Sample Final Exam Covering Chapters 1-9 (finals01)
Sample Final Exam (finals01)
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→+∞
L2
L1
v x (t )
vin ( t )
C
R1
R2
-
vout ( t )
+
L2
L1
C
vin ( t )
R2
+ − Av x (t )
v x (t )
R1
+
-
-
Answer:
The transformed circuit is
Vin ( s )
+
L1 s
L2 s
1 Cs
R2
R1
Vx ( s )
+
-
-
− AVx ( s )
There are two supernodes for which the nodal voltages are given by the source voltages. The
remaining nodal equation is
1
Sample Final Exam Covering Chapters 1-9 (finals01)
Vin ( s ) − Vx ( s ) − AVx ( s ) − Vx ( s )
+
=0
R2 L2 s
R1 Cs1 L1 s
where R1
1
Cs
L1 s =
1
Cs +
1
R1
+
1
L1s
=
R2 L2 s
R1 L1 s
R
L
s
=
and
. Simplifying the
2
2
R2 + L2 s
R1 L1Cs 2 + L1 s + R1
above equation, we get:
 ( A + 1)( R2 + L2 s ) R1 L1Cs 2 + L1 s + R1 
R1 L1Cs 2 + L1 s + R1
+
Vin ( s ) − 
 Vx ( s ) = 0
R1 L1 s
R2 L2 s
R1 L1 s


Thus, the transfer function between the input voltage and the node voltage is given by
R1 L1Cs 2 + L1 s + R1
Vx ( s )
R1 L1 s
=
Vin ( s ) ( A + 1)( R2 + L2 s ) R1 L1Cs 2 + L1 s + R1
+
R2 L2 s
R1 L1 s
=
.
R2 L2 s ( R1 L1Cs 2 + L1 s + R1 )
R1 L1 s ( A + 1)( R2 + L2 s ) + R2 L2 s ( R1 L1Cs 2 + L1 s + R1 )
The transfer function between the input voltage and the output voltage is
H A ( s) =
Vout ( s ) − AVx ( s )
− AR2 L2 s ( R1 L1Cs 2 + L1 s + R1 )
=
=
Vin ( s )
Vin ( s )
R1 L1 s ( A + 1)( R2 + L2 s ) + R2 L2 s ( R1 L1Cs 2 + L1 s + R1 )
The ideal transfer function is the limit as the op-amp gain tends to infinity:
H ( s ) = lim H A ( s ) = −
A→∞
R2 L2 ( R1 L1Cs + L1 s + R1 )
=−
R1 L1 ( R2 + L2 s )
2
L1
s + 1)
R1
L
L1 (1 + 2 s )
R2
L2 ( L1Cs 2 +
(b) [5 marks] The circuit is used as a cascade equalizer for the system
s +1
, that is, G ( jω ) H ( jω ) = 0dB, ∀ω . Let L1 = 10 H . Find the values
0.01s + 0.1s + 1
of the remaining circuit components L2 , R1 , R2 , C .
G ( s ) = −50
2
Vm ( jω )
G ( jω )
Vin ( jω )
H ( jω )
system
equalizer
Component values are obtained by setting
H ( s ) = G −1 ( s ) = −0.02
L
0.01s + 0.1s + 1
=− 2
s +1
L1
2
( L1Cs 2 +
(
2
L1
s + 1)
R1
L2
s + 1)
R2
Vout ( jω )
Sample Final Exam Covering Chapters 1-9 (finals01)
which yields ⇔ L2 = 0.2 H , R1 = 100Ω, R2 = 0.2Ω, C = 0.001F
(c) [7 marks] Sketch the Bode plot of H ( s ) and G ( s ) (magnitude only, both on the same plot).
Magnitude
Bode
H ( jω ) = −0.02
plot
of
G ( jω ) = −50
jω + 1
0.01( jω ) + 0.1( jω ) + 1
2
and
desired
0.01( jω ) 2 + 0.1( jω ) + 1
:
jω + 1
20 log10 G ( jω )
60
40
(dB)
20
10 -1
100
10 2
10 1
10 3
ω (log)
-20
-40
20 log10 H ( jω )
-60
Problem 2 (20 marks) Digital Signal Generator
A programmable digital signal generator generates a sinusoidal waveform by LTI filtering of a
staircase approximation to a sine wave x ( t ) .
x (t )
A
0.5A
T/2
-T/6
T/6
-0.5A
T/3
2T/3
5T/6
T
t
-A
3
Sample Final Exam Covering Chapters 1-9 (finals01)
(a) [9 marks] Find the Fourier series coefficients a k of the periodic signal x ( t ) . Show that the even
harmonics vanish. Express x ( t ) as a Fourier series.
Answer:
First of all, the average over one period is 0, so a0 = 0 . For k ≠ 0 ,
ak =
1
T
T
2
∫
x(t )e
− jk
2π
t
T
dt
T
−
2
−
A
=−
2T
T
3
∫e
− jk
2π
t
T
T
−
2
T
2
A
+
e
2T T∫
− jk
2π
t
T
A
dt −
T
T
3
T
6
∫e
− jk
2π
t
T
T
−
3
A
dt + ∫ e
TT
3
A
=
2T
−
− jk
2π
t
T
A
dt −
2T
0
∫e
− jk
2π
t
T
dt
T
−
6
T
6
2π
A − jk T t
dt +
e
dt
2T ∫0
6
T
6

