# Interpret periods from fourier transfrom graph of periodic impulse signals

I have obtained the fourier transform of a signal with following graph

My question (though might be too junior for the professionals in signal processing field) is: how can we extract the period of 30 from the fourier transform mag graph?

The source code is

Fs = 1;
Ts = 1/Fs;
dt = 0:Ts:30*4-Ts;

y1 = [
100 ; zeros(Fs*29,1);
100 ; zeros(Fs*29,1);
100 ; zeros(Fs*29,1);
100 ; zeros(Fs*29,1);
];

subplot(4,1,1)
plot(dt, y1, 'r')
title('impulse happen in time series', 'fontsize', 18);

X = fft(y1);
X_mag = abs(X)
subplot(4,1,2)
plot(X_mag)
title('fourier transform mag', 'fontsize', 18);


Assuming you are in the continuous-time domain, given the periodic impulse train with a period of $$T_0$$

$$x(t) = \sum_k \delta(t - k T_0)$$

then its CTFT will be another periodic impulse train in the frequency domain $$X(\omega) = \frac{2\pi}{T_0} \sum_k \delta( \omega - k\omega_0)$$

Where the period $$\omega_0$$ of the frequency impulses is related to the period of the time impulses as

$$\omega_0= \frac{2\pi}{T_0}$$

hence the time period will be given as:

$$T_0 = \frac{2\pi}{\omega_0}$$

In your case, if $$T_0 = 30$$ then $$\omega_0$$ will be $$\omega_0 = \frac{2\pi}{30} = \frac{\pi}{15}$$

Conversely, given the frequency period as $$\frac{2\pi}{30}$$ then the corresponding time period will be $$T_0 = 30$$.

Note that there will be errors in your computation depending on how you simulate the impulse train or CTFT.

• Thanks for your reply! My original intention is to extract T0 out of Fourier transform in frequency domain, it turns out given the relation of 2*Pi = T0 * W0, I come to the original question again, that is how to extract W0 in the graph. – Johnson Dec 10 '19 at 4:13