verify that the Fourier Transform of Impulse Train is another Impulse Train

Question

I'm trying to verify that the following Fourier Transform:

$$\sum \limits_{n=-\infty}^{\infty}\delta(t-nT_s) ~~~\underset{\mathcal{F}}\longleftrightarrow ~~~\Omega_s \sum \limits_{n=-\infty}^{\infty}\delta(\Omega-n\Omega_s)$$

Where:

$$\Omega_s = \frac{2\pi}{T_s}$$

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Here's my work so far:

$$y(t) = \sum \limits_{n=-\infty}^{\infty}\delta(t-nT_s)$$

taking fourier transform:

$$Y(\Omega) = \sum \limits_{n=-\infty}^{\infty}\mathcal{F} \Big[\delta(t-nT_s)\Big]\tag{1}$$

So I find two fourier transforms in the table that I think might work:

$$F(t) \underset{\mathcal{F}}\longleftrightarrow 2\pi f(-\Omega)\tag{duality}$$

$$e^{j\Omega_0 t} \underset{\mathcal{F}}\longleftrightarrow 2\pi \delta(\Omega - \Omega_0)\tag{complex exp}$$

Applying duality to the later transform i get:

$$2\pi \delta(t - t_0)\underset{\mathcal{F}}\longleftrightarrow 2 \pi ~e^{-jt_0 \Omega} $$

divide both sides of transform by $2\pi$:

$$\delta(t - t_0)\underset{\mathcal{F}}\longleftrightarrow ~e^{-jt_0 \Omega} $$

Now if I apply this Transform to (1) I get:

$$Y(\Omega) = \sum \limits_{n=-\infty}^{\infty}\mathcal{F} \Big[\delta(t-nT_s)\Big]\tag{1}$$

$$Y(\Omega) = \sum \limits_{n=-\infty}^{\infty}e^{-jnT_s\Omega}$$

How to get this to equal another impulse train? In other words, why doesn't it equal the following like every DSP textbook claims that it does:

$$\sum \limits_{n=-\infty}^{\infty}\delta(t-nT_s) ~~~\underset{\mathcal{F}}\longleftrightarrow ~~~\Omega_s \sum \limits_{n=-\infty}^{\infty}\delta(\Omega-n\Omega_s)$$

Answer 1

The Fourier series of the Dirac Delta comb with period $2\pi$ is

$$\sum_{m=-\infty }^\infty \delta(t-2m\pi)=\sum_{n=-\infty}^\infty c_n e^{int}\tag1$$

where the Fouier series coefficients, $c_n$ are given by

$$\begin{align} c_n&=\frac1{2\pi }\int_{-\pi}^{\pi} \sum_{m=-\infty}^\infty \delta(t-2m\pi)e^{-int}\,dt\\\\ &=\frac1{2\pi }\sum_{m=-\infty}^\infty\int_{-\pi}^{\pi} \delta(t-2m\pi)e^{-int}\,dt\\\\ &=\frac1{2\pi }\int_{-\pi}^{\pi} \delta(t) e^{-int}\,dt\\\\ &=\frac1{2\pi}\tag2 \end{align}$$

Using $(2)$ in $(1)$, we find that

$$\bbox[5px,border:2px solid #C0A000]{\sum_{m=-\infty}^\infty \delta(t-2m\pi )=\frac1{2\pi}\sum_{n=-\infty}^\infty e^{int}}\tag3$$

Can you fill in the details?

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EDIT:

I thought it might be instructive to mention that the relationship

$$\sum_{m=-\infty}^\infty \delta(t-2m\pi )=\frac1{2\pi}\sum_{n=-\infty}^\infty e^{int}$$

can be used to prove the Poisson Summation Formula.

Let $\phi(t)$ be a test function on the Schwartz Space of functions and let $\Phi(\omega)$ denote the Fourier Transform of $\phi(t)$ given by

$$\Phi(\omega)=\int_{-\infty}^\infty \phi(t)e^{i\omega t}\,dt$$

Then, we have in distribution

$$\int_{-\infty }^\infty \phi(t)\sum_{n=-\infty }^\infty e^{int}\,dt=\sum_{n=-\infty }^\infty \Phi(n)\tag4$$

Moreover, using $(3)$ the left-hand side of $(4)$ is also given by

$$\int_{-\infty }^\infty \phi(t)\left(2\pi \sum_{n=-\infty }^\infty \delta(t-2n\pi )\right)\,dt=2\pi \sum_{n=-\infty }^\infty \phi(2n\pi)\tag5$$

Equating $(4)$ and $(5)$ yields the Poisson Summation Formula

$$\bbox[5px,border:2px solid #C0A000]{2\pi \sum_{n=-\infty }^\infty \phi(2n\pi)=\sum_{n=-\infty }^\infty \Phi(n)}$$

