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!


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
    Oct 25 '15 at 22:06
  • $\begingroup$ The other answer was also correct. I switched the accepted one. $\endgroup$
    – clay
    Oct 26 '15 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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