1 Lecture 11: LTI FIR filter design Instructor: Dr. Gleb V. Tcheslavski Contact: gleb@ee.lamar.edu Office Hours: Room 2030 Class web site: http://ee.lamar.edu/gleb/ dsp/index.htm ELEN 5346/4304 DSP and Filter Design Fall 2008 2 Preliminary considerations M 1 H ( z ) hk z k FIR filter: (11.2.1) k 0 transfer function unit-pulse response M 1 BTW, using our “rational notation”: M 1 H ( z ) hk z k k 0 k b z k k 0 a0 ;a0 1 (11.2.2) Here M-1 is the filter order. M is the number of filter’s coefficients. Assuming that M is odd: H ( z ) h0 h1 z 1 h2 z 2 ... hM 2 z ( M 2) hM 1 z ( M 1) M 3 M 1 M 1 M 1 M 1 2 k k 2 2 2 z hM 1 hk z hk z 2 M 1 k 0 k 2 ELEN 5346/4304 DSP and Filter Design Fall 2008 (11.2.3) 3 Preliminary considerations For an FIR filter to have a linear phase: hk hM* 1k (11.3.1) That means an even symmetry of the coefficients. H ( z) z z M 1 2 M 1 2 M 3 M 3 M 1 M 1 2 2 k k h 2 2 hk z hM 1 k z M 1 2 k 0 k 0 M 3 M 1 M 1 2 k k h 2 2 h z h z k M 1 k M21 k 0 (11.3.2) Remember: FIR filters are always stable (no poles other than at the origin)! ELEN 5346/4304 DSP and Filter Design Fall 2008 4 Preliminary considerations Since the filter is stable: H (e j ) e e M 1 j 2 M 1 j 2 M 3 M 1 M 1 2 j k j k jhk jhk h 2 2 h e e e e k M21 k 0 M 3 2 M 1 h 2 h cos k h k k 2 M21 k 0 (11.4.1) Rc (e j ) Here Rc (e j ) is a real function of frequency. As a result of symmetry: H e ELEN 5346/4304 DSP and Filter Design j e j M 1 2 Fall 2008 Rc e j (11.4.2) 5 Preliminary considerations M 1 if Rc e j 0 2 H e j M 1 if R e j 0 c 2 Therefore: (11.5.1) Generalized linear phase to design ANY amplitude of frequency response: LPF, We can use Rc e HPF, BPF,… j changes sign, the phase undergoes an abrupt change of 180 When Rc e j 0 In M is even: M 2 2 Rc (e ) 2 j k 0 ELEN 5346/4304 DSP and Filter Design M 1 hk cos k hk 2 Fall 2008 (11.5.2) 6 Preliminary considerations An FIR filter will also have a linear phase in the case of odd symmetry of filter coefficients (antisymmetric impulse response), i.e.: hk hM* 1k It can be shown that (11.6.1) M 3 2 M 1 Rs (e j ) 2 hk sin k hk ;M isodd k 0 2 (11.6.2) M 2 2 M 1 Rs (e j ) 2 hk sin k hk ;M iseven k 0 2 Therefore: BTW: ELEN 5346/4304 H e j e M 1 j 2 jRs e j (11.6.3) (11.6.4) H jdifferentiator (11.6.5) H j sign Hilberttransformer (11.6.6) DSP and Filter Design Fall 2008 7 Preliminary considerations Thus: M 1 if Rs e j 0 2 2 H e j 3 M 1 if R e j 0 s 2 2 (11.7.1) Generalized linear phase Summarizing, a unit-pulse response of a GLP FIR filter must satisfy: hk h * M 1k (11.7.2) M 1 H ( z ) hk z k (11.7.3) k 0 H e j e M 1 j 2 Rc e j LPF , HPF , BPF , BSF ,... j jR e differentiators,Hilberttransformers s We only need to specify (M-1)/2 (odd M) or M/2 unique coefficients (even M). ELEN 5346/4304 DSP and Filter Design Fall 2008 (11.7.4) 8 Preliminary considerations Combining (11.7.2) and (11.7.3), we arrive at: M 1 H ( z ) hk z k k 0 z ( M 1) M 1 h k 0 * M 1 k z k M 1 m M 1 k hm* z m( M 1) m0 M 1 * m ( M 1) * * h z z H 1 z m (11.8.1) m0 hasazeroatz z0 re j hasazeroatz 1 1 j e * z0 r 1 Example: if M = 5: Has two pairs of reciprocal zeros. firls(4,[0 0.25],[1 0]); ELEN 5346/4304 DSP and Filter Design Fall 2008 Imaginary Part H ( z) h0 h1z 1 h2 z 2 h3 z 3 h4 z 4 0.5 4 0 -0.5 -1 -1 -0.5 0 Real Part 0.5 1 9 Preliminary considerations There are four types of real GLP FIR filters: Type I: symmetric hn, odd M (number of coefficients) – even order. Type II: symmetric hn, even M Type III: antisymmetric hn, odd M Type IV: antisymmetric hn, even M indicate “structural zeros” ELEN 5346/4304 DSP and Filter Design Fall 2008 10 Preliminary considerations This is where structural zeros come from… 1. For M = 4, symmetrical pulse response ( Type II FIR): H ( z) h0 h1z 1 h2 z 2 h3 z 3 evensymmetry h0 h1z 1 h1z 2 h0 z 3 H ( z) z 1 h0 h1 h1 h0 0 (11.10.1) 2. For M = 4, antisymmetrical pulse response ( Type IV FIR): H ( z) h0 h1z 1 h2 z 2 h3 z 3 oddsymmetry h0 h1z 1 h1z 2 h0 z 3 H ( z) z 1 h0 h1 h1 h0 0 (11.10.2) 3. For M = 5, antisymmetrical pulse response ( Type III FIR): H ( z) h0 h1z 1 h2 z 2 h3 z 3 h4 z 4 oddsymmetry h0 h1z 1 h2 z 2 h1z 3 h0 z 4 H ( z) z 1 h0 h1 h2 h1 h0 h2 0 0 H ( z) z 1 h0 h1 h2 h1 h0 h2 0 0 ELEN 5346/4304 DSP and Filter Design Fall 2008 (11.10.3) 11 Preliminary considerations As a result of structural zeros… Type II FIR HPF would have a zero at - specs will be violated! Type III FIR HPF would have a zero at - specs will be violated! Type IV FIR LPF would have a zero at 0 - specs will be violated! If something like tis happens, increase or decrease the FIR order to change the type of your filter. We need to be careful with selection of filter orders! ELEN 5346/4304 DSP and Filter Design Fall 2008 12 1. Least-square error minimization The desired magnitude response (something we specify): H d e j DTFT jn h e d ,n (11.12.1) n Filter coefficients could be found as: hd ,n 1 2 H d e j e jn d (11.12.2) Example: Ideal LPF hLPF ,n 1 2 jc n jc n 2 j sin c n c sin c n e e j n 1 e d 2 2 j n c n c c (11.12.3) Not causal, infinite length! We need to preserve the main features while making the filter causal. ELEN 5346/4304 DSP and Filter Design Fall 2008 13 1. Least-square error minimization The filter specified by (11.12.3) has an infinite impulse response. Therefore, we need to truncate it at some point to make an FIR filter. As a criterion for such truncation, we need to minimize the approximation error (the difference between the desired and the truncated frequency responses). Therefore, the objective is to find a finite-duration impulse response sequence, whose DTFT would approximate the desired frequency response. The magnitude of the frequency response of the “truncated filter”: H t e j ht ,n e jn U n L Where L and U are points at which the pulse response was truncated. We need to minimize the least-square integral error of approximation. ELEN 5346/4304 DSP and Filter Design Fall 2008 (11.13.1) 14 1. Least-square error minimization In particular: 1 min R 2 R h n t ,n H e H e j j t U d hd ,n ht ,n hd ,n 2 2 n L 2 L 1 h n d ,n 2 d (11.14.1) n U 1 hd ,n 2 (11.14.2) Apparently, to minimize R (LS error), we select ht,n = hd,n for n = L…U. Moreover, GLP requires symmetry