M. J. Roberts - 8/17/04
Chapter 10 -
Laplace Transform Analysis of
Signals and Systems
Selected Solutions
1. For each circuit write the transfer function between the indicated excitation and indicated
response. Express each transfer function in the standard form,
sN + bN −1sN −1 + L + b2 s2 + b1s + b0
.
H( s) = A D
s + aD −1sD −1 + L + a2 s2 + a1s + a0
(a)
Excitation: v s ( t)
Response: v o ( t)
R1
L
+
vs(t)
C
R2
vo(t)
-
H( s) =
R2
Z RLC ( s)
Z RLC ( s) + R1 sL + R2
where
(R2 + sL) sC
R
s+ 2
sL + R2
1
L
=
= 2
Z RLC ( s) =
R
1
1
+
+
s
LC
sR
C
C
1
2
2
2
s +s +
R2 + sL +
L LC
sC
1
Combine and simplify, to get
H (s) =
(b)
Excitation: is ( t)
R2
R1 LC
1
 1
R  R + R1
s2 + s 
+ 2 + 2
R1 LC
 R1C L 
Response: v o ( t)
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M. J. Roberts - 8/17/04
C2
R2
R1
+
i s(t)
C1
vo(t)
-
Use the fact that the voltage at both op-amp inputs must be zero and the fact that no
current can flow into the op-amp input terminals. Find the current t flowing toward
the op-amp through resistor, R1 , by current division. All that current must flow
through the feedback network.
(c)
Excitation: v s ( t)
Response: i1 ( t)
C2
R1
i1(t)
C1
vs(t)
R2
2. For each block diagram write the transfer function between the excitation , x( t) , and the
response, y( t) .
(a)
x(t)
∫
∫
∫
8
2
s2 Y( s) =
H( s) =
X( s)
− [8 s Y( s) + 2 Y( s)]
s
Y( s)
1
1
=
= 3
2
X( s) s( s + 8 s + 2) s + 8 s2 + 2 s
10-2
y(t)
M. J. Roberts - 8/17/04
(b)
x(t)
-1
∫
∫
-4
∫
y(t)
-10
3. Evaluate the stability of each of these system transfer functions.
100
s + 200
(a)
H( s) = −
(b)
H( s) =
80
s− 4
Pole at s = 4. Unstable.
(c)
H( s) =
6
s( s + 1)
Poles at s = 0 and s = −1. Marginally stable
(therefore unstable)
(d)
H( s) = −
15 s
s + 4s + 4
(e)
H( s) = 3
s − 10
s + 4 s + 29
(f)
s2 + 4
H( s) = 3 2
s − 4 s + 29
(g)
H( s) =
1
s + 64
(h)
H( s) =
10
s + 4 s2 + 29 s
2
Double pole at s = −2. Stable.
2
2
3
4. Find the overall transfer functions of these systems in the form of a single ratio of
polynomials in s.
(a)
X(s)
s2
10
s + 3s + 2
2
s2 + 3s + 2
10-3
Y(s)
M. J. Roberts - 8/17/04
Multiply transfer functions.
(b)
s+1
s 2 + 2s + 13
X(s)
Y(s)
1
s + 10
Add transfer functions.
(c)
s
X(s)
2
s +s+5
Y(s)
Use feedback formula.
(d)
20s
s + 200s + 290000
X(s)
2
Y(s)
1
s + 400
Use feedback formula.
5. In the feedback system below, find the overall system transfer function for these values
of forward-path gain, K.
K
H( s) =
1 + 0.1K
(a)
K = 10 6
(b)
K = 10 5
(c)
K = 10
(d)
K =1
(e)
K = −1
(f)
K = −10
H( s) =
10 6
≅ 10
1 + 10 5
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M. J. Roberts - 8/17/04
X(s)
Y(s)
K
0.1
6. In the feedback system below, plot the response of the system to a unit step, for the time
period, 0 < t < 10 , then write the expression for the overall system transfer function and
draw a pole-zero diagram, for these values of K.
X(s)
Y(s)
K
0.1
e -s
One-second
time delay
H( s) =
(a)
K = 20
H( s) = 20
K
1 + 0.1e − sK
1
1 + 2e − s
This is not the usual ratio of polynomials in s so we cannot find the inverse
transform in the usual way. Synthetically divide the denominator into the numerator
to yield the infinite series,
H( s) = 20(1 − 2e − s + 4 e −2 s − 8e −3 s + L)
Then inverse transform term-by-term.
h( t) = 20[δ ( t) − 2δ ( t − 1) + 4δ ( t − 2) − 8δ ( t − 3) + L]
The step response is the integral of the impulse response.
h −1 ( t) = 20[u( t) − 2 u( t − 1) + 4 u( t − 2) − 8 u( t − 3) + L]
Poles at 1 + 2e − s = 0 . Solving for s,
1
2
−s
The term, e , is, in general, complex because s is complex. There are multiple
solutions for s in the complex plane. The logarithm of a complex number, z = re jθ ,
is
log ( z ) = ln ( r ) + j (θ + 2 nπ ) .
e− s = −
10-5
M. J. Roberts - 8/17/04
where “log” means the generalized complex log (base e) and “ln” means the
natural log (base e) of a real number, in this case, r.
s = log(−2) = ln(2) + jπ (2 n + 1)
-
Unstable
ω
h- 1(t)
8π
8000
-1
10
t
-3
3
σ
-8 π
-8000
(b)
K = 10
(c)
K =1
(e)
K = −10
(f)
K = −20
(d)
K = −1
7. For what range of values of K is the system below stable? Plot the step responses for
K = 0, K = 4 and K = 8.
1
s2 - 4s + 4
X(s)
Y(s)
Ks
Use the feedback formula to get the overall transfer function. Then solve for the
pole locations as a function of K and then see what values of K will put the poles in
the open left half-plane.
8. Plot the impulse response and the pole-zero diagram for the forward-path and the
overall system below.
100
X(s)
Y(s)
2
s + 2s + 26
10
s + 20
H1 ( s) = 100
1
1
= 100
⇒ h1 ( t) = 20e − t sin(5 t) u( t)
2
s + 2 s + 26
(s + 1) + 25
2
Poles at s = −1 ± j 5.
H( s) = 100
s + 20
s + 22 s + 66 s + 1520
3
2
Poles at s = −22.12, 0.0612 ± j 8.29 .
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M. J. Roberts - 8/17/04
When the denominator gets to third degree in s, finding the poles becomes tedious
unless there is an obvious factorization. The MATLAB function, residue, can be
used to find the poles and to help do the partial-fraction expansion.
H( s) = −


