I want to walk through the derivation of the frequency representation of an impulse train.

The definition of the impulse train function with period $T$ and the frequency representation with sampling frequency $\Omega_s = 2\pi/T$ that I would like to derive is:

\begin{align*} s(t) &= \sum\limits_{n=-\infty}^{\infty} \delta(t - nT) \\ S(j\Omega) &= \frac{2\pi}{T} \sum\limits_{k=-\infty}^{\infty} \delta(\Omega - k\Omega_s) \\ \end{align*}

Using the exponential Fourier series representation of the impulse function and applying the Fourier transform from there results in:

\begin{align*} s(t) &= \frac{1}{T} \sum\limits_{n=-\infty}^{\infty} e^{-jn\Omega_s t} \\ S(j\Omega) &= \int_{-\infty}^\infty s(t) e^{-j\Omega t} dt \\ S(j\Omega) &= \int_{-\infty}^\infty \frac{1}{T} \sum\limits_{n=-\infty}^{\infty} e^{-jn\Omega_s t} e^{-j\Omega t} dt \\ S(j\Omega) &= \frac{1}{T} \int_{-\infty}^\infty \sum\limits_{k=-\infty}^{\infty} e^{-j(k\Omega_s + \Omega) t} dt \\ \end{align*}

To get from there to the end result, it would seem that the integration would need to be over a period of $2\pi$. Where $\Omega = -k\Omega_s$, the exponent would be $e^0$ and integrate to $2\pi$ and for other values of $\Omega$, there would be a full sine wave that would integrate to zero. However, the limits of integration are negative infinity to positive infinity. Can someone explain this? Thanks!


2 Answers 2


You correctly figured out that the occurring integrals don't converge in the conventional sense. The easiest (and definitely non-rigorous) way to see the result is by noting the Fourier transform relation

$$1\Longleftrightarrow 2\pi\delta(\Omega)$$

By the shifting/modulation property we have

$$e^{j\Omega_0t}\Longleftrightarrow 2\pi\delta(\Omega-\Omega_0)$$

So each term $e^{jn\Omega_s t}$ in the Fourier series transforms to $2\pi\delta(\Omega-n\Omega_s)$, and the result follows.

  • $\begingroup$ This is perfect and way easier than I made it out to be. thank you so much!!! $\endgroup$
    – clay
    Commented Oct 25, 2015 at 22:06
  • $\begingroup$ The other answer was also correct. I switched the accepted one. $\endgroup$
    – clay
    Commented Oct 26, 2015 at 20:56

@MattL suggested a nice, simple way to see the above result.

But if you want to see the result in the normal analysis equations you mentioned, you can do like below.

Say S(t) is a periodic train of impulses.So S(t) can be written as

$$\ S(t)= \sum_{n=-\infty}^{\infty} \delta(t-nT)$$

Now if you take the fourier series of S(t),you can write S(t) as

$$S(t) =\sum_{n=-\infty}^{\infty} C_ke^{jnw_ot} $$

Where $C_n$ are exponential fourier series coefficients and $w_o$ is the fundamental frequency .

So from exponential fourier series we know that

$$C_n= (1/T)\int_{-T/2}^{T/2} S(t)e^{-jnw_ot} dt$$

Now in the above expression substitute the value of S(t) from the first expression.

So $$C_n = (1/T)\sum_{n=-\infty}^{\infty}\int_{-T/2}^{T/2} \delta(t-nT)e^{-jnw_ot} dt$$

Now, you have to make an observation, if you observe the integral, it's from -T/2 to +T/2. During this integral period, observe that only a single impulse $\delta(t)$ exists.All the other impulse functions in the summation occur after T/2 or before -T/2. So in total the above equation for $C_n$ can be written as

$$C_n = (1/T)\sum_{n=-\infty}^{\infty} \delta(t)e^{-jnw_ot}$$

From sifting property we can write the above as

$$C_n = (1/T)e^{-jw_on(0)} = (1/T)$$

Now put this value of $C_n$ in the first S(t) equation

$$S(t) = (1/T)\sum_{n=-\infty}^{\infty} e^{jnw_ot}$$

Now find the fourier transform of above equation

$$1\Longleftrightarrow 2\pi\delta(w)$$

$$e^{jw_ot}\Longleftrightarrow 2\pi\delta(w-w_o)$$

So the fourier transform is $$S(jw) = (2\pi/T)\sum_{n=-\infty}^{\infty} \delta(w-nw_o)$$

This should help.


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