of filter coefficients, therefore L = -U. The best finite length approximation of the ideal infinite length impulse response in the LS error sense is obtained by truncation. To make the resulting FIR causal, we need to shift it by L radians. ELEN 5346/4304 DSP and Filter Design Fall 2008 15 1. Least-square error minimization Relating the point of truncation to the FIR filter order: L = (M – 1)/2 hLPFIR ,n M 1 sin c n 2 c M 1 c n 2 (11.15.1) 0.2 0.1 0 Magnitude (dB) 0.2 0.1 0 -0.1 -50 0 n Non-causal ELEN 5346/4304 DSP and Filter Design 50 -0.1 0 20 40 60 n Causal Fall 2008 80 100 Phase (degrees) 0.3 n 0.3 h h n Note: filter length M (number of filter coefficients) can be both even or odd. 0 -50 -100 0 0.2 0.4 0.6 0.8 1 Normalized Frequency ( rad/sample) 0 -2000 -4000 0 0.2 0.4 0.6 0.8 1 Normalized Frequency ( rad/sample) 16 2. Windowing effects We can view the truncation of the infinite impulse response of an ideal filter as windowing. In frequency domain, this operation is equivalent to convolution of a frequency response of the window function with the frequency response of an ideal filter. ht ,n hd ,n wn (11.16.1) from the Modulation theorem: 1 Ht e j 2 H d e j ( ) W e j d 1.2 W R: (11.16.2) |Ht()| 1 |W()| |Hd()| 0.8 Gibbs phenomenon (comes from truncated Fourier series) 0.6 0.4 0.2 Ripples of equal magnitude in PB and SB Transition band – another window artifact 0 -0.2 -1 -0.5 0 / ELEN 5346/4304 DSP and Filter Design Fall 2008 0.5 1 17 2. Windowing effects: window types As discussed previously, different window functions can be used… Fixed windows: Window MLW (4/M) PSL (dB) TB (2/M) Max SB ripple (dB) 0.9 -13 0.9 -21 Hanning 2 -31 3.1 -44 Hamming 2 -41 3.3 -53 Blackman 3 -57 5.5 -74 Rectangular Window properties ELEN 5346/4304 DSP and Filter Design Fall 2008 Filter properties 18 2. Windowing effects: window types Rectangular: (11.18.1) 1 2 n 2 2 n W 1 cos sin ,n 0,1,...M 1 2 M 1 M 1 (11.18.2) N n Hanning: Hamming: Blackman: ELEN 5346/4304 WnR 1,n 0,1,...M 1 2 n WnM 0.54 0.46cos ,n 0,1,...M 1 M 1 (11.18.3) 2 n 4 n W 0.42 0.5cos 0.08cos ,n 0,1,...M 1 (11.18.4) M 1 M 1 B n DSP and Filter Design Fall 2008 19 2. Windowing effects: window types Adjustable windows: Kaiser window: 2 2n I0 1 1 M 1 ,n 0,1,...M 1 WnK I0 x 2 I 0 ( x) 1 m 1 m! Where: m 2 Normally, 15-20 terms in the summation are sufficient. Note: if = 0, WnK = WnR. ELEN 5346/4304 DSP and Filter Design Fall 2008 (11.19.1) (11.19.2) 20 2. Windowing effects: window types Properties: TB (2/M) Max SB ripple (dB) 2 1.5 -29 3 2.0 -37 4 2.6 -45 5 3.2 -54 7 4.5 -72 8 5.1 -81 Hanning Hamming Blackman Kaiser window is the most frequently used for the FIR filter design. ELEN 5346/4304 DSP and Filter Design Fall 2008 21 2. Windowing: experimental design For the given transition band: f s p Stop-band attenuation: A 20lg s Order estimate: M Adjustable parameter: 2 A 7.95 1 14.36 f (11.21.1) (11.21.2) (11.21.3) 0.1102 ( A 8.7) A 50dB 0.5842 ( A 21)0.4 0.07886 ( A 21)21 A 50 0 A 21 (11.21.4) M controls transition band, changes ripples. Note: in practice, ripples in SB and PB are approximately