242.02
8.29
s − 0.0612
0.3785
+ 0.3785 
+

2
2
8.29 ( s − 0.0612) + 68.71 
s + 22.12
 ( s − 0.0612) + 68.71
{
}
h( t) = −0.3785e −22.12 t + 0.3785e 0.0612 t [cos(8.29 t) + 29.19 sin(8.29 t)] u( t)
9. Using the Routh-Hurwitz method, evaluate the stability of the system whose transfer
function is
s3 + 3s + 10
H( s) = 5
s + 2 s4 + 10 s3 + 4 s2 + 8 s + 20
5
4
3
1
10 8
2
4 20
−2 0
8
9
20 0
2
2
338
0 0
1 −
9
0
20
0 0
This result indicates that there are two poles in the RHP. That is confirmed by finding the
poles which lie at
s = −0.9432 + j 2.9107
s = −0.9432 − j 2.9107
s = 0.5502 + j1.207
s = 0.5502 − j1.207
s = −1.214
10. Using the Routh-Hurwitz stability test, evaluate the stability of the system whose
transfer function is of the general form,
H( s) =
N( s)
.
s + a2 s2 + a1s + a0
3
What are the relations among a2 , a1 and a0 that ensure stability?
3
2
1
a2
a0 − a1a2
1 −
a2
0
a0
10-7
a1
a0
0
0
M. J. Roberts - 8/17/04
−
a0 − a1a2
a2
For stability, a2 > 0 , −
a0 − a1a2
> 0 ⇒ a1a2 − a0 > 0 ⇒ a1a2 > a0 , a0 > 0 .
a2
11. Plot the root locus for each of the systems which have these loop transfer functions and
identify the transfer functions that are stable for all positive real values of K.
(a)
T( s) =
K
(s + 3)(s + 8)
Stable for positive K
ω
-3
σ
Ks
(s + 3)(s + 8)
(b)
T( s) =
(c)
Ks2
T( s) =
(s + 3)(s + 8)
(d)
T( s) =
K
(s + 1)(s + 4 s + 8)
2
12. Use the block diagram, of an inverting amplifier using an operational amplifier,
Vi (s)
Z f(s)
Z i(s) + Z f(s)
Ve(s)
-
A0
s
1- p
Z i(s)
Z i(s) + Z f(s)
Vo(s)
,
with A0 = 10 4 , p = −2000π , Z f = 10kΩ and Zi = 1kΩ, to find the gain and phase margins
of the amplifier.
0.5712 × 10 7
T( s) =
s + 6283
Finding the zero-dB frequency,
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M. J. Roberts - 8/17/04
T( jω 0 dB )
0.5712 × 10 7
ω
2
0 dB
+ (6283)
2
0.5712 × 10 7
=1
=
jω 0 dB + 6283
= 1 ⇒ ω 0 dB = 5.712 × 10 6 ⇒ f 0 dB = 909.1 kHz
At that frequency the phase of the loop transfer function is


π
0.5712 × 10 7
∠ T( s) = ∠
 = − or − 90°
6
2
 j 5.712 × 10 + 6283
So the phase margin is 90°. The phase approaches 90° as the frequency approaches
infinity. The loop transfer function magnitude approaches zero as the frequency
approaches infinity. The loop transfer function magnitude in dB therefore approaches
minus infinity. That means the gain margin is infinite.
13. Plot the unit step and ramp responses of unity-gain feedback systems with these
forward-path transfer functions.
(a)
H1 ( s) =
100
s + 10
H( s) =
100
s + 110
Unit step response:
100
10  1
1 
10
−110 t
H −1 ( s) =
=  −
 ⇒ h −1 ( t) = (1 − e
) u(t)
s( s + 110) 11  s s + 110 
11
Unit ramp response:
1
1 

100
10  1 110
10  1 − e −110 t 

110
H −2 ( s) = 2
=
−
+
⇒ h −2 ( t) =  t −
 u( t)
s ( s + 110) 11  s2
s
s + 110 
11 
110 


(b)
H1 ( s) =
100
s( s + 10)
(d)
H1 ( s) =
20
(s + 2)(s + 6)
(c)
H1 ( s) =
100
s ( s + 10)
2
14. Reduce these block diagrams to a single block by block-diagram reduction. Check the
answer using Mason’s theorem.
(a)
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M. J. Roberts - 8/17/04
-2
s+3
X(s)
10
s + 20
Y(s)
1
s
-2
s+3
X(s)
10s
s + 20s + 10
-20s
(s + 3) (s 2 + 20s + 10)
X(s)
Y(s)
2
Y(s)
Mason’s Theorem:
P1 ( s) =
−20
(s + 3)(s + 20)
and T1 ( s) =
10
s( s + 20)
10
and
∆1 ( s) = 1
s( s + 20)
Np
−20
∑=1 Pi (s)∆ i (s) (s + 3)(s + 20)
−20 s
. Check.
H( s) = i
=
=
2
10
∆( s)
3
20
10
+
(
)
+
+
s
s
s
(
)
1+
s( s + 20)
and
∆( s) = 1 +
(b)
s
X(s)
s
s+1
Y(s)
5
s + 20
1
s
15. Find the responses of the systems with these transfer functions to a unit-step and a
suddenly-applied, unit-amplitude, 1 Hz cosine. Also find the responses to a true unitamplitude, 1 Hz cosine (not suddenly applied) using the CTFT and compare to the
steady-state part of the total solution found using the Laplace transform.
Use the results of the chapter in which if a cosine is suddenly applied to a system
N( s)
whose transfer function is H( s) =
, the response is
D( s)
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M. J. Roberts - 8/17/04
 N (s)
y ( t ) = L−1  1  + H ( jω 0 ) cos (ω 0t + ∠ H ( jω 0 )) u ( t )
 D(s) 
where
N1 ( s)
is the partial fraction involving the transfer function poles.
D( s)
(a)
H( s) =
1
s
Unit-step response:
H −1 ( s) =
1
⇒ h −1 ( t) = t u( t) = ramp( t)
s2
Unit-amplitude 1 Hz cosine response:
Y( s) =
s
sin(2πt) u( t)
1
1
2π
2 =
2 ⇒ y( t) =
2
2
s s + (2π )
2π s + (2π )
2π
Using the CTFT:
H( jω ) =
1
jω
Y( jω ) =
δ (ω − 2π ) δ (ω + 2π ) 
1
π [δ (ω − 2π ) + δ (ω + 2π )] = π 
+

