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| − | = Practice Question on signal modulation = | + | = [[:Category:Problem_solving|Practice Question]] on signal modulation = |
Let x(t) be a signal whose Fourier transform <math>{\mathcal X} (\omega) </math> satisfies | Let x(t) be a signal whose Fourier transform <math>{\mathcal X} (\omega) </math> satisfies | ||
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=== Answer 2 === | === Answer 2 === | ||
| − | a) w<sub>c</sub> > w<sub>m</sub> | + | a) w<sub>c</sub> > w<sub>m</sub> |
| − | w<sub>c</sub> > 1000pi | + | w<sub>c</sub> > 1000pi |
| − | b)Since y(t) = x(t) e^jw<sub>c</sub>t | + | b)Since y(t) = x(t) e^jw<sub>c</sub>t |
| − | So x(t) = y(t) e^-jw<sub>c</sub>t | + | So x(t) = y(t) e^-jw<sub>c</sub>t |
| − | so to demodulate multiply by e^-jw<sub>c</sub>t | + | so to demodulate multiply by e^-jw<sub>c</sub>t |
| + | |||
| + | --[[User:Ssanthak|Ssanthak]] 12:39, 19 April 2011 (UTC) | ||
=== Answer 3 === | === Answer 3 === | ||
| − | + | a) Since c(t) only produces a phase shift, there is no potential for overlap of the signal, and no conditions are needed on ω<sub>c</sub>. Even if ω<sub>c </sub>was zero, that is fine, it just means a shift of zero (and negative would just shift it in the opposite direction.) It has to be a real number, though, right? Would we ever need to state that?<sup></sup> | |
| − | ---- | + | b) I agree with those two on the rest. |
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| + | --[[User:Kellsper|Kellsper]] 22:16, 20 April 2011 (UTC) | ||
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| + | I agree with you a) should have been, no conditions on w<sub>c</sub>.<sub></sub> | ||
| + | --[[User:Ssanthak|Ssanthak]] 09:50, 21 April 2011 (UTC) | ||
[[2011 Spring ECE 301 Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | [[2011 Spring ECE 301 Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | ||
[[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | [[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | ||
Latest revision as of 09:31, 11 November 2011
Contents
Practice Question on signal modulation
Let x(t) be a signal whose Fourier transform $ {\mathcal X} (\omega) $ satisfies
$ {\mathcal X} (\omega)=0 \text{ when }|\omega| > 1,000 \pi . $
The signal x(t) is modulated with the complex exponential carrier
$ c(t)= e^{j \omega_c t }. $
a) What conditions should be put on ωc to insure that x(t) can be recovered from the modulated signal x(t)c(t)?
b) Assuming the condition you stated in a) are met, how can one recover x(t) from the modulated signal x(t)c(t)?
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1
a) ωc > 0
b) to recover x(t) from x(t)c(t), multiply x(t)c(t) by $ e^{-j \omega_c t }. $
--Cmcmican 20:56, 7 April 2011 (UTC)
Answer 2
a) wc > wm
wc > 1000pi
b)Since y(t) = x(t) e^jwct
So x(t) = y(t) e^-jwct
so to demodulate multiply by e^-jwct
--Ssanthak 12:39, 19 April 2011 (UTC)
Answer 3
a) Since c(t) only produces a phase shift, there is no potential for overlap of the signal, and no conditions are needed on ωc. Even if ωc was zero, that is fine, it just means a shift of zero (and negative would just shift it in the opposite direction.) It has to be a real number, though, right? Would we ever need to state that?
b) I agree with those two on the rest.
--Kellsper 22:16, 20 April 2011 (UTC)
I agree with you a) should have been, no conditions on wc. --Ssanthak 09:50, 21 April 2011 (UTC)