Answer 2

The following represents a periodic impulse train with period $T_s$:

$$x(t) = \sum \limits_{n=-\infty}^{\infty} \delta(t-nT_0)$$

And the frequency of the periodic impulse train in radians/sec is:

$$\Omega_0 = \frac{2\pi}{T_s}$$

Now we want to find the Complex Fourier Series representation of x(t). If we consider one period of the impulse train, specifically between $-T_0/2$ and $T_0/2$, then a single impulse is centered on zero in this period, and the function x(t) in the $c_n$ formula:

$$c_n = \frac{1}{T_0} \int \limits_{T_0} x(t) \cdot e^{-jn\Omega_0 t}~dt$$

becomes $x(t)=\delta(t)$. And we have:

$$c_n = \frac{1}{T_0} \int \limits_{-T_0/2}^{T_0/2} \delta(t) \cdot e^{-jn\Omega_0 t}~dt$$

Applying the sifting property of the Dirac Delta function:

$$\int \limits_{a}^{b} \delta(t) \cdot f(t)~dt = \begin{cases}f(0)& a<0<b \\ 0 & \text{otherwise}\end{cases}$$

yields:

$$c_n = \frac{1}{T_0} e^{-jn\Omega_0 0}$$

$$c_n = \frac{1}{T_0}$$

Finally, we apply Complex Fourier Series formula:

$$X(\Omega) = 2\pi \sum \limits_{n=-\infty}^{\infty} c_n~ \delta(\Omega - n\Omega_0)$$

which yields:

$$X(\Omega) = 2\pi \sum \limits_{n=-\infty}^{\infty} \frac{1}{T_0}~ \delta(\Omega - n\Omega_0)$$

Previously, we noted that: $\Omega_0 = 2\pi / T_0$, thus:

$$X(\Omega) = \Omega_0 \sum \limits_{n=-\infty}^{\infty} ~ \delta(\Omega - n\Omega_0)$$

Thus, we have proven the following Fourier Transform:

$$\boxed{\sum \limits_{n=-\infty}^{\infty} \delta(t-nT_0) ~~\underset{\mathcal{F}} \longleftrightarrow ~~~ \Omega_0 \sum \limits_{n=-\infty}^{\infty} ~ \delta(\Omega - n\Omega_0)}$$

Answer 3

There is already a perfectly reasonable answer by @mark-viola. I offer mine only because once I was exposed to this approach, I cannot un-see it. And it will help readers understand it when they see it in books and papers and if they don't already know the Fourier series for the Dirac delta distribution.

My answer does not assume knowledge of the Fourier series of the Dirac delta distribution, but I will refer to some properties of the Dirichlet kernel without proof.

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1. Fourier transform of a distribution

Let $\varphi$ be a Schwartz function (a test function), and let $\mathsf{T}$ be a tempered distribution. The Fourier transform $\widehat{\mathsf{T}}$ of $\mathsf{T}$ satisfies \begin{equation} \left<\widehat{\mathsf{T}},\varphi\right> = \left<\mathsf{T},\widehat{\varphi}\right> \end{equation} for every Schwartz function $\varphi$ and its Fourier transform $\widehat{\varphi}$. We often use the integral notation for such operations, even though it is often not really a Riemann or Lebesgue integral: \begin{equation} \int\widehat{\mathsf{T}}(t)\varphi(t)dt = \int\mathsf{T}(t)\widehat{\varphi}(t)dt \end{equation}

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2. Re-arranging the Fourier transform of the comb

Let \begin{equation} \mathsf{T} = \sum_{n=-\infty}^{\infty}\delta(t - nT_s). \end{equation} We want to find an expression for $\widehat{\mathsf{T}}$. \begin{equation} \begin{split} \left<\widehat{\mathsf{T}},\varphi\right> &=~ \left<\mathsf{T},\widehat{\varphi}\right>\\ &=~ \left<\sum_{n=-\infty}^{\infty}\delta(t-nT_s),\widehat{\varphi}\right>\\ &=~ \sum_{n=-\infty}^{\infty}\left<\delta(t-nT_s),\widehat{\varphi}\right>\\ &=~ \sum_{n=-\infty}^{\infty}\widehat{\varphi}(nT_s). \end{split} \end{equation} Note that \begin{equation} \widehat{\varphi}(nT_s) = \int\varphi(k)e^{-i nT_s k}dk \end{equation} So far, we have \begin{equation} \left<\widehat{\mathsf{T}},\varphi\right> = \sum_{n=-\infty}^{\infty}\int\varphi(k)e^{-i nT_s k}dk. \end{equation} The extraordinarily good behavior of Schwartz functions ensures that we can re-write this as \begin{equation} \begin{split} \sum_{n=-\infty}^{\infty}\int\varphi(k)e^{-i nT_s k}dk &=~ \lim_{N\to\infty}\sum_{n=-N}^{N}\int\varphi(k)e^{-i nT_s k}dk\\ &=~ \lim_{N\to\infty}\int\left(\sum_{n=-N}^{N}e^{-i nT_s k}\right)\varphi(k)dk. \end{split} \end{equation}