jω
jω
jω


δ (ω − 2π ) δ (ω + 2π ) 
Y( jω ) = π 
+
− j 2π 
 j 2π
Y( jω ) =
jπ
sin(2πt)
. Check.
δ (ω + 2π ) − δ (ω − 2π )] ⇒ y( t) =
[
2π
2π
(b)
H( s) =
s
s +1
(d)
H( s) =
s2 + 2 s + 40
s2
(c)
H( s) =
s
s + 2 s + 40
2
16. For each pole-zero diagram sketch the approximate frequency response magnitude.
10-11
M. J. Roberts - 8/17/04
ω
[s]
(a)
σ
-5
ω
[s]
|H( f )|
(b)
1
σ
-2
-2
ω
[s]
4
(c)
σ
-3
-4
ω
[s]
2
(d)
σ
-4
-2
ω
[s]
10
(e)
σ
-1
-10
10-12
2
f
M. J. Roberts - 8/17/04
17. Using only a calculator, find the transfer function of a third-order ( n = 3) lowpass
Butterworth filter with cutoff frequency, ω c = 1, and unity gain at zero frequency.
The poles must be on a semicircle of radius, one, in the LHP. One pole is on the real
π
axis and the spacing between poles is
radians.
3
H ( jω ) =
1
s + 2s + 2s + 1
3
2
18. Using MATLAB, find the transfer function of an eighth-order lowpass Butterworth filter
with cutoff frequency, ω c = 1, and unity gain at zero frequency.
»[z,p,k] = buttap(8) ;
»z
z =
[]
»p
p =
-0.1951
-0.5556
-0.8315
-0.9808
-0.9808
-0.8315
-0.5556
-0.1951
+
+
+
+
-
0.9808i
0.8315i
0.5556i
0.1951i
0.1951i
0.5556i
0.8315i
0.9808i
»k
k =
1.00000000000000
H( s) =
H( s) =
1
( s + 0.1951 + j 0.9808)( s + 0.1951 − j 0.9808) 


( s + 0.5556 + j 0.8315)( s + 0.5556 + j 0.8315)


( s + 0.8315 + j 0.5556)( s + 0.8315 − j 0.5556) 
( s + 0.9808 + j 0.1951)( s + 0.9808 + j 0.1951) 


1
(s + 0.3902s + 1)(s + 1.1112s + 1)(s2 + 1.663s + 1)(s2 + 1.9616s + 1)
2
2
Multiplying the first factor by the last and then the two middle factors,
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M. J. Roberts - 8/17/04
H( s) =
H( s) =
1
(s + 2.3518s + 2.7654 s + 2.3518s + 1)(s4 + 2.7741s3 + 3.8478s2 + 2.7741s + 1)
4
3
2
1
s + 5.126 s + 13.1371s + 21.8462 s + 25.6884 s4 + 21.8462 s3 + 13.1371s2 + 5.126 s + 1
8
7
6
5
19. Find the transfer functions of these Butterworth filters.
(a)
Second-order highpass with a cutoff frequency of 20 kHz and a passband
gain of 5.
The transfer function of the normalized filter is
H norm ( s ) =
1
s + 1.414 s + 1
2
f c = 20kHz ⇒ ω c = 1.257 × 10 5
Making the transformation, s →
H (s) =
ωc
,
s
s2
s 2 + 1.777 × 10 5 s + 1.579 × 1010
The passband is high frequencies (approaching infinity). Therefore
H ( j∞ ) = 1and we need to multiply the transfer function by five to get the required
gain.
H( s) =
(b)
5 s2
s2 + 1.777 × 10 5 s + 1.579 × 1010
Third-order bandpass with a center frequency of 5 kHz, a -3 dB bandwidth of
500 Hz and a passband gain of 1.
This exercise is straightforward but the algebra is long and tedious.
H( s) =
3.1 × 1010 Ks3
s6 + 6283s5 + 2.97 × 10 9 s4 + 1.24 × 1013 s3 + 2.93 × 1018 s2 + 6.09 × 10 21 s + 9.542 × 10 26
(c)
Fourth-order bandstop with a center frequency of 10 MHz, a -3 dB
bandwidth of 50 kHz and a passband gain of 1.
More tedious algebra.
s8 + 1.579 × 1016 s6 + 9.351 × 10 31 s4 + 2.461 × 10 47 s2 + 2.429 × 10 62
H( s) = 8
s + 8.2093 × 10 5 s7 + 1.579 × 1016 s6 + 9.7223 × 10 21 s5 + 9.35 × 10 31 s4 

37 3
47 2
52
62 
+3.8377 × 10 s + 2.4607 × 10 s + 5.0499 × 10 s + 2.4285 × 10 
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M. J. Roberts - 8/17/04
20. Draw canonical system diagrams of the systems with these transfer functions.
(a)
H( s) =
1
s +1
Y( s)
1
=
⇒ X( s) = s Y( s) + Y( s) ⇒ s Y( s) = X( s) − Y( s)
X( s) s + 1
X(s)
1
s
+
-
(b)
H( s) = 4
Y(s)
s+3
s + 10
21. Draw cascade system diagrams of the systems with these transfer functions.
(a)
H( s) =
s
s +1
(b)
H( s) =
s+4
(s + 2)(s + 12)
X(s)
1
s
+
-
1
s
+
-
2
(c)
H( s) =
4
+
+
Y(s)
12
20
s( s + 5 s + 10)
2
22. Draw parallel system diagrams of the systems with these transfer functions.
(a)
H( s) =
−12
s + 3s + 10
(b)
H( s) =
2 s2
s2 + 12 s + 32
2
23. Write state equations and output equations for the circuit of Figure E23 with the inductor
current, iL ( t) , and capacitor voltage, vC ( t) , as the state variables and the voltage at the
input, v i ( t) , as the excitation and the voltage at the output, v L ( t) , as the response.
10-15
M. J. Roberts - 8/17/04
R = 10 Ω C = 1 µF
+
+
vi (t)
vC (t)
+
i L (t)
vL(t)
L = 1 mH
-
-
Figure E23 An RLC circuit
v i ( t) = R iL ( t) + vC ( t) + L i′L ( t)
1
v′C ( t) = iL ( t)
C
v L ( t ) = v i ( t ) − R i L ( t ) − vC ( t )
State equations,

