Could anyone explain to me when to consider one-sided Fourier transform and when to consider two-sided Fourier transform? Normally in a question, just an input signal is mentioned, without stating anything about the causality. Do we by default consider a system to be causal or non-causal? What's the norm?

For example, let us take this question:

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For part (a), I was considering doing a Fourier transform of \$v(t)\$, and then doing the inverse Fourier transform, with the frequencies above \$\omega_c=a\$, and below $w_c=-a$ left out. If the inverse Fourier transform is \$v'(t)\$ (say), then I guess the answer should be either \$\int_{-\infty}^{+\infty}(v'(t))^2dt\$ or \$\int_{0}^{+\infty}(v'(t))^2dt\$. Not sure which.

So, should I do the Fourier transform like



$$\hat{V}(\omega)=\int_{0}^{\infty}e^{-a|t|}e^{-jwt}dt$$ ?

  • \$\begingroup\$ Whatever that integral from zero to infinity is, it's not a Fourier Transform. And the other one isn't a Fourier Transform, either. So, if you need to do a Fourier transform, neither one of these will be it. \$\endgroup\$ – Marcus Müller Apr 21 '18 at 19:46
  • \$\begingroup\$ @MarcusMüller Ah, okay. That was a typo. \$\endgroup\$ – user186505 Apr 21 '18 at 19:49

Are you confusing the Fourier and Laplace transforms? The Laplace transform is normally defined as one-sided [1], but can be defined as two-sided. The Fourier transform is two sided (as far as I'm aware).

One way to solve this might be using Parseval's theorem, $$ \int_{-\infty}^\infty | v(t) |^2 \, \mathrm{d}t = \frac{1}{2\pi} \int_{-\infty}^\infty | V(\omega) |^2 \, \mathrm{d}\omega, $$ and changing the limits of integration from \$\infty\$ to \$a\$.

[1] https://en.wikipedia.org/wiki/Laplace_transform#Formal_definition

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