I am confused with this multiple-choice question:

Assuming that $x(n) = e^{-2n}u(n)$, what does not happen if we limit the duration of this sequences to 2014 samples?

A. The signal power leaks out into the entire frequency range

B. The energy of the obtained limited-duration signal is greater than that of the original signal $x(n)$

C. Spectral estimate is distorted

D. Spectral resolution is increased

I'm inclined to B because the summation of unlimited duration $x(n)$ would be greater than the limited one since $ x(n) > 0 $ but i don't know how to explain the other options.

Every help is appreciated. Thanks!


1 Answer 1

  • A) Fourier transform of a finite sequence is periodic and infinite. The FT of $x(t)$ is is $1/(2 + j\omega)$, which decays permanently and power zeros at infinity; for $x[n]$, an approximation (see C) of this will repeat infinitely, so non-decayed power will persist. Below is for a generic (bandlimited) signal (related):
  • B) $E = \sum_{n} |x[n]|^2$ - samples can never subtract energy, so $E_\text{long} \geq E_\text{short}$. $x[n] > 0$ isn't required.

  • C) "Distorted" here means that $\mathcal{F}\{x[n]\}$ will not be an overlap-added version of shiftings of $\mathcal{F}\{x(t)\}$ (b above), since $x[n]_{0:2013}$ is a trimmed or "incomplete" version of $x[n]_{0:\infty}$. It also means there is aliasing, since $\mathcal{F}\{x(t)\}$ has infinite bandwidth (note, there's aliasing no matter what unless $n\rightarrow\infty$ and $f_s \rightarrow \infty$).

  • D) Spectral resolution is proportional to sampling rate (time interval between samples), $f_s$, not total number of samples; I figure the assumption is that former is fixed.

  • 1
    $\begingroup$ Thank you very much! that's a good explanation to me:) $\endgroup$ Jul 31, 2021 at 12:03
  • $\begingroup$ I wonder if we sample the above signal at frequencies $2\pi*k/2014$ for k=0,1,2,...,2013. Can that signal be uniquely recovered? $\endgroup$ Aug 2, 2021 at 1:30

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