Sample Final Exam ( finals03) Fundamentals of Signals &

Sample Final Exam Covering Chapters 1-9 (finals03)
Sample Final Exam (finals03)
Covering Chapters 1-9 and part of Chapter 15 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 )
vin ( t )
C
R1
L2
vout ( t )
-
R2
+
L1
C
L2
vin ( t )
R2
+
− Av x (t )
v x ( t ) R1
+
-
Answer:
The transformed circuit is
L1 s
1 Cs
L2 s
Vin ( s )
+
R2
Vx ( s )
R1
+ − AVx ( s )
-
1
Sample Final Exam Covering Chapters 1-9 (finals03)
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 L2 s
R1 Cs1 L1 s
where R1
1
Cs
L1 s =
1
Cs +
1
R1
+
1
L1s
=
R2 L2 s
R1 L1 s
and R2 L2 s =
. Simplifying the
2
R2 + L2 s
R1 L1Cs + L1 s + R1
above equation, we get:
 ( A + 1)( R1 L1Cs 2 + L1 s + R1 ) R2 + L2 s 
R2 + L2 s
Vin ( s ) − 
+
 Vx ( s ) = 0
R2 L2 s
R1 L1 s
R2 L2 s 

Thus, the transfer function between the input voltage and the node voltage is given by
R2 + L2 s
Vx ( s )
R2 L2 s
R1 L1 s ( R2 + L2 s )
=
=
2
Vin ( s ) ( A + 1)( R1 L1Cs + L1 s + R1 ) R2 + L2 s R2 L2 s ( A + 1)( R1 L1Cs 2 + L1 s + R1 ) + R1 L1 s ( R2 + L2 s )
+
R1 L1 s
R2 L2 s
.
The transfer function between the input voltage and the output voltage is
H A ( s) =
Vout ( s ) − AVx ( s )
− AR1 L1 s ( R2 + L2 s )
=
=
Vin ( s )
Vin ( s )
R2 L2 s ( A + 1)( R1 L1Cs 2 + L1 s + R1 ) + R1 L1 s ( R2 + L2 s )
The ideal transfer function is the limit as the op-amp gain tends to infinity:
L2
s)
R1 L1 ( R2 + L2 s )
R2
H ( s ) = lim H A ( s ) = −
=−
A→∞
L
R2 L2 ( R1 L1Cs 2 + L1 s + R1 )
L2 ( L1Cs 2 + 1 s + 1)
R1
(b) [5 marks] Assume that the circuit has a DC gain of −50 , one zero at −1 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
L1 (1 +
components L2 , R1 , R2 , C .
Component values are obtained by setting
L2
s + 1)
L1
R2
s +1
H ( s ) = −50
=−
L
L2
0.01s 2 + 0.1s + 1
( L1Cs 2 + 1 s + 1)
R1
which yields L2 = 0.2 H , 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 + 1
. Bode plot:
0.01( jω ) + 0.1( jω ) + 1
2
2
Sample Final Exam Covering Chapters 1-9 (finals03)
20 log10 G ( jω )
60
40
(dB)
20
10 -1
100
10 1
10 2
10 3
10 1
10 2
103
ω (log)
-20
-40
-60
∠H ( jω )
180
135
(deg)
90
45
10 -1
10 0
ω (log)
-45
-90
-135
-180
3
Sample Final Exam Covering Chapters 1-9 (finals03)
Problem 2 (20 marks)
Consider the causal differential system described by
d 2 y (t )
dy (t )
+2
+ 2 y (t ) = 2 x(t )
2
dt
dt
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] Find the system's damping ratio
ζ
and undamped natural frequency
ω n . Give the
transfer function of the system and specify its ROC. Sketch its pole-zero plot. Is the system stable?
Justify.
Ans:
Let's take the unilateral Laplace transform on both sides of the differential equation.
 2
dy (0− ) 
−
−
−
−
s
(
s
)
sy
(0
)
Y

 + 2  sY ( s ) − y (0 )  + 2Y ( s ) = 2X ( 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 ) .
(
)
s 2 + 2s + 2 Y ( s ) = 2X ( s ) + sy (0− ) + 2 y (0− ) +
dy (0− )
dt
dy (0− )
2X ( s )
dt
Y (s) = 2
+
2
s +
s + 2s + 2
2s +
2 ( s + 2) y (0− ) +
zero-state resp.
zero-input resp.
The transfer function is H ( s ) =
2
,
s + 2s + 2
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
p1,2 = −ζω n ± jω n 1 − ζ 2 = −1 ± j 2 1 −
and the damping ratio is
ζ =
1
2
. The poles are
1
= −1 ± j .
2
Therefore the ROC is Re{s} > −1 . System is stable as jw-axis is contained in ROC. Pole-zero plot:
Im{s}
1
-1
Re{s}
-1
4
Sample Final Exam Covering Chapters 1-9 (finals03)
(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.
