1. No category

Sample Final Exam Covering Chapters 1-9 (finals04)
Sample Final Exam (finals03)
Covering Chapters 1-9 of Fundamentals of Signals & Systems
Problem 1 (20 marks)
Consider the causal op-amp circuit initially at rest depicted below. Its LTI circuit model with a voltagecontrolled source is also given below.
(a) [8 marks] Transform the circuit using the Laplace transform, and find the transfer function
H A ( s ) = Vout ( s ) Vin ( s ) . Then, let the op-amp gain A → +∞ to obtain the ideal transfer function
H ( s ) = lim H A ( s ) .
A→+∞
L1
v x (t )
C
R1
R2
vin ( t )
vout ( t )
+
L1
C
R2
vin ( t )
v x (t )
R1
+
+
-
-
− Avx (t )
Answer:
The transformed circuit:
L1 s
1 Cs
R2
Vin ( s )
+
Vx ( s )
R1
+ − AVx ( s )
-
1
Sample Final Exam Covering Chapters 1-9 (finals04)
There are two supernodes for which the nodal voltages are given by the source voltages. The
remaining nodal equation is
Vin ( s ) − Vx ( s ) − AVx ( s ) − Vx ( s )
+
=0
R2
R1 Cs1 L1 s
where R1
1
Cs
L1 s =
1
Cs +
1
R1
+
1
L1s
=
R1 L1 s
. Simplifying the above equation, we get:
R1 L1Cs 2 + L1 s + R1
 ( A + 1)( R1 L1Cs 2 + L1 s + R1 ) 1 
1
+  Vx ( s ) = 0
Vin ( s ) − 
R2
R1 L1 s
R2 

Thus, the transfer function between the input voltage and the node voltage is given by
1
Vx ( s )
R2
=
2
Vin ( s ) ( A + 1)( R1 L1Cs + L1 s + R1 ) 1
+
R1 L1 s
R2
=
.
R1 L1 s
R2 ( A + 1)( R1 L1Cs 2 + L1 s + R1 ) + R1 L1 s
The transfer function between the input voltage and the output voltage is
H A ( s) =
Vout ( s ) − AVx ( s )
− AR1 L1 s
=
=
Vin ( s )
Vin ( s )
R2 ( A + 1)( R1 L1Cs 2 + L1 s + R1 ) + R1 L1 s
The ideal transfer function is the limit as the op-amp gain tends to infinity:
H ( s ) = lim H A ( s ) = −
A→∞
R1 L1 s
=−
R2 R1 L1Cs + R2 L1 s + R2 R1
2
L1 R2 s
L
( L1Cs 2 + 1 s + 1)
R1
H (s)
has a DC gain of −50 , and that H ( s )
s
has one zero at 0 and two complex conjugate poles with ω n = 10 rd/s, ζ = 0.5 . Let L1 = 10 H .
Find the values of the remaining circuit components R1 , R2 , C .
(b) [5 marks] Assume that the transfer function H1 ( s ) =
Answer:
DC gain of H1 ( s ) =
H ( s)
=−
s
L1 R2
is given by − L1 R2 = −50 .
L1
2
( L1Cs + s + 1)
R1
Component values are obtained by setting
L1
s
R2
s
=−
H ( s ) = −50
L
0.01s 2 + 0.1s + 1
( L1Cs 2 + 1 s + 1)
R1
2
Sample Final Exam Covering Chapters 1-9 (finals04)
which yields ⇔ R1 = 100Ω, R2 = 0.2Ω, C = 0.001F
(c) [7 marks] Give the frequency response of H ( s ) and sketch its Bode plot.
Answer: Frequency response is
H ( jω ) = −50
jω
. Bode plot:
0.01( jω ) + 0.1( jω ) + 1
2
20 log10 G ( jω )
60
40
(dB)
20
10 -1
100
10 1
102
103
ω (log)
-20
-40
-60
(deg)
270
∠H ( jω )
225
180
135
90
45
10 -1
100
101
102
103
ω (log)
-45
-90
-135
-180
3
Sample Final Exam Covering Chapters 1-9 (finals04)
Problem 2 (20 marks)
Consider the causal differential system described by its direct form realization shown below,
X ( s)
+
-
-
1
s
1
s
Y ( s)
3
2 2
4
dy (0− )
= −1,
dt
unit step input signal x(t ) = u (t ) .
and with initial conditions
y (0− ) = 2 . Suppose that this system is subjected to the
(a) [8 marks] Write the differential equation of the system. Find the system's damping ratio
undamped natural frequency
ζ
and
ω n . Give the transfer function of the system and specify its ROC. Sketch
its pole-zero plot. Is the system stable? Justify.
Answer:
Differential equation:
d 2 y (t )
dy (t )
+2 2
+ 4 y (t ) = 3x(t )
2
dt
dt
Let's take the unilateral Laplace transform on both sides of the differential equation.
 2
dy (0− ) 
−
−
−
−
s
(
s
)
sy
(0
)
Y

 + 2 2  sY ( s ) − y (0 )  + 4Y ( s ) = 3X ( s )
dt 

Collecting the terms containing Y ( s) on the left-hand side and putting everything else on the righthand side, we can solve for Y ( s ) .
dy (0− )
s + 2 2 s + 4 Y ( s) = 3X ( s) + sy (0 ) + 2 2 y (0 ) +
dt
dy (0− )
( s + 2 2) y (0− ) +
3X ( s )
dt
Y (s) = 2
+
2
s +
s + 2
2
2 s +
4 2s + 4 (
2
)
zero-state resp.
−
zero-input resp.
4
−
Sample Final Exam Covering Chapters 1-9 (finals04)
The transfer function is H ( s ) =
3
s 2 + 2 2s + 4
,
and since the system is causal, the ROC is an open RHP to the right of the rightmost pole.
The undamped natural frequency is
ωn = 2
p1,2 = −ζω n ± jω n 1 − ζ 2 = − 2 ± j 2 1 −
and the damping ratio is
ζ =
1
2
. The poles are
1
=− 2± j 2.
2
Therefore the ROC is Re{s} > − 2 . System is stable as jw-axis is contained in ROC. Pole-zero plot:
Im{s}
2
− 2
Re{s}
− 2
(b) [8 marks] Compute the step response of the system (including the effect of initial conditions), its
steady-state response yss (t ) and its transient response ytr (t ) for t ≥ 0 . Identify the zero-state
response and the zero-input response in the Laplace domain.
Answer:
The unilateral LT of the input is given by
1
X ( s ) = , Re{s} > 0 ,
s
thus,
Y ( s) =
3
( s 2 + 2 2 s + 4) s
+
Re{ s}> 0
zero-state resp.
2( s + 2 2) − 1
s 2 +
2
2 s +
4
Re{ s }>−1
zero-input resp.
Let's compute the overall response:
5
=
2 s 2 + (4 2 − 1) s + 3
(s
2
)
+ 2 2s + 4 s
Sample Final Exam Covering Chapters 1-9 (finals04)
Y ( s) =
=
2 s 2 + (4 2 − 1) s + 3
(s
2
)
+ 2 2s + 4 s
A 2 + B( s + 2)
(
)
2
s+ 2 +2
, Re{s} > 0
C
s
N
+
Re{ s }> 0
Re{ s }>− 2
=
A 2 + B( s + 2)
(
)
2
s+ 2 +2
+
0.75
s
N
Re{ s}> 0
Re{ s}>− 2
Let s = − 2 to compute
2(2) + (4 2 − 1)(− 2) + 3
2(− 2)
−1 + 2
−2 2
⇒ A=
1
=
2
A+
=
1
2
0.75
A+
− 2
0.75
,
− 2
1− 2
+ 0.75 = 0.5429
2
then multiply both sides by s and let s → ∞ to get 2 = B + 0.75 ⇒ B = 1.25 :
Y ( s) =
(
0.5429 2
)
2
+
1.25( s + 2)
(
)
2
s+ 2 +2
s+ 2 +2
Re{ s }>− 2
Notice that the second term
+
0.75
s
N
Re{ s }> 0
Re{ s}>− 2
1
is the steady-state response, and thus yss (t ) = 0.75u (t ) .
s
Taking the inverse Laplace transform using the table yields
y (t ) = 0.5429e −
2t
sin( 2t )u (t ) + 1.25e −
2t
cos( 2t )u (t ) + 0.75u (t ) .
Thus, the transient response is
ytr (t ) = 0.5429e −
2t
sin( 2t )u (t ) + 1.25e−
2t
cos( 2t )u (t ) .
(c) [4 marks] Compute the percentage of the first overshoot in the step response of the system
assumed this time to be initially at rest.
Answer:
Transfer function is H ( s ) =
3
s 2 + 2 2s + 4
, Re{s} > 2 with damping ratio ζ =
6
1
2
:
Sample Final Exam Covering Chapters 1-9 (finals04)
OS = 100e
−
ςπ
1−ς 2
% = 100e
−
0.707π
0.707
% = 100e −π % = 4.3%
Problem 3 (20 marks)
The following nonlinear circuit is an ideal full-wave rectifier.
+
vin ( t )
R
- v (t ) +
αt
The voltages are vin (t ) = e
+∞
[u (t ) − u (t − T1 )] ∗ ∑ [δ (t − 2kT1 ) − δ (t − (2k − 1)T1 )] where
k =−∞
α ∈ R, α > 0 , and v (t ) = vin (t )
.
(a) [5 marks] Find the fundamental period T of the input voltage. Sketch the input and output
voltages vin (t ), v (t ) for α = 1/ T1 .
Answer:
We have T = 2T1
vin (t )
e
1
-2T1
-T1
-1
T1
2T1
t
-e
v (t )
e
1
7
-2T1
-T1
T1
2T1
t
Sample Final Exam Covering Chapters 1-9 (finals04)
(b) [8 marks] Compute the Fourier series coefficients ak of the input voltage vin (t ) for any positive
values of
α
and T1 . Write vin (t ) as a Fourier series.
Answer:
DC component :
T1
1
a0 =
2T1
T
0
1 1 αt
1
(
)
x
t
dt
=
e dt −
∫−T
∫
2T1 0
2T1
1
T
∫e
α ( t +T1 )
dt
−T1
T
1 1 αt
1 1 ατ
=
e
dt
−
e dτ = 0
2T1 ∫0
2T1 ∫0
for k ≠ 0 :
1
ak =
2T1
T1
∫
T1
1
=
2T1
T1
=
=
=
=
2π
t
2 T1
αt
∫e e
− jk
π
T1
t
dt −
0
∫e
(α − jk
π
T1
)t
0
2T1 (α − jk
0
∫
eα ( t + T1 ) e
− jk
π
T1
t
dt
− T1
π
0
π
T1
)
1
2T1 (α − jk
1
2T1
(α − jk ) t
1 α T1
T1
dt −
e ∫ e
dt
2T1
− T1
1
eα T1
dt
− T1
1
2T1
=
x (t ) e
− jk
π
T1
)
(e
(e
α T1
α T1
)
e − jkπ − 1 −
)
e − jkπ − 1 −