∫0  e

2π
− jk
t
T
−e
2π
jk
t
T

A
 dt +
2T

T
2

∫T  e

2π
− jk
t
T
−e
2π
jk
t
T
T
3
2π
2π
jk
t 

A  − jk T t
− e T  dt
 dt + ∫  e
T T


3
=
T
6
− jA
 2π
sin  k
∫
T 0
 T
T
3
j2 A

 2π
sin  k
t  dt −
∫
T T

 T
6
T
2
jA

 2π
sin  k
t  dt −
∫
T T

 T
6
jA  T 
 2π
=

 cos  k
T  2π k 
 T

t  dt

3
T
6
j2 A  T 

 2π
t +

 cos  k
T  2π k 
0
 T
T
 2π
 3 jA  T 
t +

 cos  k
T
T  2π k 
 T

6
T
2
t
T
3

 π
 2π  
 − 2 cos  k  + cos ( kπ ) − cos  k


 3
 3 

jA 
 π
 2π 
=
− cos  k  + cos  k
 − 1 + cos ( kπ ) 

2π k 
 3
 3 

=
jA
2π k
  π
 2π
cos  k 3  − 1 + 2 cos  k 3


 
Note that the coefficients are purely imaginary, which is consistent with our real, odd signal. The even
FS coefficients are for k = 2m, m = 1, 2,… :
4
Sample Final Exam Covering Chapters 1-9 (finals01)
ak = a2 m =
jA 
 2π
− cos  m

2π 2m 
 3

 4π
 + cos  m

 3


 − 1 + cos ( m2π ) 


=
jA 
π


 π

− cos  − m + mπ  + cos  m + mπ  

2π 2m 
3


 3

=
jA 
 π

 π

− cos  m − mπ  + cos  m + mπ  

2π 2m 
 3

 3

=
jA 
 π

 π

− cos  m + mπ  + cos  m + mπ   = 0

2π 2m 
 3

 3

The Fourier series representation of x ( t ) is
x(t ) =
+∞
∑ae
k =−∞
jk
k
2π
t
T
+∞
jA
= ∑
k =−∞ 2π k
k ≠0
2π

 jk T t
 π
 2π 
cos
k
cos
k
1
cos
k
π
−
+
−
+
.
(
)





e
 3
 3 


(b) [3 marks] Write x ( t ) using the real form of the Fourier series.
+∞
x (t ) = a0 + 2∑ [ Bk cos( kω 0t ) − Ck sin( kω 0t )]
k =1
Answer:
Recall that the Ck coefficients are the imaginary parts of the a k 's. Hence

−A 
 π
 2π 
− cos  k  + cos  k
 − 1 + cos ( kπ )  sin(kω 0 t ) .