equal. If this does not hold, need to select minimum of s, p. ELEN 5346/4304 DSP and Filter Design Fall 2008 22 2. Windowing: experimental design Example 11.1: design an FIR LPF using a Kaiser window for the following specs: p 0.3 ;s 0.5 ; A 40dBminSBattenuation s p 0.5 0.3 f 0.1 The transition band: 2 2 A 7.95 40 7.95 1 1 23.32 14.36 f 14.36 0.1 Order: M Parameter: 0.5842 ( A 21)0.4 0.07886 ( A 21) 3.4 We select M = 24, = 4. The filter is given by (11.15.1) with the cutoff frequency: c The transfer function: p s 2 0.4 sin c n ht ,n wn ; L n L n The resulting filter is a type II FIR which is ok for LPF. ELEN 5346/4304 DSP and Filter Design Fall 2008 23 3. Frequency sampling The desired frequency response H d e j M 1 DTFT h n 0 j n e d ,n (11.23.1) can be specified by its frequency samples at equally spaced (discrete) frequencies: 2 k k M Where = 0: = 0.5: M 1 0,1,... ,M odd 2 k 0,1,... M 2 ,M even 2 0nooffset 0.5offsetby M ELEN 5346/4304 DSP and Filter Design Fall 2008 (11.23.2) (11.23.3) (11.23.4) 24 3. Frequency sampling The sampled frequency response will be: H d ,k H d e j M 1 k hd ,ne j 2 ( k ) n M ,k 0,1,...M 1 (11.24.1) n 0 Evaluating IDFT: 1 hn M M 1 H k 0 d , k e j 2 ( k ) n M ,n 0,1,...M 1 (11.24.2) Therefore, we can compute the filter coefficients from the specified M frequency samples. Since hn is real: Hk H *M k Since hn is symmetric, we only need to specify either (M+1)/2 (M is odd) or M/2 (M is even) frequency samples to determine the pulse response. ELEN 5346/4304 DSP and Filter Design Fall 2008 (11.24.3) 25 3. Frequency sampling We can rewrite the FIR filter frequency response (11.7.4) as: M 1 j j 2 H e e ,symmetrichn hM 1 n types r j H e M 1 j H e j e 2 2 ,antisymmetrich h n M 1 n types V r (11.25.1) Sampled at the frequencies k 2 k ;k 0,1,...M 1 M (11.25.2) the response becomes: H k ELEN 5346/4304 DSP and Filter Design 2 Hr k e M Fall 2008 2 M 1 j k 2 2 M (11.25.3) 26 3. Frequency sampling 0symmetrichn antisymmetrichn Here: (11.26.1) For simplicity, we specify the following set of real frequency samples: Gk 2 1 H r k ;k 0,1,...M 1 M k H k Gk e j k e Therefore: 2 M 1 j k 2 2 M 0 0 ; 4differentcases... 1 2 ELEN 5346/4304 DSP and Filter Design Fall 2008 (11.26.2) (11.26.3) 27 3. Frequency sampling 0symmetrichn 0nooffset :k 2 k M Case 1: Hk Gk e j k e 1 hn M 1 M 1 M 1 M 2 M 1 k M 2 ELEN 5346/4304 j DSP and Filter Design 1 M M 1 j k M Gk e j k e j 2 kn M j 2 M k M 2 e j k M Gk e j k M (11.27.1) k 2 k 2 M 1 j j kn j j kn 1 U Hke Gk e e Gk e M e M Gk e M e M M k 0 k 0 k 0 k U 1 k 2 ( M k ) 2 U U j j kn j j ( M k ) n M M M GM k e e M G0 Gk e e k 1 k 1 2 1 2 1 U U j k n j k n M 2 M 2 j j 2 n Gk e e e e j 1;e j 2 n 1 G0 Gk e k 1 k 1 U 2 1 (11.27.2) G 2 G cos k n 0 k ;n 0,1,...M 1 2 k 1 M M 1 2 kn M j Fall 2008 28 3. Frequency sampling k 2 Gk 1 H r M Hk HM* k Since hn is real: 2 H r M 2 M 1 2 jM k k e k (11.28.1) (11.28.2) 2 j M M k * 2 Hr M k e M M 1 2 (11.28.3) Since Hr is real: 2 M 1 2 2 j M k Hr k e M 2 2 j M M k Hr M k e M M 1 2 2 M 2 2 2 j 2 j 2 2 j M 1 H r k Hr M k e e H M k r e M M M k k 2 2 j M k j k j H r