vC′ ( t)   0

= 
 i′L ( t)  − 1
 L
1 
0 
C vC ( t)  +  1  v ( t)


i
R
−  iL ( t)   L 
L
 vC ( t ) 
v L ( t) = [−1 − R]
 + v i ( t)
 i L ( t) 
24. Write state equations and output equations for the circuit of Figure E24 with the inductor
current, iL ( t) , and capacitor voltage, vC ( t) , as the state variables and the current at the
input, ii ( t) , as the excitation and the voltage at the output, v R ( t) , as the response.
+
ii (t)
i L (t)
R = 100 Ω
vR(t)
L = 1 mH
+
vC (t)
-
C = 1 µF
Figure E24 An RLC circuit
25. From the system transfer function,
H( s) =
s( s + 3)
,
s + 2s + 9
2
write a set of state equations and output equations using a minimum number of states.
10-16
M. J. Roberts - 8/17/04
Let Y( s) be a state, Q1 ( s) and let s Y( s) be a state, Q 2 ( s) . Then the state equations are
 s Q1 ( s)   0 1  Q1 ( s)   0 

= 
+ 2

 X( s)
s Q 2 ( s)  −9 −2 Q 2 ( s)  s + 3s
 Q1 ( s) 
Y( s) = [1 0]

Q 2 ( s) 
26. Write state equations and output equations for the system whose block diagram is in
Figure E26 using the responses of the integrators as the state variables.
∫
x(t)
∫
∫
y(t)
8
2
Figure E26 A system
27. A system is excited by the signal, x( t) = 3 u( t) , and the response is
y( t) = 0.961e −1.5 t sin( 3.122 t) u( t) . Write a set of state equations and output equations
using a minimum number of states.
X( s) =
3
s
3.122
3
Y( s) = 0.961
= 2
2
2
(s + 1.5) + 3.122 s + 3s + 12
.
Then the transfer function is
3
s
Y( s) s2 + 3s + 12
H( s) =
=
= 2
.
3
s + 3s + 12
X( s)
s
Let Y( s) be a state, Q1 ( s) and let s Y( s) be a state, Q 2 ( s) . Then the state equations are
1  Q1 ( s)  0 
 s Q1 ( s)   0

 = −
 +   X( s)

s Q 2 ( s)   12 −3Q 2 ( s)   s 
 Q1 ( s) 
Y( s) = [1 0]

Q 2 ( s) 
10-17
M. J. Roberts - 8/17/04
28. A system is described by the differential equation,
y′′ ( t) + 4 y′ ( t) + 7 y( t) = 10 cos(200πt) u( t) .
Write a set of state equations and output equations for this system with two states.
29. A system is described by the state equations and output equations,
and
 q1′ ( t)  −2 1 q1 ( t)   1 2  x1 ( t) 

= 
+



q′2 ( t)   3 0 q1 ( t)  −2 0 x 2 ( t) 
 y1 ( t)   3 5  q1 ( t) 

= 


y 2 ( t)  −2 4 q 2 ( t) 
 q1 (0 + )  0 
 x1 ( t)  −δ ( t) 
With excitation, 
=   . Find the system
= 
 and initial conditions, 
+ 
q 2 (0 )  3
x 2 ( t)   u( t) 
 y1 ( t) 
response vector, 
.
 y 2 ( t) 
s

2
 s + 2s − 3
Φ(s) = 
3

2
 s + 2 s − 3
1

s + 2s − 3 

s+2 
s 2 + 2 s − 3 
2
5
3 


 −2
+ 2 

Q( s) =  s + 3 s − 1 
9

 5
 2 + 2 − 2
 s + 3 s − 1 s 
s+2
s + 2s − 3
2
 5 −3 t 3 t 
5e −3 t + 27e t − 10 
 y1 ( t)   3 5  − 2 e + 2 e 
 u( t)

= 
 u( t) =  −3 t

t
y 2 ( t)  −2 4  5 e −3 t + 9 e t − 2 
15e + 15e − 8 
2
2

30. A system is described by the vector state equation and output equation,
q′ ( t) = Aq( t) + Bx ( t)
and
y ( t) = Cq( t) + Dx ( t) ,
10-18
M. J. Roberts - 8/17/04
−1 −3
1 0 
2 −3
1 0 
where A = 
,
B
=
,
C
=
and
D
=

0 1 
0 4 
0 0  . Define two new states,
 2 −7 






in terms of the old states, for which the A matrix is diagonal and re-write the state equations.
The eigenvalues can be found from the A matrix. They are the roots of s2 + 8 s + 13
which are s = −2.2679 and s = −5.7321. So the matrix of eigenvalues is
0 
−2.2679
Λ=
.
−5.7321
 0
The equation to solve for the transformation matrix that diagonalizes the system is
ΛT = TA . Therefore T is the matrix of eigenvectors for the matrix, A. A normalized T is
 0.8446 −0.5354 
T= 
 .
−0.3893 0.9211 
To verify that this solution is correct,
0  0.8446 −0.5354   0.8446 −0.5354 −1 −3
−2.2679
=
 0
−5.7321−0.3893 0.9211  −0.3893 0.9211  2 −7 

Using
−1.9155 1.2143  −1.9155 1.2143 
 2.2315 −5.2798  =  2.2315 −5.2798  Check.

