Finite Impulse Response (FIR) Digital Filters (II) Ideal Impulse Response Design Examples Yogananda Isukapalli 1 • FIR Filter Design Problem Given H(z) or H(ejw), find filter coefficients {b0, b1, b2, ….. bN-1} which are equal to {h0, h1, h2, ….hN-1} in the case of FIR filters. 1 z-1 z-1 z-1 z-1 x[n] h0 h1 h2 h3 hN-2 hN-1 1 1 1 1 1 y[n] Consider a general (infinite impulse response) definition: ¥ H (z) = å h[n] z-n n=-¥ 2 From complex variable theory, the inverse transform is: 1 n -1 h[n] = ò H (z)z dz 2pj C Where C is a counterclockwise closed contour in the region of convergence of H(z) and encircling the origin of the z-plane • Evaluating H(z) on the unit circle ( z = ejw ) : ¥ H (e jw ) = åh[n]e- jnw n=-¥ 1 p h[n] = ò H (e jw )e jnwdw where dz = jejw dw 2p -p 3 • Design of an ideal low pass FIR digital filter H(ejw) K -2p -p -wc 0 wc p 2p w Find ideal low pass impulse response {h[n]} 1 p h [n] = H (e jw )e jnwdw LP ò 2p -p 1 wc = Ke jnwdw 2p ò -wc Hence K h [n] = sin(nw ) n = 0, ±1, ±2, …. ±¥ LP np c 4 Let K = 1, wc = p/4, n = 0, ±1, …, ±10 The impulse response coefficients are n = 0, h[n] = 0.25 n = ±4, h[n] = 0 = ±1, = 0.225 = ±5, = -0.043 = ±2, = 0.159 = ±6, = -0.053 = ±3, = 0.075 = ±7, = -0.032 n = ±8, h[n] = 0 = ±9, = 0.025 = ±10, = 0.032 5 Non Causal FIR Impulse Response We can make it causal if we shift hLP[n] by 10 units to the right: K h [n] = sin((n -10)w ) LP (n -10)p c n = 0, 1, 2, …. 20 6 Causal FIR (N = 21) Impulse Response Notice the symmetry: h[n] = h[N-1-n] which satisfies the linear phase condition. • Filter Transfer Function 20 jw - jnw jw H LP (e ) = åhLP[n]e since z = e n=0 7 20 Implies -n H LP (z) = åhLP[n]z n=0 Filter Implementation for 21 coefficients Magnitude of filter Frequency Response for 11, 21 and 31 coefficient lengths 8 Low Pass filter Frequency Response in dB for 11 and 21 coefficients Low pass filter phase response for 21-coefficient design 9 The above figure implies jw j(w-p ) H HP (e ) = H LP (e ) n hHP[n] = hLP[n](-1) 10 Now: ¥ ¥ jw - jn(w-p ) - jnw n H HP (e ) = åhLP[n]e = åhLP[n]e (-1) n=-¥ n=-¥ ¥ n - jnw = åhLP[n](-1) e n=-¥ hHP[n] Example: Analog high pass filter specification Ideal high pass 0 10 kHz f Given: fs = 50kHz, it follows:w = WT = 2pf/fs = 0.4p rad Digital specification: 0.4p = p - wa 11 \ wc (LPF) = p - wa = p - 0.4p = 0.6p First design ideal LPF with wc= 0.6p 1 Answer : h [n] = sin(0.6pn) LP np n = 0, ±1, ±2, …. ±¥ Follows now: n hHP[n] = hLP[n](-1) (-1)n-I h [n] = sin(0.6p (n - I)) HP p (n - I) The I indicates the length of the filter Hence the above filter is a causal FIR high pass digital filter 12 13 14 15 Bandpass FIR filter design: H(ejw) K -p -wu -wo-wl 0 wl wo wu p w Shift to right: H(ejw) ejwo Shift to left: H(ejw) e-jwo Follows: hBP[n] = [2cosnwo ]hLP[n] n = 0, ±1, ±2, …. ±I where wu - wl = 2wc (lowpass) w + w w = u l o 2 16 Example: H(ejW) 1 fs = 50kHz -15 -10 0 10 15 f in kHz Digital specifications: 2p (104 ) w = = 0.4p l (5)(104 ) 2p (104 )(1.5) w = = 0.6p u (5)(104 ) wo = 0.5p w - w w = u l = 0.1p c 2 17 First design LPF: 1 hLP[n] = sin(0.1pn) n = 0, ±1, ±2, …. np Follows: hBP[n] = [2cos(0.5pn)hLP[n] n = 0, ±1, ±2, …. 18 19 20 Bandpass filter frequency response 21 Homework # 6 1. Design a length-5 FIR bandpass filter with an antisymmetric impulse response h[n], i.e. h[n]= -h[4-n], 0£n £4, satisfying the following magnitude response jp jp values : | H ( e 4 ) | = 0 . 5 and | H ( e 2 ) | = 1 . Determine the exact expression for the frequency response of the filter designed. 