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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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Read the discussion on discussion page using the discussion page tab. (Aung 11:13pm on 07/18/2008)
MAY BE: <math>X(j\omega)</math> is periodic only if x(t) is periodic
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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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(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
 
(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
  MAY BE: An exception is <math>e^{j\omega_0n}</math> which has a FT that is an impulse, however
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  MAY BE: -
x(t) = <math>e^{j\omega_0n}</math> itself is not periodic.
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<math>\int_{-\infty}^{\infty} X(j\omega) d\omega = 0 </math>
 
<math>\int_{-\infty}^{\infty} X(j\omega) d\omega = 0 </math>
 
  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.  
 
  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.  
  From this we can conclude that x(0) = 0, which holds true for odd signals.
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  From this we can conclude that x(0) = 0, which always holds true for odd signals.
  
  
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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.
 
(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.
 
  MAY BE: <math>X(e^{j0})</math> is simply <math>X(e^{j\omega})</math> evaluated at <math>\omega = 0</math>.  
 
  MAY BE: <math>X(e^{j0})</math> is simply <math>X(e^{j\omega})</math> evaluated at <math>\omega = 0</math>.  
  This only tells you that x[0] = 0, not the entire signal x[n] = 0.
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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.
  
  

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

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett