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== Exam 1, Problem 6 - Summer 2008 ==
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Read the discussion on discussion page using the discussion page tab. (Aung 11:13pm on 07/18/2008)
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a) By Inspection the linear constant-coefficient differential equation that describes the LTI system is:
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(a) The FT of <math>X(j\omega)</math> of a continuous-time signal x(t) is periodic
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MAY BE: -
  
  
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<math>y(t)= \frac{1}{7}x(t)- \frac{1}{7}\frac{dy(t)}{dt}</math>
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(b) The FT of <math>X(e^{j\omega})</math> of a continuous-time signal x[n] is periodic
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YES: <math>X(e^{j\omega})</math> is always periodic with period <math>2\pi</math>
  
  
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b) Show that the impulse response to this LTI system is given by <math>h(t)=e^{-7t}u(t)</math>
 
  
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This means that <math>x(t) = \delta{(t)} </math> and <math> y(t) = e^{-7t}u(t)</math>
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(c) If the FT of <math>X(e^{j\omega})</math> of a discrete-time signal x[n] is given as: <math>X(e^{j\omega}) = 3 + 3cos(3\omega)</math>, then the signal x[n] is periodic
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NO: The inverse transform of this signal is a set of delta functions that are not periodic.
  
  
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<math> e^{-7t}u(t) = \frac{1}{7}\delta{(t) } - \frac{1}{7}\frac{d(e^{-7t}u(t))}{dt} </math>
 
  
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(d) If the FT of <math>X(j\omega)</math> of a continuous-time signal x(t) consists of only impulses, then x(t) is periodic
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MAY BE: -
  
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Differentiating <math>\frac{1}{7}\frac{d(e^{-7t}u(t))}{dt} </math> requires use of the chain rule.
 
  
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This portion of the equation becomes:
 
  
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(e) Lets denote <math>X(j\omega)</math> the FT of a continuous-time non-zero signal x(t).  If x(t) is an odd signal, then:
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<math>\int_{-\infty}^{\infty} X(j\omega) d\omega = 0 </math>
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YES: this equation is the same as <math>\int_{-\infty}^{\infty} X(j\omega) e^{-j\omega_0t}d\omega = 0</math> where t = 0.
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From this we can conclude that x(0) = 0, which always holds true for odd signals.
  
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<math>\frac{1}{7}(-7)e^{-7t}u(t) - \frac{1}{7}\delta{(t) }e^{-7t} </math>
 
  
  
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<math> \frac{-1}{7}\delta{(t) }e^{-7t} </math> is <math> \frac{-1}{7}\delta{(t) }e^{-7t} </math> evaluated at t=0 or <math> \frac{-1}{7}\delta{(t) }(1) </math>
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(f) Lets denote <math>X(j\omega)</math> the FT of a continuous-time non-zero signal x(t).  If x(t) is an odd signal, then:
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<math>\int_{-\infty}^{\infty} |X(j\omega)|^2 d\omega = 0 </math>
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NO: using parseval's relation, we see that: <math>\int_{-\infty}^{\infty} |X(j\omega)|^2 d\omega = 0 = 2\pi \int_{-\infty}^\infty |x(t)|^2 dt </math>  
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The integral of the magnitude squared will always be positive for an odd signal.
  
  
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Plugging that back in yields:
 
  
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(g) Lets denote <math>X(e^{j0})</math> the FT of a DT signal x[n]. If <math>X(e^{j0})</math> = 0, then x[n] = 0.
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MAY BE: <math>X(e^{j0})</math> is simply <math>X(e^{j\omega})</math> evaluated at <math>\omega = 0</math>.
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This only tells you that summation of x[n] over all n is 0, not the entire signal x[n] = 0.
  
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<math> e^{-7t}u(t) = \frac{1}{7}\delta{(t) } - \frac{-7}{7}e^{-7t}u(t) - \frac{1}{7}\delta{(t) } </math>
 
  
  
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This equation simplifies to:
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(h) If the FT (<math>X(e^{j\omega})</math>) of a discrete-time signal x[n] is given as : <math>X(e^{j\omega}) = e^{-j10\omega}/(22.30e^{-j5\omega} + 11.15)</math>  then the signal x[n] is real.
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MAY BE: x[n] is real only if the properties of conjugate symmetry for real signals hold for this transform.
  
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0=0 indicating that it is correct.
 
  
  
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(i) Let x(t) be a continuous time real-valued signal for which <math>X(j\omega)</math> = 0 when <math>|\omega| > \omega_M</math> where <math>\omega_M</math> is a real and positive number. Denote the modulated signal y(t) = x(t)c(t) where c(t) = <math>cos(\omega_ct)</math> and <math>\omega_c</math> is a real, positive number. If <math>\omega_c</math> is greater than <math>2\omega_M</math>, x(t) can be recovered from y(t).
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YES: Taking the FT of c(t) we get delta functions at <math>\omega_c</math> and <math>-\omega_c</math>. 
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When convolved with the FT of the input signal <math>X(j\omega)</math>, the function <math>X(j\omega)</math> gets shifted to <math>\omega_c</math> and <math>-\omega_c</math>
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with ranges <math>(-\omega_c-\omega_M)</math> to <math>(-\omega_c+\omega_M)</math> and <math>(\omega_c-\omega_M)</math> to <math>(\omega_c+\omega_M)</math>.
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Therefore <math>(\omega_c-\omega_M) > (-\omega_c+\omega_M)</math> must hold for there to be no
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overlapping. This is equivalent to <math>2\omega_c > 2\omega_M  => \omega_c > \omega_M</math>. Since <math>\omega_c > 2\omega_M</math>, there is no overlapping
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and x(t) can be recovered.
  
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c) Find H(s) at s=jw for the LTI system with impulse response <math>h(t)=e^{-7t}u(t)</math>
 
  
  
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<math>H(s) = \int_{-\infty }^\infty h(t)e^{-st}dt</math>
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(j) Let x(t) be a continuous time real-valued signal for which <math>X(j\omega)</math> = 0 when <math>|\omega| > 40\pi</math>. Denote the modulated signal y(t) = x(t)c(t) where c(t) = <math>e^{j\omega_ct}</math> and <math>\omega_c</math> is a real, positive number. There is a constraint of <math>\omega_c</math> to guarantee that x(t) can be recovered from y(t).
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NO: The FT of c(t) is just a shifted delta function, which will simply shift the
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input signal x(t) so there is no chance of overlapping.
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<math>H(s) = \int_{-\infty }^\infty e^{-7t}u(t)e^{-st}dt</math>
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Limits change to 0 to <math> \infty </math> and u(t) drops out.
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<math>H(s) = \int_0^\infty e^(-7t)e^{-st}dt</math>
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<math>H(s) = \int_0^\infty e^{-7t-st}dt</math>
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<math>H(s) = \frac{e^{-7t-st}}{-7-s} \bigg|_0^\infty</math>
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<math>H(s) = \frac{e^{-7\infty -s\infty}}{-7-s} - \frac{e^{0}}{-7-s} </math>
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<math> \frac{e^{-7\infty -s\infty}}{-7-s} </math> Numerator goes to 0
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<math>H(s) = \frac{1}{7+s}</math>
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<math>H(jw) = \frac{1}{7+jw}</math>
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d)Find the output y(t) of the above LTI system when the input x(t) is <math> e^{j7t} </math>
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<math> y(t) = x(t) * h(t) </math> (where * indicates convolution)
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<math> x(t) = e^{j7t} </math>
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<math> h(t) = e^{-7t}u(t) </math>
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Using the properties of convolution we will choose to convolve using <math> x(t-\tau) </math> and <math> h(\tau) </math> for simplicity.
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<math> y(t) = \int_{-\infty}^\infty e^{j7(t-\tau)}e^{-7\tau}u(\tau)d\tau </math>
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Drop the u(t) and change the integration limits.
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<math> y(t) = \int_0^\infty e^{j7(t-\tau)}e^{-7\tau}d\tau </math>
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Simplify the exponentials.
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<math> y(t) = \int_0^\infty e^{-7\tau + j7t - j7\tau}d\tau </math>
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<math> y(t) = \frac{e^{-7\tau + j7\tau - j7\tau}}{-7-j7} \bigg|_0^\infty </math>
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<math> y(t) = \frac{e^{-\infty}}{-7-j7} + \frac{e^{j7t}}{7+7j} </math>
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<math> y(t) = \frac{e^{j7t}}{7+7j} </math>
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==Alternate Solutions==
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Problem 6 - Using Frequency Domain
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Latest revision as of 19:48, 30 July 2008

