Spring ECE350 2012 - Exam 3
25 April 2012
Name:
___Solutions ____
ECE 350 Signals and Systems
Spring 2012
Exam #3
One 8 x 11 sheet of notes, and a calculator are allowed during the exam.
Write all answers neatly and show your work to get full credit.
Rationalize all complex fractions.
A calculator is allowed during the exam, but should not be needed.
Write neatly and WATCH YOUR ALGEBRA!
Problem 1
Problem 2
Problem 3
Problem 4
Total
/32
/15
/28
/25
/100
ECE350 Spring 2012 - Exam 3
Name:
25 April 2012
1. (35 points) Given the following system diagram
___Solutions ____
c1(t)=cos(240t)
x1(t)
+
a1(t)
X
y1(t)
2
x1(t)
+
b(t)
d(t)
H2(j)
H1(j) a2(t) X y2(t)
y3(t)
X
c3(t)=cos(200)
c2(t)=cos(120t)
The spectrum for the real signal x1(t) is shown below:
2
X1(j)
-40
40
and the filters
H1(j)
H2(j)
1
1
-220
30
-30
220
(a) (4 points)Draw and fully label A1(j).
4
X1(j)
2
-40
40
(b) (4 points) Draw and fully label A2(j).
2
X1(j)
-30
30
(c) (5 points) Draw and fully label Y1(j).
2
1
Y1(j)
2
1
-280 -240 -200
200 240 280
(d) (5 points) Draw and fully label Y2(j).
Y2(j)
1
1
-150 -120 -90
90
120 150
ECE350 Spring 2012 - Exam 3
25 April 2012
(e) (4 points) Draw and fully label B(j).
B(j)
2
Name:
1
1
1
___Solutions ____
2
1
-280 -240 -200
-150 -120 -90
90 120 150 200 240 280
(f) (5 points) Draw and fully label D(j).
D(j)
1
1/2
-220 -200
1
-150 -120 -90
1/2
90 120 150 200 220
(g) (5 points) Draw and fully label Y3 (j).
Y3(j)
1/4
-420 -400
1/2
-350 -320 -290
1/2
1/4
1/4
1/2
-110 -80 -20 0 20 50 80
-50
1/2
1/4
110
290 320 350 400 420
ECE350 Spring 2012 - Exam 3
25 April 2012
2. (15 points) A signal is
Name:
___Solutions ____
x(t ) cos2 (30t )
(a) (9 points) What is the equation describing X(j)? Which properties did you use to
decide this? Draw X(j).
2 methods:
1) trig identities: cos(2) = 2(cos2()-1) cos2() = (1+cos(2))/2 X(j) =
() + /2 (+60) + /2 (-60)
2) modulation
*
-30
30
-30
30
=
/2
/2
-60
0
60
(b) (6 points) According to Nyquist, what is the smallest sampling frequency, s, that
may be used to sample x(t) so that full reconstruction is possible?
Nyquist states that s > 2m
s > 2*60 = 120, so 120.00001
Here because the signal has impulses at the larges frequencies, the strictly greater
matters.
ECE350 Spring 2012 - Exam 3
25 April 2012
3. (28 points) For the transfer function
Name:
___Solutions ____
1 s
s s 20
a) (5 points) Sketch the pole-zero plot for this transfer function
H ( s)
2
Complex Plane
Im
6
4
2
0
Re
-2
-4
-6
-6
-4
-2
0
2
4
6
b) (4 points) How many and what are the possible ROCs for this transfer function?
Sketch the possible ROCs.
Three possible ROCs: Re{s}<-4, < -4 Re{s} < 5, Re{s}>5
Complex Plane
Im
Complex Plane
Im
Complex Plane
Im
5
5
5
0
Re 0
Re 0
-5
-5
ROC
-5
0
5
-5
ROC
-5
0
Re
5
ROC
-5
0
5
c) (6 points) For each ROC in (b) above, determine stability and causality of the
impulse response.
ROC
Re{s}<-4
-4 < Re{s} < 5
Re{s}>5
Stability
not stable
stable
not stable
Causality
not causal
not causal
causal
ECE350 Spring 2012 - Exam 3
25 April 2012
d) (8 points) If h(t) is stable, find h(t).
H (s)
Name:
___Solutions ____
4/9 5/9
1 s
s s 20
s 5
s4
2
h(t) = -5/9e-4tu(t) +4/9e5tu(-t)
e) (5 points) Sketch each term of h(t) from (d).
-5/9*exp(-4*t)u(t)
4/9*exp(5*t)u(-t)
0
0.35
-0.05
0.3
-0.1
0.25
-0.15
0.2
-0.2
0.15
-0.25
0.1
-0.3
0.05
-0.35
-0.4
-5
0
t
0
-5
5
0
t
5
-5/9*exp(-4*t)u(t) + 4/9*exp(5*t)u(-t)
0.3
0.2
0.1
0
-0.1
-0.2
-0.3
-0.4
-5
-4
-3
-2
-1
0
t
1
2
3
4
5
ECE350 Spring 2012 - Exam 3
25 April 2012
4. (20 points) For the transfer function
Name:
___Solutions ____
s 2 25
s 2 s 1.25
a) (5 points) Sketch the pole-zero plot for this transfer function.
H ( s)
Complex Plane
Im
6
4
2
0
Re
-2
-4
-6
-6
-4
-2
0
2
4
6
b) (8 points) Sketch the magnitude of the Fourier Transform associated with H(s).
limit approaches a
constant
maxima near =1
-5
-1
1
5
touches zero
ECE350 Spring 2012 - Exam 3
Name:
___Solutions ____
25 April 2012
c) (7 points) Draw the magnitude of the straight line Bode plot for this transfer
function.
Does it agree with your answer in (b)? Why/why not?
s 2
25 1
2
5
s 25
H (s) 2
s s 1.25
s 2
s
1.25
1
1.25 1.25
Both numerator and denominator are second order with complex (or imaginary) roots.
Therefore keep in second order form.\
The numerator bends upward 40dB/decade at =5.
The denominator bends downward at -40dB/decade at =sqrt(1.25)~1.1
20log10(25/1.25)=26
20
0
-20
-26
-40
1
5
10
ECE350 Spring 2012 - Exam 3
25 April 2012
Name:
___Solutions ____
Bode Diagram
Magnitude (dB)
50
0
-50
-100
- 150
45
Phase (deg)
0
-45
-90
-135
-180
0
1
10
10
Frequency (rad/sec)
It is the same but in part (b) we didnt know what the gain of the constant
attenuation would be for frequencies greater than =5. The bode plot tells us that
it is a lower gain than the attenuation at low frequencies.
d) (5 points) What type of filter is this?
This is a low pass filter. It passes low frequencies. The high frequencies are passed
with a constant gain, but it is much lower than the gain of the low frequencies. In
part (b) we didnt know what the gain of the constant attenuation would be for
frequencies greater than =5. The bode plot tells us that it is a lower gain than the
attenuation at low frequencies.