I need to find Fourier transform of

$$\sum_{n=-\infty }^\infty e^{-\lvert t-2n\rvert }$$

I found the solution online. But I think the choice of limits in equation (1) of the solution is wrong. It should be probably from $-\infty$ to $2n$ for the first integral and from $2n$ to $\infty$ for the second integral. But I am confused about what the limits would be as the both integration and summation is involved. and I am not sure whose limits to change (probably limits of integration must be changed only and that of summation must be kept as it is)

Given Solution:

enter image description here

Please tell me what the limits will be and the explanation. Also if possible please solve the problem.

Original answer:

$$ \frac{(1-e^{-2(1+jw)})}{(1-e^{-2})(1+jw)}-\frac{(e^{-2})(1-e^{-2(1+jw)})}{(1-e^{-2})(1-jw)}$$

  • $\begingroup$ Is that an authorized solution manual (of which book?)? I was wondering because the solution in the manual is wrong. $\endgroup$
    – Matt L.
    Feb 15, 2017 at 18:38
  • $\begingroup$ It is a question from Oppenheim Signals and Systems. I am not sure if it is the authorised solution manual. You can find it here slideshare.net/Gilgitian/… $\endgroup$
    – Soumee
    Feb 16, 2017 at 9:00
  • $\begingroup$ OK, in any case, the solution appears to be wrong. $\endgroup$
    – Matt L.
    Feb 16, 2017 at 10:35

2 Answers 2


First of all, the signal is periodic, so its spectrum must be discrete. Hence, your final solution ("Original answer") cannot be correct.

I would go the following way:

  1. Calculate $X(f)=\mathcal{F}\left\{\exp(-\lvert t\rvert)\right\}$
  2. Apply the time-shifting property of the Fourier Transform to transform all time-shifts of the function.
  3. See that you can factor out $X(f)$ from the resulting sum and then see that the terms in the sum correspond to a comb function where the diracs are $\frac{1}{2}$ apart from each other.

Edit: Below is some code to plot the function:

Fs = 100  # some sampling frequency for discretizing the signal to plot it
T = 10    # the time range
t = np.arange(-T, T, 1/Fs)  # the time samples

# define the basis function
f = lambda t: np.exp(-abs(t))

# plot the basis function and the periodic extension
plt.plot(t, f(t))
plt.plot(t, sum(f(t-2*n) for n in range(-10, 10)))

enter image description here

Edit: Regarding your comment:

$$ \begin{align} \mathcal{F}\left\{e^{−\lvert t\rvert}\right\}&=X_1(j\omega)=\frac{2}{1+\omega^2}\\ \mathcal{F}\left\{\sum_{n}e^{-\lvert t-2n\rvert}\right\}&=\sum_nX(j\omega)\exp\left(j2n \omega\right)\\ &=X(j\omega)\sum_n\exp(j2n \omega) \end{align} $$

Here's the trick: $$\sum_n\exp(j2n \omega)=\frac{1}{2}\sum_n\delta\left(\omega-\frac{2\pi}{2}n\right)$$

Have a look at the wikipedia article and also this article for example on why this holds.

  • $\begingroup$ Sorry if my question is bit naive. I cant understand why $\sum_{n=-\infty }^\infty e^{-|t-2n|}$ is periodic. For what values of t will x(t+T) be equal to x(t) ? Can you suggest an online graph plotter as I want to see The nature of this function. I don't understand how to plot this as it involves two variables t and n. Thank you. $\endgroup$
    – Soumee
    Feb 15, 2017 at 10:01
  • $\begingroup$ @user3237657 The function only depends on t, as n is defined within the sum and is hence not a free variable. I have added some code to plot the function in python and the resulting output image. $\endgroup$ Feb 15, 2017 at 10:17
  • $\begingroup$ sir I have done what you said:F.T of $$ e^{-|t|}= X_1(jw)= \frac{2}{1+w^2}$$ $$ x(t)=...+e^{-|t+4|}+e^{-|t+2|}+e^{-|t|}+e^{-|t-2|}+e^{-|t-4|}+...$$ <----F.T----> $$...+\frac{e^{jw4}}{1+w^2}+\frac{e^{jw2}}{1+w^2}+\frac{e^0}{1+w^2}+\frac{e^{jw4}}{1+w^2}+\frac{e^{jw4}}{1+w^2}+... $$ $$=\sum_{n=-\infty}^\infty \frac{e^{-jw2n}}{1+w^2}$$ $\endgroup$
    – Soumee
    Feb 15, 2017 at 11:01
  • $\begingroup$ But this is the wrong answer I guess. Since this is a periodic function, Fourier transform should consist of impulses(comb function as you said). Where have i gone wrong?... $\endgroup$
    – Soumee
    Feb 15, 2017 at 11:08
  • $\begingroup$ @user3237657 Your derivation is correct. The thing which is missing is that the sum of exponentials equals the Dirac Comb. I've edited my answer with some more detail. $\endgroup$ Feb 15, 2017 at 12:01

One straightforward way of solving this problem is by applying Poisson's Sum Formula:

$$\sum_{n=-\infty}^{\infty}g(t-nT)=\frac{1}{T}\sum_{k=-\infty}^{\infty}G(k\omega_0)e^{jk\omega_0t},\quad \omega_0=\frac{2\pi}{T}\tag{1}$$

where $G(\omega)$ is the Fourier transform of $g(t)$.