 

q′2 ( t) = Tq1′ ( t)
we get
Using
 0.8446 −0.5354 
q′2 ( t) = 
q1′ ( t) .
−0.3893 0.9211 
q′2 ( t) = TA1T−1q2 ( t) + TB1x ( t) = A 2q2 ( t) + B 2 x ( t)
we get
−1
 0.8446 −0.5354 −1 −3 0.8446 −0.5354 
 0.8446 −0.5354 1 0 
q′2 ( t) = 
q
(
)
+
t
−0.3893 0.9211 0 1 x ( t)


 2
−0.3893 0.9211  2 −7 −0.3893 0.9211 



or
0 
−2.2679
 0.8446 −0.5354 
q′2 ( t) = 
q 2 ( t) + 

 x ( t)
−5.7321
 0
−0.3893 0.9211 
31. For the original state equations and output equations of Exercise 30 write a differentialequation description of the system.
The original state equations are
10-19
M. J. Roberts - 8/17/04
and
 q1′ ( t)  −1 −3 q1 ( t)  1 0  x1 ( t) 

= 
+



q′2 ( t)   2 −7 q 2 ( t)  0 1 x 2 ( t) 
 y1 ( t)  2 −3 q1 ( t)  1 0  x1 ( t) 

= 
+
 .


y 2 ( t)  0 4 q 2 ( t)  0 0 x 2 ( t) 
From the output equation,
or
 y1′ ( t)  2 −3 q1′ ( t)  1 0  x1′ ( t) 

= 
+



y′2 ( t)  0 4 q′2 ( t)  0 0 x′2 ( t) 
 y1′ ( t)  2 −3−1 −3 q1 ( t)  1 0  x1 ( t)  1 0  x1′ ( t) 

= 
+
 + 





y′2 ( t)  0 4  2 −7 q 2 ( t)  0 1 x 2 ( t)  0 0 x′2 ( t) 
 y1′ ( t)  −8 15  q1 ( t)  2 −3 x1 ( t)  1 0  x1′ ( t) 

= 
+
+
.



y′2 ( t)   8 −28 q 2 ( t)  0 4 x 2 ( t)  0 0 x′2 ( t) 
Solving the output equations for the states,
 q1 ( t)  2 −3  y1 ( t)  1 0  x1 ( t) 

= 
− 

 

q 2 ( t)  0 4  y 2 ( t)  0 0 x 2 ( t) 
−1
or
1
 q1 ( t)   2

= 
q 2 ( t)   0

3
1
8  y1 ( t)  − 
1 y 2 ( t)   2

0
4

0  x1 ( t) 

.
0 x 2 ( t) 
Then
 1
 y1′ ( t)  −8 15  2

= 

y′2 ( t)   8 −28  0

3
1
8  y1 ( t)  − 
1 y 2 ( t)   2

0
4

 x ( t)  2 −3 x ( t)  1 0  x′ ( t) 
0 1
1
1

 + 0 4 
+



x 2 ( t)  0 0 x′2 ( t) 
0 x 2 ( t)  

 y1′ ( t)  −4 3  y1 ( t)   6 −3 x1 ( t)  1 0  x1′ ( t) 
4 

= 
+
+



y
(
)
t
′

 2   4 −4 y 2 ( t)  −4 4 x 2 ( t)  0 0 x′2 ( t) 
3
y ( t) + 6 x1 ( t) − 3 x 2 ( t) + x1′ ( t)
4 2
y′2 ( t) = 4 y1 ( t) − 4 y 2 ( t) − 4 x1 ( t) + 4 x 2 ( t)
y1′ ( t) = −4 y1 ( t) +
32. Find the s-domain transfer functions for the circuits below and then draw block diagrams
for them as systems with Vi ( s) as the excitation and Vo ( s) as the response.
10-20
M. J. Roberts - 8/17/04
R = 10 kΩ
+
+
vi (t)
vo(t)
C = 1 µF
L = 5 mH
−
−
(a)
H( s) = 100
s
s + 100 s + 2 × 10 8
2
V(s)
o
100
V(s)
i
1
s
+
1
s
100
+
+
8
2 10
L = 5 mH
C = 1 µF
R = 10 kΩ
+
+
vi (t)
vo(t)
−
(b)
R = 10 kΩ
−
L = 5 mH
+
+
vi (t)
C = 1 µF
vo(t)
−
−
(c)
10-21
M. J. Roberts - 8/17/04
R = 10 kΩ
R = 10 kΩ
+
+
vi (t)
vo(t)
C = 1 µF
C = 1 µF
−
−
(d)
33. Determine whether the systems with these transfer functions are stable, marginally
stable or unstable.
(a)
(c)
(e)
s( s + 2)
s2 + 8
s2
H( s) = 2
s + 4s + 8
s
H( s) = 3
s + 4 s2 + 8 s
H( s) =
s( s − 2)
s2 + 8
s2
H( s) = 2
s − 4s + 8
H( s) =
(b)
(d)
34. Find the expression for the overall system transfer function of the system below.
K
s+10
X(s)
Y(s)
β
H (s) =
K
s + 10 + β K
Pole at s = −10 − βK
(a)
Let β = 1. For what values of K is the system stable?
Pole at s = −10 − K
System is stable for K > −10
(b)
Let β = −1. For what values of K is the system stable?
(c)
Let β = 10. For what values of K is the system stable?
35. Find the expression for the overall system transfer function of the system below.
For what positive values of K is the system stable?
X(s)
K
(s+1)(s+2)
10-22
Y(s)
M. J. Roberts - 8/17/04
36. Find the expression for the overall system transfer function of the system below.
Using MATLAB plot the paths of the poles of the overall system transfer function as a
function of K. For what positive values of K is the system stable?
X(s)
H( s) =
K
(s+1)(s+2)(s+3)
Y(s)
K
K
K
Y( s)
= 2
=
= 3
2
X( s) ( s + 1)( s + 2)( s + 3) + K ( s + 3s + 2)( s + 3) + K s + 6 s + 11s + 6 + K
Although this denominator can be factored it is probably easier just to numerically
explore the pole locations versus the value of K. A MATLAB program was written to graph
the pole locations for a range of K’s and it was found that when K = 60 the poles just touch
the ω axis, indicating marginal stability. When K = 60, the poles are at
s = −6 , ± j 3.3166 .
Therefore for 0 < K < 60 the system is stable.
37. Thermocouples are used to measure temperature in many industrial processes. A
thermocouple is usually mechanically mounted inside a “thermowell”, a metal sheath
which protects it from damage by vibration, bending stress, or other forces. One effect
of the thermowell is that its thermal mass slows the effective time response of the
thermocouple/thermowell combination compared witht the inherent time response of the
thermocouple alone. Let the actual temperature on the outer surface of the thermowell in
Kelvins be Τs ( t) and let the voltage developed by the thermocouple in response to
temperature be v t ( t) . The response of the thermocouple to a one-Kelvin step change in
the thermowell outer-surface temperature from T1 to T1 + 1 is
t


−


v t ( t) = K T1 + 1 − e 0.2  u( t) 




where K is the thermocouple temperature-to-voltage conversion constant.
µV
. Design an active filter which processes
K
the thermocouple voltage and compensates for its time lag making the overall system have a
response to a one-Kelvin step thermowell-surface temperature change that is itself a step of
voltage of 1mV.
(a)
Let the conversion constant be K = 40
The unit step response of the thermocouple-thermowell combination is
v t ( t) = K (1 − e −5 t ) u( t) .
The impulse response is the derivative of the step response,
h t ( t) = 5Ke −5 t u( t) .
10-23
M. J. Roberts - 8/17/04
The transfer function is the transform of the impulse response,
H t ( s) =
5K
.
s+5
The desired overall frequency response is
H( s) =
1mV
.
K
(Here K is the kelvin not the thermocouple gain, K.) Therefore the transfer function of the
compensating active filter is
s+5
H( s) 10 −3
H f ( s) =
=
= 2 × 10 −4
.
K
5
K
H t ( s)
s+5
This is a system with a real zero creating a corner frequency, ω c = 5.
synthesized by the circuit of Figure S37 .
Rf
Ri
+
C
This can be
+
vi (t)
vo (t)
-
-
Figure S37 Thermocouple-thermowell compensator
Choose resistor and capacitor values to locate the poles and zeros in the proper places.
(b)
Suppose that the thermocouple also is subject to electromagnetic interference (EMI)
from nearby high-power electrical equipment. Let the EMI be modeled as a sinusoid with
an amplitude of 20 µV at the thermocouple terminals. Calculate the response of the
thermocouple-filter combination to EMI frequencies of 1 Hz, 10 Hz and 60 Hz. How big is
the apparent temperature fluctuation caused by the EMI in each case?
At 1 Hz:
H f ( j 2π ) = 5( j 2π + 5) = 40.14 ∠51°
So the response to the 20 µV excitation is about 800 µV which is equivalent to
about 0.8 K.
At 10 Hz:
At 60 Hz:
38. A laser operates on the fundamental principle that a pumped medium amplifies a
travelling light beam as it propagates through the medium. Without mirrors a laser
becomes a single-pass travelling wave amplifier (Figure E38-1).
10-24
M. J. Roberts - 8/17/04
Pumped Laser Medium
Light In
Light Out
Figure E38 -1 A one-pass travelling-wave light amplifier
This is a system without feedback. If we now place mirrors at each end of the pumped
medium, we introduce feedback into the system.
Mirror
Pumped Laser Medium
Mirror
Figure E38-2 A regenerative travelling-wave amplifier
When the gain of the medium becomes large enough the system oscillates creating a
coherent output light beam. That is laser operation. If the gain of the medium is less that
that required to sustain oscillation, the system is known as a regenerative travelling-wave
amplifier (RTWA).
Let the electric field of a light beam incident on the RTWA from the left be the
excitation of the system, E inc ( s) , and let the electric fields of the reflected light, E refl ( s) , and
the transmitted light, E trans ( s) , be the responses of the system (Figure E38-3) .
ri
E refl (s)
E circ (s)
jt i
E inc (s)
gfp
jt o
ro
ri
grp
jt i
Figure E38-3 Block diagram of an RTWA
Let the system parameters be as follows:
Electric field reflectivity of the input mirror, ri = 0.99
Electric field transmissivity of the input mirror, ti = 1 − ri2
Electric field reflectivity of the output mirror, ro = 0.98
Electric field transmissivity of the output mirror, t o = 1 − ro2
10-25
E trans(s)
M. J. Roberts - 8/17/04
Forward and reverse path electric field gains, g fp ( s) = grp ( s) = 1.01e −10
Find an expression for the frequency response,
frequency range, 3 × 1014 ± 5 × 10 8 Hz.
−9
s
E trans ( f )
, and plot its magnitude over the
E inc ( f )
E circ ( s) = jti E inc ( s) + g fp ( s) ro grp ( s) ri E circ ( s)
E circ ( s) =
jti
E ( s)
1 − g fp ( s) ro grp ( s) ri inc
E trans ( s) = jto g fp ( s) E circ ( s)
E trans ( s) = −
The transfer function is
H( s) =
ti to g fp ( s)
E ( s)
1 − ro ri g 2fp ( s) inc
ti to g fp ( s)
E trans ( s)
=−
E inc ( s)
1 − ro ri g 2fp ( s)
and the frequency response is
H( jω ) = −
ti to g fp ( jω )
.
1 − ro ri g 2fp ( jω )
|H(j2πf )|
3
f
2.999995e+14
3.000005e+14
39. A classical example of the use of feedback is the phase-locked loop used to demodulate
frequency-modulated signals (Figure E39) .
x(t)
+
-
y VCO(t)
Phase
Detector
xLF (t)
Loop
Filter, H LF (s)
VoltageControlled
Oscillator
Figure E39 A phase-locked loop
10-26
y(t)
M. J. Roberts - 8/17/04
The excitation, x( t) , is a frequency-modulated sinusoid. The phase detector detects the
phase difference between the excitation and the signal produced by the voltage-controlled
oscillator. The response of the phase detector is a voltage proportional to phase difference.
The loop filter filters that voltage. Then the loop filter response controls the frequency of
the voltage-controlled oscillator. When the excitation is at a constant frequency and the
loop is “locked” the phase difference between the two phase-detector excitation signals is
zero. (In an actual phase detector the phase difference is 90° at lock. But that is not
significant in this analysis since that only causes is a 90° phase shift and has no impact on
system performance or stability.) As the frequency of the excitation, x( t) , varies, the loop
detects the accompanying phase variation and tracks it. The overall response signal, y( t) , is
a signal proportional to the frequency of the excitation.
The actual excitation, in a system sense, of this system is not x( t) , but rather the
phase of x( t) , φ x ( t) , because the phase detector detects differences in phase, not voltage.
Let the frequency of x( t) be fx ( t) . The relation between phase and frequency can be seen
by examining a sinusoid. Let x( t) = A cos(2πf 0 t) . The phase of this cosine is 2πf 0 t and,
for a simple sinusoid ( f 0 constant), it increases linearly with time. The frequency is f 0 , the
derivative of the phase. Therefore the relation between phase and frequency for a
frequency-modulated signal is
1 d
fx ( t ) =
(φ (t)) .
2π dt x
Let the frequency of the excitation be 100 MHz. Let the transfer function of the
Hz
. Let the transfer function of the loop filter be
voltage-controlled oscillator be 10 8
V
H LF ( s) =
1
.
s + 1.2 × 10 5
V
. If the frequency of the excitation
radian
signal suddenly changes to 100.001MHz, plot the change in the output signal, ∆ y( t) .
Let the transfer function of the phase detector be 1
Let φ diff ( t) be phase difference between x( t) and yVCO ( t) . Then the following are
the relations among the signals,
X LF ( s) = Φ diff ( s)
Y( s) =
X LF ( s)
s + 1.2 × 10 5
FVCO ( s) = 10 4 Y( s)
ΦVCO ( s) = 2π
FVCO ( s)
s
Φ diff ( s) = Φ x ( s) − ΦVCO ( s)
10-27
M. J. Roberts - 8/17/04
Combining equations,
X LF ( s)
5
Φ diff ( s)
XVCO ( s) = s + 1.2 × 10 =
s
s( s + 1.2 × 10 5 )
FVCO ( s) = 10 4
ΦVCO ( s) = 2π
10 8
Φ diff ( s)
s + 1.2 × 10 5
Φ diff ( s)
Φ diff ( s)
s + 1.2 × 10 5 = 2π × 10 4
s
s( s + 1.2 × 10 5 )
Φ diff ( s) = Φ x ( s) − 2π × 10 8
Φ diff ( s)
s( s + 1.2 × 10 5 )

2π × 10 8 
Φ diff ( s) 1 +
= Φ x ( s)
5 
 s( s + 1.2 × 10 ) 
Φ diff ( s)
=
Φ x ( s)
s( s + 1.2 × 10 5 )
1
=
2π × 10 8
s( s + 1.2 × 10 5 ) + 2π × 10 8
1+
s( s + 1.2 × 10 5 )
Y( s)
s
=
Φ x ( s) s( s + 1.2 × 10 5 ) + 2π × 10 8
Fx ( s) =
sΦ x ( s)
2π
2π
Y( s)
=
.
Fx ( s) s( s + 1.2 × 10 5 ) + 2π × 10 8
In steady state with no frequency modulation and a frequency of 100 MHz, y( t) = 1.
The response to a step frequency change of 1 kHz, ∆fx ( t) = 1000 u( t) , is
∆ Y( s) =
2π
1000
2000π
= 2
5
8
s( s + 1.2 × 10 ) + 2π × 10 s
s ( s + 1.2 × 10 5 ) + 2π × 10 8 s
10 −5
5.033 × 10 −7
1.05 × 10 −5
−
∆ Y( s) =
+
s
s + 1.145 × 10 5
s + 5487
(
)
∆ y( t) = 10 −5 + 5.033 × 10 −7 e −1.145 ×10 t − 1.05 × 10 −5 e −5487 t u( t)
5
10-28
M. J. Roberts - 8/17/04
∆y( t)
1e-05
0.001
t
40. Plot the root locus for each of the systems which have these loop transfer functions and
identify the transfer functions that are stable for all positive real values of K.
(a)
T( s) =
K ( s + 10)
(s + 1)(s2 + 4 s + 8)
Unstable for some positive real K.
ω
-1
-10
σ
K ( s2 + 10)
(b)
T( s) =
(c)
T( s) =
K
s + 37 s + 332 s + 800
(d)
T( s) =
K (s − 4)
s+4
(e)
T( s) =
K (s − 4)
(s + 4)2
(f)
T( s) =
K ( s + 6)
(s + 5)(s + 9)(s2 + 4 s + 12)
(s + 1)(s2 + 4 s + 8)
3
2
41. The circuit below is a simple approximate model of an operational amplifier with the
inverting input grounded.
(a)
Define the excitation of the circuit as the current of a current source applied to the
non-inverting input and define the response as the voltage developed between the noninverting input and ground. Find the transfer function and graph its frequency response.
This transfer function is the input impedance.
Z i ( s) = Ri = 1MΩ
(b)
Define the excitation of the circuit as the current of a current source applied to the
output and define the response as the voltage developed between the output and ground with
10-29
M. J. Roberts - 8/17/04
the non-inverting input grounded. Find the transfer function and graph its frequency
response. This transfer function is the output impedance.
Z i ( s) = Ro = 10Ω
(c)
Define the excitation of the circuit as the voltage of a voltage source applied to the
non-inverting input and define the response as the voltage developed between the output and
ground. Find the transfer function and graph its frequency response. This transfer
function is the voltage gain.
1
1
sCx
= A0 Vi ( s)
Vo ( s) = VX ( s) = A0 Vi ( s)
1
sRx Cx + 1
+ Rx
sCx
H( s) =
A
Vo ( s)
1
1
= 0
= 1.25 × 10 7
1
s + 125
Vi ( s) Rx Cx s +
Rx Cx
The corner frequency is approximately 20 Hz.
Rx
Ro
+
+
Ri
-
vi (t)
-
Output
+
A0v i (t)
Cx
vx(t)
v x(t)
-
Ri = 1MΩ , Rx = 1kΩ , Cx = 8 µF , Ro = 10 Ω , A0 = 10 6
42. Change the circuit of Exercise 41 to the circuit below. This is a feedback circuit which
establishes a positive closed-loop voltage gain of the overall amplifier. Repeat steps (a),
(b) and (c) of Problem #6 for the feedback circuit and compare the results. What are
the important effects of feedback for this circuit?
10-30
M. J. Roberts - 8/17/04
Rx
Ro
+
+
Ri
+
vi (t)
A0v i (t)
-
-
Output
Cx
Rs
vx(t)
v x(t)
-
Rf
Ri = 1MΩ , Rx = 1kΩ , Cx = 8 µF , Ro = 10 Ω , A0 = 10 6 , R f = 10 kΩ , Rs = 5 kΩ
(a)
Z i ( s) =
Vi ( s) + Vs ( s) Ri I i ( s) + Vs ( s)
V ( s)
=
= Ri + s
I i ( s)
I i ( s)
I i ( s)
Vs ( s)Gs + [Vs ( s) − Vo ( s)]G f = I i ( s)
Vx ( s) = A0
Ri
1
Vi ( s) = A0
I ( s)
sRx Cx + 1 i
sRx Cx + 1
(
)
Vo ( s) Go + G f − Vs ( s)G f − Vx ( s)Go = 0
Combining the last two equations,
(
)
Vo ( s) Go + G f − Vs ( s)G f − A0
Ri
I ( s)Go = 0
sRx Cx + 1 i
Then
Gs + G f
 −G
f

