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(a)The Fourier transform of X(jw) 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)
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MAY BE: X(jw) is periodic only if x(t) is periodic
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(b)The Fourier transform of X(ejw) of a continuous-time signal x[n] 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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  YES: X(ejw) is always periodic with period 2pi
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  MAY BE: -
  
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(c)If the FT of X(ejw) of a discrete-time signal x[n] is given as: X(ejw) = 3 + 3cos(3w), then the signal x[n] is periodic
 
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MAY BE:
 
  
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(d)If the FT of X(jw) of a continuous-time signal x(t) consists of only impulses, then x(t) is periodic
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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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  MAY BE: e^jw0n has a FT that is an impulse
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  YES: <math>X(e^{j\omega})</math> is always periodic with period <math>2\pi</math>
  
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(e)Lets denote X(jw) the FT of a continuous-time non-zero signal x(t).  If x(t) is an odd signal, then  
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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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(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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(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:
 
<math>\int_{-\infty}^{\infty} X(j\omega) d\omega = 0 </math>
 
<math>\int_{-\infty}^{\infty} X(j\omega) d\omega = 0 </math>
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  YES: this eqation 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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  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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(f)Lets denote X(jw) the FT of a continuous-time non-zero signal x(t).  If x(t) is an odd signal, then  
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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:
 
<math>\int_{-\infty}^{\infty} |X(j\omega)|^2 d\omega = 0 </math>
 
<math>\int_{-\infty}^{\infty} |X(j\omega)|^2 d\omega = 0 </math>
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  NO: from 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> The integral of the magnitude squared will allways be positive for an odd signal.
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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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(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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(g)Lets denot X(ejw) the FT of a DT signal x[n]. If X(ej0) = 0, then x[n] = 0.
 
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MAY BE: X(ej0) is simply X(ejw) evaluated at w = 0. This does not tell you anything about the original signal x[n] other than x[0] = 0.
 
  
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(h)
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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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(i)Let x(t) be a continuous time real-valued signal for which X(jw) = 0 when |w| > wm where wm is a real and positive number. Denote the modulated signal y(t) = x(t)c(t) where c(t) = cos(wct) and wc is a real, positive nubmer. IF wc is greater than 2wm, x(t) can be recovered from y(t).
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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:  
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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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(j)Let x(t) be a continuous time real-valued signal for which X(jw) = 0 when |w| > 40pi. Denote the modulated signal y(t) = x(t)c(t) where c(t) = e{jwct} and wc is a real, positive nubmer. There is a constraint of wc to guarantee that x(t) can be recovered from y(t).
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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 input signal x(t) so there is no chance of overlapping.
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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.

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

Dr. Paul Garrett