The question is to find the CTFT of $$x(t) = e^{-t}u(t)\cdot \sum_{n=-\infty}^{\infty} \delta\left(t-\frac{n}{2}\right)$$

Now I know that multiplication in time means convolution in the Fourier Domain but how can I find the CTFT of $\sum_{n=-\infty}^{\infty} \delta(t-\frac{1}{2}n)$ since I am not allowed to directly use FT pairs?


3 Answers 3


Using the Fourier Transform property related to multiplication:

$$\mathcal{F}\{x(t)\} = \frac{1}{2\pi} \left(\mathcal{F}\{e^{-t}u(t)\} \ast \mathcal{F}\left\{\sum_{n=-\infty}^{\infty} \delta(t-\frac{1}{2}n)\right\}\right)$$

as you have correctly identified.

$\sum_{n=-\infty}^{\infty} \delta(t-\frac{1}{2}n)$ is a periodic signal whose period if $\frac 12$ units of time. So, first find its F.S.E. coefficients:

$$a_k = 2 \int_{-1/4}^{1/4} \delta(t)e^{-j4\pi kt} dt = 2$$

where I chose one period as $[-\frac{1}{4}, \frac{1}{4}]$

As such,

\begin{align*} \mathcal{F}\left\{\sum_{n=-\infty}^{\infty} \delta(t-\frac{1}{2}n)\right\} &= \sum_{k=-\infty}^{\infty}2\pi\cdot 2\cdot \delta\left(\omega - \frac{2\pi}{\frac 12}k\right) \quad \quad \textbf{FT is Linear}\\ & = \sum_{k=-\infty}^{\infty}4\pi \delta\left(\omega - 4\pi k\right) \end{align*}


$$\mathcal{F}\{e^{-t}u(t)\} = \int_{0}^{\infty} e^{-t}e^{-j\omega t}dt = \frac{1}{1+j\omega}$$


\begin{align*} \mathcal{F}\{x(t)\} &= \frac{1}{2\pi} \left[ \left( \frac{1}{1+j\omega}\right) \ast \left( \sum_{k=-\infty}^{\infty}4\pi \delta\left(\omega - 4\pi k\right)\right)\right]\\ & = \sum_{k=-\infty}^{\infty}\frac{2}{1+j(\omega - 4\pi k)} \end{align*}

Note that a convolution with a shifted impulse just shifts the function!

Sidenote: There is another approach that yields a closed form solution which I might type up in a short while as a separate answer perhaps.

  • $\begingroup$ Thank you for your answer but I couldn't really understand why we needed to find the Fourier Series coefficients first? Shouldn't we be able to take the FT directly? $\endgroup$
    – user67157
    Commented Jul 6, 2023 at 10:31
  • $\begingroup$ @AcerbicNarcissist If you try to take the Fourier Transform directly you will get a summation that diverges. It's a bit of a tricky concept so I won't get too much into it but when we write the Dirac Comb as a weighted sum of complex exponentials we can take its Fourier Transform without any mathematical ambiguity. Also for such questions you don't always have to follow the approach given here, see my other answer. $\endgroup$ Commented Jul 6, 2023 at 10:51

The results shown in the two answers provided by Ahsan Yousaf don't agree. This answer is about explaining why the two solutions are different, and how to arrive at the correct solution. Note that this is not about some trivial error in the computations, but it is about a fundamental property of the Fourier transform which is well-known but which has been ignored for many years in the usage of such a popular method as the impulse invariance method for designing digital filters from analog impulse responses. The necessary correction was only found and published in the year 2000 (independently by Mecklenbräucker and Jackson). See also the corresponding wikipedia page.

First, let's see the difference between the two solutions. The solution expressed as an infinite sum given in this answer is correct (orange in the top figure below). The closed form solution in this answer is incorrect (blue in the top figure below). The reason why it is incorrect is exactly the same as why the impulse invariance method has been used incorrectly for so many years.

The error is the following: the Fourier transform converges to the arithmetic mean of the left-hand and right-hand limits at jump discontinuities. That's why when sampling a continuous-time signal and computing its Fourier transform, we need to choose the samples such that their values equal the arithmetic mean of the left-hand and right-hand limits at discontinuities. If we don't sample at discontinuities of the continuous-time function there is no problem. However, we often sample a causal function at $t=0$, and that's also where we frequently find a discontinuity (as an example, think of a simple first-order lowpass filter).

Applying the above to the derivation of a closed-form expression for the Fourier transform of the given function results in the following solution: \begin{align*} \mathcal{F}\left\{e^{-t}u(t)\sum_{n=-\infty}^{\infty}\delta\left(t-\frac{n}{2}\right)\right\} & = \mathcal{F}\left\{\frac12\delta(t)+\sum_{n=1}^{\infty}e^{-n/2}\delta\left(t-\frac{n}{2}\right)\right\}\\ & = \frac12 + \sum_{n=1}^{\infty}e^{-n(1+j\omega)/2} \\ & = \frac12 + \frac{e^{-(1+j\omega)/2}}{1-e^{-(1+j\omega)/2}} \\ & = \frac12\frac{1+e^{-(1+j\omega)/2}}{1-e^{-(1+j\omega)/2}} \tag{1} \end{align*} Note that for the sample at $n=0$ we used the arithmetic mean of the left-hand and right-hand limits of the function $e^{-t}u(t)$ at $t=0$ (which is $1/2$).

