Sample Final Exam (finals00)
Sample Final Exam Covering Chapters 1-9 (finals00)
Sample Final Exam (finals00) Covering Chapters 1-9 of Fundamentals of Signals & Systems Problem 1 (20 marks) Consider the transfer function:
H (s) =
s 2 − 2s + 1 (0.01s 2 + 0.1 3s + 1)( s + 2)
.
(a) [6 marks] Find the poles and zeros of H ( s ) (specify how many there are at ∞ ). Give all possible regions of convergence of the transfer function H ( s) . Answer:
p1 = −2
z1 = 1,
Poles are p2 = −5 3 + 5 j = −8.66 + 5 j
Zeros are z2 = 1
z3 = ∞
p3 = −5 3 − 5 j = −8.66 − 5 j There are 3 possible ROCs: ROC1: Re{s} < −5 3 = −8.66 ROC2: −5 3 < Re{s} < −2 ROC3: Re{s} > −2
(b) [10 marks] Give the region of convergence of H ( s ) that corresponds to the impulse response of a stable system, and sketch it on a pole-zero plot. Is the stable system causal? Explain. Compute the impulse response h( t ) of this stable system and give its Fourier transform H ( jω ) . Answer: System is stable for the ROC that contains the jω -axis: ROC3. This system is also causal as ROC3 is an open RHP and the transfer function is rational.
Im{s} 5 -8.66
-2
1 -5
The partial fraction expansion of H ( s) yields: 1
ROC3
Re{s}
Sample Final Exam Covering Chapters 1-9 (finals00)
H (s) =
s 2 − 2s + 1 (0.01s 2 + 0.1 3s + 1)( s + 2)
=
100( s 2 − 2s + 1) ( s 2 + 10 3s + 100)( s + 2)
100( s − 1) 2
=
( s + 5 3 − j 5)( s + 5 3 + j 5)( s + 2) 1 1 1 = 72.58 + j 225.2 + 72.58 − j 225.2 + 12.976
s + 5 3 − j 5
s + 5 3 + j5 s+2 C A B A=
100( s − 1) 2 ( s + 5 3 + j 5)( s + 2) s =−5
= 3+ j5
100(−5 3 − 1 + 5 j ) 2 (10 j )(−5 3 + 2 + 5 j )
=
10(−5 3 − 1 + 5 j ) 2 j (−5 3 + 2) − 5
= 43.512 + j135.24
B = A∗ = 43.512 − j135.24 C=
100( s − 1) 2 ( s + 10 3s + 100) s =−2 2
=
900
=
(4 − 20 3 + 100)
900 = 12.976 69.359
Using the table and simplifying, we find the following impulse response: ROC
Re{s} < −1 : h(t ) = (43.512 + j135.24)e( −5 = 2e−5
3t
3 +5 j )t
{
u (t ) + (43.512 − j135.24)e( −5
3 −5 j ) t
}
Re (43.512 + j135.24)e j 5t u (t ) + 12.976e −2t u (t )
= 2(142.07) e−5 = 284.14 e−5
3t
3t
cos(5t + 1.2595)u (t ) + 12.976e −2t u (t )
cos(5t + 1.2595)u (t ) + 12.976e −2t u (t )
= 2 e−5 3t [43.512 cos(5t ) − 135.24sin(5t )]u (t ) + 12.976e −2t u (t ) 100
80
60
40
20
0
-20
-40
0
0.5
1
1.5
2
2.5
Fourier transform of h(t ) is H ( jω ) :
H ( jω ) =
( jω ) 2 − 2 jω + 1 (0.01( jω ) 2 + 0.1 3 jω + 1)( jω + 2) 2
3
3.5
4
4.5
5
u (t ) + 12.976e−2t u (t )
Sample Final Exam Covering Chapters 1-9 (finals00)
(c) [4 marks] Give the direct form realization (block diagram) of H ( s) . Answer:
100 −200
+ X ( s)
-
3
s W ( s)
-
1 s
-
sW ( s)
1 s
s 2W ( s)
1 s
W ( s)
+ 100
+
+
Y ( s)
10 3 + 2 20 3 + 100
200
Problem 2 (5 marks) True or False? (a) The Fourier transform X ( jω ) of the product of a real signal x ( t ) and an impulse real. Answer: False.
δ (t − 1)
is
(b) The system defined by y (t ) = x(t ) is time-invariant. Answer: False. 2
(c) The Fourier series coefficients a k of a purely imaginary periodic signal x ( t ) have the following ∗
property: ak = − a− k . Answer: True. (d) The causal linear discrete-time system defined by y[ n − 2] + 2 y[ n − 1] + y[ n] = x[ n] is stable. Answer: False. (e) The signal x[ n] = e Answer: True.
2 j n 5
is not periodic.
