Question

4.1. Consider an FIR filter of length 5 with a symmetric impulse response. i.e. hinl = h14-n, input consisting of a sum of three cosine sequences of angular frequencies: 0.3 rad/samples, 0.7 rad/samples, respectively, is applied to this filter. Determine the impulse response coercicm passes only the midfrequency component of the input. =h[4 - n), 0 <n< 4. An rad/samples, 0.4 rad/samples, and ponse coefficients so that the filter

Answer 1

0.72 . Givon 5-Length FIR filter . h cm) = (4-n) Oensys and given the Impulse Responses symmetrical. Let noo hlo) = h (4) = a. . n=1 h( = h (3) = b n=2 h(2) = n(2) = C The Impulse response coebbelents are · h(n) = f a, b, a, b, a} mo, Hlevw) = a + beviot ce-j2w7be-j3w Ha éjyw 20 (heres syns Mimg cannedsvifo.3 n) + cos Corun) + coslo.tn CamScanner

3 (7) = n(evia) a = 'Wo frequency of 3(m). the Input signal. 7. Given that w=0.3 rad/samples, w=0.7 rad sam .عام gets not passes through the filter, and W=0.4 rad) samples Input passes through- w= 0.4 rad) samples -mid frequency: TH Cevi) |w = 0.3 = 0 , 1 Cevio) ] = out = 0 that the pass band gain Let of us assume filter = 1 Hace deo) Ivo = 0.4 = 1. CamScanner

Hceso) = (at beswt te jaw the j3w+bauw). I Hie uw) = (at ae duw) of (below & be o3o) + ce daw - e-izola e izog ae diw) te 2w ( be jo, be sis) t ce2 è và e té = COS X e vzw (za cos zw + 26 cos w + c) # Cewis) = 1 Acemia) | <¢. or) luce vio) levo 2 COS 26 + 2bcoSwtc. Stanned with CamScanner

WE 0,3 l Cevie) = 0 We 0.3 da cos 2W + ab cos w+ cm-0.2 = 0 2a cos Co. 6) + 26 cos (0.3) +C = 0 1.65 a + 1091 6 +C =0 © w=0.4 1 Cevio) so + 2a cos (e x0.7) + 26 cos (0.7) +C =0 0.33a + 1.52b + c = 6 - ©w=ooy (H Cevie) 1 = 2. +6 = 4 1 da +2b Cos (0.4 cos (2x0.4) 1.39 at 1.846+ c = 1 — Cs solving with the d® & ©© CamScanner

1.65 a + 1.91b+c= O both = -0.332 -1526 substitute these in 4. gly equations. 1,65 a +1.91 b - 0.33 -1-52b = 0 1.32 a + 0.36bro - © 1-39 a 1.84 6-0.332-1.5 2b = 1. 1-06 a + 0.32b = 1 - a b = -1-32 a 0.39 653.38 (-a) 5 substitute in 6 1.06 a 0.32 X 338 *a = 1 -0.02166 = 1 a = - 46.29 16= 156.4602 Scanned with CamScanner

c = -0.33 (A) -1.52 (b) C = -0.33 (46.29) – 1-52 ( 156646) C= 15.2757 - 237.8). C = - 222.54 hcm = [ a, b, c, b, a] I-46.29 156.46, -222-54, 156.46, for the ... are the coefhcients where a, b, c Scanned with Carfilterner