EE 341 Quiz 3 Solutions
1. (a) n = 200k/500 = 400. (b) The new carrier frequency is n 110k = 44 MHz. To obtain a carrier
frequency of 100 MHz, use fLO = 56 or 144 MHz. (c) The BPF should have a center frequency of 100
MHz and a bandwidth of BF M = 2(f + B
EE 341 Quiz 2 Solutions
Fall 2008
1. (a) We use the same approach as in homework problem 5.2-2. The instantaneous frequency is
fi (t) = fc + 5000 cos 1000t + 30, 000 cos 3000t. Then the frequency deviation is f = 35 kHz. (i)
BEM = 2(f + B ) = 2(35 + 1.5)
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EE 341 Quiz 2 Solutions
1. (a) i (t) = c + 8000 cos 2000t, B = 2000/2 = 1000 Hz, f = 8000/2 = 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 2000t.
For exa
EE 3+1 HWF Solo-Hons
4.3-8 The signal at point a is [A + m{t)] cos (wet). The signal at point b is:
:r (r) = [A + m [t)]2co52 (wet) = W (1 + cos (2am
The lowpass lter suppresses the term containing cos (Email. Hence. the signal at point e is:
2 m 2
E6 34-! f-l W 2 Soluhoas
3.7-4 Application of duality property [Eq. (3.26)] to pair 3 of Table 3.1 yields
20
__ <=> eam
a2 + (21st)2
Applying the the time-scaling property with a = 5-1;, we obtain
G22: F g 2ne'l2"
The signal energy is given by
Go
E6 34-! f-l W 2 Soluhoas
3.7-4 Application of duality property [Eq. (3.26)] to pair 3 of Table 3.1 yields
20
__ <=> eam
a2 + (21st)2
Applying the the time-scaling property with a = 5-1;, we obtain
G22: F g 2ne'l2"
The signal energy is given by
Go
EE 341 Final Exam
Fall 2008
December 19, 2008
Name:
Problem Maximum
Number
Points
1
12
2
18
3
15
4
15
5
15
Total
75
Your
Score
You are permitted to use three 8.5 11 formula sheets (both sides). No homework solutions (a penalty
will be assessed for noncom