Problem 3 (15 marks) Consider the radar system depicted below where the radar's emitting antenna emits a pulse
x(t ) = e −100t sin(103 t )u (t ) that gets reflected by a boat and is received by the radar's receiving 3
Sample Final Exam Covering Chapters 1-9 (finals00)
antenna as xr (t ) = 0.1e
100 t0
e −100t sin(103 t − 103 t0 )u (t − t0 ) . An LTI filter H ( jω ) processes xr (t )
(
to generate another pulse y (t ) = e
−10( t −t0 )
)
+ e −20( t −t0 ) u (t − t0 ) that is used to measure the time of
arrival of the received pulse.
x (t )
x r (t )
y (t )
H ( jω ) t (a) [10 marks] Find the frequency response H ( jω ) of the filter. Is the filter BIBO stable? Justify your answer. Write the causal LTI differential equation that would implement this filter. Answer: We have X r ( jω ) H ( jω ) = Y ( jω ) , where
X r ( jω ) = 0.1X ( jω )e− jω t0 = X ( jω ) =
100e − jωt0 ( jω + 100) 2 + (103 ) 2
103 ( jω + 100) 2 + (103 ) 2
and
1 1 e − jωt0 (2 jω + 30) + = Y ( jω ) = e − jωt0 jω + 10 jω + 20 ( jω + 10)( jω + 20) Hence
e − jω t0 (2 jω + 30) (2 jω + 30) ( jω + 100) 2 + (103 ) 2 Y ( jω ) ( jω + 10)( jω + 20) H ( jω ) = = = X r ( jω ) 100( jω + 10)( jω + 20) e − jωt0 100 2 3 2 ( jω + 100) + (10 ) This filter is NOT BIBO stable because it is not proper (degree of num>degree of denom.) Differential equation:
4
Sample Final Exam Covering Chapters 1-9 (finals00)
100
d 2 y (t ) dy (t ) d 3 x(t ) d 2 x(t ) dx(t ) + 3000 + 20000 y ( t ) = 2 + 430 + 2026000 + 30300000 x(t ) 2 3 2 dt dt dt dt dt
(b) [5 marks] Knowing that electromagnetic waves propagate at the speed of light c = 3 × 10 m/s , and that t0 = 10 µ s , how far is the boat from the radar? 8
Answer: The time delay results from the pulse having traveled twice the distance d between the boat and the radar, thus
ct0 3 × 108 ms ⋅ 10−5 s d= = = 1500m 2 2 Problem 4 (20 marks) Consider the causal first-order circuit initially at rest depicted below. It can be used to implement a first-order lead or lag, depending on the values of its components. Its ideal circuit model with a voltage-controlled source is also given below. (a) [8 marks] Transform the ideal circuit using the Laplace transform, and use nodal analysis to find the transfer function H A ( s ) = Vout ( s ) Vin ( s ) . Then, let A → +∞ to obtain the ideal transfer function
H ( s ) = lim H A ( s ) . Note that the second op-amp stage is just an inverter such that A→+∞
vout (t ) = −v y (t ) .
C2 C1
v x (t )
vin ( t )
100kΩ
R2
100kΩ
-
R1
-
+
v y (t )
+
vout ( t )
C2 C1
vin ( t )
+
R1
R2
v x (t )
+ -
-
Answer:
5
v y (t ) = − Av x (t )
Sample Final Exam Covering Chapters 1-9 (finals00) The transformed circuit is 1 C2 s 1 C1 s
Vin ( s )
+
R2
Vx ( s )
R1
+ -
-
− AVx ( s )
There are two supernodes for which the nodal voltages are given by the source voltages. The remaining nodal equation is
Vin ( s ) − Vx ( s ) R
1 1 C1s
where the notation R1
1 C1s
=
+
− AVx ( s ) − Vx ( s ) R1
1 C1s
=0
R1 denotes the equivalent impedance of the parallel connection of R1C1s + 1
the resistor and capacitor. Simplifying the above equation, we get:
( A + 1)( R2 C2 s + 1) R1C1 s + 1 R1C1 s + 1 + Vin ( s ) − Vx ( s ) = 0 R1 R2 R1 Thus, the transfer function between the input voltage and the node voltage is given by
Vx ( s ) = Vin ( s )
R1C1 s + 1
( A + 1) R1 R1 [( A + 1)C2 + C1 ]s + +1 R2
.