2. An FIR filter of length 5 is defined by a symmetric impulse response i.e. h[n]= h[4-n], 0£n £4,. Let the input to this filter be a sum of 3 cosine sequences of angular frequencies: 0.2 rad/samples, 0.5 rad/samples, and 0.8 rad/samples, respectively. Determine the impulse response coefficients so that the filter passes only the midfrequency component of the input. 3. The frequency response H(ejω) of a length-4 FIR filter with a real and antisymmetric impulse response has the following specific values: H(ejπ)=8, and H(ejπ/2)= -2+j2. Determine H(z). 22 Frequency transformations of ideal filters. High pass Filter The ideal low filter frequency response is shown below. hLP[n] jw represents the non-causal impulse response of HLP(e ) jw HLP(e ) -2p 0 2p -2p-wc -p -wc wc p 2p-wc w Shifting the low pass filter frequency by p, we get j(w-p) HLP(e ) -2p 0 2p -p-wc -p -p+wc p-wc p p+wc w 23 The above figure is the frequency response of an ideal high pass jw filter with a cutoff frequency of p-wc (HHP(e ) ).Hence, jw j(w-p ) H HP (e ) = H LP (e ) From the shifting property of the Fourier transform, we get jpn hHP [n] = hLP [n]e n hHP [n] = (-1) hLP [n] 1 where h [n] = sin(nw ) LP pn c Hence, to design an ideal high pass filter with cutoff frequency of wa, first design a ideal low pass filter with a cutoff frequency of (p-wa). Then use the above transformation to obtain the impulse response of the ideal high pass filter. 24 Band pass Filter The frequency response for an ideal band pass filter is shown below. The center frequency is wo. jw HBP(e ) K -p -wu -wo wl wl wo wu p jw Low pass frequency response HLP(e ) -p -wc wc p w Looking at the frequency responses of a band pass filter and a low pass filter, we can observe that a band pass filter is obtained by shifting the low pass filter to the left and to the right by wo and adding the two shifted responses. 25 Shifting the low pass filter by wo to the left and to the right and adding , we get j(w-wo) j(w+wo) HLP(e )+HLP(e ) -p -wc-wo -wo wc-wo -wc+wo wo wc+wo p Comparing the ideal band pass filter and the above figure, we can observe that wl = -wc + wo wu = wc + wo w - w w = u l c 2 w + w w = u l o 2 26 jw The frequency response of the band pass filter HBP(e ) is jw j(w-wo ) j(w+wo ) H BP (e ) = H LP (e ) + H LP (e ) From the shifting property of the Fourier transform , we have jwon - jwon hBP [n] = hLP [n]e + hLP [n]e jwon - jwon hBP [n] = hLP [n]{e + e } hBP [n] = hLP [n]{2cos(nwo )} To design a band pass filter, first design a low pass filter with cutoff frequency wc given by w - w w = u l c 2 And use the above transformation to get the impulse response of the band pass filter . 27 Band stop Filter jw The frequency response of an ideal band stop filter (HBS(e )) is jw HBS(e ) -p 0 p -wu -wo -wl wl wo wu w Looking at the frequency responses of band stop and band pass filters, we observe that the band stop filter is obtained by subtracting the band pass from 1. jw jw H BS (e ) = 1- H BP (e ) 28 ¥ ¥ - jnw - jnw å hBS [n]e = 1- å hBP [n]e n=-¥ n=-¥ If we work out the first few terms, we get j2w jw - jw - j2w ...... + hBS [-2]e + hBS [-1]e + hBS [0] + hBS [1]e + hBS [2]e j2w jw - jw - j2w = ...... + hBP [-2]e + hBP [-1]e +1- hBP [0] + hBP [1]e + hBP [2]e Hence, the impulse response coefficients for a band stop filter are obtained as hBS [0] = 1- hBP [0] = 1- 2hLP [0] hBS [n] = -hBP [n] = -hLP [n]{2cos(nwo } n ¹ 0 29 Low pass K h [n] = sin(nw ) n = 0,±1,±2,...,±I LP np c H(ejw) K -p -wc wc p High pass n hHP[n] = (-1) hLP[n], n = 0,±1,±2,...,±I H(ejw) K 30 -p -(p -wc) (p- wc) p Band pass hBP[n] = [2cos(nwo )]hLP[n], n = 0,±1,±2,...,±I w + w w - w = 2w , w = u l u l c 0 2 H(ejw) K -p -wu -wo wl wl wo wu p Band stop hBS [0] = K - hBP[0], hBS [n] = -hBP[n], n = ±1,±2,...,±I wu + wl wu - wl = 2wc , w0 = 2 H(ejw) K -p -wu -wo wl wl wo wu p 31 Examples 1.