Read the discussion on discussion page using the discussion page tab. (Aung 11:13pm on 07/18/2008)

(a) The FT of $ X(j\omega) $ of a continuous-time signal x(t) is periodic 

MAY BE: -


(b) The FT of $ X(e^{j\omega}) $ of a continuous-time signal x[n] is periodic 

YES: $ X(e^{j\omega}) $ is always periodic with period $ 2\pi $


(c) If the FT of $ X(e^{j\omega}) $ of a discrete-time signal x[n] is given as: $ X(e^{j\omega}) = 3 + 3cos(3\omega) $, then the signal x[n] is periodic 

NO: The inverse transform of this signal is a set of delta functions that are not periodic.


(d) If the FT of $ X(j\omega) $ of a continuous-time signal x(t) consists of only impulses, then x(t) is periodic 

MAY BE: -


(e) Lets denote $ X(j\omega) $ the FT of a continuous-time non-zero signal x(t). If x(t) is an odd signal, then: $ \int_{-\infty}^{\infty} X(j\omega) d\omega = 0 $ 

YES: this equation is the same as $ \int_{-\infty}^{\infty} X(j\omega) e^{-j\omega_0t}d\omega = 0 $ where t = 0. 
From this we can conclude that x(0) = 0, which always holds true for odd signals.


(f) Lets denote $ X(j\omega) $ the FT of a continuous-time non-zero signal x(t). If x(t) is an odd signal, then: $ \int_{-\infty}^{\infty} |X(j\omega)|^2 d\omega = 0 $ 

NO: using parseval's relation, we see that: $ \int_{-\infty}^{\infty} |X(j\omega)|^2 d\omega = 0 = 2\pi \int_{-\infty}^\infty |x(t)|^2 dt  $ 
The integral of the magnitude squared will always be positive for an odd signal.


(g) Lets denote $ X(e^{j0}) $ the FT of a DT signal x[n]. If $ X(e^{j0}) $ = 0, then x[n] = 0. 

MAY BE: $ X(e^{j0}) $ is simply $ X(e^{j\omega}) $ evaluated at $ \omega = 0 $. 
This only tells you that summation of x[n] over all n is 0, not the entire signal x[n] = 0.


(h) If the FT ($ X(e^{j\omega}) $) of a discrete-time signal x[n] is given as : $ X(e^{j\omega}) = e^{-j10\omega}/(22.30e^{-j5\omega} + 11.15) $ then the signal x[n] is real. 

MAY BE: x[n] is real only if the properties of conjugate symmetry for real signals hold for this transform.


(i) Let x(t) be a continuous time real-valued signal for which $ X(j\omega) $ = 0 when $ |\omega| > \omega_M $ where $ \omega_M $ is a real and positive number. Denote the modulated signal y(t) = x(t)c(t) where c(t) = $ cos(\omega_ct) $ and $ \omega_c $ is a real, positive number. If $ \omega_c $ is greater than $ 2\omega_M $, x(t) can be recovered from y(t). 

YES: Taking the FT of c(t) we get delta functions at $ \omega_c $ and $ -\omega_c $.  
When convolved with the FT of the input signal $ X(j\omega) $, the function $ X(j\omega) $ gets shifted to $ \omega_c $ and $ -\omega_c $
with ranges $ (-\omega_c-\omega_M) $ to $ (-\omega_c+\omega_M) $ and $ (\omega_c-\omega_M) $ to $ (\omega_c+\omega_M) $. 
Therefore $ (\omega_c-\omega_M) > (-\omega_c+\omega_M) $ must hold for there to be no 
overlapping. This is equivalent to $ 2\omega_c > 2\omega_M  => \omega_c > \omega_M $. Since $ \omega_c > 2\omega_M $, there is no overlapping 
and x(t) can be recovered.


(j) Let x(t) be a continuous time real-valued signal for which $ X(j\omega) $ = 0 when $ |\omega| > 40\pi $. Denote the modulated signal y(t) = x(t)c(t) where c(t) = $ e^{j\omega_ct} $ and $ \omega_c $ is a real, positive number. There is a constraint of $ \omega_c $ to guarantee that x(t) can be recovered from y(t). 

NO: The FT of c(t) is just a shifted delta function, which will simply shift the 
input signal x(t) so there is no chance of overlapping.
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