The transfer function between the input voltage and the output voltage is
H A ( s) =
Vout ( s ) − AVx ( s ) =− = Vin ( s ) Vin ( s )
A( R1C1 s + 1) ( A + 1) R1 R1 [( A + 1)C2 + C1 ]s + +1 R2
The ideal transfer function is the limit as the op-amp gain tends to infinity:
H ( s ) = lim H A ( s ) = A→∞
R2 ( R1C1 s + 1) R1 ( R2 C2 s + 1)
(b) [7 marks] Suppose that the circuit is used as an equalizer in the following communication system and R1 = R2 = 1k Ω . A rectangular pulse is transmitted over an LTI communication channel with
1 . The pulse is distorted at the receiving end of the channel, and jω + 1 you are asked to design (i.e., specify the capacitances C1 , C2 of the capacitors) a first-order LTI frequency response G ( jω ) =
equalizer that would reshape the pulse. The specification is that the magnitude of the combined frequency response of the channel and the equalizer H ( jω )G ( jω ) must have a DC gain of 0dB and a -3dB bandwidth of 100 radians/s. Use Bode plots (magnitude only) to guide your designs. 6
Sample Final Exam Covering Chapters 1-9 (finals00)
Vm ( jω )
Vin ( jω )
G ( jω )
communication channel
H ( jω )
Vout ( jω )
equalizer
Answer: Bode plot of
G ( jω ) =
1 and desired H ( jω ) : jω + 1 20 log10 H ( jω )
(dB)40
0
1
100
ω (log)
20 log10 G ( jω )
20 log10 G ( jω ) H ( jω )
-20dB/dec The circuit must be a first-order lead with a transfer function of the form
H (s) = where
ατ s + 1 , τ s +1
τ = 0.01 and α = 100 . Frequency response:
jω + 1 0.01 jω + 1 There are two break frequencies for the Bode plot: ω1 = 1, ω 2 = 100 . The DC gain is 0dB. Identifying (1000C1 s + 1) with the parameters of H ( s ) = , we find for our design C1 = 1000 µ F , C2 = 10 µ F . (1000C2 s + 1) H ( jω ) =
(c) [5 marks] Compute the response vout ( t ) of the equalizer circuit to a 1-Volt, 2-second pulse
vm (t ) = u (t ) − u (t − 2) transmitted through the communication channel. Compute the energy in the residual distortion error e(t ) := vm (t ) − vout (t ) . Answer:
Vout ( s ) = H ( s )G ( s )Vm ( s ) =
1 e −2 s 1 −2 s 1 − = − −e s (0.01s + 1) s (0.01s + 1) s s + 100
Taking the inverse transform, we get
(
)
(
)
vout (t ) = 1 − e −100t u (t ) − 1 − e −100(t −2) u (t − 2) the energy in the residual distortion error is given by 7
1 1 s − s + 100
Sample Final Exam Covering Chapters 1-9 (finals00)
+∞
∫
2
e(t ) dt =
−∞
+∞
∫ 0
2
2
+∞
0
2
−e −100t + e −100( t −2) u (t − 2) dt = ∫ e −200t dt +
∫
2
e 200 − 1 e−200t dt
2 2 +∞ 1 1 200 e −200t − e − 1 e −200t 0 2 200 200 2 1 1 200 e −400 − 1 − =− e − 1 0 − e −400 200 200 2 1 1 e −400 − 1 + =− 1 − e −200 200 200 1 1 −200 ≅ 0.01 = − e 100 100
=−
or using Parseval's relation: +∞
∫
−∞
1 e(t ) dt = 2π
1 = 2π
2
+∞
∫
−∞
2
1 − e − jω 2 1 dω = 0.01 jω + 1 2π
+∞
+∞
∫
(1 − cos 2ω ) 2 + ( sin 2ω )
−∞
0.0001ω 2 + 1
2
dω
+∞
2 − 2 cos 2ω 1 1 − cos 2ω ∫−∞ 0.0001ω 2 + 1 dω = π −∞∫ 0.0001ω 2 + 1 dω
Problem 5 (10 marks) The following circuit with a Zener diode is an ideal clamping circuit. The input voltage is shown below
vin (t ) = e
−t
Volts
t vin (t ), vin (t ) < 1 . vin (t ) ≥ 1 1,
and the output voltage is v (t ) =
+
+
-
vin ( t )
+ -
Vz1=1 V
v (t ) -
[10 marks] Sketch the output voltage v ( t ) and compute its the Fourier transform V ( jω ) . 8
Sample Final Exam Covering Chapters 1-9 (finals00) Answer: The circuit doesn't saturate!
v (t )
1 1
1 V ( jω ) =
∞
∫ v(t )e
− jω t
dt
−∞ 0
=
∫
+∞
∫e
et e − jωt dt +
−∞
dt
− (1+ jω ) t
dt
e
0
0
=
− t − jω t
∫e
(1− jω ) t
+∞
∫e
dt +
−∞
0
0 +∞ 1 1 e(1− jω ) t − e − (1+ jω )t 0 −∞ 1 − jω 1 + jω 1 1 = + 1 − jω 1 + jω 2 = 1+ω2
=
Problem 6 (20 marks) −
−
The following circuit has initial conditions on the capacitor vC ( 0 ) and inductor i L (0 ) .
C
vs (t )
L
+
+
R1
-
v (t )
R2
-
(a) [4 marks] Transform the circuit using the unilateral Laplace transform. Answer:
-
Li L ( 0 − )
Ls
+-
-
Vs ( s)
+
1 vc ( 0 − ) s
R1
R2
I2 ( s)
I1 ( s) 9
+
+
1 Cs
V( s)
-
Sample Final Exam Covering Chapters 1-9 (finals00)
(b) [8 marks] Find the unilateral Laplace transform V ( s) 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) + LiL (0− ) = 0 ⇒ I1 ( s) =
1 L ( R1 + R2 + Ls )I2 ( s) − iL (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
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 ) = R2 I2 ( s ) =
R1 R2 CsVs ( s ) (1 + R1Cs ) LR2 iL (0 − ) − R1 R2 Cvc (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 = 1 Ω, L = 0.035425 H, C = 0.564576 F . What type of filter is it (lowpass, bandpass or highpass)?
