Revision as of 18:19, 13 April 2011

Homework 10 Solutions, ECE301 Spring 2011 Prof. Boutin

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Question 1

a) We can write

$ y_1(t)=e^{j \theta_c}x(t)e^{j\omega_c t} $

Notice that this is exactly as modulating by $ e^{j\omega_c t} $ but now we are multiplying with a complex exponential independent of $ t $ (phase shift). We can recover the signal $ x(t) $ for any $ \omega_c $, and hence there are no conditions put on $ \omega_c $.

b) In order to recover signal $ x(t) $, we multiply $ y_1(t) $ by $ e^{-j(\omega_c+\theta_c)} $.

c) We can write

$ y_2(t)=x(t)\left(\frac{e^{j\omega_c t}-e^{-j\omega_c t}}{2j}\right) $

Taking the FT of $ y_2(t) $, we get:

$ \begin{align} \mathcal{Y}_2(\omega)&=\frac{1}{2\pi(2j)}\mathcal{X}(\omega)*[2\pi\delta(\omega-\omega_c)-2\pi\delta(\omega+\omega_c)] \\ &=\frac{1}{2j}\mathcal{X}(\omega-\omega_c)-\frac{1}{2j}\mathcal{X}(\omega+\omega_c) \end{align} $

Now, to insure that we can recover signal $ x(t) $ we need to avoid having the two images of $ X(\omega) $ overlap. Hence we need $ \omega_c>\omega_M $. But $ \omega_M=2000\pi/2=1000\pi $. Hence in order for $ x(t) $ to be recoverable we need:

$ \omega_c>1000\pi $

d)In order to recover signal $ x(t) $ we multiply $ y_2(t) $ by $ \sin(\omega_c t) $ first. The signal after multiplying with $ \sin(\omega_c t) $ is:

$ \begin{align} r(t)&=y_2(t)\sin(\omega_c t) \\ &=\sin^2(\omega_c t)x(t) \\ &=\frac{1}{2}x(t)-\frac{1}{2}\cos(2\omega_c t)x(t) \end{align} $

Thus in order to recover $ x(t) $ we need to filter out the second term of $ r(t) $ and amplify the remainder by a factor of 2 (you may want to draw the FT of $ r(t) $ to verify this). To achieve that, we pass $ r(t) $ through a low pass filter with a cut-off frequency $ \omega_{cut}=\omega_M=1000\pi $ and gain 2. The frequency response of this low pass filter is:

$ \mathcal{H}(\omega)=\left\{\begin{array}{ll} 2, & \mbox{ for } |\omega|<1000\pi\\ 0, & \mbox{ elsewhere} \end{array}\right. $

Note that the cut-off frequency of the low pass filter can actually be anywhere between $ \omega_M $ and $ 2\omega_c-\omega_M $.

Question 2

a) The Fourier series coefficients of $ c(t) $ are:

$ \begin{align} a_k&=\frac{\sin(\frac{2\pi k 10^{-3}}{4\times 10^{-3}})}{\pi k} \\ &=\frac{\sin{\frac{\pi k}{2}}}{\pi k} \end{align} $

and using the synthesis equation of the Fourier series we get:

$ c(t)=\sum_{k=-\infty}^{\infty} a_k e^{j\frac{2\pi k}{T}t}=\sum_{k=-\infty}^{\infty} a_k e^{j2000\pi kt} $

Taking the FT of the latter sum, we get:

$ \mathcal{C}(\omega)=2\pi\sum_{k=-\infty}^{\infty}a_k \delta(\omega-2000\pi k) $

Now, let $ y(t)=x(t)c(t) $, then:

$ \begin{align} \mathcal{Y}(\omega)&=\frac{1}{2\pi}\mathcal{X}(\omega)*\mathcal{C}(\omega) \\ &=\frac{1}{2\pi}\mathcal{X}(\omega)*2\pi\sum_{k=-\infty}^{\infty}a_k \delta(\omega-2000\pi k) \\ &=\sum_{k=-\infty}^{\infty}a_k \mathcal{X}(\omega-2000\pi k) \end{align} $

In order to recover $ x(t) $ we need to avoid aliasing and hence $ 2000\pi>2\omega_M $. Then $ \omega_M<1000\pi $.

b) We need to find $ a_0 $ since the image at DC is multiplied by it:

$ a_0=\lim_{k\to 0}\frac{\sin{\frac{\pi k}{2}}}{\pi k}=\frac{\frac{\pi k}{2}}{\pi k}=\frac{1}{2} $

Now, to recover $ x(t) $ we need to filter out the images other than the image at DC and and multiply it by $ \frac{1}{a_0}=2 $. Hence we use a low pass filter with the following frequency response:

$ \mathcal{H}(\omega)=\left\{\begin{array}{ll} 2, & \mbox{ for } |\omega|<\omega_{M}\\ 0, & \mbox{ elsewhere} \end{array}\right. $

Note that the cut-off frequency of the low pass filter can actually be anywhere between $ \omega_M $ and $ 2000\pi-\omega_M $.

Question 4

a)

$ \begin{align} X(s)&=\int_{-\infty}^{\infty}e^{-5t}u(t+3)e^{-st} dt \\ &=\int_{-3}^{\infty}e^{-(5+s)t} dt \\ &=-\frac{1}{s+5}e^{-(5+s)t}|^{\infty}_{-3} \\ &=\frac{e^{3(5+s)}}{s+5} \end{align} $

where the ROC is:

$ \mathcal{R}e\{5+s\}>0 $

$ 5+\mathcal{R}e\{s\}>0 $

$ \mathcal{R}e\{s\}>-5 $

b)

$ \begin{align} X(s)&=\int_{-\infty}^{\infty}e^{-5t}u(-t+5)e^{-st} dt \\ &=\int_{-\infty}^{5}e^{-(5+s)t} dt \\ &=-\frac{1}{s+5}e^{-(5+s)t}|_{-\infty}^{5} \\ &=-\frac{e^{-5(5+s)}}{s+5} \end{align} $

where the ROC is:

$ \mathcal{R}e\{5+s\}<0 $

$ 5+\mathcal{R}e\{s\}<0 $

$ \mathcal{R}e\{s\}<-5 $

c)

$ \begin{align} X(s)&=\int_{-\infty}^{\infty}e^{-5t}[u(t)-u(t-3)]e^{-st} dt \\ &=\int_{0}^{3}e^{-(5+s)t} dt \\ &=-\frac{1}{s+5}[e^{-3(5+s)}-1]\\ &=\frac{1}{s+5}-\frac{e^{-3(5+s)}}{s+5} \end{align} $

where we did not need to constraint our result to a ROC (which is expected since the signal has finite duration).

d)

$ \begin{align} X(s)&=\int_{-\infty}^{\infty}e^{-2|t|}e^{-st}dt \\ &=\int_{-\infty}^{0}e^{2t}e^{-st}dt + \int_{0}^{\infty}e^-{2t}e^{-st}dt \\ &=\int_{-\infty}^{0}e^{(2-s)t}dt + \int_{0}^{\infty}e^-{(2+s)t}dt \\ &=\frac{1}{2-s}e^{(2-s)t}|^{0}_{-\infty}-\frac{1}{2+s}e^{-(2+s)t}|_{0}^{\infty} \\ &=\frac{1}{2-s}+\frac{1}{2+s} \end{align} $

where for the first term to converge we need $ \mathcal{R}e\{2-s\}>0 $, which is equivilant to $ \mathcal{R}e\{s\}<2 $, and for the other term to converge we need $ \mathcal{R}e\{2+s\}>0 $, or equivalently $ \mathcal{R}e\{s\}>-2 $. Thus the ROC is the intersection of those two constraints and hence ROC is $ -2<\mathcal{R}e\{s\}<2 $.

HW10

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