EE 341 Quiz 2 Solutions 1. (a) ωi(t) = ωc + 8000π cos 2000πt, B

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EE 341 Quiz 2 Solutions
1. (a) ωi (t) = ωc + 8000π cos 2000πt, B = 2000π/2π = 1000 Hz, ∆f = 8000π/2π = 4000 Hz. BEM =
2(∆f + B) = 2(4000 + 1000) = 10 kHz.
(b) This signal can be either an FM or a PM signal. If it is a PM signal, then kp m(t) = 4 sin 2000πt.
For example, kp = 4 and m(t) = sin 2000πt. If it is an FM signal, then kf m(t) = 8000π cos 2000πt. For
example, kf = 8000π and m(t) = cos 2000πt.
(c) The average transmit power is PT = 1002 /2 = 5000 W.
2. (a) Using the trigonometric relationship, 2 cos 2000πt cos 300πt = cos 1700πt + cos 2300πt,
x(t) = 20 cos 2000πt + 10 cos 2000πt cos 300πt = [20 + 10 cos 300πt] cos 2000πt
(1)
so that m(t) = 10 cos 300πt.
(b) Using the expression for x(t) in the problem statement, the transmit power is PT = [52 +202 +52 ]/2 =
225 W. Alternatively, using the form in (1), we showed in class that PT = [A2 + Sm ]/2 where A = 20 and
Sm = 102 /2 = 50 is the average message power. PT = [202 + 50]/2 = 225 W.
3. (a) m(t) = 4 cos 400πt, M (f ) = 2[δ(f −200)+δ(f +200)], Xdsb (f ) = 2[δ(f −1200)+δ(f +1200)+δ(f −
800) + δ(f − 800)]. From the plot of Xusb (f ) = 2[δ(f − 1200) + δ(f + 1200)] in Fig P3, x(t) = 4 cos 2400πt.
(b) The time domain expression for the USB is x(t) = m(t) cos ωc t − mh (t) sin ωc t. The Hilbert transform
of 4 cos 400πt is 4 cos(400πt − π/2) = 4 sin 400πt. Then the USB signal is given by
x(t) = 4 cos 400πt cos 2000πt − 4 sin 400πt sin 2000πt = 4 cos 2400πt.
4. (a) If the input to a coherent demodulator is [A + m(t)] cos ωc t, then the output is A + m(t). By
adding a dc block, as in Figure P4(a), the output of the dc block would be m(t).
(b) As we showed in class and as given in the text, if the local oscillator is out of phase with the sender’s
oscillator, then there can be a significant reduction in the power of the received signal (this is also true for
the frequency). To address this, one needs an additional circuit to synchronize the phase (and frequency)
which is expensive. Using an envelope detector is cheaper.
(c) In Figure P4(c), two AM modulators are used to generate a DSB-SC signal.
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