Ans:
The unilateral LT of the input is given by
1
X ( s ) = , Re{s} > 0 ,
s
thus
Y ( s) =
2
(
)
s 2 + 2s + 2 s
2s + 3
2 s 2 + 3s + 2
=
s 2 +
2 s +
2
s 2 + 2s + 2 s
+
Re{ s}>−1
zero-input resp.
Re{ s}> 0
zero-state resp.
(
)
Let's compute the overall response:
Y ( s) =
=
2s 2 + 3s + 2
(s
2
)
+ 2s + 2 s
A + B( s + 1)
( s + 1) + 1
A + B( s + 1)
( s + 1) + 1
2
Re{ s }> 0
get 2 = B + 1 ⇒ B = 1 :
Y ( s) =
1
Ns
+
Re{ s}>−1
Let s = −1 to compute
C
s
N
+
2
Re{ s }>−1
=
, Re{s} > 0
Re{ s}> 0
1 1
1
= A+
⇒ A = 0 , then multiply both sides by s and let s → ∞ to
−1 1
−1
( s + 1)
( s + 1) + 1
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
y (t ) = e −t cos(t )u (t ) + u (t ) .
Thus, the transient response is ytr (t ) = e
−t
cos(t )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:
5
Sample Final Exam Covering Chapters 1-9 (finals03)
2
1
:
, with damping ratio ζ =
s + 2s + 2
2
Transfer function is H ( s ) =
OS = 100e
−
ςπ
1−ς 2
% = 100e
2
−
0.707π
0.707
% = 100e −π % = 4.3%
Problem 3 (15 marks)
2T1
. This signal is the input to
T
−5 t
an LTI system with impulse response h(t ) = te u (t ) and output y (t ) .
Consider the rectangular waveform x ( t ) of period T and duty cycle
x (t )
1
-T
-T1
T
T1
t
y (t )
x (t )
h (t )
(a) [5 marks] Compute the frequency response H ( jω ) of the LTI system using the Fourier integral.
Give expressions for its magnitude H ( jω ) and phase ∠H ( jω ) as functions of
The frequency response of the system is given by
H ( jω ) =
+∞
+∞
0
0
−5t − jω t
∫ te e dt =
∫ te
− (5 + jω ) t
dt
+∞
1
1
te − (5+ jω ) t  −
=
0
−(5 + jω )
−(5 + jω )
0
1
− (5 + jω ) t +∞


e
0
−(5 + jω ) 2 
1
=
(5 + jω ) 2
=
Magnitude:
6
+∞
∫e
0
− (5 + jω ) t
dt
ω.
Sample Final Exam Covering Chapters 1-9 (finals03)
H ( jω ) =
Phase:
1
25 + ω 2
∠H ( jω ) = 2 atan 2 ( −ω ,5 )
(b) [3 marks] Find the Fourier series coefficients
duty cycle.
ak of the input voltage x (t ) for T = 1 s and a 40%
The period given corresponds to a signal frequency of 1Hz, and the 40% duty cycle means that
T1 =
2T
.
10
From the lecture notes, we get:
ak =
2
2k
sinc( )
5
5
(c) [3 marks] Compute the Fourier series coefficients bk of the output signal
described in (b) above), and compute its power spectrum.
We have
y (t ) (for the input
ω0 = 2π :
1
1
= 2 sinc( 2k )
bk = H ( jkω 0 )ak = 2 sinc( 2k )
2
5
5 (5 + jkω 0 )
5
5 (5 + jk 2π ) 2
The power spectrum of the output signal is
bk
2
=
4sinc 2 ( 25k )
25(25 + (k 2π ) 2 ) 2
(d) [4 marks] Compute the total average power P of the output signal z (t ) of the following ideal
lowpass filter H lp ( s ) , whose input is y (t ) of the above system. The frequency response of H lp ( s ) is
shown below, and its cutoff frequency is
ω c = 5π rd/s .
