Latest revision as of 05:15, 7 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) $

First of all, we have

$ \delta(ax) = \left\{ \begin{array}{ll} \infty, & x=0\\ 0, & \text { else}. \end{array} \right. $

Now, given the fact that $ \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 \\ e^{j\pi t}&= \frac{1}{2\pi} \int\limits_{-\infty}^{\infty}{\mathcal X}(\omega)e^{j\omega t}d\omega \end{align} $

By comparing the left hand side and right hand side of the equation, we have

$ \frac{1}{2\pi}{\mathcal X}(\omega) = \delta(\omega - \pi) $

Therefore,

$ {\mathcal X}(\omega) = 2\pi \delta(\omega - \pi) $

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