Determine below signal is power or energy and derive its power or energy accordingly. $$x(t)=\sum_{n=-\infty }^\infty e^{-\lvert t-n\rvert } \tag 1$$

we have : $$E_x=\int_{-\infty}^{\infty}(\sum_{n=-\infty }^\infty e^{-\lvert t-n\rvert })^2dt \tag 2$$ $$P_x =\underset{T\to +\infty }{\mathop{\lim }}\,\frac{1}{2T}\int_{-T}^{T}(\sum_{n=-\infty }^\infty e^{-\lvert t-n\rvert })^2dt \tag 3$$

but squared summation makes it difficult to calculate the integrals , I didn't know how to continue.

  • 1
    $\begingroup$ I don't think this can converge. You are adding an infinite number of signals with finite energy. Since all individual values are positive there is no cancelation or destructive interference so I I'm pretty sure that the energy of the sum is equal or larger larger than the sum of the energies which is already infinite. $\endgroup$
    – Hilmar
    Commented Oct 19, 2022 at 12:11

3 Answers 3


The signal is clearly periodic with period $T=1$. Consequently, its energy is infinite, but we can compute its power:


For $t\in [0,1]$ we can write


So we have

$$\begin{align}\sum_{n=-\infty}^{\infty}e^{-|t-n|}&=\sum_{n=1}^{\infty}e^{t-n}+\sum_{n=-\infty}^{-1}e^{n-t}+e^{-t}\\&=\left(e^t+e^{-t}\right)\sum_{n=1}^{\infty}e^{-n}+e^{-t}\\&=\frac{e^t+e^{-t}}{e-1}+e^{-t},\qquad 0\le t\le 1\tag{3}\end{align}$$

Using $(3)$ we can compute $P_x$ by solving an elementary integral:

$$P_x=\int_0^1\left[\frac{e^t+e^{-t}}{e-1}+e^{-t}\right]^2dt\approx 4.0053\tag{4}$$

We can cross-check our result by using another way to compute $P_x$. From Parseval's identity for Fourier series we know that


where $c_k$ are the complex Fourier coefficients of $x(t)$. Using Poisson's sum formula we get

$$\sum_{n=-\infty}^{\infty}g(t-nT)=\frac{1}{T}\sum_{k=-\infty}^{\infty}G\left(\frac{2\pi k}{T}\right)e^{j2\pi kt/T}\tag{6}$$

where $G(\omega)$ is the Fourier transform of $g(t)$. In our example we have $g(t)=e^{-|t|}$ and $T=1$. We also have


Consequently, the Fourier coefficients of $x(t)$ are

$$c_k=G(2\pi k)=\frac{2}{1+(2\pi k)^2}\tag{8}$$

And from $(5)$ we obtain

$$P_x=4\sum_{k=-\infty}^{\infty}\frac{1}{\big[1+(2\pi k)^2\big]^2}=4+8\sum_{k=1}^{\infty}\frac{1}{\big[1+(2\pi k)^2\big]^2}\approx 4.0053\tag{9}$$

which matches with the result $(4)$. Note that the sum in $(9)$ can be computed quite accurately with only a few terms because the values in the sum decay very fast with increasing $k$.

  • $\begingroup$ How is $x(t)=\sum_{n=-\infty }^\infty e^{-\lvert t-n\rvert }$ periodic? It looks monotonically increasing to me. $\endgroup$
    – TimWescott
    Commented Oct 21, 2022 at 23:17
  • $\begingroup$ @TimWescott: $x(t)$ is a sum of periodically shifted versions of the function $e^{-|t|}$, so it must be periodic. It's also straightforward to show that $x(t)=x(t+k)$ for integer $k$, meaning it's periodic with period $T=1$. $\endgroup$
    – Matt L.
    Commented Oct 22, 2022 at 14:12
  • $\begingroup$ OK -- I see that now. $\endgroup$
    – TimWescott
    Commented Oct 23, 2022 at 3:47

First look at the case $t=0$: $$x(0) = \sum_{n=-\infty}^{\infty}e^{-|-n|}$$ Let's fiddle around a bit: $$x(0)=e^0 + 2\cdot\sum_{n=1}^{\infty}e^{-n}$$ Looks familiar? This is the power series, so we can ditch the sum and rewrite as: $$x(0)=1+2\left(\frac{1}{e-1}\right)$$ So far, so good, this is a constant. But what about the other values of $t$? First let's look at the integer values of $t$: $$x(m)=\sum_{n=-\infty}^{\infty}e^{-|m-n|}$$ This is obviously just a time shift of the $t=0$ case, with the symmetry point/origin at $m=n$, so $$x(m)=x(0)=1+2\left(\frac{1}{e-1}\right)$$ So, we found that $x(t)$ is periodic in some way. That screams power signal, so we have that part of the question answered. What about the values of $t$ in between integers and the actual power? Without loss of generality we look only at the interval $[-0.5,0.5]$, wich is one period of the signal: $$P_x=\int_{-0.5}^{0.5}\left(\sum_{n=-\infty}^{\infty}e^{-|t-n|}\right)^2dt$$ From here, take a look at Matts answer, where he resolves the three different cases for the absolute in the exponent. I had a mistake in my solution, so I deleted this part of my answer.

  • $\begingroup$ thanks for your answer dear Max, in "Making use of the absolute in the exponent" part how did you write $e^{-abs(t-n)}$ in this way: $e^{t-abs(-n)}$ $\endgroup$ Commented Oct 19, 2022 at 15:21
  • $\begingroup$ By limiting t to positive values via the Integration borders. $\endgroup$
    – Max
    Commented Oct 19, 2022 at 15:24
  • $\begingroup$ I'm afraid that this result is wrong. Your simplification of $-|t-n|$ doesn't work that way, and the sum $\sum_{n=-\infty}^{\infty}e^{-n}$ cleary doesn't converge. $\endgroup$
    – Matt L.
    Commented Oct 20, 2022 at 13:21
  • $\begingroup$ $\sum_{n=-\infty}^{\infty}e^{-n}$ does not converge, but $\sum_{n=-\infty}^{\infty}e^{-|-n|}$ does! I forgot the abs-brackets. And maybe my argumentation was off, but I think in the infinite sum, you can pull out the $t$ like that. If the term was naked, you were right, but in the sum, I think you can do this. $\endgroup$
    – Max
    Commented Oct 20, 2022 at 14:05
  • $\begingroup$ I just saw you answer, I get it now. I'll edit my answer. Thanks! $\endgroup$
    – Max
    Commented Oct 20, 2022 at 14:25

Let my try a shortcut on direct calculation, just providing hints. Being unwary about convergence: $x(t)$ looks periodic to me, since: $$x(t+1) = \sum_{n=-\infty}^{n=\infty}e^{-|t+1-n|}= \sum_{n=-\infty}^{n=\infty}e^{-|t-(n-1)|}= \sum_{m=-\infty}^{m=\infty}e^{-|t-m|}$$ with $m=n-1$ variable change. Therefore you can restrict a lot of computations to what happens in one period.

The total energy $E_x$ is unlikely to be bounded, as each one-period is positive. Asymptotic bounds could be obtained. For power $P_x$, it should be doable to bound each term $e^{-|t-m|}$ on integer intervals $[N,N+1[$.


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