| Line 35: | Line 35: | ||
We cannot compute the Fourier transform directly because | We cannot compute the Fourier transform directly because | ||
| − | <math>{\mathcal X}(\omega) = \int\limits_{-\infty}^{\infty}x(t)e^{-j\omega t}dt</math> | + | <math>{\mathcal X}(\omega) = \int\limits_{-\infty}^{\infty}x(t)e^{-j\omega t}dt = \int\limits_{-\infty}^{\infty}e^{-j(\omega -\pi) t}dt</math> |
cannot be integrated. | cannot be integrated. | ||
| Line 41: | Line 41: | ||
Instead, we can find out <math>{\mathcal X}(\omega) </math> using inverse Fourier transform. | Instead, we can find out <math>{\mathcal X}(\omega) </math> using inverse Fourier transform. | ||
| − | <math>x(t) = \frac{1}{2\pi} \int\limits_{-\infty}^{\infty}{\mathcal X}(\omega)e^{j\omega t}d\omega</math> | + | <math> |
| + | \begin{align} | ||
| + | x(t) &= \frac{1}{2\pi} \int\limits_{-\infty}^{\infty}{\mathcal X}(\omega)e^{j\omega t}d\omega \\ | ||
| + | |||
| + | |||
| + | &= \sum_{n=-\infty}^{\infty} | ||
| + | |||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | <math></math> | ||
Revision as of 15:38, 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}) $
Question 2
We cannot compute the Fourier transform directly because
$ {\mathcal X}(\omega) = \int\limits_{-\infty}^{\infty}x(t)e^{-j\omega t}dt = \int\limits_{-\infty}^{\infty}e^{-j(\omega -\pi) t}dt $
cannot be integrated.
Instead, we can find out $ {\mathcal X}(\omega) $ using inverse Fourier transform.
$ \begin{align} x(t) &= \frac{1}{2\pi} \int\limits_{-\infty}^{\infty}{\mathcal X}(\omega)e^{j\omega t}d\omega \\ &= \sum_{n=-\infty}^{\infty} \end{align} $
