# Should I consider systems to be causal or non-causal by default in questions like these?

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:

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

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

or

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

• 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. – Marcus Müller Apr 21 '18 at 19:46
• @MarcusMüller Ah, okay. That was a typo. – user186505 Apr 21 '18 at 19:49

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 $a$.