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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?

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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.

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  • $\begingroup$ Yes, neither periodic comb is localized in its respective independent variable (time / frequency). $\endgroup$ – Peter K. Feb 6 '15 at 14:00
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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.

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    $\begingroup$ Right, but the main fallacy is to think that a Dirac comb is localized in time. It isn't because it is periodic. So the uncertainty theorem doesn't say anything useful about a Dirac comb. $\endgroup$ – Matt L. Feb 6 '15 at 13:28
  • $\begingroup$ @MattL., that's not how I understand the original question. I think he's actually arguing that the dirac train is entirely delocalised in its native domain and therefore should Fourier transform to something very localised. $\endgroup$ – Jazzmaniac Feb 6 '15 at 13:59
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    $\begingroup$ OK, looks like there's a misunderstanding what the OP means by 'another line'. I thought this refers to a flat spectrum (just like the spectrum of a Dirac impulse he referred to before). But you thought this refers to a spectral line, i.e. one single frequency. At least now I understand how your answer could answer the OP's question. $\endgroup$ – Matt L. Feb 6 '15 at 14:34
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    $\begingroup$ @MattL., I actually thought he means the usual graphical representation of Dirac distributions when he writes "line". In any case, he will have to clarify as the question can be really read in at least two different ways. $\endgroup$ – Jazzmaniac Feb 6 '15 at 15:26
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    $\begingroup$ well, the "standard" definition is a physical statement relating momentum and position uncertainties (specifically standard deviations) and has an $\hbar$ in there. and even so, in this case, you have to define what is meant by "$\Delta t$" and "$\Delta f$". that constant (which you specify as $\frac{1}{4 \pi}$) can't be too far from unity (in the log scale), but it need not be $\frac{1}{4 \pi}$ except due to a specific definition for "$\Delta t$" and "$\Delta f$". $\endgroup$ – robert bristow-johnson Feb 9 '15 at 21:11
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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:

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

  1. 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.

  1. 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.

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    $\begingroup$ @Jazzmaniac, that's a falsehood. when have i ever been condescending toward mathematicians? (me thinks you're projecting a bit.) BTW, it's been 38 years since i have had 2 semesters of functional analysis at the graduate level. don't remember everything, but i sure do remember what a metric space is, a normed metric space (i think they were sometimes called "Banach spaces"), and inner product spaces (sometimes called "Hilbert spaces"), and what a functional is (maps from one of these to a number). and i know what linear spaces are. about $\delta(t)$, i don't mind them being naked. $\endgroup$ – robert bristow-johnson Feb 7 '15 at 12:53
  • $\begingroup$ You go on with a wrong argument that suggests mathematicians don't get 1 when they integrate over a Dirac distribution. Well, you can't demonstrate any better that you haven't understood the Dirac distribution, even if you have taken a class on functional analysis. It doesn't need electrical engineers like you to "fix" mathematics. And I will keep pointing that out to you until you stop talking about mathematicians like that. It's entirely your choice. $\endgroup$ – Jazzmaniac Feb 7 '15 at 15:30
  • $\begingroup$ that's a falsehood, too, @Jazzmaniac. i am saying that, consistent with what mathematicians tell us, the Dirac delta function is not really a function (even though we electrical engineers don't worry about that distinction and deal with it as if it were a function) because if it were a function that was zero almost everywhere, the integral would be zero. why do you keep misrepresenting me? what is the ax you're grinding? $\endgroup$ – robert bristow-johnson Feb 9 '15 at 20:59
  • $\begingroup$ @robertbristow-johnson "electrical engineers play a little fast and loose with the Dirac delta function." Paul Dirac was an electrical engineer. Claude Shannon was also an electrical engineer. I admonish you from making such general and inaccurate statements. You claim to be an electrical engineer and clearly understand distribution theory. $\endgroup$ – Mark Viola May 4 '16 at 3:49
  • $\begingroup$ nearly every undergraduate electrical engineering textbook on Linear System Theory or Signals and Systems or some similar name, will introduce and treat the Dirac Delta as a limiting case of a "nascent delta". e.g. : $$ \delta(t) = \lim_{a \to 0}\frac{1}{a \sqrt{\pi}} \mathrm{e}^{-t^2/a^2} $$ or some other unit area pulse function that you can make skinny. i would not be surprized that in published papers, folks like Shannon or Dirac (didn't know that) would stick with the conservative facts: $$ \int f(t) \delta(t-\tau) \ dt = f(\tau) $$ and $$ \delta(t)=0 \quad \forall \ t \ne 0$$. $\endgroup$ – robert bristow-johnson May 4 '16 at 5:34
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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.

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