Sample Final Exam Covering Chapters 1-9 (finals02)
Sample Final Exam (finals00)
Covering Chapters 1-9 of Fundamentals of Signals & Systems
Problem 1 (20 marks)
The unit step response of an LTI system was measured to be
s (t ) = 2 e−
π
sin(t − )u (t ) + u (t ) − tu (t ) .
6
(a) [10 marks] Find the transfer function H ( s ) of the system. Specify its ROC. Sketch its pole-zero
3t
plot.
Answer:

H ( s ) = sS ( s ) = sL  2e−

3t
sin(t −
π

)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 ) 


3
s+ 3
1 1
= s
−
+
− 2
2
2
s
s 
(
3)
1
(
+
+
+
+
s
s
3)
1

=
− 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
1
Sample Final Exam Covering Chapters 1-9 (finals02)
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
Im{s}
1
-1.7
-1.39
1.17
Re{s}
-1
(b) [4 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) [6 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
2
1
s
W ( s)
−4
+ +
+
Y ( s)
Sample Final Exam Covering Chapters 1-9 (finals02)
Problem 2 (20 marks)
−
−
The following circuit has initial conditions on the capacitor vC ( 0 ) and inductor i L (0 ) .
C
vs (t )
R2
+
L
R1
-
+
v (t )
-
(a) [4 marks] Transform the circuit using the unilateral Laplace transform.
Answer:
Vs ( s)
+
R2
+-
Ls
R1
I2 ( s)
I1 ( s)
+
-
1
vc ( 0 − )
s
-
1
Cs
+
V( s)
Li L ( 0 − )
-
(b) [8 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
⇒ I2 ( s) = − Vs ( s) +
vc (0− ) + (1 +
)I1 ( 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
3
Sample Final Exam Covering Chapters 1-9 (finals02)
R1CsVs ( s)
(1 + R1Cs) Li L (0− ) − R1Cvc (0− )
I2 ( s) =
+
LR1Cs2 + ( L + R1 R2 C ) s + R1 + R2 LR1Cs2 + ( L + R1 R2 C ) s + R1 + R2
And finally the output voltage is
V ( s ) = LsI2 ( s ) − LiL (0 − ) =
LR1Cs 2 Vs ( s )
(1 + R1Cs ) L2 siL (0 − ) − R1CLsvc (0− )
+
− LiL (0− )
LR1Cs + ( L + R1 R2 C ) s + R1 + R2
LR1Cs 2 + ( L + R1 R2 C ) s + R1 + R2
2
=
LR1Cs 2 Vs ( s )
{(1 + R1Cs ) L2 s − [ L2 R1Cs 2 + L ( L + R1 R2 C ) s + L( R1 + R2 )]}iL (0 − ) − R1CLsvc (0− )
+
LR1Cs 2 + ( L + R1 R2 C ) s + R1 + R2
LR1Cs 2 + ( L + R1 R2 C ) s + R1 + R2
=
LR1Cs 2 Vs ( s )
−[ LR1 R2 Cs + L( R1 + R2 )]iL (0− ) − R1CLsvc (0 − )
+
LR1Cs 2 + ( L + R1 R2 C ) s + R1 + R2
LR1Cs 2 + ( L + R1 R2 C ) s + R1 + R2
(c) [8 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
H ( s ) :=
=
Bode Plot:
V( s )
=
Vs ( s )
s2
=
L + R1 R2 C
R + R2
s2 +
s+ 1
s2 +
LR1C
LR1C
s2
1
891
+ 109
891
1
891
s+
1
891
s2
s2
1
s2
=
=
1
s + 1)
s 2 + 110 s + 1000 ( s + 10)( s + 100) 1000 ( 101 s + 1)( 100
20 log 10 H ( jω )
20
(dB) 10-1 100
101
102
ω (log)
-20
-40
-60
-80
-100
1 + 109
891
4
Sample Final Exam Covering Chapters 1-9 (finals02)
∠H ( jω )
(rd)
5π/4
π 10-1 100
101
102
3π/4
103
ω (log)
π/2
π/4
0
Problem 3 (15 marks)
Consider the causal differential system described by
d 2 y (t )
dy (t )
dx(t )
+2
+ 4 y (t ) = 4
+ 4 x(t )
2
dt
dt
dt
and with initial conditions
dy (0− )
= −2,
dt
y (0− ) = −4 . Suppose that this system is subjected to the
input signal
x(t ) = u (t ) .
Find the system's damping ratio
ζ
and undamped natural frequency
the system and specify its ROC. Compute the steady-state response
response
ω n . Give the transfer function of
yss (t ) and the transient
ytr (t ) for t ≥ 0 .
Answer:
Let's take the unilateral Laplace transform on both sides of the differential equation.
 2
dy (0− ) 
−
−
Y
−
−
s
(
s
)
sy
(0
)

