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A signal is given by

$$x(t)=\begin{cases} e^{-t} &t\geq 0\\ 0 &t < 0 \\\end{cases}$$

  1. Find the Fourier transform of the signal.

$$X(\omega)=\int_{0}^{\infty}e^{-t}e^{-j\omega t} dt$$ and I get $$X(\omega)=\left(\frac{1}{1+j\omega}\right)$$

  1. Find total energy of the signal using time domain representation

$$X(\omega)=\int_{0}^{\infty}|e^{-t}|^2 dt$$

and I get $$E = \frac{1}{2}$$ Is this correct?

  1. Find total energy of the signal using its frequency domain representation $$E(\omega)=\int_{-\infty}^{\infty}\left|\frac{1}{1+j\omega}\right|^2 d\omega$$

I think this is the approach I have to use in order to calculate the energy right? And since this contains an imaginary part, how can I integrate this? :)

  1. Find the percentage of energy contained in the frequency range 200 - 500 Hz
    If I take the magnitude in the 3rd Q in order to integrate it, I don't get a function of $\omega$
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    $\begingroup$ A hint for 3.: $|1 + j \omega| = \sqrt{1 + \omega^2}$. Also this. $\endgroup$ – user66081 Dec 21 '15 at 2:45
  • $\begingroup$ So the integral simplifies into $$\left(\frac{1}{1+\omega ^2}\right)$$ and that is evaluated $\pi$ ??? $\endgroup$ – Blogger Dec 21 '15 at 3:14
  • $\begingroup$ Another thing I need to clarify is, what is the point of putting the absolute sign in these integrals? How can we integrate a function with imaginary terms in it if there are no absolute sign in them? $\endgroup$ – Blogger Dec 21 '15 at 3:15
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The Laplace transform $$ \mathscr{L}\{f\}(\omega)=\int_{0}^{\infty}e^{-\omega t}f(t)dt $$ may be examined on a vertical line in the complex plane $\omega = u+iv$ for $-\infty < v < \infty$ for any fixed $u$ for which $\int_{0}^{\infty}e^{-ut}|f(t)|dt < \infty$. This function of $u$ $$ \mathscr{L}\{f\}(u+iv)=\int_{0}^{\infty}e^{-ivt}(e^{-ut}f(t))dt $$ is related to the Fourier transform of $e^{-ut}f(t)$ where $f$ is extended to be $0$ on $(-\infty,0)$. By Parseval's equality for the Fourier transform, $$ \int_{0}^{\infty}|f(t)e^{-ut}|^2dt = \frac{1}{2\pi}\int_{-\infty}^{\infty} |\mathscr{L}\{f\}(u+iv)|^2 dv $$ This is true for all $u$ for which $F_{u}(t) = f(t)e^{-ut}$ is square integrable in $t$ on $[0,\infty)$.

In your case, $$ \mathscr{L}\{f\}(u+iv) = \int_{0}^{\infty}e^{-t}e^{-(u+iv)t}dt = \frac{1}{1+u+iv} $$ Your function is square integrable for $u=0$, which gives energy \begin{align} \frac{1}{2}=\int_{0}^{\infty}e^{-2t}dt & = \frac{1}{2\pi}\int_{-\infty}^{\infty}\left|\frac{1}{1+iv}\right|^2 dv \\ & =\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1}{1+v^2}dv \\ & = \left.\frac{1}{2\pi}\tan^{-1}(v)\right|_{v=-\infty}^{\infty} \\ & =\frac{1}{2\pi}\pi=\frac{1}{2}. \end{align} The reconstruction of $f$ is $$ f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathscr{L}\{f\}(iv)e^{ivt}dv $$ The signal $e^{ivt}$ is a periodic function which goes through one complete cycle every $2\pi/|v|$ seconds. So the frequency of $e^{ivt}$ is $$ \frac{1\mbox{cycle}}{(2\pi/|v|) \mbox{seconds}}=\frac{|v|}{2\pi}\frac{\mbox{cycles}}{\mbox{second}} $$ You want the components of the signal for which $$ 200 < \frac{|v|}{2\pi} < 500. $$ The energy corresponding to these components is obtained by integrating the energy density function over these frequencies. $$ \frac{1}{2\pi}\int_{-500(2\pi)}^{-200(2\pi)}\frac{1}{1+v^2}dv + \frac{1}{2\pi}\int_{200(2\pi)}^{500(2\pi)}\frac{1}{1+v^2}dv \\ = \frac{1}{\pi}\int_{400\pi}^{1000\pi}\frac{1}{1+v^2}dv \\ = \frac{1}{\pi}\{\tan^{-1}(1000\pi)-\tan^{-1}(400\pi)\} $$

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    $\begingroup$ This explains it way beyond and that's why I like stackexchange \m/ Thank you @TrialAndError this is what I was looking for and was bothering me! $\endgroup$ – Blogger Dec 21 '15 at 3:45
  • $\begingroup$ @Blogger : You're welcome. $\endgroup$ – DisintegratingByParts Dec 21 '15 at 3:56

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