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Homework 5 Solutions, ECE301 Spring 2011 Prof. Boutin
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Question 1
$ \begin{align} \mathcal{X}(\omega)&= \int_{-\infty}^{\infty} e^{-3|t|}e^{-j\omega t} dt \\ &= \int_{-\infty}^0 e^{3t}e^{-j\omega t} dt + \int_0^{\infty} e^{-3t}e^{-j\omega t} dt \\ &= \int_{-\infty}^0 e^{(-j\omega +3)t} dt + \int_0^{\infty} e^{-(j\omega +3)t} dt \\ &= \frac{1}{-j\omega +3} \left[e^{(-j\omega +3)t}\right]_{-\infty}^0 - \frac{1}{j\omega +3} \left[e^{-(j\omega +3)t}\right]^{\infty}_0 \\ &= \frac{1}{-j\omega +3} \left[1-0 \right] - \frac{1}{j\omega +3} \left[0-1\right] \\ &= \frac{1}{-j\omega +3} + \frac{1}{j\omega +3} \\ &= \frac{j\omega +3}{\omega^2 +9}+\frac{-j\omega +3}{\omega^2 + 9} \\ &= \frac{6}{\omega^2+9} \end{align} $
To verify our answer using the table, we first write:
$ x(t)=e^{-3|t|}=e^{3t}u(-t)+e^{-3t}u(t) $.
From the table:
$ \mathfrak{F}\left\{e^{-3t}u(t)\right\}=\frac{1}{j\omega+3} $.
Using time-reversal property from the table:
$ \mathfrak{F}\left\{e^{3t}u(-t)\right\}=\frac{1}{-j\omega+3} $.
Now, since the Fourier transform (FT) is linear, we have that:
$ \begin{align} \mathcal{X}(\omega)&=\mathfrak{F}\left\{e^{-3t}u(t)\right\}+\mathfrak{F}\left\{e^{3t}u(-t)\right\} \\ &=\frac{1}{j\omega+3}+\frac{1}{-j\omega+3} \\ &=\frac{6}{\omega^2+9} \end{align} $.
Question 2
$ \begin{align} \mathcal{X}(\omega)&= \int_{-\infty}^{\infty} \sin^2\left(\pi t +\frac{\pi}{12}\right)e^{-j\omega t}dt \\ &= \int_{-\infty}^{\infty} \left[\frac{1}{2} - \frac{1}{2}\cos\left(2\pi t + \frac{\pi}{6}\right) e^{-j\omega t} \right]dt \\ &= \frac{1}{2} \int_{-\infty}^{\infty} e^{-j\omega t} dt - \frac{e^{j\frac{\pi}{6}}}{4} \int_{-\infty}^{\infty} e^{j2\pi t} e^{-j\omega t} dt -\frac{e^{-j\frac{\pi}{6}}}{4} \int_{-\infty}^{\infty}e^{-j2\pi t}e^{-j\omega t} dt \\ &= \frac{1}{2} \int_{-\infty}^{\infty} e^{-j\omega t} dt - \frac{e^{j\frac{\pi}{6}}}{4} \int_{-\infty}^{\infty} e^{j(2\pi -\omega)t} dt -\frac{e^{-j\frac{\pi}{6}}}{4} \int_{-\infty}^{\infty}e^{-j(2\pi + \omega)t} dt \\ &=\pi \delta(\omega) - \frac{\pi e^{j\frac{\pi}{6}}}{2} \delta(\omega - 2\pi) - \frac{\pi e^{-j\frac{\pi}{6}}}{2} \delta(\omega + 2\pi), \end{align} $
where we have used the following property:
$ \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-j\omega t} dt = \delta(\omega) $
Now to verify our answer we write:
$ \begin{align} x(t)&=\sin^2\left(\pi t - \frac{\pi}{12}\right) \\ &= \frac{1}{2}-\frac{1}{2}\cos\left(2\pi t + \frac{\pi}{6}\right) \\ &= \frac{1}{2}-\frac{1}{2}\cos\left[2\pi\left( t + \frac{1}{12}\right)\right] \end{align} $
Now using (from the table) the time-shift property, FT of a constant, and FT of a cosine we get:
$ \begin{align} \mathcal{X}(\omega) &= \pi\delta(\omega) - \frac{\pi e^{j\frac{\omega}{12}}}{2} \delta(\omega - 2\pi)- \frac{\pi e^{j\frac{\omega}{12}}}{2} \delta(\omega + 2\pi) \\ &= \pi\delta(\omega) - \frac{\pi e^{j\frac{\pi}{6}}}{2} \delta(\omega - 2\pi)- \frac{\pi e^{-j\frac{\pi}{6}}}{2} \delta(\omega + 2\pi) \end{align} $
where we have used the fact that $ f(t)\delta(t-t_0)=f(t_0)\delta(t-t_0) $.