R1 R2 C s R1 R2 Cs R1 + R2 V( s) H ( s) = = = LR1C 2 ( L + R1 R2 C ) Vs ( s ) LR1Cs 2 + ( L + R1 R2 C ) s + R1 + R2 s + s +1 R1 + R2 R1 + R2 Frequency response:
10
Sample Final Exam Covering Chapters 1-9 (finals00)
R1 R2 C jω R1 + R2 0.2823 jω H ( jω ) = = LR1C ( L + R1 R2 C ) 0.01( jω ) 2 + 0.3( jω ) + 1 ( jω ) 2 + ( jω ) + 1 R1 + R2 R1 + R2 from which we find
ω n = 10rd/s, ζ = 1.5
and the poles are
p1 = −15 + 10 (1.5) 2 − 1 = −3.82 and
p1 = −15 − 10 (1.5) 2 − 1 = −26.18 . Gain is -11dB. Bode plot:
(dB)
20 log 10 H ( jω )
20 10-1
100
3.8 101 26 3.5
ω (log)
102
-20 -40 -60
(rd)
3π/4 π/2 10-1
∠H ( jω )
100 3.8 101 26
102
103
ω (log)
π/4 0 −π/4 −π/2 This is a bandpass filter. 11
Sample Final Exam Covering Chapters 1-9 (finals00)
Problem 7 (10 marks) Suppose we want to control the level of liquid in a tank that is part of a continuous chemical process. We can do this by varying the opening of the control valve based on a feedback measurement of the tank level. The control valve input signal x ( t ) is taken to be a delta variation of valve opening from its nominal opening. The output signal y (t ) is the tank level.
x ( t ): delta valve opening level sensor control valv e
y ( t ): level
tank
The open-loop tank dynamics are modeled by the following nonminimum phase transfer function
x (t )
y (t )
G ( s)
G ( s) =
−( s − 2) s( s + 1)
(a) [5 marks] Compute and sketch the unit step response valve opening x ( t ) .
Y (s) =
−( s − 2) A B C = + + 2 2 s ( s + 1) ( s + 1) s s
we find
Y (s) =
3 −3 2 + + ( s + 1) s s 2
inverse transform:
y (t ) = 3e− t u (t ) − 3u (t ) + 2tu (t )
12
y (t ) of the tank level for a unit step in delta
Sample Final Exam Covering Chapters 1-9 (finals00)
Plot: 10
8
6
y (t )
4
2
0
-2
-4
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
y (t ) perfectly. We want to test the feedback controller K ( s) = λ ( s + α ) to control the level. That is, we want the tank level y (t ) to track the desired level ydes (t ) .
(b) [5 marks] Assume that we can measure the tank level
y des ( t )
+ e (t ) -
y(t ) K ( s)
Find the closed-loop transfer function step response?
G ( s)
H ( s): = Y ( s) Ydes ( s) . What is the final value y (+∞) of the
Answer: Closed-loop transfer function:
13
Sample Final Exam Covering Chapters 1-9 (finals00)
K ( s)G ( s) 1 + K ( s )G ( s ) −λ ( s + α )( s − 2) s( s + 1) = −λ ( s + α )( s − 2) 1+ s( s + 1) −λ ( s + α )( s − 2) = s( s + 1) − λ ( s + α )( s − 2) −λ ( s + α )( s − 2) = 2 s + s − λ ( s 2 + (α − 2) s − 2α ) −λ ( s + α )( s − 2) = 2 (1 − λ ) s + [1 − λ (α − 2) ] s + 2αλ
H (s) =
With controller parameters
H (s) =
λ = 0.5
and
α = 3:
−0.5( s + 3)( s − 2) −( s + 3)( s − 2) = s2 + s + 6 0.5s 2 + 0.5s + 3
Unit step response: NOT REQUIRED!!!
Y (s) =
−( s + 3)( s − 2) s( s 2 + s + 6)
A B( s + 0.5) + C 6 + s s2 + s + 6 A B( s + 0.5) + C 6 = + s 23 2 ( s + 0.5)2 + ( ) 2 ⇔ =
23 = 2.398, 2
− s 2 + s + 6 = ( s 2 + s + 6) + Bs( s + 0.5) + Cs 6 ⇔ B = −2,
0.5(−2) + C 6 = 1 ⇔ C =
2 6
=
2 3
Hence,
y (t ) = u (t ) − 2e −0.5t cos(
23 2 −0.5t 23 t )u (t ) + e t )u (t ) sin( 2 3 2
and the final value is y (+∞) = H (0) = 1 .