The given function is obtained with $g(t)=e^{-|t|}$ and $T=2$. We have


and from $(1)$ we get

$$\sum_{n=-\infty}^{\infty}e^{-|t-2n|}=\sum_{k=-\infty}^{\infty}\frac{1}{1+k^2\pi^2}e^{jk\pi t}\tag{3}$$

With $$\mathcal{F}\{e^{jk\pi t}\}=2\pi\delta(\omega-k\pi)$$

the Fourier transform of $(3)$ is obtained as

$$2\pi \sum_{k=-\infty}^{\infty}\frac{1}{1+k^2\pi^2}\delta(\omega-k\pi)\tag{4}$$

The "given solution" in your question is flawed because of the very reason you noticed (the integration limits are wrong). It is surprising that also the solution in the solution manual is wrong. The basic form of that solution is correct (a weighted Dirac comb), but the weights are wrong.

Of course you can also compute the Fourier transform by direct application of the Fourier integral:

$$\begin{align}\int_{-\infty}^{\infty}\sum_{n=-\infty}^{\infty}e^{-|t-2n|}e^{-j\omega t}dt&=\sum_{n=-\infty}^{\infty}\int_{2n}^{\infty}e^{-(t-2n)}e^{-j\omega t}dt+\sum_{n=-\infty}^{\infty}\int_{-\infty}^{2n}e^{(t-2n)}e^{-j\omega t}dt\\&=\sum_{n=-\infty}^{\infty}e^{2n}\int_{2n}^{\infty}e^{-(1+j\omega)t}dt+\sum_{n=-\infty}^{\infty}e^{-2n}\int_{-\infty}^{2n}e^{(1-j\omega)t}dt\\&=\sum_{n=-\infty}^{\infty}e^{2n}\frac{1}{1+j\omega}e^{-2n(1+j\omega)}+\sum_{n=-\infty}^{\infty}e^{-2n}\frac{1}{1-j\omega}e^{2n(1-j\omega)}\\&=\frac{1}{1+j\omega}\sum_{n=-\infty}^{\infty}e^{-2nj\omega}+\frac{1}{1-j\omega}\sum_{n=-\infty}^{\infty}e^{-2nj\omega}\\&=\frac{2}{1+\omega^2}\sum_{n=-\infty}^{\infty}e^{-2nj\omega}\\&=\frac{2\pi}{1+\omega^2}\sum_{n=-\infty}^{\infty}\delta(\omega-n\pi)\\&=2\pi\sum_{n=-\infty}^{\infty}\frac{1}{1+n^2\pi^2}\delta(\omega-n\pi)\end{align}$$

where the last two equalities follow from the Fourier series representation of the Dirac comb and from the sifting property of the Dirac impulse, respectively.

  • $\begingroup$ I understood your solution from step (1) to (4). But I have a few doubts: First: In wikipedia it has been given that: $$ \sum_{n=-\infty}^\infty s(t+nP)=\sum_{k=-\infty}^\infty \frac{1}{P}.\hat s(\frac{k}{P})e^{j2\pi \frac{k}{P}t}$$ It also follows that $$ \sum_{n=-\infty}^\infty g(t-nT)=\sum_{k=-\infty}^\infty \frac{1}{T}.\hat g(\frac{k}{T})e^{-j2\pi \frac{k}{T}t}$$ $\endgroup$
    – Soumee
    Feb 16, 2017 at 5:48
  • $\begingroup$ Note that the sign of the exponent has changed. Furthur if we substitute $T=\frac{2\pi}{w_o}$ in RHS of the equation, we get: $$ \sum_{n=-\infty}^\infty g(t-nT)=\sum_{k=-\infty}^\infty \frac{1}{T}.\hat g(\frac{kw_o}{2\pi})e^{-jkw_ot}$$ So there are two things which has come up extra in your equation (1): a negative sign in the exponent and $\frac{1}{2\pi}$ in argument of $\hat g()$ , which will alter the result. Please clarify this doubt. $\endgroup$
    – Soumee
    Feb 16, 2017 at 5:56
  • $\begingroup$ @user3237657: First of all, note that by a simple index substitution you can easily see that $$\sum_{n=-\infty}^{\infty}g(t-nT)=\sum_{n=-\infty}^{\infty}g(t+nT)$$ because it doesn't matter in which order you add up the shifted versions of $g(t)$. So the left-hand side doesn't change. Further, your conclusion that changing the index on the LHS would change the sign of the exponent on the RHS is wrong. So you can change the sign of $n$ on the LHS and nothing changes on the RHS. $\endgroup$
    – Matt L.
    Feb 16, 2017 at 6:49
  • $\begingroup$ @user3237657: Your other doubt has to do with the way the Fourier transform of $g(t)$ is written. I chose to use $\omega=2\pi f$ as the independent variable, whereas on the wikipedia page they choose $f$ as the independent variable. Both conventions are used frequently. $\endgroup$
    – Matt L.
    Feb 16, 2017 at 6:50
  • $\begingroup$ Sir Please clarify another doubt:(I am extending this comment to the next block as we are allowed to type a limited number of characters per comment) . Sir Maximilian Matthé's solutions also seems to be correct. He has done something like this: $$X(jw)=\int_{-\infty}^\infty x(t)e^{-jwt}dt$$ $$=\int_{-\infty}^\infty \sum_{n=-\infty}^\infty e^{-|t-2n|}e^{-jwt}dt$$ $$=\sum_{n=-\infty}^\infty \frac{2}{1+w^2}e^{-jw2n}$$ $\endgroup$
    – Soumee
    Feb 16, 2017 at 7:32

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