).5. h[].5..5 H() (db) H() π π.4. π.4 π

17 c J. Fessler, May 7, 4, 3:8 (studet versio) 8.7 Further illustratios of the tradeoffs. The error has to go somewhere: passbad(s), stopbad(s), ad/or trasitio bad(s). 3 W() 3 W() 3 W() /π.4.5 /π.4.5 /π.374 H(), M=3.6 H(), M=3.78 H(), M= /π /π /π 3 W() 3 W() 3 W() /π.4.5 /π.4.55 /π H(), M=3 H(), M=7 H(), M= /π /π /π 8..7 Compariso of FIR methods skim

18 8.8 c J. Fessler, May 7, 4, 3:8 (studet versio) 8.3 Desig of IIR Filters from Aalog Filters Why IIR? With IIR desigs we ca get the same approximatio accuracy (of the magitude respose) as FIR but with a lower order filter. The tradeoff is oliear phase. Aalog filter desig is a mature field. There are well kow methods for selectig RLC combiatios to approximate some desired frequecy respose H d (F ). Geerally, the more (passive) compoets used, the closer oe ca approximate H d (F ), (to withi compoet toleraces). Essetially all aalog filters are IIR, sice the solutios to liear differetial equatios ivolve ifiite-duratio terms of the form t m e λt u(t). Oe way to desig IIR digital filters is to piggyback o this wealth of desig experiece for aalog filters. What do aalog filters look like? Ay RLC etwork is describe by a liear costat coefficiet differetial equatio of the form N k= α k d k dt k y a(t) = M k= β k d k dt k x a(t) with correspodig system fuctio (Laplace trasform of the impulse respose) M k= H a (s) = β ks k N k= α ks. k Each combiatio of N, M, {α k }, {β k } correspods to some arragemet of RLCs. Those implemetatio details are uimportat to us here. The frequecy respose of a geeral aalog filter is H a (F ) = H a (s). s=jπf Overview Desig N, M, {α k }, {β k } usig existig methods. Map from s plae to z-plae somehow to get a k s ad b k s, i.e., a ratioal system fuctio correspodig to a discrete-time liear costat coefficiet differece equatio. Ca we achieve liear phase with a IIR filter? We should be able to use our aalysis tools to aswer this. Recall that liear phase implies that H(z) = z N H ( z ), so if z is a pole of the system fuctio H(z), the z is also a pole. So ay fiite poles (i.e., other tha or ) would lead to istability. There are o causal stable IIR filters with liear phase. Thus we desig for the magitude respose, ad see what phase respose we get IIR by approximatio of derivatives skim s = z T 8.3. IIR by impulse ivariace skim h[] = h a (T ) z = e st

19 c J. Fessler, May 7, 4, 3:8 (studet versio) IIR filter desig by biliear trasformatio Suppose we have used existig aalog filter desig methods to desig a IIR aalog filter with system fuctio M k= H a (s) = β ks k N k= α ks. k For a samplig period T s, we ow make the biliear trasformatio s = T s z + z. This trasformatio ca be motivated by the trapezoidal formula for umerical itegratio. See text for the derivatio. Defie the discrete-time system fuctio H(z) = H a (s) s= z. T s + z This trasformatio yields a ratioal system fuctio, i.e., a ratio of polyomials i z. This H(z) is a system fuctio whose frequecy respose is related to the frequecy respose of the aalog IIR filter. Example. Cosider a st-order aalog filter with a sigle pole at s = α Picture where α >, with system fuctio H a (s) = α + s. How would you build this? Usig the followig RL voltage divider, where V out (s) = i(t) R R+sL V i(s) = H a (s) = R/L R/L+s. vi(t) + R vout(t) L Applyig the biliear trasformatio to the above (Laplace domai) system fuctio yields: + z H(z) = H a (s) s= z = = T s + z α( + z ) + = ( z ) T s = α + z T s + z α + /T s + z /T s α /T s + α z = α + /T s + z (α + /T s ) + (α /T s )z + z [ pz = α + /T s p + + /p ] pz, where p = αt ) s (, ). What is h[]? h[] = + αt α+/t s ( p δ[] +( + /p)p u[]. s Im(z) p Re(z) Where did zero at z = come from?

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