z (t )
y (t )
H lp ( s )
Hlp ( jω )
1
ωc
−ωc
7
ω
Sample Final Exam Covering Chapters 1-9 (finals03)
Answer:
The filter keeps only the dc components, the first harmonic components at ±2π rd/s and the second
harmonic components at ±4π rd/s . Thus,
2
2
P = b0 + 2 b1 + 2 b2
=
4sinc 2 ( 25k )
25(25 + (kω 0 ) 2 ) 2
2
=
2
+2
k =0
4sinc 2 ( 25k )
25(25 + (kω 0 ) 2 ) 2
2
+2
k =1
4sinc 2 ( 25k )
25(25 + (kω 0 ) 2 ) 2
k =2
4sinc ( )
4sinc ( )
4
+2
+2
3
2 2
25
25(25 + (2π ) )
25(25 + (4π ) 2 ) 2
2
5
4
5
4sin 2 ( 25π )
4sin 2 ( 45π )
4
= 3 +2
+2
25
25( 25π ) 2 (25 + (2π ) 2 ) 2
25( 45π ) 2 (25 + (4π ) 2 ) 2
=
sin 2 ( 25π )
sin 2 ( 45π )
4
2
2
+
+
= 0.000256 + 0.0000441 + 0.00000052 = 0.0003
253
(2π ) 2 (25 + (4π ) 2 ) 2
π 2 (25 + (2π ) 2 ) 2
Problem 4 (20 marks)
System identification
Suppose that you perform two input-output experiments on an LTI system. In the first experiment, the
system has nonzero initial conditions at t = 0 , but the input is set to zero. You record the output in this
case to be:
y zi (t ) = e − t cos(t − π 3)u (t )
In the second experiment, the system has the same initial conditions as in the first experiment, but the
input is given by:
x(t ) = e −2t u (t ) ,
and you 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.
Answer:
First, we have to remove the effect of the initials conditions from the output:
8
Sample Final Exam Covering Chapters 1-9 (finals03)
y zs (t ) = y (t ) − y zi (t ) = e− t [ cos t + sin t ] u (t ) − e−2t u (t ) − e −t cos(t − π 3)u (t )
1 j (t −π 3)
(e
+ e− j (t −π 3) )u (t )
2
1
= e − t [ cos t + sin t ] u (t ) − e −2t u (t ) − e −t (e− jπ 3 e jt + e jπ 3 e− jt )u (t )
2
−t
−2 t
−t
= e [ cos t + sin t ] u (t ) − e u (t ) − e Re(e − jπ 3 e jt )u (t )
= e − t [ cos t + sin t ] u (t ) − e −2t u (t ) − e −t
= e − t [ cos t + sin t ] u (t ) − e − t [cos(− π 3) cos t + sin(π 3) sin t ]u (t ) − e−2t u (t )
= e − t [ cos t + sin t ] u (t ) − e − t [0.5cos t +
3
sin t ]u (t ) − e −2t u (t )
2


3
) sin t  u (t ) − e −2t u (t )
= e − t 0.5cos t + (1 −
2


Take the Laplace transforms of the input and output signals using the table:
X (s) =
Yzs ( s ) =
1
, Re{s} > −2
( s + 2)
0.5s + 0.5
0.134
1
+
−
2
2
2
2
s+2
( s + 1) + 1 ( s + 1) + 1
N
Re{
s }>−2
Re{ s}>−1
=
Re{ s }>−1
0.5s + 0.634
1
−
2
2
s+2
( s + 1) + 1
N
Re{
s }>−2
Re{ s }>−1
=
(0.5s 2 + 1.634s + 1.268) − ( s 2 + 2s + 2)
, Re{s} > −1
( s 2 + 2 s + 2)( s + 2)
=−
0.5s 2 + 0.366s + 0.732
, Re{s} > −1
( s 2 + 2 s + 2)( s + 2)
Then, the transfer function is obtained as follows:
H (s) =
Yzs ( s )
=
X (s)
−
0.5s 2 + 0.366 s + 0.732
0.5s 2 + 0.366 s + 0.732
1 s 2 + 0.732 s + 1.464
( s 2 + 2 s + 2)( s + 2)
=−
=−
1
2
s 2 + 2s + 2
s 2 + 2s + 2
( s + 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 .
9
Sample Final Exam Covering Chapters 1-9 (finals03)
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) [4 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
y
(
t
)
=
−
0.5
− 0.366
− 0.732 x(t )
2
2
dt
dt
dt
dt
(c) [6 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:
X ( s)
Y ( s)
W ( s)
1
2
s + 2s + 2
−0.5s − 0.366s − 0.732
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 ) = −0.5s 2W ( s ) − 0.366 sW ( s ) − 0.732W ( s ) .
The direct form realization of the system is given below:
−0.5
−0.366
s 2W ( s)
X ( s)
+
-
-
sW ( s)
1
s
1
s
2
2
10
+ +
W ( s)
−0.732
+
Y ( s)
Sample Final Exam Covering Chapters 1-9 (finals03)
Problem 5 (15 marks)
(a) [12 marks] Compute the step response z[n] of the following causal discrete-time LTI system
initially at rest:
x[n] = u[n]
y[n]
y[n] − 0.2 y[n − 1] − 0.48 y[n − 2]
z[n]
z[n] = y[n] − y[n −1]
= x[n] − 0.2x[n − 1]
S1
S2
Answer:
First, by interchanging the first and second block, we can see that the problem is equivalent to
computing the impulse response of S1 :
x[n] = u[n]
w[n]
w[n] = x[n] − x[n − 1]
z[n] − 0.2z[n −1] − 0.48z[n − 2]
= w[n] − 0.2w[n −1]
S1
S2
The intermediate signal is just the unit impulse:
w[n] = u[n] − u[n − 1] = δ [n] .
