Sample Final Exam (finals03)
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
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