 3
 3 
k =1 π k 

+∞
x(t ) = ∑
(c) [2 marks] Design an ideal lowpass filter that will produce the perfect sinusoidal waveform
y (t ) = sin
2π
t at its output with x(t ) as its input. Sketch its frequency response and specify its gain
T
K and cutoff frequency ω c .
Answer:
Hlp ( jω )
K
−ωc
ωc
ω
5
Sample Final Exam Covering Chapters 1-9 (finals01)
The cutoff should be between the fundamental and the second harmonic, say
−π
should be K =
A

π 
 2π
 − cos  3  + cos  3
 




 − 1 + cos (π ) 


−1
=
−π
A
ωc =
 1 1

 − 2 − 2 − 2
3π
. The gain
T
−1
=
π
3A
.
(d) [6 marks] Now suppose that the first-order lowpass filter whose differential equation is given below
is used to filter x(t ) .
d
y (t ) + y (t ) = Bx(t )
dt
T
. Give the Fourier series representation of the output
where the time constant is chosen to be τ =
2π
y (t ) . Compute the total average power in the fundamental components P1tot and in the third
harmonic components P3tot . Find the value of the dc gain B such that the output w(t ) produced by
the fundamental harmonic of the real Fourier series of x(t ) has unit amplitude.
τ
Answer:
B
τ s +1
B
H ( jω ) =
τ jω + 1
H (s) =
+∞
2π jk 2π t
B
jA
y (t ) = ∑ ak H ( jk
)e
= ∑
T
k =−∞
k =−∞ jk + 1 2π k
+∞
k ≠0
2π

 jk T t
 π
 2π 
cos
k
cos
k
1
cos
k
−
+
−
+
π
(
)





e
 3
 3 


Power:
P1tot
B jA 
π 
 2π
=2
− cos   + cos 

j + 1 2π 
3
 3


 − 1 + cos (π ) 


2
B jA  1 1
B j3 A

=2
− − − 2 = 2

j + 1 2π  2 2
j + 1 2π

2
2
2
= B2
P3tot
3A
9 A2 B 2
=
2π
4π 2
B jA
− cos (π ) + cos ( 2π ) − 1 + cos (π ) 
=2
j3 + 1 2π 
2
B jA
=2
[ 0] = 0
j 3 + 1 2π
DC gain B:
We found that the gain at
ω0
should be
6
2
Sample Final Exam Covering Chapters 1-9 (finals01)
π
3A
B
= H ( jω 0 ) =
⇔B=
π
(τω0 )
τ jω 0 + 1
2
=
B
(τω 0 )
2
+1
+1
3A
Problem 3 (15 marks)
Compute the response y[ n] of an LTI system described by its impulse response h[ n] shown below
and whose magnitude from a time instant to the next decays by 0.9 on its support. The input signal
x[n] is the rectangular pulse shown below shown below. Give a mathematical expression for h[n] .
h[n]
1
x[n]
1
-1
-1
0 1 2 3 4
-2
n
1
0
3
2
4
Answer:
Mathematical expression for impulse response: h[ n] = (−0.9)
( n + 2)
u[n + 2]
We break down the problem into 3 intervals for n.
For n < −2 : h[ n − k ] is zero for k>=0, hence g[ k ] = h[k ]x[ n − k ] = 0 ∀k and y[ n] = 0 .
For −2 ≤ n ≤ 1 : Then g[ k ] = h[ k ] x[ n − k ] ≠ 0 for k = 0,… , n + 2 . We get
n+ 2
n+2
n+2
k =0
k =0
k =0
y[n] = ∑ g[k ] = ∑ (−0.9) n − k + 2 = (−0.9) n + 2 ∑ (−0.9) − k
− n −3

10 k
n + 2  1 − ( −0.9)
= (−0.9) ∑ (− ) = (−0.9) 
−1 
9
k =0
 1 − (−0.9) 
(−0.9) n + 2 − (−0.9) −1 (−0.9) n +3 − 1
=
=
−1.9
1 − (−0.9) −1
n+ 2
n+ 2
For n ≥ 2 : Then g[ k ] = h[ k ] x[ n − k ] ≠ 0 for k = 0,… ,4 . We get
3
3
3
k =0
k =0
k =0
y[n] = ∑ g[k ] = ∑ (−0.9) n − k + 2 = (−0.9) n + 2 ∑ (−0.9) − k
 1 − (−0.9) −4 
= (−0.9) n + 2 
−1 
 1 − (−0.9) 
(−0.9) n + 2 − (−0.9) n − 2
=
1 − (−0.9) −1
7
n
Sample Final Exam Covering Chapters 1-9 (finals01)
In summary, the output signal of the LTI system is

