# The Fourier transform of a “comb function” is a comb function?

Let $f(x) = \sum_{n=-\infty}^{\infty} \delta(x - n)$, where $\delta$ is the Dirac delta function. This function $f$ (a "comb function") is important in signal processing because evenly sampling a function $g$ can be viewed as multiplying $g$ pointwise with $f$. This idea is the first step towards understanding how to approximate the Fourier transform of $g$, given evenly spaced samples of $g$. The next step is to note that $\hat{gf} = \hat{g}*\hat{f}$, where $\hat{f}$ is the Fourier transform of $f$ and $*$ denotes convolution.

Is it true that $\hat f$ is also a comb function? If that statement isn't quite right, what is the correct statement?

I would like to see: 1) an informal, intuitive, nonrigorous but easy derivation of the Fourier transform of $f$ . 2) A rigorous version of the same calculation. (How would you even define $f$ rigorously, is it a distribution?)

(I'd also be interested in recommendations of math textbooks that cover this topic, including the Nyquist sampling theorem, even if it's only an exercise or series of exercises in an analysis textbook.)

• The following should be helpful: mathoverflow.net/questions/14568/… – Justthisguy Aug 20 '16 at 22:12
• Intuitively: The Fourier coefficient at frequency $\omega$ is nonzero iff the sinusoid $e^{ix\omega}$ lines up with the "teeth" of the comb. – Rahul Aug 22 '16 at 23:29
• en.wikipedia.org/wiki/Dirac_comb#Fourier_series – rikhavshah Aug 22 '16 at 23:56
• If I remember correctly from fifty years ago, the only function that is its own fourier transform is the bell curve. For the discrete case, it would be the binomial coefficients (which approach the bell curve when there are many of them). Important in antenna design and optics apodizing. – richard1941 Aug 24 '16 at 21:12
• @richard1941 it is not true that the Bell curve is the only function that is its own fourier transform. As you can see here, the Bell curve times a Hermite polynomial of order $4 n$ is also invariant under Fourier transformation. – Fabian Aug 29 '16 at 21:19

## 3 Answers

Assume an arbitrary periodic function $f(t)$ with period $T$. Consider the Fourier series representation of $f(t)$, in which $\omega_0=\frac{2\pi}{T}$: $$f(t)=\sum_{n=-\infty}^{+\infty}c_n e^{i n \omega_0 t}$$ Take the Fourier transform of the sides: \begin{align} \mathcal{F}\{f(t)\}=&\mathcal{F}\{\sum_{n=-\infty}^{+\infty}c_n e^{i n \omega_0 t}\}\\ =&\sum_{n=-\infty}^{+\infty}c_n\mathcal{F}\{ e^{i n \omega_0 t}\}\\ F(\omega)=&2\pi\sum_{n=-\infty}^{+\infty}c_n\delta(\omega-n\omega_0) \end{align} This means that the Fourier transform of a periodic signal is an impulse train where the impulse amplitudes are $2\pi$ times the Fourier coefficients of that signal.

With $f(t)=\delta(t)$, the Fourier series coefficients are $c_n=\frac{1}{T}$ for all $n$.

Hence, $$\mathcal{F}\{\sum_{n=-\infty}^{+\infty}\delta(t-nT)\}=\frac{2\pi}{T} \sum_{n=-\infty}^{+\infty}\delta(\omega-n\omega_0)$$ or in comb notation: $$\boxed{\mathcal{F}\{\text{comb}_T(t)\}=\omega_0\ \text{comb}_{\omega_0}(\omega)}$$

where $$\text{comb}_A(x)\triangleq\sum_{n=-\infty}^{+\infty}\delta(x-nA)$$

• Thanks, this is great. Do you recommend any books for this topic? Can you elaborate on how you compute the Fourier transform of $e^{j n \omega_0 t}$? – littleO Aug 28 '16 at 0:17
• The content can be found on most relevant books. However, if you look at signal processing textbooks you may find similar topics. The Fourier transform of $e^{jn\omega_0t}$ can be found simply by recalling the "frequency shifting" property of Fourier transform, that is : $\mathcal{F}\{e^{jn\omega_0t}g(t)\}=G(\omega-n\omega_0)$ and the fact that $\mathcal{F}\{1\}=2\pi\delta(\omega)$. – msm Aug 28 '16 at 1:41
• In order to get to $\mathcal{F}\{1\}=2\pi\delta(\omega)$ itself, one needs to accept $\mathcal{F}\{\delta(t)\}=1$ and then use the "duality" property of Fourier transform that is : $\mathcal{F}\{F(-t)\}=2\pi f(\omega)$. If you are interested in why $\mathcal{F}\{\delta(t)\}=1$ as well, assume a form of pulse such as Gaussian, triangle, rectangle, etc. in the time domain and limit the pulse width to zero (i.e. $\delta(t)$). You will get a frequency representation that expands more and more and tends to a constant in limit. – msm Aug 28 '16 at 2:57
• Shouldn't the Fourier coefficients for $\operatorname{comb}_T (t)$ be $\frac{2}{T}$, since we have $$\frac{1}{T} \int_0^T \operatorname{comb}_T (t) e^{-jn\omega_0 t} \operatorname{dt} = \frac{1}{T} \int_0^T \big( \delta(t) + \delta(t-T) \big) \operatorname{dt} \ ?$$ This leads to an extra factor $2$ in your Fourier Transform of the comb function. – Mussé Redi Jun 2 '18 at 15:22
• I'm using the convention $$f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega)e^{i\omega t} \operatorname{dt} \ .$$ Which convention are you using? – Mussé Redi Jun 2 '18 at 15:28

Intuitive Explanation

The Comb is a sum of Time Shifted Dirac Delta.

The Fourier Transform of a Dirac Delta is known to be a constant.

The Fourier Transform of a Time Shifted Function is known to be Fourier Transform of the function multiplied by a complex exponential factor which is $\exp(-i 2 \pi f T)$

Just apply this points to the Comb Function considered as a sum of Time Shifted Dirac Delta with distance $kT$ and you get a sum of Frequency Shifted exponential functions, each of which multiplied by a constant.

Finally use Euler’s Formula to consider complex exponentials as a periodic sinusoidal function and observe that you have constructive interference only in frequencies which are integer multiple of $\frac{1}{T}$

Up to constants and normalizations, this Dirac comb is its own Fourier transform: the assertion of this evaluated on (for example) a Schwartz function, $\sum_{n\in \mathbb Z} f(n) \;=\; \sum_{n\in\mathbb Z} \widehat{f}(n)$, is the Poisson summation formula.