k 1 H r e e 1 (11.28.4) M k e M M ELEN 5346/4304 DSP and Filter Design Fall 2008 29 3. Frequency sampling Therefore: Gk GM k (11.29.1) The number of coefficients (frequency samples) to be specified: M 1 2 ;odd M U M 1;evenM 2 (11.29.2) - forced zero at z = -1 Since the unit-pulse response is already evaluated, we can estimate the corresponding frequency response as its zero-padded DFT: Hˆ e j DFT hn 0 and check whether the specifications are satisfied. ELEN 5346/4304 DSP and Filter Design Fall 2008 30 3. Frequency sampling 0symmetrichn 1 2ffset :k 2 (k 0.5) M Case 2: The desired frequency samples: Gk 0.5 2 1 H r M k 1 k 2 (11.30.1) 2 U 2 1 1 hn Gk 0.5 2sin k n ;n 0,1,...M 1 M k 0 2 2 M M 1 2 ;odd M U M 1;evenM 2 ELEN 5346/4304 DSP and Filter Design Fall 2008 (11.30.2) (11.30.3) 31 3. Frequency sampling 1antisymmetrichn 0noffset :k 2 k M Case 3: The desired frequency samples: k 2 k Gk 0.5 1 H r M (11.31.1) 2 M 1 2 2 k 1 hn Gk sin n ;M isodd M k 1 2 M M 2 2 1 2 n 1 hn 1 GM 2 2 Gk sin M k 1 M ELEN 5346/4304 DSP and Filter Design Fall 2008 1 k n ;M iseven 2 (11.31.2) (11.31.3) 32 3. Frequency sampling 1antisymmetrichn Case 4: 0.5ffset :k 2 k 0.5 M The desired frequency samples: 2 k 0.5 Gk 0.5 1 H r M 2 U 2 1 1 hn Gk 0.5 cos k n ;n 0,1,...M 1 M k 0 2 2 M k M 3 2 ;odd M U M 1;evenM 2 ELEN 5346/4304 DSP and Filter Design (11.32.1) (11.32.2) (11.32.3) Fall 2008 33 3. Frequency sampling 1.2 We specify the desired response in the SB and the PB only… actual specified 1 The stopband attenuation in this case (only SB and PB samples are given) is approximately -20 dB. |H()| 0.8 0.6 0.4 0.2 1.2 0 actual specified 1 0 0.1 0.2 0.3 fractional frequency 0.4 0.5 |H()| 0.8 Alternatively, we can add transition sample(s) and make the frequency response smoother. 0.6 0.4 0.2 0 0 ELEN 5346/4304 0.1 0.2 0.3 fractional frequency DSP and Filter Design 0.4 0.5 The stopband attenuation would be: for one TB sample: approximately -40 dB. for two TB sample: approximately -60 dB. for three TB sample: approximately -80 dB. Fall 2008 34 3. Frequency sampling The transition band sample(s) “optimized” experimentally are: # of TB samples Sample(s) value(s) SB attenuation, dB T1 0.3789795 – if = 0 0.3570496 – if = 0.5 ~ 40 T2 [0.1065 0.5886] ~ 60 T3 [0.025779 0.251635 0.723071] [0.030957 0.27557 0.744348] Note: only one of them is correct. ~ 80 T4 [0.006061 0.09324 0.4082 0.82097] ~ 100 Effects of adding TB sample(s) are increased SB attenuation and wider TB. These samples must be optimized for the situation. Selection of the offset can help too. Note: we can specify any desired frequency response at the samples but not in between! Ultimately, we can view all frequency samples as the TB samples, which leads to an optimal (equiripple) design. ELEN 5346/4304 DSP and Filter Design Fall 2008 35 4. Optimal equiripple design We are interested in a LP filter satisfying the following requirements: 1 p H r e j 1 p ;PB s H r e j (11.35.1) ;SB s Considering four FIR types: Type 1: symmetric pulse response, odd M. hn hM 1n (11.35.2) The real-valued frequency response: M 3 2 M 1 2 M 1 M 1 H r e hM 1 2 hn cos n k n ak cos k 2 2 k 0 n 0 2 j hM 1 ,k 0 2 ak M 1 2 h , k 1, 2,... M 1 2 2 k ELEN 5346/4304 DSP and Filter Design Fall 2008 (11.35.3) (11.35.4) 36 4. Optimal equiripple design hn hM 1n Type 2: symmetric pulse response, even M. (11.36.1) The real-valued frequency response: M 2 2 M 2 M 2 1 M 1 H r e j 2 hn cos n k n bk cos k (11.36.2) 2 2 2 k 1 n 0 M 1 2 (11.36.3) cos k (11.36.4) bk 2hM ,k 1, 2,... 