Why is the Fourier transform of a Dirac comb a Dirac comb?

Question

This doesn't make sense to me, because the Heisenberg inequality states that $\Delta t\Delta \omega$ ~ 1.

Therefore when you have something perfectly localized in time, you get something completely distributed in frequency. Hence the basic relationship $\mathfrak{F}\{\delta(t)\} = 1$ where $\mathfrak{F}$ is the Fourier transform operator.

But for the Dirac comb, applying the Fourier transform, you receive another Dirac comb. Intuitively, you should also get another line.

Why does this intuition fail?

Answer 1

I believe that the fallacy is to believe that a Dirac comb is localized in time. It isn't because it is a periodic function and as such it can only have frequency components at multiples of its fundamental frequency, i.e. at discrete frequency points. It can't have a continuous spectrum, otherwise it wouldn't be periodic in time. Just like any other periodic function, a Dirac comb can be represented by a Fourier series, i.e. as an infinite sum of complex exponentials. Each complex exponential corresponds to a Dirac impulse in the frequency domain at a different frequency. Summing these Dirac impulses gives a Dirac comb in the frequency domain.

Answer 2

Your intuition fails because you're starting with wrong assumptions. Heisenberg's uncertainty doesn't say what you think it says. As you already say in your question, it's an inequality. To be precise, it's

$$\Delta t \cdot \Delta f \geq\frac{1}{4\pi}$$

There is no reason why the uncertainty product has to be close to its lower bound for all signals. In fact, the only signals that achieve this lowest bound are Gabor atoms. For all other signals, expect it to be larger and possibly even infinite.

Answer 3

electrical engineers play a little fast and loose with the Dirac delta function, which the mathematicians insist is not a function (or, at least, not a "regular" function, but is a "distribution"). the mathematical fact is that if $f(t)=g(t)$ "almost everywhere" (which means at every value of $t$ except for a countable number of discrete values), then $$ \int f(t)dt = \int g(t)dt $$.

well the functions $f(t)=0$ and $g(t)=\delta(t)$ are equal everywhere except at $t=0$, yet we electrical engineers insist that their integrals are different. but if you set aside this little (and, in my opinion, non-practical) difference, the answer to your question is:

1. the Dirac comb function $$ \mathrm{III}_T(t) \triangleq \sum\limits_{k=-\infty}^{+\infty} \delta(t - kT) $$ is a periodic function of period $T$ and therefore has a Fourier series: $$ \mathrm{III}_T(t) = \sum\limits_{n=-\infty}^{+\infty} c_n \ e^{j 2 \pi n t/T} $$

2. if you blast out the coefficients, $c_n$, of the Fourier series you get:

$$ \begin{align} c_n & = \frac{1}{T}\int\limits_{t_0}^{t_0+T} \mathrm{III}_T(t) e^{-j 2 \pi n t/T} dt \\ & = \frac{1}{T}\int\limits_{-T/2}^{T/2} \delta(t) e^{-j 2 \pi n t/T} dt \quad \quad (k=0)\\ & = \frac{1}{T}\int\limits_{-T/2}^{T/2} \delta(t) e^{-j 2 \pi n 0/T} dt \\ & = \frac{1}{T} \quad \quad \forall n \\ \end{align} $$

3. so the Fourier series for the Dirac comb is

$$ \mathrm{III}_T(t) = \sum\limits_{n=-\infty}^{+\infty} \frac{1}{T} \ e^{j 2 \pi n t/T} $$

which means you're just summing up a bunch of sinusoids of equal amplitude.

4. the Fourier Transform of a single complex sinusoid is:

$$ \mathfrak{F} \left\{ e^{j 2 \pi f_0 t} \right\} = \delta(f-f_0) $$

and there is this property of linearity regarding the Fourier Transform. the rest of the proof is an exercise left to the reader.

Answer 4

I shall try to give an intuition. The way we could probably think is : "One Dirac delta gives us a 1 in frequency domain. Now I give infinite number of Dirac deltas. Shouldn't I get a higher DC?" Now let us see whether by adding all those frequency components mentioned in the Dirac comb in the frequency domain(FD), we get another Dirac comb in time domain(TD). We are adding continuous waveforms and getting deltas at discrete points. Sounds weird.

Coming back to the FD. We have a Dirac comb with spacing $ \omega_0 $. To put it in words, we have deltas at $ 0,\pm\omega_0,\pm2\omega_0,\pm3\omega_0 $ and so on. We thus have a DC and infinite number of cosines, namely $ \cos(\omega_0 t), \cos(2\omega_0 t), \cos(3\omega_0 t) $ and so on.

Let's consider points in time domain corresponding to $ t = \frac{2n\pi}{\omega_0}$. All the above cosine waves will give us value 1. Hence they all add up and give us non zero value at those points. Now what about any other t? We need to get convinced that they will all add up to zero.

Now deviating slightly, let's consider a waveform $cos(kn) ; n = 0,1,2,3,4...\infty$. We know that unless k can be expressed as a fraction multiplied by $\pi$, it's aperiodic. What does that mean? There is not a single repeating sample. Each of the samples are unique. Looking it from another perspective, we have infinite number of samples which are unique and part of a cosine wave. This means taking all the infinite points, we will be able to construct a single CONTINUOUS cosine wave completely once. What if $cos(kn)$ is periodic? We already know that the sum of samples will be zero periodically based on value of k. Hence, sum of all the samples of $cos(kn)$ will give us zero for any value of k, except $k = 2\pi$'s multiple.

Returning back to our original problem : We now take an arbitrary $t=t_0 \neq 2r\pi$. Now we have $ \cos(0\omega_0 t_0)[dc] + \cos(\omega_0 t_0) + \cos(2\omega _0 t_0) + \cos(3\omega_0 t_0) $....as the value at $t=t_0$. But we have already proved this infinite sum =0 for any t except $ t=\frac{2n\pi}{\omega _0} $, where all these cosines add up to give dirac deltas.