# Why does the Fourier Transform of the impulse look so different from the Fourier Transform of the impulse train?

The fourier transform of the impulse functions is:

$$\delta(t) \longleftrightarrow 1$$

The shifted delta:

$$\delta(t-nT) \longleftrightarrow e^{-j \Omega nT}$$

But the fourier transform of the impulse train is:

$$\sum_{n=-\infty}^{\infty} \delta(t-nT) \longleftrightarrow \frac{2\pi}{T} \sum_{k=-\infty}^{\infty} \delta \left(\Omega - k\tfrac{2\pi}{T} \right)$$

How could I go about proving the impulse train fourier transform using the definition of the fourier transform of the shifted delta? It seems confusing... Thank you!

• Hi Darklink! have you also considered the extension of Fourier transform into periodic functions ? In other words, have you considered the periodic nature of the impulse train ? – Fat32 Sep 16 '18 at 19:34
• I think many (most?) signals and systems textbooks contain a derivation of this result... have you seen any? – MBaz Sep 16 '18 at 21:29
• Ive seen derivations using other ways, but I'd like to find an intuitive way to deriving it using the definition of the fourier transform of the shifted delta function. – Darklink9110 Sep 16 '18 at 23:40
• Suggest going to Wikipedia and looking up Dirac comb. – robert bristow-johnson Sep 17 '18 at 4:48
• If you're looking for an intuitive rather than a mathematical explanation, you can read this article. dsprelated.com/showarticle/1014.php – Qasim Chaudhari Sep 18 '18 at 0:13

This answer shows that $\sum\limits_{n=-\infty}^{\infty} e^{j n \Omega T}$ is actually the Fourier Series of $\frac{2\pi}{T} \sum_{k=-\infty}^{\infty} \delta(\Omega - k\tfrac{2\pi}{T})$.

$$\delta(t-nT) \frac{2\pi}{T} e^{j n \Omega T}$$ which means $$\sum\limits_{n = -\infty}^{\infty} \delta(t-nT) \longleftrightarrow \sum\limits_{n = -\infty}^{\infty} e^{j n \Omega T}$$ Now consider the exponential Fourier series of $\frac{2\pi}{ T} \sum_{k=-\infty}^{\infty} \delta(\Omega - k\Omega_s)$, because this function is periodic, so we can say $$\frac{2\pi}{ T} \sum_{k=-\infty}^{\infty} \delta(\Omega - k\Omega_s) = \sum\limits_{n=-\infty}^{\infty} c_n e^{j n \Omega T}$$ where $$c_n = \frac{1}{\Omega_s}\int\limits_{-\frac{\Omega_s}{2}}^{\frac{\Omega_s}{2}} \frac{2\pi}{ T} \sum_{k=-\infty}^{\infty} \delta(\Omega - k\Omega_s) e^{j n \Omega T} d (\Omega T)$$ and $\Omega_s = \frac{2\pi}{T}$ by definition.

Take $\alpha = \Omega T$, $$c_n = \frac{1}{\Omega_s} \frac{2\pi}{ T} \sum_{k=-\infty}^{\infty} \int\limits_{-\frac{T\Omega_s}{2}}^{\frac{T\Omega_s}{2}}\delta(\frac{\alpha }{T} - k\Omega_s) e^{-j n \alpha} d \alpha$$ The term $\frac{1}{\Omega_s} \frac{2\pi}{T} = 1$ and take now $$\beta = \frac{\alpha}{T}$$ $$c_n = \sum_{k=-\infty}^{\infty} \int\limits_{-\frac{\Omega_s}{2}}^{\frac{\Omega_s}{2}}\delta(\beta - k\Omega_s) e^{-j n \beta T} d \beta \tag{1}$$ We know that $$\int_{a}^b \delta(x - x_0) f(x) \ dx = f(x_0) \quad \text{if } a < x_0 < b$$ and zero elsewhere. So the integral in $(1)$ contains one value within $[-\frac{\Omega_s}{2},\frac{\Omega_s}{2}]$ i.e. for $k = 0$, so $$c_n = \sum_{k=-\infty}^{\infty} \int\limits_{-\frac{\Omega_s}{2}}^{\frac{\Omega_s}{2}}\delta(\beta - k\Omega_s) e^{-j n \beta T} d \beta = e^{-j n (0) T} = 1$$ So $c_n = 1$ for all $n$. We have just proved that $$\frac{2\pi}{ T} \sum_{k=-\infty}^{\infty} \delta(\Omega - k\Omega_s) = \sum\limits_{n=-\infty}^{\infty} e^{j n \Omega T}$$ So whether you say $\sum\limits_{n=-\infty}^{\infty} e^{j n \Omega T}$ or $\frac{2\pi}{ T} \sum_{k=-\infty}^{\infty} \delta(\Omega - k\Omega_s)$, it is the same thing.

Since you asked for an intuitive answer, I can offer this:

The Fourier transform of any repeated function is the transform of the function itself as an envelope of impulses spaced at the repetition rate.

Why is this?

This is because any frequencies that are not integer harmonics of the base repetition rate cannot exist; if they did the resulting function could not repeat in that repetition interval!

The unique characteristic of the integer harmonics is that these are the only frequency components that will start at the exact same position at the start of the next repetition interval of the fundamental frequency.

To further add to this: I always found the following properties in my slide below interesting and useful, especially since they work both ways: when I don't have an intuitive view in one domain, I can often swap time and frequency as I do have an intuitive view in the other. For example the first property: if it is periodic in one domain, it must be discrete in the other (and vice versa). Above we talked about a waveform repeating in time, being discrete in frequency. The opposite view is a waveform that is discrete in time; it repeats in frequency (such as the spectrum after A/D conversion).

As @MattL points out in the comments, there are cases where this won’t occur exactly as stated, such as shifting the discrete waveform in frequency (or in time) by an irrational amount (see example he have in the comments)—- in this case the waveform in the other domain appears to repeat, but it never actually does in a strict mathematical sense.

• It should be noted that not every discrete signal is periodic in the other domain. E.g., $X(f)=\delta(f-1)+\delta(f-\sqrt{2})$ is discrete in the frequency domain but not periodic in the time domain. – Matt L. Sep 17 '18 at 14:59
• @MattL. Isn’t that a frequency translated sinusoid? I see, because of the irrational location it can never repeat in time even though it is a sinusoid on a carrier midway between 1 and $\sqrt(2)$? Am I following? If so, that’s a great example and case- thanks Matt! – Dan Boschen Sep 17 '18 at 15:11
• Yes, you got it, it's because of the irrational frequency ratio. – Matt L. Sep 17 '18 at 15:23
• I would leave the general (inaccurate) rule as it is, and just add a remark that there are special cases where it doesn't work out as expected. And these cases occur when the shifts (in frequency or time) don't have a greatest common divisor. – Matt L. Sep 17 '18 at 16:01
• Yes, they must lie on a uniform grid, but arbitrarily many can be zero, e.g., $\delta(f-1)+\delta(f-100)$ would be fine. And the value zero must also lie on the same grid. – Matt L. Sep 18 '18 at 18:47