END OF EXAMINATION 14
A =1
Sample Final Exam (finals00) Covering Chapters 1-9 of Fundamentals of Signals & Systems Problem 1 (20 marks) Consider the transfer function:
H (s) =
s 2 − 2s + 1 (0.01s 2 + 0.1 3s + 1)( s + 2)
.
(a) [6 marks] Find the poles and zeros of H ( s ) (specify how many there are at ∞ ). Give all possible regions of convergence of the transfer function H ( s) . Answer:
p1 = −2
z1 = 1,
Poles are p2 = −5 3 + 5 j = −8.66 + 5 j
Zeros are z2 = 1
z3 = ∞
p3 = −5 3 − 5 j = −8.66 − 5 j There are 3 possible ROCs: ROC1: Re{s} < −5 3 = −8.66 ROC2: −5 3 < Re{s} < −2 ROC3: Re{s} > −2
(b) [10 marks] Give the region of convergence of H ( s ) that corresponds to the impulse response of a stable system, and sketch it on a pole-zero plot. Is the stable system causal? Explain. Compute the impulse response h( t ) of this stable system and give its Fourier transform H ( jω ) . Answer: System is stable for the ROC that contains the jω -axis: ROC3. This system is also causal as ROC3 is an open RHP and the transfer function is rational.
Im{s} 5 -8.66
-2
1 -5
The partial fraction expansion of H ( s) yields: 1
ROC3
Re{s}
Sample Final Exam Covering Chapters 1-9 (finals00)
H (s) =
s 2 − 2s + 1 (0.01s 2 + 0.1 3s + 1)( s + 2)
=
100( s 2 − 2s + 1) ( s 2 + 10 3s + 100)( s + 2)
100( s − 1) 2
=
( s + 5 3 − j 5)( s + 5 3 + j 5)( s + 2) 1 1 1 = 72.58 + j 225.2 + 72.58 − j 225.2 + 12.976
s + 5 3 − j 5
s + 5 3 + j5 s+2 C A B A=
100( s − 1) 2 ( s + 5 3 + j 5)( s + 2) s =−5
= 3+ j5
100(−5 3 − 1 + 5 j ) 2 (10 j )(−5 3 + 2 + 5 j )
=
10(−5 3 − 1 + 5 j ) 2 j (−5 3 + 2) − 5
= 43.512 + j135.24
B = A∗ = 43.512 − j135.24 C=
100( s − 1) 2 ( s + 10 3s + 100) s =−2 2
=
900
=
(4 − 20 3 + 100)
900 = 12.976 69.359
Using the table and simplifying, we find the following impulse response: ROC
Re{s} < −1 : h(t ) = (43.512 + j135.24)e( −5 = 2e−5
3t
3 +5 j )t
{
u (t ) + (43.512 − j135.24)e( −5
3 −5 j ) t
}
Re (43.512 + j135.24)e j 5t u (t ) + 12.976e −2t u (t )
= 2(142.07) e−5 = 284.14 e−5
3t
3t
cos(5t + 1.2595)u (t ) + 12.976e −2t u (t )
cos(5t + 1.2595)u (t ) + 12.976e −2t u (t )
= 2 e−5 3t [43.512 cos(5t ) − 135.24sin(5t )]u (t ) + 12.976e −2t u (t ) 100
80
60
40
20
0
-20
-40
0
0.5
1
1.5
2
2.5
Fourier transform of h(t ) is H ( jω ) :
H ( jω ) =
( jω ) 2 − 2 jω + 1 (0.01( jω ) 2 + 0.1 3 jω + 1)( jω + 2) 2
3
3.5
4
4.5
5
u (t ) + 12.976e−2t u (t )
Sample Final Exam Covering Chapters 1-9 (finals00)
(c) [4 marks] Give the direct form realization (block diagram) of H ( s) . Answer:
100 −200
+ X ( s)
-
3
s W ( s)
-
1 s
-
sW ( s)
1 s
s 2W ( s)
1 s
W ( s)
+ 100
+
+
Y ( s)
10 3 + 2 20 3 + 100
200
Problem 2 (5 marks) True or False? (a) The Fourier transform X ( jω ) of the product of a real signal x ( t ) and an impulse real. Answer: False.
δ (t − 1)
is
(b) The system defined by y (t ) = x(t ) is time-invariant. Answer: False. 2
(c) The Fourier series coefficients a k of a purely imaginary periodic signal x ( t ) have the following ∗
property: ak = − a− k . Answer: True. (d) The causal linear discrete-time system defined by y[ n − 2] + 2 y[ n − 1] + y[ n] = x[ n] is stable. Answer: False. (e) The signal x[ n] = e Answer: True.
2 j n 5
is not periodic.
Problem 3 (15 marks) Consider the radar system depicted below where the radar's emitting antenna emits a pulse
x(t ) = e −100t sin(103 t )u (t ) that gets reflected by a boat and is received by the radar's receiving 3
Sample Final Exam Covering Chapters 1-9 (finals00)
antenna as xr (t ) = 0.1e
100 t0
e −100t sin(103 t − 103 t0 )u (t − t0 ) . An LTI filter H ( jω ) processes xr (t )
(
to generate another pulse y (t ) = e
−10( t −t0 )
)
+ e −20( t −t0 ) u (t − t0 ) that is used to measure the time of
arrival of the received pulse.
x (t )
x r (t )
y (t )
H ( jω ) t (a) [10 marks] Find the frequency response H ( jω ) of the filter. Is the filter BIBO stable? Justify your answer. Write the causal LTI differential equation that would implement this filter. Answer: We have X r ( jω ) H ( jω ) = Y ( jω ) , where
X r ( jω ) = 0.1X ( jω )e− jω t0 = X ( jω ) =
100e − jωt0 ( jω + 100) 2 + (103 ) 2
103 ( jω + 100) 2 + (103 ) 2
and
1 1 e − jωt0 (2 jω + 30) + = Y ( jω ) = e − jωt0 jω + 10 jω + 20 ( jω + 10)( jω + 20) Hence
e − jω t0 (2 jω + 30) (2 jω + 30) ( jω + 100) 2 + (103 ) 2 Y ( jω ) ( jω + 10)( jω + 20) H ( jω ) = = = X r ( jω ) 100( jω + 10)( jω + 20) e − jωt0 100 2 3 2 ( jω + 100) + (10 ) This filter is NOT BIBO stable because it is not proper (degree of num>degree of denom.) Differential equation:
4
Sample Final Exam Covering Chapters 1-9 (finals00)
100
d 2 y (t ) dy (t ) d 3 x(t ) d 2 x(t ) dx(t ) + 3000 + 20000 y ( t ) = 2 + 430 + 2026000 + 30300000 x(t ) 2 3 2 dt dt dt dt dt
(b) [5 marks] Knowing that electromagnetic waves propagate at the speed of light c = 3 × 10 m/s , and that t0 = 10 µ s , how far is the boat from the radar? 8
Answer: The time delay results from the pulse having traveled twice the distance d between the boat and the radar, thus
ct0 3 × 108 ms ⋅ 10−5 s d= = = 1500m 2 2 Problem 4 (20 marks) Consider the causal first-order circuit initially at rest depicted below. It can be used to implement a first-order lead or lag, depending on the values of its components. Its ideal circuit model with a voltage-controlled source is also given below. (a) [8 marks] Transform the ideal circuit using the Laplace transform, and use nodal analysis to find the transfer function H A ( s ) = Vout ( s ) Vin ( s ) . Then, let A → +∞ to obtain the ideal transfer function
H ( s ) = lim H A ( s ) . Note that the second op-amp stage is just an inverter such that A→+∞
vout (t ) = −v y (t ) .
C2 C1
v x (t )
vin ( t )
100kΩ
R2
100kΩ
-
R1
-
+
v y (t )
+
vout ( t )
C2 C1
vin ( t )
+
R1
R2
v x (t )
+ -
-
Answer:
5
v y (t ) = − Av x (t )
Sample Final Exam Covering Chapters 1-9 (finals00) The transformed circuit is 1 C2 s 1 C1 s
Vin ( s )
+
R2
Vx ( s )
R1
+ -
-
− AVx ( s )
There are two supernodes for which the nodal voltages are given by the source voltages. The remaining nodal equation is
Vin ( s ) − Vx ( s ) R
1 1 C1s
where the notation R1
1 C1s
=
+
− AVx ( s ) − Vx ( s ) R1
1 C1s
=0
R1 denotes the equivalent impedance of the parallel connection of R1C1s + 1
the resistor and capacitor. Simplifying the above equation, we get:
( A + 1)( R2 C2 s + 1) R1C1 s + 1 R1C1 s + 1 + Vin ( s ) − Vx ( s ) = 0 R1 R2 R1 Thus, the transfer function between the input voltage and the node voltage is given by
Vx ( s ) = Vin ( s )
R1C1 s + 1
( A + 1) R1 R1 [( A + 1)C2 + C1 ]s + +1 R2
.
