Question

3. Consider the real-valued causal linear-phase FIR impulse response, h, hl0, hl]. h40. Write an equation for the group delay of this filter as a function of w radians/sample, for w E [0, T). (Credit will be earned only for a simplified expression.) 시n] ー0.0001, 一0.0004, 一0.0003, 0.0009, 0.0018, 一0.0004. 一0.0043, 一0.0031, 0.0054, 0.0103. = 0.0010, -0.0183, -0.0133, 0.0195, 0.0380, 一0.0016, -0.0677, -0.0569. 0.0923, 0.2999. 0.3981, 0.2999, 0.0923, - 0.0569,0.0677, -0.0016, 0.0380, 0.0195, -0.0133,0.0183, 一0.0010, 0.0 103, 0.0054,一0.0031,一0.0043,一0.0004, 0.00 18, 0.0009,-0.0003,-0.0004,-0.0001

Answer 1

Total 41 samples are present in h[n]. h[0]=-0.0001 h[21]=0.2999 h[1]=-0.0004 h[22] =0.0923 h[2]=- 0.0003 h[23] =-0.0569 h[3]=0.0009 h[24] =-0.0677 h[4] = 0.0018 h[25]=- 0.0016 h[5]=-0.0004 h[26]=0.0380 h[6]=-0.0043 h[27]=0.0195 h[7]=-0.0031 h[28] =-0.0133 h[8] =0.0054 h[29]=-0.0183 h[9]=0.0103 h[30]=-0.0010 h[10]=-0.0010 h[11]=-0.0183 h[31]=0.0103 h[12]=- 0.0133 h[32]=0.0054 h[13]=0.0195 h[33]=-0.0031 h[14]=0.0380 h[34]=-0.0043 h[15]=-0.0016 n[35] =- 0.0004 h[16]=- 0.0677 h[36] =0.0018 h[17]=-0.0569 h[37]=0.009 h[18]=0.0923 h[38] =- 0.0003 h[19]=0.2999 h[39] =- 0.0004 h[20]=0.3981 h[40]=-0.0001

The real valued causal (FIR) filter whose real valued impulse response h[n zero is for n2 N The impulse response is given by, 715-00 H(el) = n[n]e=son [1]e=;® +h[2]e-2jø +........+ H(ekº)={,[N-11e*'" (N-1) ....+ h[n –1]e=i[N-7) 22 (N-1) Here, N is odd then h[n] is symmetric about So, IN-1) un = h - h 2+nah (N- n Now, H(ej@) will be ja N-1) 12 Hele) -1) { le + file . [-17 twth[e of*) +hpolero( #(2)"(N=1) 2010)* +212]471 +2[9:11 -j@(N-1)

Here, A(0) is A(@) = 2*[D]cos o (9,7) + 2H[1]cos (~=? -.) +...+([=1] As it is a real value function and the phase of H(ed) will be, CH(evº) - (N-1) Write down the formula for the group-delay in terms of o. d(ZH (ejm)) group-delay =-- do - (N-1) do --/-(N-1) 2 _(N-1) Calculate the group-delay. (N-1) group-delay =