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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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--[[User:Cmcmican|Cmcmican]] 20:56, 7 April 2011 (UTC) | --[[User:Cmcmican|Cmcmican]] 20:56, 7 April 2011 (UTC) | ||
| − | + | :<span style="color:green">Instructor's comment: Good! For the test, make sure you have a clear mental picture of what is happening in the frequency domain when you do this and why it works. -pm </span> | |
=== 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 | ||
| − | + | 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) | |
| + | :<span style="color:green">Instructor's comment: It is a bit confusing if you talk about a "cutoff f", because the variable "f" is used to denote frequency in hertz. In this course, we are using <math>\omega</math> for which the units are radians per time unit. -pm</span> | ||
=== Answer 3 === | === Answer 3 === | ||
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 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)
- Instructor's comment: Good! For the test, make sure you have a clear mental picture of what is happening in the frequency domain when you do this and why it works. -pm
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)
- Instructor's comment: It is a bit confusing if you talk about a "cutoff f", because the variable "f" is used to denote frequency in hertz. In this course, we are using $ \omega $ for which the units are radians per time unit. -pm
Answer 3
Write it here.