The transfer function between the input voltage and the output voltage is
H A ( s) =
Vout ( s ) − AVx ( s ) =− = Vin ( s ) Vin ( s )
A( R1C1 s + 1) ( A + 1) R1 R1 [( A + 1)C2 + C1 ]s + +1 R2
The ideal transfer function is the limit as the op-amp gain tends to infinity:
H ( s ) = lim H A ( s ) = A→∞
R2 ( R1C1 s + 1) R1 ( R2 C2 s + 1)
(b) [7 marks] Suppose that the circuit is used as an equalizer in the following communication system and R1 = R2 = 1k Ω . A rectangular pulse is transmitted over an LTI communication channel with
1 . The pulse is distorted at the receiving end of the channel, and jω + 1 you are asked to design (i.e., specify the capacitances C1 , C2 of the capacitors) a first-order LTI frequency response G ( jω ) =
equalizer that would reshape the pulse. The specification is that the magnitude of the combined frequency response of the channel and the equalizer H ( jω )G ( jω ) must have a DC gain of 0dB and a -3dB bandwidth of 100 radians/s. Use Bode plots (magnitude only) to guide your designs. 6
Sample Final Exam Covering Chapters 1-9 (finals00)
Vm ( jω )
Vin ( jω )
G ( jω )
communication channel
H ( jω )
Vout ( jω )
equalizer
Answer: Bode plot of
G ( jω ) =
1 and desired H ( jω ) : jω + 1 20 log10 H ( jω )
(dB)40
0
1
100
ω (log)
20 log10 G ( jω )
20 log10 G ( jω ) H ( jω )
-20dB/dec The circuit must be a first-order lead with a transfer function of the form
H (s) = where
ατ s + 1 , τ s +1
τ = 0.01 and α = 100 . Frequency response:
jω + 1 0.01 jω + 1 There are two break frequencies for the Bode plot: ω1 = 1, ω 2 = 100 . The DC gain is 0dB. Identifying (1000C1 s + 1) with the parameters of H ( s ) = , we find for our design C1 = 1000 µ F , C2 = 10 µ F . (1000C2 s + 1) H ( jω ) =
(c) [5 marks] Compute the response vout ( t ) of the equalizer circuit to a 1-Volt, 2-second pulse
vm (t ) = u (t ) − u (t − 2) transmitted through the communication channel. Compute the energy in the residual distortion error e(t ) := vm (t ) − vout (t ) . Answer:
Vout ( s ) = H ( s )G ( s )Vm ( s ) =
1 e −2 s 1 −2 s 1 − = − −e s (0.01s + 1) s (0.01s + 1) s s + 100
Taking the inverse transform, we get
(
)
(
)
vout (t ) = 1 − e −100t u (t ) − 1 − e −100(t −2) u (t − 2) the energy in the residual distortion error is given by 7
1 1 s − s + 100
Sample Final Exam Covering Chapters 1-9 (finals00)
+∞
∫
2
e(t ) dt =
−∞
+∞
∫ 0
2
2
+∞
0
2
−e −100t + e −100( t −2) u (t − 2) dt = ∫ e −200t dt +
∫
2
e 200 − 1 e−200t dt
2 2 +∞ 1 1 200 e −200t − e − 1 e −200t 0 2 200 200 2 1 1 200 e −400 − 1 − =− e − 1 0 − e −400 200 200 2 1 1 e −400 − 1 + =− 1 − e −200 200 200 1 1 −200 ≅ 0.01 = − e 100 100
=−
or using Parseval's relation: +∞
∫
−∞
1 e(t ) dt = 2π
1 = 2π
2
+∞
∫
−∞
2
1 − e − jω 2 1 dω = 0.01 jω + 1 2π
+∞
+∞
∫
(1 − cos 2ω ) 2 + ( sin 2ω )
−∞
0.0001ω 2 + 1
2
dω
+∞
2 − 2 cos 2ω 1 1 − cos 2ω ∫−∞ 0.0001ω 2 + 1 dω = π −∞∫ 0.0001ω 2 + 1 dω
Problem 5 (10 marks) The following circuit with a Zener diode is an ideal clamping circuit. The input voltage is shown below
vin (t ) = e
−t
Volts
t vin (t ), vin (t ) < 1 . vin (t ) ≥ 1 1,
and the output voltage is v (t ) =
+
+
-
vin ( t )
+ -
Vz1=1 V
v (t ) -
[10 marks] Sketch the output voltage v ( t ) and compute its the Fourier transform V ( jω ) . 8
Sample Final Exam Covering Chapters 1-9 (finals00) Answer: The circuit doesn't saturate!
v (t )
1 1
1 V ( jω ) =
∞
∫ v(t )e
− jω t
dt
−∞ 0
=
∫
+∞
∫e
et e − jωt dt +
−∞
dt
− (1+ jω ) t
dt
e
0
0
=
− t − jω t
∫e
(1− jω ) t
+∞
∫e
dt +
−∞
0
0 +∞ 1 1 e(1− jω ) t − e − (1+ jω )t 0 −∞ 1 − jω 1 + jω 1 1 = + 1 − jω 1 + jω 2 = 1+ω2
=
Problem 6 (20 marks) −
−
The following circuit has initial conditions on the capacitor vC ( 0 ) and inductor i L (0 ) .
C
vs (t )
L
+
+
R1
-
v (t )
R2
-
(a) [4 marks] Transform the circuit using the unilateral Laplace transform. Answer:
-
Li L ( 0 − )
Ls
+-
-
Vs ( s)
+
1 vc ( 0 − ) s
R1
R2
I2 ( s)
I1 ( s) 9
+
+
1 Cs
V( s)
-
Sample Final Exam Covering Chapters 1-9 (finals00)
(b) [8 marks] Find the unilateral Laplace transform V ( s) 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) + LiL (0− ) = 0 ⇒ I1 ( s) =
1 L ( R1 + R2 + Ls )I2 ( s) − iL (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
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 ) = R2 I2 ( s ) =
R1 R2 CsVs ( s ) (1 + R1Cs ) LR2 iL (0 − ) − R1 R2 Cvc (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 = 1 Ω, L = 0.035425 H, C = 0.564576 F . What type of filter is it (lowpass, bandpass or highpass)?