The top figure below shows the incorrect closed-form solution of this answer (blue) with the correct solution given in this answer (orange). In the bottom figure, the correct closed-form solution given by $(1)$ is shown, together with the equivalent solution expressed by an infinite sum.

enter image description here

Proof that the solution $(1)$ and the infinite series solution given in this answer are identical:

First, note that $(1)$ can be rewritten as

\begin{align*} \frac12\frac{1+e^{-(1+j\omega)/2}}{1-e^{-(1+j\omega)/2}} & =\frac12\textrm{coth}\,\left(\frac{1+j\omega}{4}\right) \end{align*}

With the partial fraction expansion of $\textrm{coth}\,z$ this equals \begin{align*} \frac12\textrm{coth}\,\left(\frac{1+j\omega}{4}\right) & = \frac{2}{1+j\omega}+\frac{1+j\omega}{4}\sum_{k=1}^{\infty}\frac{1}{\left(\frac{1+j\omega}{4}\right)^2+\pi^2k^2} \\ & = \boxed{\frac{2}{1+j\omega}+4(1+j\omega)\sum_{k=1}^{\infty}\frac{1}{\left(1+j\omega\right)^2+16\pi^2k^2}}\tag{2} \end{align*}

The infinite series solution in this answer can be rewritten as follows:

\begin{align*} \sum_{k=-\infty}^{\infty}\!\frac{2}{1+j(\omega-4\pi k)} & = \frac{2}{1+j\omega} + 2\!\sum_{k=1}^{\infty}\left[\frac{1}{1+j(\omega-4\pi k)}\!+\!\frac{1}{1+j(\omega+4\pi k)}\right] \\ & = \frac{2}{1+j\omega} + 4\!\sum_{k=1}^{\infty}\frac{1+j\omega}{\left[(1+j\omega)-j4\pi k\right]\left[(1+j\omega)+j4\pi k\right]} \\ & = \boxed{\frac{2}{1+j\omega} + 4(1+j\omega)\sum_{k=1}^{\infty}\frac{1}{(1+j\omega)^2+16\pi^2 k^2}}\tag{3} \end{align*}

which is identical to the formula $(2)$, Q.E.D.

  • 2
    $\begingroup$ + 1 Very nice. I actually had no idea about this correction. Thanks for bringing it to our attention! $\endgroup$ Commented Jul 6, 2023 at 18:48
  • $\begingroup$ I didn't go through the math, but it seems to me that this should be straight forward since the dirac impulse train was offset by 1/2 meaning that there is no sampling of the edge of $e^{-t} u(t)$. I don't see any jump discontinuity getting sampled. $\endgroup$ Commented Jul 6, 2023 at 20:27
  • $\begingroup$ In your Eq. (1), I don't see where that $\frac12 \delta(t)$ comes from. $\endgroup$ Commented Jul 6, 2023 at 20:31
  • $\begingroup$ @robertbristow-johnson: Sampling of the discontinuity happens at $n=0$. That's also where $\frac12\delta(t)$ comes from. $\endgroup$
    – Matt L.
    Commented Jul 6, 2023 at 20:59
  • 1
    $\begingroup$ Oh, you're right. I just looked at it wrong. Multiplying by 1/2 is not the same as adding 1/2. $\endgroup$ Commented Jul 6, 2023 at 22:28

Alternate Solution:

Write $x(t)$ as:

\begin{align*} x(t) &= [e^{-t}u(t)]\cdot \sum_{n=-\infty}^{\infty} \delta(t-\frac{1}{2}n)\\ & = \sum_{n=0}^{\infty}e^{-t}\delta(t-\frac 12 n)\\ & = \sum_{n=0}^{\infty}e^{-\frac 12 n}\delta(t-\frac 12 n) \end{align*}

Taking the FT:

\begin{align*} \mathcal{F}\{x(t)\} = \mathcal{F}\left\{\sum_{n=0}^{\infty}e^{-\frac 12 n}\delta(t-\frac 12 n) \right\} &= \sum_{n=0}^{\infty}e^{-\frac 12 n}\mathcal{F}\{\delta(t-\frac 12 n)\} \quad \quad \textbf{FT is Linear}\\ & = \sum_{n=0}^{\infty}e^{-\frac 12 n}e^{-j\frac 12 n} \quad \quad \textbf{Shift Property}\\ & = \sum_{n=0}^{\infty}\left[e^{-\frac{1+j\omega}{2}}\right]^n\\ & = \frac{1}{1-e^{-\frac{1+j\omega}{2}}} \end{align*}

  • $\begingroup$ Compare the two solutions in your answers and you'll note that they are not identical. Can you spot the error? (It's not a simple sign error or the like, but a more fundamental thing). $\endgroup$
    – Matt L.
    Commented Jul 6, 2023 at 10:49
  • $\begingroup$ @MattL. I have reviewed my solutions again but cannot seem to find any error in my methods or computation. Could you point out what it is? $\endgroup$ Commented Jul 6, 2023 at 11:10
  • $\begingroup$ But you do see that they're not identical, right? $\endgroup$
    – Matt L.
    Commented Jul 6, 2023 at 11:17
  • $\begingroup$ Yes, but I still cannot find a mistake. I would be grateful if you could point out please. $\endgroup$ Commented Jul 6, 2023 at 11:29
  • $\begingroup$ Ok, it's a bit of a longer story, I'll write up an answer explaining it. $\endgroup$
    – Matt L.
    Commented Jul 6, 2023 at 12:30

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