0,

 (−0.9) n +3 − 1
,
y[n] = 
1.9
−

 (−0.9) n + 2 − (−0.9) n −2
,

1 − (−0.9) −1

n < −2
−2 ≤ n < 1
2≤n
Problem 4 (10 marks)
The unit step response of an LTI system was measured to be
s (t ) = 2 e−
3t
π
sin(t − )u (t ) + u (t ) − tu (t ) .
6
(a) [6 marks] Find the transfer function H ( s ) of the system. Specify its ROC. Sketch its pole-zero
plot.
8
Sample Final Exam Covering Chapters 1-9 (finals01)

H ( s ) = sS ( s ) = sL  2e−

3t
π

sin(t − )u (t ) + u (t ) − tu (t ) 
6


 −
= sL  2 e

π
− j ( t− )
 j ( t −π6 )
6
−
e
e
3t 

2j




= s L  2 e−



 3
1
3
1
− j )e jt − (
+ j )e − jt
(
3t
2
2
2
 2
2j



 3
1
− j ) e−
 (
2
= s L  2



 3 −
e
 (
2
= s L 



= sL 

(
3 e−
3t
3t + jt
−
3t + jt
−(


 u (t ) + u (t ) − tu (t ) 





3
1
+ j ) e−
2
2




 u (t ) + u (t ) − tu (t ) 






3t − jt
j
3 −
e
2
sin t − e−
3t
3t − jt
1
) − j e−
2
j
3t + jt
)




 u (t ) + u (t ) − tu (t ) 






1
− j e−
2
3t − jt




 u (t ) + u (t ) − tu (t ) 






cos t u (t ) + u (t ) − tu (t ) 


s+ 3
3
1 1
= s
−
+
− 2
2
2
s
s 
+
+
s
(
3)
1
(
3)
1
s
+
+

=
=
− s 3 + [( s + 3) 2 + 1]( s − 1)
s[( s + 3) 2 + 1]
− s 3 + [ s 2 + 2 3s + 4]( s − 1)
s[( s + 3) 2 + 1]


