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Unformatted text preview: ECE 210/211 Analog Signal Processing Spring 2011
Basar, Eden, SchuttAine, Trick University of Illinois Exam 2 Thursday, March 17, 201 1  7:008:15 PM 1
Name:
A
. Seem” 9 AM 10 AM 1 PM 2 PM
(Clrcle one)
r
. Class: ECE 210 ECE 211
(Circle one)
L Please clearly PRINT your name TN CAPITAL LETTERS and circle your section in the boxes
above. This is a closed book and closed notes exam. Calculators are not allowed. Please show all your
work. Backs of pages may be used for scratch work if necessary. All answers should include
units wherever appropriate. Problem 1 (25 points) Problem 2 (25 points) Problem 3 (25 points) Problem 4 (25 points) Total (100 points) Problem 1 (25 points) (21) 39 + 3H
Coscot y(t) 3F What is the frequency co of the input current that will cause y(t) to be zero? (0 :
(b) Compute the phasors F of the following signals.
_ . F =
f[(t) — cos 5t + s1n 5t 1
F2 = (C) i) What impedance ZL will absorb the maximum average power in the circuit shown above? Z: L ii) What is the maximum average power P absorbed by ZL. (d) An LTl system has frequency response H(co) as follows: H(<o) Ifthe input f(t) is 3 + 10 Cos % t+ 7 Sin(§ t+ 30°], the output y(t) is Problem 2 (25 points) The circuit shown below illustrates a radiofrequency (RF) transmitter driving a transmission line
and an antenna. The source v0 is a steady state sinusoidal source operating at a frequency (0 = 106 s'1 . At the end of the transmission is a variable capacitor at the base of an antenna
represented by a resistor R = 100 Q in parallel with a 1 pH inductor. Tuning capacitor V0“) 0
1 pH \—V——/b\—V—/ \—\/——/ TRANSMITTER TRANSMITTER ANTENNA
LINE (a) Find the total impedance Z 0fthe circuit between ab, in terms of C. Z: (b) The capacitor C is to be tuned until the current i is in phase with the voltage source
v0(t) = v0 Cos wt. Calculate the value of C. Problem 3 (25 points) (a) For the following circuit, ﬁnd the frequency response function H((o) = Simplify your
answer.
IQ
1 Q y(t)
f(t) 0 2F
4H HUD) = (b) A linear system with input f(t) and output y(t) is described by the frequency response Y ' . .
— = ((0) = J0) .Determme the followmg: F 5+jw (i) Amplitude of y(t) when f(t) = J? sin(5t+ Amplitude = (ii) Output y(t) when input is f(t) = 7 + 5 cos (5t)V. Problem 4 (25 points) f(t) In the exponential form the fourier coefﬁcients for l the given pulse are F0 =1, F, =—l—, F, =0, F3 =——l, 2 7r ” 37! (a) (10 pts) For the given square wave express
the fourier coefﬁcients G" in terms of the F" coefﬁcients above. Also, compute
an and b” for n = 0, l, 2, and 3. (b) (15pm) Repeat part (a) for the given triangular
Wave. Compute X0 and express X, in terms I of F” fornzi l, i 2, also compute an and
b” for n = 0, l, 2, and 3. f(r). period T = 1’1 MCoefﬁcients ll)” 00 '11“) I
Zn=—oo File] 0 F,, = % fr f(r)e‘j"'”"’a'1 Exponential £29 + 220:, a,, cos(nw(,!) + b" sin(na)0!) Trigonometric bn : _ F—n) Cu = 2iFni
0n 2 1F" Compact for real f(!) 52° + Z°° c” cos(nw,,! + 9,.) Ii=l “5'9 51 Summary of different representations of periodic signal [(1) having period T = and funda ( Mma‘ fre<Juency a)". The formula for F,, in the upper right corner will be derived in Section 6.2. Condition: Constant K f(1)<—> F,,,g(1) <—> Gm... Scaling Kf(() 6 KB, Addition f(,) + gm + H 1:" + G" + 1 Time shift Delay 10 f(, _ ,0) H Fne—jnwnn, Derivative Continuous f(!) $4 H jnwan Hermitian Real f(t) f(—I) = f(!)
ft—I) = —f(t) F—n=F‘ I1 'Even function f(!) = 329 + 22:1 an cos(na)(,!) f(1) 2 22:1 b" sin(na),,!)
P E % IT mmzdr = $21.00 tFnP J Odd function 'Average power Table 6.3 Properties of Fourier series. ...
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 Spring '08
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 Signal Processing, maximum average power, Basar

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