Read the discussion on discussion page using the discussion page tab. (Aung 11:13pm on 07/18/2008) --The reasoning for (a) (d) (g) are wrong. Think again. (Aung 11:11 pm on 07/18/08)
(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.
Alternate Solution - Using Frequency Domain
File:ECE301Summer2008asanAlternate Sol. Problem 6 Exam 1.pdf
