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− | = Practice Question on signal modulation = | + | = 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) Multiply by cos(w<sub>c</sub>t) then pass it through a Low Pass Filter with a gain of 2 and a cutoff f of w<sub>c</sub> | + | b) Multiply by cos(w<sub>c</sub>t) then pass it through a Low Pass Filter with a gain of 2 and a cutoff f of w<sub>c</sub> |
+ | |||
+ | <sub></sub> H(w) = 2 [u(w+w<sub>c</sub>)-u(w-w<sub>c</sub>)] | ||
+ | --[[User:Ssanthak|Ssanthak]] 12:49, 19 April 2011 (UTC) | ||
− | |||
=== Answer 3 === | === Answer 3 === |
Revision as of 07:49, 19 April 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 sinusoidal carrier
c(t) = cos(ωct).
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 conditions 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 > ωm = 1,000π must be met to insure that x(t) can be recovered.
b) To demodulate, first multiply again by cos(ωct). Then feed the resulting signal through a low pass filter with a gain of 2 and a cutoff frequency of ωc.
--Cmcmican 20:56, 7 April 2011 (UTC)
Answer 2
a) wc > wm
wc > 1000pi
b) Multiply by cos(wct) then pass it through a Low Pass Filter with a gain of 2 and a cutoff f of wc
H(w) = 2 [u(w+wc)-u(w-wc)] --Ssanthak 12:49, 19 April 2011 (UTC)
Answer 3
Write it here.