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No-Tune-9435 t1_j5vzhjj wrote

Reply to comment by blutfink in what information an impulse response graph provides about headphones? by MEGA_AEOIU792

You didn’t answer my question at all though. If we want to represent an FR from 20-20khz, what time window would be appropriate for the impulse response? Or have you actually not done this math before?

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TaliskerBay22 t1_j5w0hju wrote

Look here is an example, the time domain lasts for a ms and the Fourier response extends from 0 to some tens of KHz. Plug it in Matlab and tweak it as you like

% Define time-domain signal
dt = 0.000001; % time step (s)
t = 0:dt:0.001; % time vector (s)
x = cos(50000.*t-0.0004).*exp(-(t-0.0003).^2/0.00006^2); % time-domain signal
% Perform FFT
X = fft(x); % FFT of time-domain signal
f = (0:length(X)-1)/(dt*length(X)); % frequency vector (Hz)
% Plot results
figure;
subplot(2,1,1);
plot(t,x);
xlabel('Time (s)');
ylabel('Amplitude');
title('Time-domain signal');
subplot(2,1,2);
Amplitude=abs(X);
semilogy(f(1:length(X)/20),Amplitude(1:length(X)/20));
xlabel('Frequency (Hz)');
ylabel('Amplitude');
title('Frequency-domain signal (FFT)');

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blutfink t1_j5w0fke wrote

In the extreme case? Infinitesimally short time. The FT of a Dirac pulse is a flat, constant response from 0 Hz to infinity Hz (or, in the case of DFT, Nyquist frequency).

As I said, the FT of a Gaussian is a Gaussian. Really sharp in the time domain means really wide in the frequency domain. If you do not understand why that is, your intuition about FTs will be flawed.

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blutfink t1_j5w2dc2 wrote

Maybe this helps: The impulse response in OPs post image basically does not have to be windowed at all, it’s decently close to zero at the edges. Plug it into a FT and you’ll get a decently accurate FR.

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