(4 − 2 3)
4
(2 3 − 1)  s 2 +
s−

(2 3 − 1)
(2 3 − 1) 
(2 3 − 1) s 2 + (4 − 2 3) s − 4

=
=
s[( s + 3) 2 + 1]
s[( s + 3) 2 + 1]
ROC: Re{s} > 0 .
p1 = 0
z1 = −1.3875,
Poles are p2 = − 3 + j , Zeros are zeros of s + 0.21748s − 1.62331 : z2 = 1.1700
2
z3 = ∞
p3 = − 3 − j
9
Sample Final Exam Covering Chapters 1-9 (finals01)
Im{s}
1
-1.7
-1.39
1.17
Re{s}
-1
(b) [2 marks] Is the system causal? Is it stable? Justify your answers.
Answer:
System is causal: ROC is an open RHP and transfer function is rational.
This system isn't stable as ROC doesn't include the imaginary axis (or because rightmost pole 0 has a
nonnegative real part.)
(c) [2 marks] Give the direct form realization (block diagram) of H ( s) .
Answer:
H (s) =
(2 3 − 1) s 2 + (4 − 2 3) s − 4
s3 + 2 3s 2 + 4s
2 3 −1
4−2 3
+
X ( s)
s 3W ( s)
- - -
1
s
s 2W ( s)
sW ( s)
1
s
2 3
4
0
10
1
s
W ( s)
+ +
−4
+
Y ( s)
Sample Final Exam Covering Chapters 1-9 (finals01)
Problem 5 (5 marks)
True or False?
(a) The Fourier transform Z ( jω ) of the convolution of a real even signal x ( t ) and a real odd signal
y (t ) is imaginary even.
Answer: False.
(b) The system defined by y (t ) = tx (t ) is time-invariant.
Answer: False.
(c) The Fourier series coefficients bk of the periodic signal y (t ) = x(10t ) are given by bk =
1
ak ,
10
FS
where x(t ) ↔ ak .
Answer: False.
(d) The Fourier transform X ( jω ) of the product of a real signal x ( t ) and an impulse
Answer: True.
(e) The fundamental period of the signal x[ n] = sin(
δ (t )
3π
2π
n) cos( n) is 60.
5
3
Answer: False. (N=30)
Problem 6 (15 marks)
−
−
The following circuit has initial conditions on the capacitor vC ( 0 ) and inductor i L (0 ) .
C
vs (t )
L
+
-
+
R2
R1
v (t )
-
(a) [3 marks] Transform the circuit using the unilateral Laplace transform.
Answer:
-
-
Vs ( s)
+
Ls
+-
+
1
Cs
Li L ( 0 − )
1
vc ( 0 − )
s
+
R2
R1
I2 ( s)
I1 ( s)
11
-
V ( s)
is real.
Sample Final Exam Covering Chapters 1-9 (finals01)
(b) [6 marks] Find the unilateral Laplace transform of
v (t ) .
Answer:
Let's use mesh analysis.
For mesh 1:
1
1
I1 ( s) − vc (0− ) − R1[I1 ( s) − I2 ( s)] = 0
Cs
s
1
1
1
vc (0− ) + (1 +
)I1 ( s)
⇒ I2 ( s) = − Vs ( s) +
R1
R1s
R1Cs
Vs ( s) −
For mesh 2:
R1[I1 ( s) − I2 ( s)] − ( R2 + Ls)I2 ( s) + Li L (0− ) = 0
1
L
⇒ I1 ( s) = ( R1 + R2 + Ls)I2 ( s) − i L (0− )
R1
R1
Substituting, we obtain
I2 ( s) = −
LM
N
1
1
1
1
L
Vs ( s) +
vc (0− ) + (1 +
)
( R1 + R2 + Ls)I2 ( s) − i L (0 − )
R1
R1s
R1Cs R1
R1
OP
Q
[ R12 Cs − (1 + R1Cs)( R1 + R2 + Ls)]I2 ( s) = − R1CsVs ( s) + R1Cvc (0− ) − (1 + R1Cs) Li L (0− )
−[ LR1Cs 2 + ( L + R1 R2 C ) s + R1 + R2 ]I2 ( s) = − R1CsVs ( s) + R1Cvc (0− ) − (1 + R1Cs) Li L (0− )
Solving for I2 ( s) , we get
I2 ( s) =
R1CsVs ( s)
(1 + R1Cs) Li L (0− ) − R1Cvc (0− )
+
LR1Cs2 + ( L + R1 R2 C ) s + R1 + R2 LR1Cs2 + ( L + R1 R2 C ) s + R1 + R2
And finally the output voltage is
V ( s ) = R2 I2 ( s ) =
R1 R2 CsVs ( s )
R2 (1 + R1Cs ) LiL (0− ) − R1 R2 Cvc (0 − )
+
LR1Cs 2 + ( L + R1 R2 C ) s + R1 + R2
LR1Cs 2 + ( L + R1 R2 C ) s + R1 + R2
(c) [6 marks] Draw the Bode plot (magnitude and phase) of the frequency response from the input
voltage Vs ( jω ) to the output voltage V ( jω ) . Assume that the initial conditions on the capacitor and
the inductor are 0. Use the numerical values:
R1 = 1 Ω, R2 =
109
1
Ω, L =
H, C = 1 F .
891
891
Answer:
For the values given, the transfer function from the source voltage to the output voltage is
12
Sample Final Exam Covering Chapters 1-9 (finals01)
V( s)
H ( s ) :=
=
Vs ( s )
=
s2 +
