Find Magnitude Responce at $\omega=\dfrac{\pi}{2M}$ where $M$ is the number of channel in QMF I have frequency response of low pass filter

    M=8  %%%channel 
    x=8*512 
    d1=fir1(N-1,wc/pi,w); 
    % L=length(d1); 
    % for k=1:L 
    %     b2(k)=((-1)^k)*d1(k); 
    % end 
    %%%%%%Frequency response 
    h1=fft(d1,x);
    a1=max(abs(h1)); 
    m1=20*log10(abs(h1/ a1));
    c=0:x-1 
    plot(f1/x,m1)


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