Revision as of 21:47, 18 July 2008 by Mbarga (Talk)

(a) The FT of $ X(j\omega) $ of a continuous-time signal x(t) is periodic

MAY BE: $ X(j\omega) $ is periodic only if x(t) is periodic

(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

MAY BE:

(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^{j\omega_0n} $ has a FT that is an impulse

(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 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 x[0] = 0, not the entire signal x[n] = 0.


(h)


(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

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang