FFT of Truncated(2048) Ideal Impulse Response

h = [h(N/2+1:N),h(1:N/2)]; % zero-centered form (almost)
hw = wzp .* h; % Apply window to ideal impulse response
Hw = fft(hw);  % Frequency response we really get

% Compute total stopband attenuation:
ierr = norm(Hw(N/2+2:N))/norm(Hw)
     = 0.0378 = 3.8 percent

For spectral displays, rotate buffer so negative frequencies are on the left and dc is at $k=N/2$ :

Hwp = [Hw(N/2+2:N), Hw(1:N/2+1)]; % Neg. freqs on left
Hwpn = abs(Hwp); Hwpn = Hwpn/max(Hwpn);
plot(f,20*log10(Hwpn)); grid;
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