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I have a signal that I want to sample using delta functions. The signal is: $x(t) = W^2sinc^2(Wt)$

and after the sampling we will have the signal $z(t)$. We know the form of the signal in the frequency domain, and it is:

$$Z(f) = W \sum \limits _{k=-\infty}^\infty \Lambda \left( \frac{f-kf_s}{W} \right)$$

With $\Lambda(x)$ the triangular function, which is:

$$ \Lambda(x) = \left\{ \begin{array}{l l} 1-|x| & \quad , |x|<1\\ 0 & \quad \text{otherwise} \end{array} \right. $$

The thing is that I want to calculate the energy of $z(t)$, within the frequencies $[-W,W]$. Using the Parseval theorem we can find this energy through the fourier transform, $Z(f)$:

$$ \mathcal{E}_z = \int \limits _{-W}^W |Z(f)|^2 df $$

But I am not sure how to handle the infinite summation within the integral. I thought:

$$ \begin{align*} \mathcal{E}_z &= \int \limits _{-W}^W |Z(f)|^2 df = \int \limits _{-W}^W\sum \limits _{k=-\infty}^\infty \left (W \Lambda \left( \frac{f-kf_s}{W} \right) \right) ^ 2df \\ &= 2 \int \limits _0^W \sum \limits _{k=-\infty}^\infty (W - |f + kf_s|) ^ 2 df = 2 \sum \limits _{k=-\infty}^\infty \int \limits _0^W (W - |f + kf_s|) ^ 2 df \end{align*} $$

Saying, $u = W - |f + kf_s|$, but I don't know if I have to change the limits of the integral, and if I do it, in the end I have a summation practically infinite. And this is not correct, because this integral is the area of a triangle (with some distortions depending on $f_s$ and aliasing phenomena).

How can I constrain the infinit sum according to the limits of my interation?

I don't want to give me a straight solution, I 'd rather be given a way to tackle these kind of problems. I just gave a specific example to be clear.

jasonsaras
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  • So would a discussion of when it's "allowed" to interchange the order of integration and summation in similar circumstances be useful to you? – hardmath Jan 16 '14 at 15:08
  • Are you sure that your initial statement that the sampled signal energy is given by $E_z = \int \limits _{-W}^W |Z(f)|^2 df $ is actually correct? As far as I'm aware, during sampling, higher frequency components alias into the bandwidth interval of the sampling, so it's not just the integral of the spectral energy density over the bandlimit. As a counterexample, consider a sine wave whose frequency is beyond the bandlimit. According to your integral, the sampled signal energy will be zero, but this is clearly false, since the sampled signal will be nonzero. – DumpsterDoofus Jan 16 '14 at 15:12
  • @hardmath, Yes it could be. In this situation, the limits of intergration is affecting ths summation, because Z(f) is a series of triangles (due to the existance of summation), and the integral keeps only the triangle around zero and perhaps a piece of the two triangles besides it (if the sampling frequency is below Nyquist frequency). A solution can be given geometrically, but I want to know the analytic solution -jasonsaras – jasonsaras Jan 16 '14 at 15:12
  • @DumpsterDoofus Consider that we have the $z(t)$ signal and we have aliasing, meaning that the triangles of the spectrum is overlaping. If I take $z(t)$ through a low-pass filter cutting-off frequencies larger than W, we will have a spectrum with a triangle centering in zero, and two pieces of the two others triangles besides this, due to the aliasing. I want to calculate the energy of this spectrum. Correct me if I am wrong. – jasonsaras Jan 16 '14 at 15:19
  • Actually, now that I reread your question I'm a little confused. You say "We know the form of the signal in the frequency domain", but are you referring to the original signal or the sampled signal? For the original signal, you simply have $Z(w)=\sqrt{\frac{\pi }{2}} W \Lambda \left(\frac{w}{2 W}\right)$, it's not a sum of triangle functions. For the sampled signal, if $2f_s\geq W$ (I think? might be a wrong factor of 2) then you will have perfect reconstruction, and the sampled signal will be identical in the frequency domain to the true signal. Otherwise, the edges will fold in. – DumpsterDoofus Jan 16 '14 at 16:32
  • BTW, you can compute the signal in the frequency domain in Mathematica via FourierTransform[W^2 Sinc[W x]^2, x, w], although it outputs it in a form which is hard to recognize as a triangle pulse (although when plotted it looks fine). – DumpsterDoofus Jan 16 '14 at 16:36
  • So there are two cases: if $2f_s\geq W$ (or maybe it's $4f_s\geq W$, I'm too tired to check) then the sampled signal energy will be identical to the original. Otherwise, the edges will fold in and you can compute it numerically. There might be an analytic way to do this too, but it might take some thinking. – DumpsterDoofus Jan 16 '14 at 16:40
  • Never mind, you're using $Z(w)$ to refer to the sampled signal in the frequency domain, which is correct. I'll think more about this later when I have time. – DumpsterDoofus Jan 16 '14 at 17:00
  • On second thought, isn't the total energy of the sampled signal simply the sum of the squares of the samples, or $f_s^{-1}\sum_{k=-\infty}^\infty W^4\mbox{sinc}^4(W k f_s)$? That has a closed analytic form. – DumpsterDoofus Jan 16 '14 at 18:30

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