 + 2  sY ( s ) − y (0 )  + 4Y ( s ) = 4sX ( s ) + 4X ( 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 ) .
5
Sample Final Exam Covering Chapters 1-9 (finals02)
(s
2
)
+ 2s + 4 Y ( s ) = 4sX ( s ) + 4X ( s ) + sy (0− ) + 2 y (0− ) +
dy (0− )
dt
dy (0− )
4( s + 1)X ( s)
dt
+
Y ( s) = 2
2
s
2
s
4
s
2
s
4
+
+
+
+
( s + 2) y (0− ) +
zero-state resp.
The transfer function is
zero-input resp.
4( s + 1)
s + 2s + 4
2
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
and the damping ratio is
ζ = 0.5 .
The poles are
p1,2 = −ζω n ± jω n 1 − ζ 2 = −1 ± j 3 .
Therefore the ROC is
Re{s} > −1 .
The unilateral LT of the input is given by
1
X ( s ) = , Re{s} > 0 ,
s
thus
Y ( s) =
4 ( s + 1)
( s + 2s + 4
)s
2
+
−4 s − 10
−4s 2 − 6s + 4
=
s 2 +
2 s +
4
s 2 + 2s + 4 s
Re{ s}>−1
zero-input resp.
Re{ s}> 0
zero-state resp.
(
)
Let's compute the overall response:
Y ( s) =
=
−4 s 2 − 6 s + 4
(s
2
)
+ 2s + 4 s
, Re{s} > 0
A 3 + B ( s + 1)
( s + 1) + 3
2
+
Re{ s}> 0
Re{ s}>−1
=
A 3 + B( s + 1)
( s + 1) + 3
2
Re{ s }>−1
C
s
N
+
1
Ns
Re{ s }> 0
6
1
1
A+
=
⇒ A = − 3 , then multiply both sides by s and let s → ∞
−3
−1
3
to get −4 = B + 1 ⇒ B = −5 :
Let s = −1 to compute
6
Sample Final Exam Covering Chapters 1-9 (finals02)
Y ( s) =
− 3( 3) − 5( s + 1)
( s + 1) + 3 2
+
Re{ s}>−1
Notice that the second term
1
Ns
Re{ s }> 0
1
is the steady-state response, and thus yss (t ) = u (t ) .
s
Taking the inverse Laplace transform using the table yields
ytr (t ) =  − 3e−t sin( 3t ) − 5e−t cos( 3t )  u (t ) .
Problem 4 (10 marks)
Consider the following second-order, causal difference LTI system S initially at rest:
S:
2 y[n] − 1.8 y[n − 1] + 0.4 y[n − 2] = x[n] − x[n − 2]
(a) [4 marks] What is the characteristic polynomial of S ? What are its zeros? Is the system stable?
Justify your answer.
Answer:
Let's rewrite the difference equation as
S:
y[n] − 0.9 y[n − 1] + 0.2 y[n − 2] = 0.5 x[n] − 0.5 x[n − 2]
p ( z ) = z − 0.9 z + 0.2 = ( z − 0.4)( z − 0.5)
The zeros are z1 = 0.4, z2 = 0.5 .
2
(b) [6 marks] Compute the impulse response of S for all n.
Answer:
The homogeneous response is given by
y[n] = A(0.4) n + B(0.5) n , n > 0 .
ha [n] − 0.9ha [n − 1] + 0.2ha [n − 2] = δ [n]
The initial conditions for the homogeneous equation for n>0
are
ha [−1] = 0 and ha [0] = δ [0] = 1 .
Now we can compute the coefficient A and B:
y[−1] = A(0.4) −1 + B(0.5) −1 = 2.5 A + 2 B = 0
y[0] = A + B = 1
Hence
A = −4, B = 5
and the intermediate impulse response is
ha [n] =  −4(0.4) n + 5(0.5) n  u[n]
Finally, the impulse response is
h[n] = 0.5ha [n] − 0.5ha [n − 2] =  −2(0.4) n + 2.5(0.5) n  u[n] −  −2(0.4) n − 2 + 2.5(0.5) n − 2  u[n − 2]
7
Sample Final Exam Covering Chapters 1-9 (finals02)
Problem 5 (5 marks)
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 real odd.
Answer: False.
t
(b) The system defined by y (t ) =
∫ x(τ )dτ
is time-invariant.
0
Answer: False.
(c) The Fourier series coefficients a k of a real and even periodic signal x ( t ) have the following
∗
property: a k = a− k .
Answer: True.
(d) The Fourier transform of the convolution of a real even signal with the impulse
Answer: False.
(e) The fundamental period of the signal x[ n] = sin(
δ (t − 1)
is real.
3π
n)e jπ n is 14.
7
Answer: True.
Problem 6 (15 marks)
(a) [12 marks] Consider the periodic signal x ( t ) depicted below. Give a mathematical expression for
x (t ) . Find its fundamental frequency ω 0 . Compute its Fourier series coefficients ak . Express x (t )
as a Fourier series.
x (t )
4
1
-3
-5 -4
-2
-1
-1 1
3
2
-4
Answer:
8
4
5
6
t
Sample Final Exam Covering Chapters 1-9 (finals02)
This signal can be written as:
x(t ) =
+∞
∑ −((t − 6m) − 1) [u (t + 3 − 6m) − u(t − 6m)] + ((t − 6m) − 4) [u (t − 6m) − u (t − 3 − 6m)]
m =−∞
Its fundamental period and frequency are T = 6,
ω0 =
2π π
= . The average value over one period
6
3
is given by:
6
1
a0 = ∫ x(t )dt = 0
60
The FS coefficients ak for k ≠ 0 are given by
π
− jk t
1
ak = ∫ x(t )e 3 dt
TT
π
0
=
π
3
− jk t
− jk t
1
1
(1 − t )e 3 dt + ∫ (t − 4)e 3 dt
∫
6 −3
60
0
π
− jk t 

