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Given <math>\int\limits_{-\infty}^{\infty}\delta(x)dx=1 </math> | Given <math>\int\limits_{-\infty}^{\infty}\delta(x)dx=1 </math> | ||
− | then <math>\int\limits_{-\infty}^{\infty}\delta(ax)dx \overset{\underset{\mathrm{y=ax}}{}}{=} \int\limits_{-\infty}^{\infty}\delta(y) | + | then <math>\int\limits_{-\infty}^{\infty}\delta(ax)dx \overset{\underset{\mathrm{y=ax}}{}}{=} \int\limits_{-\infty}^{\infty}\frac{1}{a}\delta(y)dy</math> |
+ | |||
+ | Therefore, <math>\delta(ax) = \frac{1}{a}\delta(x) </math> | ||
Revision as of 12:06, 4 September 2011
Homework 1, ECE438, Fall 2011, Prof. Boutin
Question 1
In ECE301, you learned that the Fourier transform of a step function $ x(t)=u(t) $ is the following:
$ {\mathcal X} (\omega) = \frac{1}{j \omega} + \pi \delta (\omega ). $
Use this fact to obtain an expression for the Fourier transform $ X(f) $ (in terms of frequency in hertz) of the step function. (Your answer should agree with the one given in this table.) Justify all your steps.
Answer: Recall the relation between frequency in hertz $ f $ and frequency in radius $ \omega $
$ \omega =2\pi f $
Pull in the relation into the fact, we obtain
$ {\mathcal X}(2\pi f) = \frac{1}{j 2\pi f} + \pi \delta (2\pi f ). $
Then we justify the following equality.
$ \delta(ax) = \frac{1}{a}\delta(x) $
Given $ \int\limits_{-\infty}^{\infty}\delta(x)dx=1 $
then $ \int\limits_{-\infty}^{\infty}\delta(ax)dx \overset{\underset{\mathrm{y=ax}}{}}{=} \int\limits_{-\infty}^{\infty}\frac{1}{a}\delta(y)dy $
Therefore, $ \delta(ax) = \frac{1}{a}\delta(x) $