R1 R2 C s R1 R2 Cs R1 + R2 V( s) H ( s) = = = LR1C 2 ( L + R1 R2 C ) Vs ( s ) LR1Cs 2 + ( L + R1 R2 C ) s + R1 + R2 s + s +1 R1 + R2 R1 + R2 Frequency response:
10
Sample Final Exam Covering Chapters 1-9 (finals00)
R1 R2 C jω R1 + R2 0.2823 jω H ( jω ) = = LR1C ( L + R1 R2 C ) 0.01( jω ) 2 + 0.3( jω ) + 1 ( jω ) 2 + ( jω ) + 1 R1 + R2 R1 + R2 from which we find
ω n = 10rd/s, ζ = 1.5
and the poles are
p1 = −15 + 10 (1.5) 2 − 1 = −3.82 and
p1 = −15 − 10 (1.5) 2 − 1 = −26.18 . Gain is -11dB. Bode plot:
(dB)
20 log 10 H ( jω )
20 10-1
100
3.8 101 26 3.5
ω (log)
102
-20 -40 -60
(rd)
3π/4 π/2 10-1
∠H ( jω )
100 3.8 101 26
102
103
ω (log)
π/4 0 −π/4 −π/2 This is a bandpass filter. 11
Sample Final Exam Covering Chapters 1-9 (finals00)
Problem 7 (10 marks) Suppose we want to control the level of liquid in a tank that is part of a continuous chemical process. We can do this by varying the opening of the control valve based on a feedback measurement of the tank level. The control valve input signal x ( t ) is taken to be a delta variation of valve opening from its nominal opening. The output signal y (t ) is the tank level.
x ( t ): delta valve opening level sensor control valv e
y ( t ): level
tank
The open-loop tank dynamics are modeled by the following nonminimum phase transfer function
x (t )
y (t )
G ( s)
G ( s) =
−( s − 2) s( s + 1)
(a) [5 marks] Compute and sketch the unit step response valve opening x ( t ) .
Y (s) =
−( s − 2) A B C = + + 2 2 s ( s + 1) ( s + 1) s s
we find
Y (s) =
3 −3 2 + + ( s + 1) s s 2
inverse transform:
y (t ) = 3e− t u (t ) − 3u (t ) + 2tu (t )
12
y (t ) of the tank level for a unit step in delta
Sample Final Exam Covering Chapters 1-9 (finals00)
Plot: 10
8
6
y (t )
4
2
0
-2
-4
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
y (t ) perfectly. We want to test the feedback controller K ( s) = λ ( s + α ) to control the level. That is, we want the tank level y (t ) to track the desired level ydes (t ) .
(b) [5 marks] Assume that we can measure the tank level
y des ( t )
+ e (t ) -
y(t ) K ( s)
Find the closed-loop transfer function step response?
G ( s)
H ( s): = Y ( s) Ydes ( s) . What is the final value y (+∞) of the
Answer: Closed-loop transfer function:
13
Sample Final Exam Covering Chapters 1-9 (finals00)
K ( s)G ( s) 1 + K ( s )G ( s ) −λ ( s + α )( s − 2) s( s + 1) = −λ ( s + α )( s − 2) 1+ s( s + 1) −λ ( s + α )( s − 2) = s( s + 1) − λ ( s + α )( s − 2) −λ ( s + α )( s − 2) = 2 s + s − λ ( s 2 + (α − 2) s − 2α ) −λ ( s + α )( s − 2) = 2 (1 − λ ) s + [1 − λ (α − 2) ] s + 2αλ
H (s) =
With controller parameters
H (s) =
λ = 0.5
and
α = 3:
−0.5( s + 3)( s − 2) −( s + 3)( s − 2) = s2 + s + 6 0.5s 2 + 0.5s + 3
Unit step response: NOT REQUIRED!!!
Y (s) =
−( s + 3)( s − 2) s( s 2 + s + 6)
A B( s + 0.5) + C 6 + s s2 + s + 6 A B( s + 0.5) + C 6 = + s 23 2 ( s + 0.5)2 + ( ) 2 ⇔ =
23 = 2.398, 2
− s 2 + s + 6 = ( s 2 + s + 6) + Bs( s + 0.5) + Cs 6 ⇔ B = −2,
0.5(−2) + C 6 = 1 ⇔ C =
2 6
=
2 3
Hence,
y (t ) = u (t ) − 2e −0.5t cos(
23 2 −0.5t 23 t )u (t ) + e t )u (t ) sin( 2 3 2
and the final value is y (+∞) = H (0) = 1 .
END OF EXAMINATION 14
A =1