2 k H r e j cos Or: M 1 2 b 2 k 0 k b0 0.5 b1 Where: bk 2 bk bk 1 ;k 1, 2,... bM 2 ELEN 5346/4304 DSP and Filter Design 1 2bM 2 Fall 2008 M 2 2 (11.36.5) 37 4. Optimal equiripple design Type 3: antisymmetric pulse response, odd M. hn hM 1n (11.37.1) The real-valued frequency response: M 3 2 M 1 M 1 H r e j 2 hn sin n k n 2 2 n 0 ck 2hM 1 ,k1, 2,... 2 k M 1 2 M 1 2 c k 1 k sin k (11.37.2) (11.37.3) M 3 2 H r e j sin ck cos k Or: (11.37.4) k 0 c M 3 c M 1 Where: 2 2 ck 1 ck 1 2ck ;2 k M 5 2 c0 0.5c2 c1 ELEN 5346/4304 DSP and Filter Design Fall 2008 (11.37.5) 38 4. Optimal equiripple design Type 4: antisymmetric pulse response, even M. hn hM 1n (11.38.1) The real-valued frequency response: M 2 2 M 2 M M 1 H r e j 2 hn sin n k n d k sin k 2 2 k 1 n 0 M 2 (11.38.3) cos k (11.38.4) dk 2hM ,k1, 2,... 2 k H r e j sin Or: M 2 2 d 2 k 0 k (11.38.2) d M 3 2d M Where: 2 2 d k 1 d k 2d k ;2 k M 3 2 d0 0.5d1 d1 ELEN 5346/4304 DSP and Filter Design Fall 2008 (11.38.5) 39 4. Optimal equiripple design Therefore, we can express the frequency response for all FIR types as: H r e j Q e j P e j P e Where: Filter type 1: 2: hn hM 1n M odd hn hM 1n M even 3: hn hM 1n M odd 4: hn hM 1n M even ELEN 5346/4304 DSP and Filter Design j L k 0 k (11.39.1) cos k (11.39.2) Q(ej) P(ej) M 1 1 cos 2 sin sin Fall 2008 2 2 ak cos k k 0 M 2 2 k 0 M 3 k 0 M 2 k 0 2 bk cos k ck cos k 2 dk cos k 40 4. Optimal equiripple design The desired frequency response: 1inPB H dr e 0inSB j We can compute (11.40.1) H r e j and compare it to H dr e j . Adjusting P() and selecting filter type, we can obtain the desired frequency characteristic. We introduce the weighting function on the approximation error: 1inSB W e s p inPB j If s < p – bigger error in the passband is allowed. ELEN 5346/4304 DSP and Filter Design Fall 2008 (11.40.2) 41 4. Optimal equiripple design The weighted approximation error: E e j W e j H dr e j H r e j W e j H dr e j Q e j P e j H dr e j j j j ˆ e j Hˆ e j P e j W e Q e P e W dr j Q e (11.41.1) Where the modified weighting function and the modified desired frequency response are: Wˆ e j W e j Q e j Hˆ dr e j ELEN 5346/4304 DSP and Filter Design H dr e j Q e Fall 2008 j (11.41.2) (11.41.3) 42 4. Optimal equiripple design For the given error function E(e j), the Chebyshev (mini-max) approximation problem is to determine the filter parameters k that minimize the maximum absolute error over the frequency bands of interest: L j j ˆ ˆ min max E e min max W e H dr e k cos k over k S over k S k 0 j (11.42.1) Usually, the frequency bands of interest are specified as: S PB SB (11.42.2) a disjoint union of the passband and the stopband: 0 - p and s - (for a LPF). That is we “ignore” the transition band. More precisely speaking, error over the TB will not be optimized. The solution of this problem can be found via the alternation theorem. ELEN 5346/4304 DSP and Filter Design Fall 2008 43 4. Optimal equiripple design The alternation theorem L A necessary and sufficient conditions for P e k cos k to be k 0 j ˆ unique and best approximation to H e in S is that the error function j dr E(e j) exhibit at least L + 2 extremal frequencies in S. That is, there must exist at least L + 2 frequencies {i}, such that: 1 2 ... L2 E e ji E e ji1 E e ji max E e j ,i 1, 2,..., L 2 S I.e., error alternates in sign between two successive extremal frequencies. ELEN 5346/4304 DSP and Filter Design Fall 2008 (11.43.1) (11.43.2) (11.43.3) 44 4. Optimal equiripple design: Example Example 11.2: design a LPF. Since both the desired frequency response Hdr and the weighting function W are piecewise constants: dE e j d j dH e d r j j j W e H dr e H r e 0 d d (11.44.1) Consequently, the frequencies {i} that correspond to the peaks of the error function E(e j) also correspond to the peaks of Hr(e j), i.e., where the frequency response meets the specified error tolerance. Since Hr(e j) is a polynomial of degree L: Hr e j L k 0 L k k L cos k k nk cos k cos k 0 n 0 n k (11.44.2) k 0 Hr(e j) has at most L-1 local minima and maxima and, therefore, at most L+1 extremal frequencies (bandages) since we add = 0 and = . Furthermore, the band-edge frequencies p and s are also extrema of the error function. Therefore, E(e j) has at most L+3 extremal frequencies. ELEN 5346/4304 DSP and Filter Design Fall 2008 45 4. Optimal equiripple design: Example However, from the alternation theorem, there are at least L+2 extremal frequencies in E(e j). Thus the error function for the LPF design will have either L+2 or L+3 (extra ripple filters) extremal frequencies. At the specified extremal frequencies n: alternations n Wˆ e jn Hˆ dr e jn P e jn 1 ;n 0,1,...L 1 (11.45.1) Here represents the maximum value of the error function E(e j). 1 n Hˆ dr e jn n 0,1,...L 1 (11.45.2) 1 ˆ e jn n 0,1,...L 1 k cos n k H dr Wˆ e jn k 0 (11.45.3) P e jn Wˆ e jn or alternatively: n L ELEN 5346/4304 DSP and Filter Design Fall 2008 46 4. Optimal equiripple design: Example We can consider the {k} and as the unknown (to be determined) design parameters for the given extremal frequencies. Therefore: 1 cos 0 cos 20 1 cos 1 cos 21 1 cos L 1 cos 2L 1 cos L0 cos L1 cos LL 1 1 Wˆ e 1 L 1 0 Hˆ dr e j0 jL1 1 j Hˆ dr e 1 (11.46.1) L jL1 jL1 ˆ ˆ H e W e dr 1 Wˆ e j0 Note: initially, both the design parameters and the extremal frequencies are unknown. The above system can be solved by the iterative algorithm (Remez exchange algorithm): 1) Guess the set of L + 1 extremal frequencies; 2) Solve for {k} and ; 3) Determine the error function as in (11.39.1); 4) Determine the new set of L + 1 extremal frequencies and go to 2)… Keep running until E(e j) - convergence. ELEN 5346/4304 DSP and Filter Design Fall 2008 47 4. Optimal equiripple design Alternatively, we can