Hence, the problem reduces to:
z[n] − 0.2 z[n − 1] − 0.48 z[n − 2] = δ [n] − 0.2δ [n − 1] .
We first compute ha [ n] :
ha [n] − 0.2ha [n − 1] − 0.48ha [n − 2] = δ [n]
The zeros of the characteristic polynomial are 0.8 and −0.6 , so
p( z ) = ( z − 0.8)( z + 0.6) = z 2 − 0.2 z − 0.48
homogeneous response for n>0:
ha [n] = A(0.8) n u[n] + B(−0.6) n u[n] .
Initial conditions ha [ −1] = 0, ha [0] = 1 lead to
ha [−1] = 0 = 1.250 A − 1.667 B
ha [0] = 1 = A + B
⇒ A = 0.571, B = 0.429
Thus,
ha [n] = 0.571(0.8) n u[n] + 0.429(−0.6) n u[n] ,
and finally:
11
z[n]
Sample Final Exam Covering Chapters 1-9 (finals03)
z[n] = ha [n] − 0.2ha [n − 1]
= 0.571(0.8) n u[n] + 0.429(−0.6) n u[n] − 0.1143(0.8) n −1 u[n − 1] − 0.0857(−0.6) n −1 u[n − 1]
= δ [n] + [0.571 − 0.1143(0.8) −1 ](0.8) n u[n − 1] + [0.429 − 0.0857( −0.6) −1 ](−0.6) n u[n − 1]
= δ [n] + 0.428(0.8) n u[n − 1] + 0.572(−0.6) n u[n − 1]
= 0.428(0.8) n u[n] + 0.572(−0.6) n u[n]
(b) [3 marks] Compute the impulse response of the system in (a)
Answer:
The impulse response is the first difference of the step response, so:
h[n] = z[n] − z[n − 1]
= 0.428(0.8) n u[n] + 0.572(−0.6) n u[n] − 0.428(0.8) n −1 u[n − 1] − 0.572(−0.6) n −1 u[n − 1]
Problem 6 (10 marks)
Consider the following sampling system where the sampling frequencies are
x p (t )
x (t )
x
p (t ) =
H lp1 ( jω )
w(t )
p (t ) =
1
k =−∞
KH lp 2 ( jω )
x
+∞
∑ δ (t − kT )
wp (t )
ω s1 =
2π
2π
, ωs2 =
.
T1
T2
y (t )
+∞
∑ δ (t − kT )
2
k =−∞
The spectrum X ( jω ) of the input signal
x(t ) , and the frequency responses of the two ideal lowpass
filters, are shown below. The gain of the second lowpass filter is K > 0 .
X ( jω )
1
−W
1
W
ω
ω c1
−ω c1
(a) [2 marks] For what range of sampling frequencies
first sampler (from
H lp 2 ( jω )
H lp1 ( jω )
x(t ) to x p (t ) )?
12
ω s1
1
ω
−ω c 2
ωc 2
ω
is the sampling theorem satisfied for the
Sample Final Exam Covering Chapters 1-9 (finals03)
Answer:
The sampling theorem is satisfied for
ω s1 > 2W
for the first sampler.
(b) [2 marks] The cutoff frequencies of the lowpass filters are given by
ω c1 = 3W , ω c 2 = W . Assume
that the lowest admissible sampling frequency is chosen for the first sampler. For what range of
sampling frequencies ω s 2 is the sampling theorem satisfied for the second sampler (from w(t ) to
wp (t ) )?
Answer:
The sampling theorem is satisfied for
ω s 2 > 6W
.
(c) [6 marks] Choosing the lowest sampling frequencies in the ranges that you found for the two
samplers, sketch the spectra X p ( jω ) , W ( jω ) , W p ( jω ) , and Y ( jω ) . Find the gain K of the
second filter that leads to y (t ) = x (t ) .
Answer:
For the spectra, we set
ω s1 = 2W
and
ω s 2 = 6W
, so that T1 =
π
W
, T2 =
π
3W
:
X p ( jω )
1/T1
−2ωs1 -3W
−ω s1
-W
W
ωs1
3W
2ωs1
ω
3W
4W
ω
3W
2ωs1
ω
W ( jω )
1/T1
-3W -2W
-W
W
2W
W p ( jω )
1/T1T2=3W2 /π2
−2ωs1 -3W
−ω s1
-W
W
13
ωs1
Sample Final Exam Covering Chapters 1-9 (finals03)
Y ( jω )
3W 2
π2
-W
Finally, K =
π2
3W 2
K
W
.
END OF EXAMINATION
14
ω