Question 3
$ \begin{align} E_{\infty} \left\{x(t)\right\} &= \frac{1}{2\pi} \int_{-\infty}^{\infty} \left| |\omega|^3 [u(\omega + 8)-u(\omega-5)]\right|^2 d\omega \\ &=\frac{1}{2\pi} \int_{-\infty}^{\infty} \omega^6[u(\omega+8)-u(\omega-5)]d\omega \\ &=\frac{1}{2\pi} \int_{-8}^{5}\omega^6 d\omega \\ &=\frac{1}{14\pi} \omega^7\Bigg|^5_{-8} \\ &=\frac{1}{14\pi}[5^7-(-8)^7] \\ &=\frac{2175277}{14\pi} \end{align} $
Question 4
$ \begin{align} y(t)&=x(-3t+2) \\ &=x\left[-3\left(t-\frac{2}{3}\right)\right] \end{align} $
From this we can deduce that we have time scaling first by a factor of -3, and then we have a time delay by 2/3.
Now,
$ \mathfrak{F}\left\{x(-3t)\right\}=\frac{1}{3}\mathcal{X}(-\frac{\omega}{3}) $,
using the time scaling property of the FT.
Then,
$ \mathcal{Y}(\omega)=\frac{e^{-j\frac{2}{3}\omega}}{3}\mathcal{X}(-\frac{\omega}{3}) $
Question 5
a)
$ \mathcal{H}(\omega)=\mathfrak{F}\left\{e^{-3t}u(t)\right\}=\frac{1}{3+j\omega} $
b)
$ \mathcal{X}(\omega)=\mathfrak{F}\left\{e^{-2(t-2)}u(t-2)\right\}=\frac{e^{-2j\omega}}{2+j\omega} $
$ \begin{align} \mathcal{Y}(\omega)&=\mathcal{H}(\omega) \mathcal{X}(\omega) \\ &= \frac{e^{-2j \omega}}{(3+j\omega)(2+j\omega)} \\ &= \frac{e^{-2j\omega}}{6+5j\omega-\omega^2} \end{align} $
Now we need to find the partial fraction expansion of $ \mathcal{Y}(\omega) $.
Let
$ \frac{1}{(3+j\omega)(2+j\omega)}=\frac{A}{3+j\omega}+\frac{B}{2+j\omega} $
By multiplying both sides by $ (3+j\omega)(2+j\omega) $ and simplifying we get:
$ 2A+3B+(A+B)j\omega=1 $
After comparing, we get $ A=-1 $ and $ B=1 $.
Then,
$ \mathcal{Y}(\omega)=-\frac{e^{-2j \omega}}{(3+j\omega)}+\frac{e^{-2j \omega}}{(2+j\omega)} $
Now, using FT pairs and the time shift property, we have:
$ y(t)=e^{-2(t-2)}u(t-2)-e^{-3(t-2)}u(t-2) $
Question 6
a)
Taking the FT of both sides of the differential equation we get:
$ \begin{align} -\omega^2\mathcal{Y}(\omega)&=3j\omega\mathcal{Y}(\omega)-2\mathcal{Y}(\omega)+\mathcal{X}(\omega) \\ (-3j\omega-\omega^2+2)\mathcal{Y}(\omega)&=\mathcal{X}(\omega) \\ \mathcal{Y}(\omega)&=\frac{\mathcal{X}(\omega)}{2-3j\omega-\omega^2} \\ \end{align} $
Then,
$ \mathcal{H}(\omega)=\frac{1}{2-3j\omega-\omega^2} $.
b)
The frequency response of the system can be written as:
$ \mathcal{H}(\omega)=\frac{1}{(j\omega-1)(j\omega-2)} $.
Note that this can be done by either direct inspection or by finding the roots of the quadratic equation in the denominator.
Now, we need to find the partial fraction expansion of $ \mathcal{H}(\omega) $.
Let
$ \frac{1}{(j\omega-1)(j\omega-2)}=\frac{A}{j\omega-1}+\frac{B}{j\omega-2} $
By multiplying both sides by $ (j\omega-1)(j\omega-2) $ and simplifying we get:
$ -2A-B+(A+B)j\omega=1 $
After comparing, we get $ A=-1 $ and $ B=1 $.
Then,
$ \mathcal{H}(\omega)=-\frac{1}{j\omega-1}+\frac{1}{j\omega-2}=\frac{1}{1-j\omega}-\frac{1}{2-j\omega} $
Now, using linearity of FT, FT pairs, and time reverse property, we get:
$ h(t)=e^{t}u(-t)-e^{2t}u(-t) $ .
