1
$\begingroup$

I am solving old exam problems in preparation of my exam in signals, and I'm having trouble with a question.

For a continuous aperiodic signal with the spectrum $X(\omega)=\exp(-\omega^2)$ I am to derive the energy of the signal. I have made it this far, in agreement with the solution manual:

$W = \frac{1}{2\pi}\int_{-\infty}^{\infty} \exp(-(\sqrt{2}\omega)^2)\,\mathrm{d}\omega$

The assignment then suggests substituting in the error function

$\mathrm{erf}(z) = \frac{1}{\sqrt{\pi}} \int_{-z}^z \exp(-y^2)\,\mathrm{d}y$

and exploiting that $\mathrm{erf}(z) \rightarrow 1$ towards infinity, which makes good sense, but the result in the solution manual before reducing the expression is

$W = \frac{1}{2\pi} \sqrt{\frac{\pi}{2}} \mathrm{erf}(\sqrt{2}\omega)\Big\rvert_{\omega \rightarrow \infty}$

I don't understand where the $\frac{1}{2}$ in the square root comes from, so I would appreciate if anyone could help me understand this.

$\endgroup$
1
$\begingroup$

The factor just comes from the substitution $y=\sqrt{2}\omega$ in the integral defining the error function:

$$\begin{align}E_x=\frac{1}{2\pi}\int_{-\infty}^{\infty}|X(\omega)|^2d\omega&=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-2\omega^2}d\omega\\&=\frac{1}{2\pi}\frac{1}{\sqrt{2}}\int_{-\infty}^{\infty}e^{-(\sqrt{2}\omega)^2}d(\sqrt{2}\omega)\\&=\frac{1}{2\pi}\frac{1}{\sqrt{2}}\int_{-\infty}^{\infty}e^{-y^2}dy\\&=\frac{1}{2\pi}\sqrt{\frac{\pi}{2}}\lim_{x\to\infty}\textrm{erf}(x)\\&=\frac{1}{2\sqrt{2\pi}}\end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.