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Homework 1 Solution, ECE438, Fall 2015, Prof. Boutin
Note: Please pay attention to the difference between $ X \ $ and $ {\mathcal X} $.
A complex exponential
$ x(t)=e^{j2 \pi 800 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 800) \\ &=\delta(f - 800), \end{align} $
where the last line follows from the scaling property of the Dirac delta distribution.
A sine
$ x(t)=\sin (2\pi 600 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 600) - \frac{2 \pi}{2 j} \delta(2\pi f + 2 \pi 600) \\ &=\frac{1}{2j}\delta(f-600) - \frac{1}{2j}\delta(f+600) , \end{align} $
where the last line follows from the scaling property of the Dirac delta distribution.
A cosine
$ x(t)=\cos (2\pi 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 1) + \frac{2 \pi}{2 } \delta(2\pi f + 2 \pi 1) \\ &=\frac{1}{2}\delta(f-1) + \frac{1}{2}\delta(f+1) , \end{align} $
where the last line follows from the scaling property of the Dirac delta distribution.
A periodic function
From the table, we have the transform pair of a periodic function as
$ \sum_{k=-\infty}^{\infty} a_k e^{j\omega_0t} \leftrightarrow 2\pi \sum_{k=-\infty}^{\infty} a_k \delta(\omega-k\omega_0) $
where $ a_k $ are the Fourier Series coefficients of the function.
We can represent this in terms of $ f $ using the definition $ \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} $
To find Fourier series coefficients coefficients, we use the formula
$ a_k=\frac{1}{T_0}\int_{\tau}^{\tau+T_0} x(t)e^{-j\omega_0 kt}dt $
Note that we can integrate over any one period of the function. Performing this integration gives
$ \begin{align} a_k&=\frac{1}{T_0}\int_{\tau}^{\tau+T_0} x(t)e^{-j\omega_0 kt}dt \\ &=\frac{1}{4}\int_{-2}^{2} x(t)e^{-j \frac{\pi kt}{2}}dt \\ &=\frac{1}{4}\int_{-1}^{1} e^{-j \frac{\pi kt}{2}}dt \\ &=\frac{1}{4}\frac{-2}{j\pi k} \left [e^{-j\frac{\pi kt}{2}} \right ]_{-1}^1 \\ &= \frac{-1}{j2\pi k} \left [e^{-j\frac{\pi k}{2}} - e^{j\frac{\pi k}{2}}\right ] \\ &= \frac{\sin \left ( \frac{\pi}{2} k \right )}{\pi k} \end{align} $
Then we have that
$ X(f)= \sum_{k=-\infty}^{\infty} \frac{\sin \left ( \frac{\pi }{2}k \right )}{\pi k} \delta \left (f-\frac{k}{4} \right ) $
An impulse train
$ x(t)=\sum_{n=-\infty}^{\infty} \delta (t-5n) $
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-5n) \leftrightarrow &\frac{2 \pi}{T} \sum_{k=-\infty}^{\infty} \delta \left ( 2\pi f- \frac{2\pi 5}{T} \right ) \\ &=\frac{1}{T} \sum_{k=-\infty}^{\infty} \delta \left (f- \frac{5}{T} \right ) \mbox{, using the scaling property of the delta} \end{align} $
Discussion
You may discuss the homework below.
- Should the final answer read:
$ \begin{align} \sum_{n=-\infty}^{\infty} \delta (t-5n) \leftrightarrow &\frac{2 \pi}{5} \sum_{k=-\infty}^{\infty} \delta \left ( 2\pi f- \frac{2\pi k}{5} \right ) \\ &=\frac{1}{5} \sum_{k=-\infty}^{\infty} \delta \left (f- \frac{k}{5} \right ) \mbox{, using the scaling property of the delta} \end{align} $
If not, why? Confused about the 5/T in answer solution which I thought should be k/5.
- You are correct. There is a typo in the early state of the solutions, and it was carried over by copy and paste. Thanks for pointing this out! -pm
