| Line 19: | Line 19: | ||
\end{align} | \end{align} | ||
</math><br> | </math><br> | ||
| − | Where the last line | + | Where the last line follows from the [https://www.projectrhea.org/rhea/index.php/Homework_3_ECE438F09 scaling] property of the delta function. |
---- | ---- | ||
===A sine=== | ===A sine=== | ||
| − | <math> x(t)=sin(t) </math> | + | <math> |
| + | \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}</math> <br> | ||
| + | |||
| + | <math> | ||
| + | \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} | ||
| + | </math> | ||
---- | ---- | ||
===A cosine=== | ===A cosine=== | ||
| − | <math>x(t)=cos(t)</math> | + | <math>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}</math> |
| + | |||
| + | <math> | ||
| + | \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} | ||
| + | </math> | ||
| + | |||
---- | ---- | ||
===A periodic function=== | ===A periodic function=== | ||
Revision as of 16:36, 8 September 2014
Contents
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) $
Discussion
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