1
891
R2
s
L
L + R1 R2 C
R + R2
s2 +
s+ 1
LR1C
LR1C
109 s
1 + 109
+ 109
891
891
s
+
1
1
891
891
109 s
s + 110s + 1000
109 s
=
( s + 10)( s + 100)
s
= 0.109 1
1
( 10 s + 1)( 100
s + 1)
=
2
Bode Plot:
20 log 10 H ( jω )
20
(dB) 10-1 100
101
102
ω (log)
-20
-40
-60
-80
-100
13
Sample Final Exam Covering Chapters 1-9 (finals01)
∠H ( jω )
(rd)
3π/4
π/2
π/4
0
10-1 100
101
102
103
−π/4
ω (log)
−π/2
Problem 7 (15 marks)
A classical technique to control the position of an inertial load (e.g., a robot arm) driven by a
permanent-magnet DC motor is to vary the armature current based on a potentiometer measurement
of the robot's joint angle θ (t ) , as shown in the diagram below.
dc motor
θ (t )
potentiometer
ia ( t )
K (s)
em (t ) θ (t )
m
+
−
B
−1
+
v (t )
J, b
−
+10Volts
θ d (t )
load (robot
arm)
Before closing the feedback loop, we first look at the open-loop dynamics of this system in (a) and (b).
The torque τ ( t ) in Newton-metres applied to the load by the motor is proportional to the armature
current in Amps:
τ (t ) = Aia (t ) .
14
Sample Final Exam Covering Chapters 1-9 (finals01)
The robot arm is modeled as an inertia J with viscous friction represented by the coefficient b . The
equation of movement for the load is
d 2θ (t )
dθ (t )
J
+b
= τ (t ) .
2
dt
dt
(a) [2 marks] Assume that the system is initially at rest. Find the open-loop transfer function
G ( s): = Θ( s) I a ( s) relating the joint angle Θ( s) to the motor's armature current input I a ( s) .
Answer:
G ( s) =
A
Θ( s)
=
I a ( s) s( Js + b)
(b) [6 marks] Assume that A = 1 Nm A , J = 1 Nm
rd
s2
and b = 1 Nm
unit step response sω ( t ) of the load's angular velocity
current. What is the ±5% settling time
Answer:
Gω ( s ) =
ω (t ) =
rd
s
. Compute the open-loop
dθ (t )
for a unit step in armature
dt
t s (an approximation is OK)?
s Θ( s )
1
=
I a ( s) s + 1
The unit step response sω ( t ) is given by
1
Θ( s )
1
1
1
1
Sω ( s ) = Gω ( s ) =
=
=
= −
s
I a ( s ) s ( s + 1) s ( s + 1) s s + 1
Taking the inverse Laplace transform, we get
sω (t ) = (1 − e −t )u (t )
The ±5% settling time is approximately equal to three time constants: t s ≈ 3 s .
(c) [7 marks] Based on the above figure, find the gain B of the joint angle sensor (potentiometer) in
−1
Volts/radian. Assume, as shown in the figure above, that there is a gain B in the feedback path
canceling out the sensor gain, and thus the link angle θ ( t ) is measured perfectly, i.e.,
θm (t) = θ (t) . We want to analyze the feedback controller K ( s) = λ ( s + α ) to control the robot's
joint angle. That is, we want θ ( t ) to track the desired angle θd (t ) .
Answer:
Sensor gain is B =
10
π
V rd
15
Sample Final Exam Covering Chapters 1-9 (finals01)
θ d (t )
θ (t )
+
K ( s)
-
G ( s)
Find the closed-loop transfer function H ( s ) := Θ( s ) Θ d ( s ) . Find expressions for the closed-loop
damping ratio
ζ
and undamped natural frequency
λ . For α = 4 , find a value of λ
ωn
in terms of the controller parameters
that will result in a closed-loop damping ratio of
Closed-loop transfer function:
K ( s )G ( s )
1 + K ( s )G ( s )
λ (s + α )
s ( s + 1)
=
λ (s + α )
1+
s ( s + 1)
λ (s + α )
=
s ( s + 1) + λ ( s + α )
λ (s + α )
= 2
s + (1 + λ ) s + αλ
H (s) =
We get
For
ω n = αλ
α =4
and
and given
ζ =
ζ =
1+ λ
2 αλ
1
ζ =
2
.
, we have:
1
=
1+ λ
2 4λ
1 (1 + λ ) 2
⇒ =
⇒ 8λ = 1 + 2λ + λ 2
2
16λ
⇒ 1 − 6λ + λ 2 = 0
2
⇒ λ1,2 =
6 ± 36 − 4
= 3± 2 2
2
both values are positive (for stability) and hence acceptable.
END OF EXAMINATION
16
ζ =
1
2
.
α
and