( ( − 1) − 1) + ( − 1)
k
k
eα T1
2T1 (α − jk
(e
+ 1)
2T1 (α − jk
−1
( ( − 1) − 1 )
k
2T1α − j 2 k π
Fourier series:
vin (t ) =
(
)e
+∞
(eαT1 + 1) ( −1) − 1
k =−∞
2T1α − j 2kπ
∑
k
T1
)
eα T1
2T1α − j 2 k π
α T1
π
8
jk
π
T1
t
π
T1
)
(1 − e
− α T1
e jkπ
)
(1 − e
− α T1
e jkπ
)
Sample Final Exam Covering Chapters 1-9 (finals04)
(c) [5 marks] Compute the Fourier series coefficients bk of v ( t ) again for any positive values of
α
and T1 .
Answer: Here the fundamental period is T = T1 .
DC component :
T
T
1 1
1 1 αt
b0 = ∫ x(t )dt = ∫ e dt
T1 0
T1 0
=
T1
1
1
 eαT1 − 1
eα t  =
0
α T1
α T1
for k ≠ 0 :
bk =
=
1
T1
T1
∫ x (t ) e
− jk
2π
t
T1
dt =
0
1
T1 (α − jk
2π
)
T1
(e
α T1
1
T1
T1
∫e
(α − jk
2π
)t
T1
dt
0
)
−1 =
eα T1 − 1
α T1 − j 2 k π
(d) [2 marks] Compute the Fourier series coefficients of the output voltage signal v (t ) for the case
α →0
with T1 held constant. What time-domain signal v (t ) do you obtain in this case?
Answer:
When α → 0 we get a constant signal for v (t ) .
lim b0 = lim
α →0
α →0
1
 eα T1 − 1 = 1
α T1
1−1
eα T1 − 1
=
=0
α →0 α T − j 2kπ
− j 2kπ
1
lim bk = lim
α →0
Problem 4 (15 marks)
System identification
Suppose we know that the input of a differential LTI system is
x (t ) = te −2 t u(t ) ,
and we measured the output to be
y (t ) = e − t cos t + sin t u(t ) − e −2 t u(t ) .
y (t )
x (t )
H ( s)
(a) [10 marks] Find the transfer function H ( s) of the system and its region of convergence. Is the
system causal? Is it stable? Justify your answers.
9
Sample Final Exam Covering Chapters 1-9 (finals04)
Answer:
First take the Laplace transforms of the input and output signals using the table:
X ( s) =
Y ( s) =
1
, Re{s} > −2
( s + 2) 2
s +1
1
1
+
−
2
2
2
2
(s + 1
)
+
(s + 1
)
+
s+2
1 1
Re{s}>−1
=
;
Re{s}>−2
Re{s}>−1
s+2
1
−
2
2
(s + 1
)
+
s+2
1
Re{s}>−1
;
Re{s}>−2
( s 2 + 4 s + 4) − ( s 2 + 2 s + 2)
, Re{s} > −1
( s2 + 2 s + 2)( s + 2)
2( s + 1)
= 2
, Re{s} > −1
( s + 2 s + 2)( s + 2)
=
Then, the transfer function is simply
2( s + 1)
Y ( s) ( s + 2 s + 2)( s + 2) 2( s + 1)( s + 2)
s 2 + 3s + 2
=2 2
=
= 2
H ( s) =
1
( s + 2 s + 2)
X ( s)
s + 2s + 2
2
( s + 2)
2
To determine the ROC, first note that the ROC of Y ( s) should contain the intersection of the ROC's of
H ( s) and X ( s) . There are two possible ROC's for H ( s) : (a) an open left half-plane to the left of
Re{s} = −1 , (b) an open right half-plane to the right of Re{s} = −1 . But since the ROC of X ( s) is
an open right half-plane to the right of s = −2 , the only possible choice is (b). Hence, the ROC of
H ( s) is Re{s} > −1 .
The system is causal as the transfer function is rational and the ROC is a right half-plane. It is also
stable as both complex poles p1,2 = −1 ± j are in the open left half-plane.
(b) [2 marks] Find an LTI differential equation representing the system.
Answer:
It can be derived from the transfer function obtained in (a):
d 2 y (t )
dy (t )
d 2 x (t )
dx (t )
+
2
+
2
(
)
=
2
+6
+ 4 x (t )
y
t
2
2
dt
dt
dt
dt
(c) [3 marks] Find the direct form realization of the transfer function H ( s) .
Answer:
The transfer function can be split up into two systems as follows:
10
Sample Final Exam Covering Chapters 1-9 (finals04)
X ( s)
Y ( s)
W ( s)
1
2
s + 2s + 2
2s + 6s + 4
2
The input-output system equation of the first subsystem is
s 2W ( s) = −2 sW ( s) − 2W ( s) + X ( s) ,
and for the second subsystem we have
Y ( s) = 2 s2W ( s) + 6 sW ( s) + 4W ( s) .
The direct form realization of the system is given below:
2
6
s 2W ( s)
X ( s)
+
-
-
sW ( s)
1
s
1
s
W ( s)
+ +
4
Y ( s)
+
2
2
Problem 5 (15 marks)
(a) [10 marks] Compute the Fourier transform X ( jω ) of the following aperiodic signal x(t ) and give
its magnitude and phase.
x (t )
1
T/2
T
t
-1
Answer:
11
Sample Final Exam Covering Chapters 1-9 (finals04)
X ( jω ) =
T
∫ x (t ) e
− jω t
dt
0
T /2
=
∫
e − jω t dt −
=
1
− jω
=
1
jω
1
=
jω
Magnitude:
e − jω t dt
4
ω