1
=
(1 − t )e 3  −

π
 −3 6 jk π
−6 jk 
3
3
1
0
∫e
π
− jk t
3
−3
3
π
π
3
− jk t 
− jk t

1
3
dt +
(t − 4)e
 +
∫0 e 3 dt
π 
π
 0 6 jk
−6 jk
3
3
1
0
 − jk π3 t 
1
1
1
1
1 − 4e jkπ  +
 −e − jkπ + 4  +
=
e
 +
2 
2
− jk 2π
− jk 2π
 π
 π 

−3
−6  jk 
6  jk 
 3
 3
1
3
3
5 − 5e jkπ  − 2 2 1 − e jkπ  − 2 2 1 − e − jkπ 
=
− jk 2π
2k π
2k π
j5
3
1 − e jkπ  − 2 2 1 − e jkπ 
=
k 2π
k π
3
5 

k
= − 2 2 + j
 1 − (−1) 
k 2π 
 k π
The Fourier series representation of x ( t ) is
x(t ) =
+∞
∑
k =−∞
π
ak e
jk t
3
=
+∞
3
5

∑
− 2 2 + j
k 2π
k π
k =−∞ 
k ≠0
(b) [3 marks] Compute the Fourier transform of x ( t ) .
Answer:
9
π
jk t

k
3
.
 1 − (−1)  e

3
 − jk π3 t 
e


0
Sample Final Exam Covering Chapters 1-9 (finals02)
X ( jω ) =
∞
∑ 2π a δ (ω − kω
k
k =−∞
=
∞

0
)
5
6
∑  − k π + j k  1 − (−1)
2
k =−∞
k ≠0
k
π
 δ (ω − k )
3
Problem 7 (15 marks)
(a) [8 marks] Compute the 95% rise time of the unit step response s (t ) of
H (s) =
0.01s + 1
, Re{s} > −10 . Sketch s (t ) , indicating the important features.
0.1s + 1
Answer:
H (s) =
Notice that
0.01s + 1 ατ s + 1
=
, Re{s} > −10, α = 0.1, τ = 0.1
0.1s + 1
τ s +1
α > 0.05 . The step response is
−
t
s(t ) = αu(t ) + (1 − α )(1 − e τ )u(t ) .
s( t )
1
0.95
α
0.05
t
tr
The 95% rise time is given by the difference between the times when the response reaches the value
0.95 and the value 0.05. Since
−
α > 0.05 ,the output is already larger than 0.05 at t = 0+ .
t95%
0.95 = α + (1 − α )(1 − e τ )
⇒ t95% = −τ (ln 0.05 − ln(1 − α )) = [2.9957 + ln(1 − α )]τ = [2.9957 + ln(0.9)]0.1 = 0.289
⇒ ts = t95% = 0.289
(b) [7 marks] For the stable, causal second-order system H ( s ) =
3
, find the value of
0.02 s + as + 2
2
the real parameter a that will cause the system to have a 10% first overshoot in its step response.
Answer:
10
Sample Final Exam Covering Chapters 1-9 (finals02)
H (s) =
3
1.5
=
=
2
0.02 s + as + 2 0.01s + 0.5as + 1
1.5
2ζ
ω n −2 s 2 +
s +1
2
⇒ ω n = 10, a =
4ζ
ωn
ωn
= 0.4ζ
For a 10% first overshoot:
10 = 100e
−
ςπ
1−ς 2
ς2 =
ςπ
−
1−ς 2
ςπ
1−ς 2
= 0.1
= ln 0.1 = −2.3026
(2.3026) 2
π
⇒e
−
2
(1 − ς 2 )
 (2.3026) 2  (2.3026) 2
⇒ ς 1 +
=
π2
π2


(2.3026) 2
2
(2.3026) 2
2
π
⇒ς =
=
= 0.3495
(2.3026) 2 π 2 + (2.3026) 2
1+
2
2
π
⇒ ς = 0.59
Therefore a = 0.4ζ = 0.4(0.59) = 0.24 .
END OF EXAMINATION
11