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 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. $
