I have frequency response of low pass filter

    M=8  %%%channel 
    % L=length(d1); 
    % for k=1:L 
    %     b2(k)=((-1)^k)*d1(k); 
    % end 
    %%%%%%Frequency response 
    m1=20*log10(abs(h1/ a1));

I have to find magnitude responce at $\omega=\dfrac{\pi}{2M}$ i.e., $h_1$ at $\omega=\dfrac{\pi}{2M}$

  • $\begingroup$ So what's your problem with finding the magnitude at $\omega=\pi/2M$? $\endgroup$ – Matt L. Aug 23 '15 at 19:44

Your code has too many unassigned variables for us to execute it. Your FFT freq axis, in radians/sample, goes from 0 –to- 2pi. The equivalent mathematical FFT integer freq axis index goes from 0 –to- 4095. You want to know h1 at w = pi/(2M) radians/sample. Let's use variable k to represent the mathematical FFT integer index corresponding to your w.

So we can set two ratios equal to each other: k/4096 = w/2pi. Next we write k/4096 = (pi/(2M))/2pi. This gives us k = 4096/(4M) = 4096/32 = 128 as the math index corresponding to your w. Because of Matlab's unpleasant indexing method the magnitude response you want to examine is your h1(k+1) = h1(129) FFT sample. [By the way, as long as your M is an integer power of two, k will be an integer. If M is not an integer power of two, you'll have to use either a 'floor()' or 'ceil()' command to compute the mathematical FFT integer index k.]

  • $\begingroup$ if i use freqz instead of fft then what is magnitude responce at ω = π/2M.Code Given is d1=fir1(N-1,wc/pi,w); [h1,fre1]=freqz(d1,1,x) a1=max(abs(h1)); plot(fre1/pi,a1) $\endgroup$ – Rajat Sharma Aug 24 '15 at 19:11
  • $\begingroup$ If you insist on using the freqz() command instead of the fft() command, use the command [h1,fre1]=freqz(d1,1,x, 'whole') to plot the full freq magnitude response of your d1 coefficients. The '2' on the right side of the freq axis is equivalent to 2pi radians/sample (Fs samples/second). $\endgroup$ – Richard Lyons Aug 25 '15 at 11:49

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