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| − | ( | + | Read the discussion on discussion page using the discussion page tab. (Aung 11:13pm on 07/18/2008) |
| − | + | -------------------------------------------------------------------------------------------------------------- | |
| − | ( | + | (a) The FT of <math>X(j\omega)</math> of a continuous-time signal x(t) is periodic |
| − | + | MAY BE: - | |
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| − | ( | + | (b) The FT of <math>X(e^{j\omega})</math> of a continuous-time signal x[n] is periodic |
| − | + | YES: <math>X(e^{j\omega})</math> is always periodic with period <math>2\pi</math> | |
| − | (e)Lets denote X( | + | |
| + | |||
| + | (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 | ||
| + | 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 | ||
| + | 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> | ||
| − | YES: this | + | 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 always holds true for odd signals. | ||
| − | (f)Lets denote X( | + | |
| + | |||
| + | (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> | ||
| − | NO: | + | 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> |
| + | The integral of the magnitude squared will always be positive for an odd signal. | ||
| + | |||
| + | |||
| + | |||
| + | (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>. | ||
| + | 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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| − | (h) | + | (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. |
| + | 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( | + | (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). |
| − | YES: | + | YES: Taking the FT of c(t) we get delta functions at <math>\omega_c</math> and <math>-\omega_c</math>. |
| + | 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> | ||
| + | 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>. | ||
| + | Therefore <math>(\omega_c-\omega_M) > (-\omega_c+\omega_M)</math> must hold for there to be no | ||
| + | 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 | ||
| + | and x(t) can be recovered. | ||
| − | (j)Let x(t) be a continuous time real-valued signal for which X( | + | (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). |
| − | 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. | + | 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. | ||
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.
