(New page: =Homework 10 Solutions, ECE301 Spring 2011 Prof. Boutin= Students should feel free to make comments/corrections or ask questions directly on this page. ==Q...)
 
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==Question 1==
 
==Question 1==
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a) We can write
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:<math>y_1(t)=e^{j \theta_c}x(t)e^{j\omega_c t}</math>
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 +
Notice that this is exactly as modulating by <math>e^{j\omega_c t}</math> but now we are multiplying with a complex exponential independent of <math>t</math> (phase shift).
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We can recover the signal <math>x(t)</math> for any <math>\omega_c</math>, and hence there are no conditions put on <math>\omega_c</math>.
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b) In order to recover signal <math>x(t)</math>, we multiply <math>y_1(t)</math> by <math class="inline">e^{-j(\omega_c+\theta_c)}</math>.
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 +
c) We can write
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:<math>y_2(t)=x(t)\left(\frac{e^{j\omega_c t}-e^{-j\omega_c t}}{2j}\right)</math>
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Taking the FT of <math>y_2(t)</math>, we get:
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:<math>\begin{align}
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\mathcal{Y}_2(\omega)&=\frac{1}{2\pi(2j)}\mathcal{X}(\omega)*[2\pi\delta(\omega-\omega_c)-2\pi\delta(\omega+\omega_c)] \\
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&=\frac{1}{2j}\mathcal{X}(\omega-\omega_c)-\frac{1}{2j}\mathcal{X}(\omega+\omega_c)
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\end{align}
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</math>
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Now, to insure that we can recover signal <math>x(t)</math> we need to avoid having the two images of <math>X(\omega)</math> overlap. Hence we need <math>\omega_c>\omega_M</math>. But <math>\omega_M=2000\pi/2=1000\pi</math>. Hence in order for <math>x(t)</math> to be recoverable we need:
 +
 +
:<math>\omega_c>1000\pi</math>
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 +
d)In order to recover signal <math>x(t)</math> we multiply <math>y_2(t)</math> by <math>\sin(\omega_c t)</math> first. The signal after multiplying with <math>\sin(\omega_c t)</math>  is:
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 +
:<math>\begin{align}
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r(t)&=y_2(t)sin(\omega_c t) \\
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&=sin^2(\omega_c t)x(t) \\
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&=\frac{1}{2}x(t)-\frac{1}{2}cos(2\omega_c t)x(t)
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\end{align}
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</math>
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 +
Thus in order to recover <math>x(t)</math> we need to filter out the second term of <math>r(t)</math> and amplify it by a factor of 2. To do this, we pass <math>r(t)</math> through a low pass filter with a cut-off frequency <math>\omega_cut=\omega_M=1000\pi</math> and gain 2. The frequency response of this low pass filter is:
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 +
:<math>\mathcal{H}(\omega)=\left\{\begin{array}{ll}
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2, & \mbox{  for  }  |\omega|<1000\pi\\
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0, & \mbox{  elsewhere}
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\end{array}\right.
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</math>
 
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[[HW10 ECE301 Spring2011 Prof Boutin| HW10]]
 
[[HW10 ECE301 Spring2011 Prof Boutin| HW10]]

Revision as of 17:49, 13 April 2011

Homework 10 Solutions, ECE301 Spring 2011 Prof. Boutin

Students should feel free to make comments/corrections or ask questions directly on this page.

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 it by a factor of 2. To do this, 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. $

HW10

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