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We have been given x(t) and h(t) with their functions and we are required to compute the energy spectra of y $S_y(f)=S_X(f)|H(f)|^2$ Where $S_y$ is the energy spectrum of $y(t)$ and $S_x$ is the energy spectrum of $x(t)$

I have found $S_x(f)$ and $H(f)=\frac {2a} {a^2 +r\pi^2f^2}$ with a being a generic number in $R>0$ .

What im a bit confused about is. When it asks for $|H(f)^2|$ is it asking for the literal function $H(f)=\frac {2a} {a^2 +r\pi^2f^2}$ squared, or do i have to find the energy of $|H(f)|$?

I dont need computations, i just need a bit of clarification on this part

Severjan Lici
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1 Answers1

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If you have signal $x(t)$ and filter $h(t)$ with respective Fourier transforms $X(\omega)$ and $H(\omega)$, then the filtered signal $y(t)$ is obtained by $$ y(t)=(x \ast h)(t) \Longleftrightarrow Y(\omega)=X(\omega)H(\omega). $$ If you want the energy spectrum, or PSD, you take absolute values and square in order to get from 'complex amplitude' to 'energy': $$ |Y(\omega)|^2=|X(\omega)|^2|H(\omega)|^2, $$ which is a result of $Y(\omega)Y^\ast(\omega)=|Y(\omega)|^2$.

You can write the result a bit differently as $$ S_y(\omega)=S_x(\omega)|H(\omega)|^2. $$ The phrase 'take absolute values and square' is no more and no less what to do. So if your $H(\omega)$ is positive and real valued and therefore $H(\omega)=|H(\omega)|$ (which seems to apply to your case), then you can just take the square of it.

Your $H(\omega)$ has the same form as $|H(\omega)|^2$ of a first order lowpass filter. Check your results.

Jeroen Boschma
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