Latest revision as of 05:31, 22 September 2014

Homework 1 Solution, ECE438, Fall 2014, Prof. Boutin

Note: Please pay attention to the difference between $ X \ $ and $ {\mathcal X} $.

A complex exponential

$ x(t)=e^{j2 \pi f_0 t} $

From table, $ {\mathcal X} (\omega)= 2\pi \delta(\omega - \omega_0) $, therefore
$ \begin{align} X(f) & = {\mathcal X} (2 \pi f)\\ &= 2\pi \delta(2\pi f - 2\pi f_0) \\ &=\delta(f - f_0), \end{align} $
where the last line follows from the scaling property of the Dirac delta distribution.

A sine

$ x(t)=\sin (2\pi f_0 t) $

From table, $ {\mathcal X} (\omega)= \frac{\pi}{i} \left[\delta (\omega - \omega_0) - \delta (\omega + \omega_0)\right] $, therefore
$ \begin{align} X(f) & = {\mathcal X} (2 \pi f)\\ &= \frac{2 \pi}{2j} \delta (2\pi f - 2\pi f_0) - \frac{2 \pi}{2 j} \delta(2\pi f + 2 \pi f_0) \\ &=\frac{1}{2j}\delta(f-f_0) - \frac{1}{2j}\delta(f+f_0) , \end{align} $
where the last line follows from the scaling property of the Dirac delta distribution.

A cosine

$ x(t)=\cos (2\pi f_0 t) $

From table, $ {\mathcal X} (\omega)= \pi \left[\delta (\omega - \omega_0) + \delta (\omega + \omega_0)\right] $, therefore
$ \begin{align} X(f) & = {\mathcal X} (2 \pi f)\\ &= \frac{2 \pi}{2 } \delta (2\pi f - 2\pi f_0) + \frac{2 \pi}{2 } \delta(2\pi f + 2 \pi f_0) \\ &=\frac{1}{2}\delta(f-f_0) + \frac{1}{2}\delta(f+f_0) , \end{align} $
where the last line follows from the scaling property of the Dirac delta distribution.

A periodic function

$ x(t)=\sum_{k=-\infty}^{\infty} a_k e^{jk2\pi f_0 t} $
From the table, we have the transform pair:
$ \sum_{k=-\infty}^{\infty} a_k e^{j\omega_0t} \leftrightarrow 2\pi \sum_{k=-\infty}^{\infty} a_k \delta(\omega-k\omega_0) $
Therefore, using the definition that $ \omega=2\pi f $:
$ \begin{align} \sum_{k=-\infty}^{\infty} a_k e^{j2\pi f_0t} \leftrightarrow &2\pi \sum_{k=-\infty}^{\infty} a_k \delta(2\pi f-k2\pi f_0) \\ &=\sum_{k=-\infty}^{\infty} a_k \delta(f-k f_0) \mbox{, by the scaling property of the delta} \end{align} $

An impulse train

$ x(t)=\sum_{n=-\infty}^{\infty} \delta (t-nT) $
From the table, we have the transform pair:
$ \sum_{n=-\infty}^{\infty} \delta (t-nT) \leftrightarrow \frac{2 \pi}{T} \sum_{k=-\infty}^{\infty} \delta \left ( \omega - \frac{2\pi k}{T} \right ) $
Therefore, using the definition that $ \omega=2\pi f $:
$ \begin{align} \sum_{n=-\infty}^{\infty} \delta (t-nT) \leftrightarrow &\frac{2 \pi}{T} \sum_{k=-\infty}^{\infty} \delta \left ( 2\pi f- \frac{2\pi k}{T} \right ) \\ &=\frac{1}{T} \sum_{k=-\infty}^{\infty} \delta \left (f- \frac{k}{T} \right ) \mbox{, using the scaling property of the delta} \end{align} $

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