Latest revision as of 10:34, 7 March 2011

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) $.

Instructor's comment: Actually, this is only true for $ t\neq 0 $, since $ x(0)=1 $. However, changing the value of a signal at a single point does not change its Fourier transform. -pm

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) $

Instructor's note: In order to use this property, one first needs to prove it. Unfortunately, proving it is quite tricky, as was noted in class. Therefore, the above method cannot be used in this homework. Here is a correct solution:

$ \begin{align} \sin^2\left(\pi t +\frac{\pi}{12}\right)&= \left( \frac{1}{2j}e^{j (\pi t +\frac{\pi}{12})}-\frac{1}{2j}e^{-j (\pi t +\frac{\pi}{12})} \right)^2 \\ &= \left( \frac{1}{2j}e^{j \frac{\pi}{12}}e^{j\pi t}-\frac{1}{2j}e^{-j \frac{\pi}{12}}e^{-j \pi t} \right)^2 \\ & = \frac{1}{2} - \frac{e^{j\frac{\pi}{6}}}{4} e^{j 2\pi t} -\frac{e^{-j\frac{\pi}{6}}}{4} e^{-j2\pi t} \\ \Rightarrow {\mathcal F} \left( \sin^2\left(\pi t +\frac{\pi}{12}\right)\right) &= \int_{-\infty}^\infty \left( \frac{1}{2} - \frac{e^{j\frac{\pi}{6}}}{4} e^{j 2\pi t} -\frac{e^{-j\frac{\pi}{6}}}{4} e^{-j2\pi t} \right) e^{-j\omega t} dt \\ &= \int_{-\infty}^\infty \frac{1}{2} e^{-j\omega t} dt - \int_{-\infty}^\infty \frac{e^{j\frac{\pi}{6}}}{4} e^{j 2\pi t} e^{-j\omega t} dt -\int_{-\infty}^\infty \frac{e^{-j\frac{\pi}{6}}}{4} 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 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} {\mathcal F} \left(1\right) - \frac{e^{j\frac{\pi}{6}}}{4} {\mathcal F} \left( e^{j 2\pi t} \right) -\frac{e^{-j\frac{\pi}{6}}}{4} {\mathcal F} \left( e^{-j2\pi t} \right). \end{align} $

So all we need is the Fourier transform of $ e^{j \omega_0 t} $, which cannot be found by integration. Since we are not allowed to use the table of properties and pairs, we will have to "guess it" and justify our answer.

Claim: $ {\mathcal F} \left(e^{j \omega_0 t} \right) = 2 \pi \delta(\omega-\omega_0) $

Proof: $ {\mathcal F}^{-1} \left( 2 \pi \delta(\omega-\omega_0) \right) = \frac{1}{2\pi} \int_{-\infty}^{\infty} 2 \pi \delta(\omega-\omega_0) e^{-j\omega t} d\omega = e^{-j\omega_0 t} $.

Therefore

$ {\mathcal F} \left( \sin^2\left(\pi t +\frac{\pi}{12}\right)\right)= \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) $

-pm

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) $.

Instructor's note: Alternatively, you can view x(t) as a product of two sinusoidal. By the multiplication property, the Fourier transform of x(t) is thus the convolution of the Fourier transform of a sinusoidal with itself. -pm

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}) $

Instructor's note: It is probably easier to do this problem using the definition of the Fourier transform.

$ \begin{align} {\mathcal F} \left( x(-3t+2) \right) &= \int_{-\infty}^\infty x(-3t+2) e^{-j\omega t} dt, \\ &= \int_{\infty}^{-\infty} x(u) e^{-j\omega \frac{(u-2)}{-3}} \frac{du}{-3}, \text{ (letting }u=-3t+2), \\ &= \frac{1}{3} \int_{-\infty}^\infty x(u)e^{-\frac{j\omega u}{-3}}e^{\frac{-2j\omega}{3}} du,\\ &= \frac{e^{\frac{-2j\omega}{3}}}{3}\int_{-\infty}^\infty x(u)e^{-\frac{j\omega u}{-3}} du, \\ &= \frac{e^{\frac{-2j\omega}{3}}}{3} \mathcal{X}(\frac{\omega}{-3}). \end{align} $

Question 5

a)

$ \mathcal{H}(\omega)=\mathfrak{F}\left\{e^{-3t}u(t)\right\}=\int_{-\infty}^\infty e^{-3t}u(t) e^{-j\omega t} dt = \int_{0}^\infty e^{-3t} e^{-j\omega t} dt =\left. \frac{e^{(-3-j\omega) t}}{ -3 -j \omega }\right|_0^\infty =\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) $ .

HW5

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