compute analytically [Rabiner ‘75]: 0 Hˆ dr e j 1Hˆ dr e j ... L 1Hˆ dr e j 0 0 Wˆ e j0 1 1 Wˆ e j1 ... L1 (1) L 1 Wˆ e jL1 L 1 (11.47.1) L 1 where: 1 k n 0 cos k cos n (11.47.2) nk Therefore, the initial guess of the L + 2 extremal frequencies allows us to compute the maximum value of the error . Thus: P e j k x k ,x cos L k 0 ELEN 5346/4304 DSP and Filter Design Fall 2008 (11.47.3) 48 4. Optimal equiripple design Since we know that the polynomial at the points xn = cos n has the values: P e jn Hˆ e jn dr (1) n ,n 0,1,...L 1 jn ˆ W e (11.48.1) the Lagrange interpolation formula can be used, which leads to L P e j jk P e k k 0 L x x k 0 k x cos xk cos k L 1 k n 0 xk xn Here: nk ELEN 5346/4304 DSP and Filter Design x xk Fall 2008 (11.48.2) k (11.48.3) (11.48.4) (11.48.5) 49 4. Optimal equiripple design Having the solution for P, we can compute the error function: E e j Wˆ e j Hˆ dr e j P e j (11.49.1) on a dense set of frequency points. Usually, 16M frequency points are sufficient. If the error exceeds the estimated tolerance , we select a new set of frequencies corresponding to the L+2 largest peaks of error function and the procedure starts from (11.45.1). New set of critical frequencies will lead to increased . As a result, the algorithm produces the optimal solution with equal ripples in the PB and the SB for the given M. The order of the filter can be estimated as: Mˆ ELEN 5346/4304 DSP and Filter Design 20lg s p 13 14.6 s p 2 Fall 2008 1 (11.49.2) 50 Summary Historically, the window-based algorithm was the first proposed method for FIR design. Its major disadvantage is the lack of precise control of the critical frequencies, such as s and p. These values, in general, depend on the type of the window and the filter length M. The frequency sampling method is attractive since it specifies an arbitrary frequency response at the uniformly spaced frequencies and the transition band is a multiple of 2/M. However, no control for the response in between the samples. The Chebyshev approximation method provides a total control of the filter specifications and may lead to an equiripple design. This way, the approximation error is spread evenly across the PB and SB, which leads to an optimal filter. This method is usually preferred. ELEN 5346/4304 DSP and Filter Design Fall 2008 51 Remarks and conclusions Since orders of FIR filters can be quite high, these filters can introduce a considerable group delay, which is approximately M/2. The following Matlab functions are handy for FIR design: fir1 – FIR filter design using the windowing method fir2 – FIR filter design via the frequency sampling method firpm – Parks-McClellan optimal equiripple FIR design firpmord - Parks-McClellan optimal equiripple FIR order estimator ELEN 5346/4304 DSP and Filter Design Fall 2008 52 Appendix: the ideal transfer functions LPF: HPF: BPF: BSF: ELEN 5346/4304 DSP and Filter Design hLP ,n sin c n , n n (11.52.1) hHP,n 1 c ,n 0 sin c n , n 0 n (11.52.2) hBP ,n sin c 2 n sin c1n , n 0 n n (11.52.3) hBS ,n c 2 c1 1 ,n 0 sin c1n sin c 2 n , n 0 n n (11.52.4) Fall 2008