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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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(Formatting to follow after dinner...)
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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: -
  
a) By Inspection th linear constant-coefficient differential equation that describes the LTI system is:
 
  
y(t)=1/7x(t)-(1/7)(dy(t)/dt)
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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>
  
  
  
b) Show that the impulse response to this LTI system is given by h(t)=e^(-7t)u(t)
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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.
  
This means that x(t) = delta(t) and y(t) = e^(-7t)u(t)
 
  
e^(-7t)u(t) = (1/7)delta(t) - (1/7)d(e^(-7t)u(t)/dt
 
  
Differentiating (1/7)d(e^(-7t)u(t)/dt requires use of the chain rule.
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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: -
  
This portion of the equation becomes:
 
  
(1/7)(-7)(e^(-7t))u(t) - (1/7)delta(t)(e^(-7t))
 
  
-(1/7)delta(t)(e^(-7t)) is -(1/7)delta(t)(e^(-7t)) evaluated at t=0 or -(1/7)delta(t)*1
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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.
  
Plugging that back in yields:
 
  
e^(-7t)u(t) = (1/7)delta(t) - (1/7)(-7)(e^(-7t))u(t) - (1/7)delta(t)
 
  
This equation simplifies to:
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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.
  
0=0 indicating that it is correct.
 
  
  
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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.
  
c) Find H(s) at s=jw for the LTI system with impuls response h(t)=e^(-7t)u(t)
 
  
H(s) = Integral(-infinity to infinity)h(t)*e^(-st)dt
 
  
H(s) = Integral(-infinity to infinity)e^(-7t)u(t)*e^(-st)dt
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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.
  
Limits change to 0 to infinity and u(t) drops out.
 
  
H(s) = Integral(0 to infinity)e^(-7t)*e^(-st)dt
 
  
H(s) = Integral(0 to infinity)e^(-7t-st)dt
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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.
  
H(s) = e^(-7t-st)/(-7-s) evaluated from 0 to inifinity
 
  
H(s) = e^(-7*infinity-s*infinity)/(-7-s) - e^(0)/(-7-s)
 
  
H(s) = e^(-7*infinity-s*infinity)/(-7-s) goes to 0
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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  
H(s) = 1/(7+s)
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input signal x(t) so there is no chance of overlapping.
 
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H(jw) = 1/(7+jw)
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d)Find the output y(t) of the above LTI system when the input x(t) is e^(j7t)
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y(t) = x(t) convolved with h(t)
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x(t) = e^(j7t)
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h(t) = e^(-7t)u(t)
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Using the properties of convolution we will choose to convolve using x(t-tau) and h(tau) for simplicity.
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y(t) = Integral(-infinity to infinity)e^(j7(t-tau))e^(-7tau)u(tau)dtau
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Drop the u(t) and change the integration limits.
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y(t) = Integral(0 to infinity)e^(j7(t-tau))e^(-7tau)dtau
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Simplify the exponentials.
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y(t) = Integral(0 to infinity)e^(-7tau + j7t - j7tau)dtau
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y(t) = (e^(-7tau + j7tau - j7tau))/(-7-j7) evaluated from 0 to inifity
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y(t) = e^(-infinity)/(-7-j7) + e^(j7t)/(7+7j)
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y(t) = e^(j7t)/(7+7j)
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Latest revision as of 20:48, 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.

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang