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Homework 1 Solution, ECE438, Fall 2014, Prof. Boutin
A complex exponential
$ x(t)=e^{j2 \pi f_0 t} $
From table, $ e^{j\omega_0t} \leftrightarrow 2\pi \delta(\omega - \omega_0) $, therefore
$ \begin{align} e^{j2\pi f_0 t } \leftrightarrow &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 delta function.
A sine
$ \begin{align} x(t)=sin(2\pi f_0 t) =\frac{1}{2j} e^{j2\pi f_0 t} - \frac{1}{2j} e^{-j2\pi f_0 t} \end{align} $
$ \begin{align} \mathcal{F} \left \{ sin (2 \pi f_0 t) \right \} &= \mathcal{F} \left \{ \frac{1}{2j} e^{j2\pi f_0 t} - \frac{1}{2j} e^{-j2\pi f_0 t} \right \} \\ &= \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) \mbox{, using the transform of the complex exponential} \\ &= \frac{1}{2j}\delta(f-f_0) - \frac{1}{2j}\delta(f+f_0) \mbox{, by the scaling property of the delta} \end{align} $
A cosine
$ x(t)=cos(2\pi f_0 t) = \frac{1}{2}e^{j2\pi f_0t} + \frac{1}{2}e^{-j2\pi f_0 t} $
$ \begin{align} \mathcal{F} \left \{ cos (2 \pi f_0 t) \right \} &= \mathcal{F} \left \{ \frac{1}{2} e^{j2\pi f_0 t} + \frac{1}{2} e^{-j2\pi f_0 t} \right \} \\ &= \frac{2 \pi}{2} \delta (2\pi f - 2\pi f_0) + \frac{2 \pi}{2 } \delta(2\pi f + 2 \pi f_0) \mbox{, using the transform of the complex exponential} \\ &= \frac{1}{2}\delta(f-f_0) + \frac{1}{2}\delta(f+f_0) \mbox{, by the scaling property of the delta} \end{align} $
A periodic function
$ x(t)=x(t-T) $
An impulse train
$ x(t)=\sum_{n=-\infty}^{\infty} \delta (t-nT) $
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