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| − | The Fourier transform of X(jw) of a continuous-time signal x(t) is periodic | + | (a)The Fourier transform of X(jw) of a continuous-time signal x(t) is periodic |
| + | MAY BE: X(jw) is periodic only if x(t) is periodic | ||
| − | + | (b)The Fourier transform of X(ejw) of a continuous-time signal x[n] is periodic | |
| + | YES: X(ejw) is always periodic with period 2pi | ||
| + | (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 | ||
| + | MAY BE: | ||
| + | (d)If the FT of X(jw) of a continuous-time signal x(t) consists of only impulses, then x(t) is periodic | ||
| + | MAY BE: e^jw0n has a FT that is an impulse | ||
| − | + | (e)Lets denote X(jw) 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> | ||
| + | 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. | ||
| − | + | (f)Lets denote X(jw) 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> | ||
| + | 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. | ||
| + | (g)Lets denot X(ejw) the FT of a DT signal x[n]. If X(ej0) = 0, then x[n] = 0. | ||
| + | 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. | ||
| + | (h) | ||
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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). | ||
| + | YES: | ||
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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). | |
| − | + | 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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Revision as of 20:17, 18 July 2008
(a)The Fourier transform of X(jw) of a continuous-time signal x(t) is periodic
MAY BE: X(jw) is periodic only if x(t) is periodic
(b)The Fourier transform of X(ejw) of a continuous-time signal x[n] is periodic
YES: X(ejw) is always periodic with period 2pi
(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
MAY BE:
(d)If the FT of X(jw) of a continuous-time signal x(t) consists of only impulses, then x(t) is periodic
MAY BE: e^jw0n has a FT that is an impulse
(e)Lets denote X(jw) 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 eqation 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 holds true for odd signals.
(f)Lets denote X(jw) 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: from 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 allways be positive for an odd signal.
(g)Lets denot X(ejw) the FT of a DT signal x[n]. If X(ej0) = 0, then x[n] = 0.
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.
(h)
(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).
YES:
(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).
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.
