| Line 14: | Line 14: | ||
Pull in the relation into the fact, we obtain | Pull in the relation into the fact, we obtain | ||
| − | <math>{\mathcal X}(2\pi f) = \frac{1}{j 2\pi f} + \pi \delta (2\pi f ).</math> | + | <math>{\mathcal X}(2\pi f) = \frac{1}{j 2\pi f} + \pi \delta (2\pi f ). (*)</math> |
Then we justify the following equality. | Then we justify the following equality. | ||
| Line 22: | Line 22: | ||
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}\frac{1}{a}\delta(y)dy</math> | + | then <math>\int\limits_{-\infty}^{\infty}\delta(ax)dx \overset{\underset{\mathrm{y=ax}}{}}{=} \int\limits_{-\infty}^{\infty}\frac{1}{a}\delta(y)dy = \frac{1}{a}</math> |
Therefore, <math>\delta(ax) = \frac{1}{a}\delta(x) </math> | Therefore, <math>\delta(ax) = \frac{1}{a}\delta(x) </math> | ||
| + | Therefore, (*) can be further simplified to | ||
| + | |||
| + | <math>X(f) = \frac{1}{j 2\pi f} + \frac{1}{2}\delta (f), where X(f) := {\mathcal X}{j 2\pi f}</math> | ||
---- | ---- | ||
Revision as of 13:11, 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 = \frac{1}{a} $
Therefore, $ \delta(ax) = \frac{1}{a}\delta(x) $
Therefore, (*) can be further simplified to
$ X(f) = \frac{1}{j 2\pi f} + \frac{1}{2}\delta (f), where X(f) := {\mathcal X}{j 2\pi f} $
