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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.
The signal is clearly periodic with period $T=1$. Consequently, its energy is infinite, but we can compute its power:
$$P_x=\frac{1}{T}\int_T\big|x(t)\big|^2dt=\int_{0}^{1}\left[\sum_{n=-\infty}^{\infty}e^{-|t-n|}\right]^2dt\tag{1}$$
For $t\in [0,1]$ we can write
$$|t-n|=\begin{cases}n-t,&n>0\\t,&n=0\\t-n,&n<0\end{cases}\tag{2}$$
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
$$P_x=\sum_{k=-\infty}^{\infty}|c_k|^2\tag{5}$$
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
$$G(\omega)=\mathcal{F}\big\{e^{-|t|}\big\}=\frac{2}{1+\omega^2}\tag{7}$$
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 $k$.
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.
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 $[N,N+1[$.