e

T
− jω


1  − jω T
2
− 1 +
e
−
e


 jω 

T
− jω

− 2e 2 + 1

T
− jω
2
 − jω T
e

2
T
− jω


1
2
−
1
e

 =
jω


sin 2 (ω
X ( jω ) =
Phase:
∫
T /2
0
= j
T
4
ω
T
− jω
 − jω T4 jω T4

4
−
e
(
e
e
)



2
T
T − jω 2
)e
4
sin 2 (ω
T
)
4
T π

 −ω 2 + 2 , ω > 0

T π
∠ X ( jω ) =  −ω − , ω < 0
2 2

0,
ω =0

(b) [5 marks] Write the Fourier series coefficients ak of the following rectangular waveform y (t ) in
terms of X ( jω ) that you obtained in (a) and compute them.
y (t )
A
-T
-0.5T
0.5T
-A
Answer:
12
T
t
Sample Final Exam Covering Chapters 1-9 (finals04)
1
2π
)
X ( jk
T
T
2π T
1
4
2π T − jk T 2
sin 2 (k
)e
=A j
2π
T
T 4
k
T
2 A 2 π − jkπ
2A
sin ( k )e
(( −1) k − 1)(−1) k
= j
= j
2
kπ
kπ (−2)
ak = A
We have
= jA
((−1) k − 1)
kπ
Problem 6 (10 marks)
Just answer true or false.
(a) The Fourier transform Z ( jω ) of the product of a real even signal x ( t ) and a real odd signal
y (t ) is imaginary.
Answer: True.
(b) The system defined by y (t ) = x (t + 1) is causal.
Answer: False.
(c) The Fourier series coefficients a k of a purely imaginary even periodic signal x ( t ) have the
∗
following property: ak = ak .
Answer: False.
(d) The causal linear discrete-time system defined by y[ n − 2] + 0.4 y[ n − 1] − 0.45 y[ n] = x[ n − 1] is
stable.
Answer: False.
(e) The fundamental period of the signal x[ n] = sin(
3π
n) is 10.
5
Answer: True.
END OF EXAMINATION
13