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3. The Dirichlet kernel

The expression inside the parentheses is the Dirichlet kernel: \begin{equation} \begin{split} \sum_{n=-N}^{N}e^{-i nT_s k} &=~ \sum_{n=-N}^{N}e^{i nT_s k}\\ &=~ \sum_{n=-N}^{N}e^{in(-T_s k)}\\ &=~ \underbrace{\frac{\sin\left((N+\frac{1}{2})T_s k\right)}{\sin\left(\frac{1}{2}T_s k\right)}}_{D_N(T_s k)}. \end{split} \end{equation} Now we have \begin{equation} \left<\widehat{\mathsf{T}},\varphi\right> = \int D_N(T_s k)\varphi(k)dk. \end{equation} We do a simple change of variable that will pay off soon. Let $r = T_s k$, so thast $k = r/T_s$ and $dk = dr/T_s$. then \begin{equation} \left<\widehat{\mathsf{T}},\varphi\right> = \int D_N(r)\varphi(r/T_s)\frac{dr}{T_s}. \end{equation}

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4. The Dirichlet kernel's superpower

The Dirichlet kernel's utility for us is the following: for Schwartz functions (and for some other functions that are not quite so well-behaved), \begin{equation} \lim_{N\to\infty}\int_{-\pi}^{\pi} D_N(r)\psi(r)dt = 2\pi\psi(0). \end{equation} $D_N$ is also $2\pi$-periodic.

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5. Integrals over $2\pi$-wide intervals

We break the integral over $\mathbb{R}$ into integrals over $[-\pi,\pi]$, $[\pi,3\pi]$, $[-3\pi,-\pi]$, and so on. The middle of each such interval is $2n\pi$ for some $n\in\mathbb{Z}$. \begin{equation} \begin{split} \int D_N(r)\varphi(r/T_s)\frac{dr}{T_s} &=~ \sum_{n=-\infty}^{\infty}\int_{(2n+1)\pi}^{(2n+3)\pi} D_N(r)\varphi(r/T_s)\frac{dr}{T_s}\\ &=~ \sum_{n=-\infty}^{\infty}\int_{-\pi}^{\pi} D_N(r + 2n\pi)\varphi\left(\frac{r + 2n\pi}{T_s}\right)\frac{dr}{T_s} \end{split} \end{equation}

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6. Using the Dirichlet kernel's periodicity

Since $D_N$ is $2\pi$-periodic, we can re-write the last line: \begin{equation} \sum_{n=-\infty}^{\infty}\int_{-\pi}^{\pi} D_N(r + 2n\pi)\varphi\left(\frac{r + 2n\pi}{T_s}\right)\frac{dr}{T_s} = \sum_{n=-\infty}^{\infty}\int_{-\pi}^{\pi} D_N(r)\varphi\left(\frac{r + 2n\pi}{T_s}\right)\frac{dr}{T_s}. \end{equation} The extraordinarily good behavior of Schwartz functions endows series of their integrals very good behavior, and we move the limiting operation as $N\to\infty$ inside the sum: \begin{equation} \begin{split} \lim_{N\to\infty}\sum_{n=-\infty}^{\infty}\int_{-\pi}^{\pi} D_N(r)\varphi\left(\frac{r + 2n\pi}{T_s}\right)\frac{dr}{T_s} &=~ \sum_{n=-\infty}^{\infty}\lim_{N\to\infty}\int_{-\pi}^{\pi} D_N(r)\varphi\left(\frac{r + 2n\pi}{T_s}\right)\frac{dr}{T_s}\\ &=~ \sum_{n=-\infty}^{\infty}2\pi\varphi\left(\frac{2n\pi}{T_s}\right)\frac{1}{T_s}\\ &=~ \frac{2\pi}{T_s}\sum_{n=-\infty}^{\infty}\left<\delta\left(t-\frac{2n\pi}{T_s}\right),\varphi(t)\right> \end{split} \end{equation}

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7. Conclusion

We have shown that, in the sense of distributions, \begin{equation} \textrm{the Fourier transform of}~~~\sum_{n=-\infty}^{\infty}\delta(t-nT_s)~~~\textrm{is}~~~ \frac{2\pi}{T_s}\sum_{n=-\infty}^{\infty}\delta\left(k-\frac{2n\pi}{T_s}\right). \end{equation} where $k$ is the conjugate variable.