(
1

−G f Vs ( s)  
R
 I ( s)

i
=
Go + G f Vo ( s)   A0 sR C + 1 Go  i

x x

)(
)
∆ = Gs + G f Go + G f − G 2f = GsGo + GsG f + G f Go
RiG f Go
−G f
1
Go + G f + A0
Vs ( s) 1
sRx Cx + 1
Ri
=
=
G
G
G
A
+
GsGo + GsG f + G f Go
I i ( s) ∆ 0
o
f
sRx Cx + 1 o
10-31
M. J. Roberts - 8/17/04
1 1
R f Ro
1
1
Ri Rs
+
+ A0
RsR f + RsR0 + A0
sRx Cx + 1
Vs ( s) Ro R f
sRx Cx + 1
=
=
1 1
1 1
1 1
I i ( s)
R f + Ro + Rs
+
+
Rs Ro Rs R f R f Ro
Ri
Zi ( s) = Ri +
Ri Rs
sRx Cx + 1
R f + Ro + Rs
RsR f + RsR0 + A0
Substituting in numbers,
Zi ( s) = 10 6 +
50 × 10 6 + 50, 000 + 10 6
5 × 10 9
8 × 10 −3 s + 1
15, 010
At low frequencies,
1012
Z i ( s) ≅
3
This input impedance is much higher than in the open-loop case.
(b)
Ground the input terminal for this calculation.
Z o ( s) =
Vo ( s) Vs ( s) + V f ( s)
=
I o ( s)
I o ( s)
[V (s) − V (s)]G + [V (s) − V (s)]G
o
s
f
o
[V (s) − V (s)]G
s
o
f
x
o
= I o ( s)
+ Vs ( s)(Gs + Gi ) = 0
Since the non-inverting input is grounded we can write
Vx ( s) = − Vs ( s) A0
1
sRx Cx + 1
Combining equations and solving
 A0Go

Vo ( s) G f + Go + 
− G f  Vs ( s) = I o ( s)
 sRx Cx + 1

(
)
(
)
− Vo ( s)G f + G f + Gs + Gi Vs ( s) = 0
10-32
M. J. Roberts - 8/17/04
A0Go

− G f Vo ( s)  I o ( s) 
sRx Cx + 1
= 


G f + Gs + Gi Vs ( s)   0 

G f + Go

 −G f
 A0Go

∆ = G f + Go G f + Gs + Gi + G f 
− Gf 
 sRx Cx + 1

(
)(
)
(
)
∆ = G f (Gs + Gi ) + Go G f + Gs + Gi +
1 I o ( s)
Vo ( s) =
∆ 0
Z o ( s) = 10
A0Go
− G f G f + Gs + Gi
I o ( s)
=
sRx Cx + 1
∆
G f + Gs + Gi
G f + Gs + Gi
Z o ( s) =
Z o ( s) = Ro
A0GoG f
sRx Cx + 1
(
)
G f (Gs + Gi ) + Go G f + Gs + Gi +
A0GoG f
sRx Cx + 1
RsRi + R f Ri + R f Rs
Ro ( Ri + Rs ) + RsRi + R f Ri + R f Rs + RsRi
(
A0
sRx Cx + 1
)
10 6 Rs + R f + R f Rs
(
)
10(10 6 + Rs ) + 10 6 Rs + R f + R f Rs + 10 6 Rs
10 6
8 × 10 −3 s + 1
Substituting in numbers,
Z o ( s) ≅ 10
1
15.05 × 10 9
≅
15
5 × 10 6
5 × 10
0
1
15.06 × 10 9 +
.
+
8 × 10 −3 s + 1
8 × 10 −3 s + 1
At low frequencies
Z o ( s) ≅ 0.2 × 10 −6
This output impedance is much lower than in the open-loop case.
(c)
(
)
Vo ( s) Go + G f − Vx ( s)Go − Vs ( s)G f = 0
(
)
Vs ( s) Gs + G f + Gi − Vo ( s)G f − Vex ( s)Gi = 0
where Vex ( s) is the overall excitation voltage with respect to ground
10-33
M. J. Roberts - 8/17/04
Vx ( s) = (Vex ( s) − Vs ( s)) A0

Go + G f

 −G f

 0

−Go
0
1
∆ = − A0
1
sRx Cx + 1



Vo ( s)  

0
−G f


 
Gs + G f + Gi Vx ( s)  = 
Vex ( s)Gi

1
V ( s)  
1

A0
 s  Vex ( s) A0 sR C + 1 
sRx Cx + 1 

x x

(
)(
)
G f Go
− Go + G f Gs + G f + Gi + G 2f
sRx Cx + 1
1
Vo ( s) =
∆
−Go
−G f
0
Vex ( s)Gi
0 Gs + G f + Gi
1
1
Vex ( s) A0
A0
1
sRx Cx + 1
sRx Cx + 1


−Go
1
G + G f + Gi Go
+ G f  − Vex ( s) A0
− Vex ( s)Gi  A0
sRx Cx + 1 s
 sRx Cx + 1

Vo ( s) =
∆
(
)


−Go
1
G + G f + Gi Go
Gi  A0
+ G f  + A0
sRx Cx + 1 s
 sRx Cx + 1

Vo ( s)
H( s) =
=
G f Go
Vex ( s)
A0
+ Go + G f Gs + G f + Gi − G f2
sRx Cx + 1
(
(
(
)(
)
)
)
A0
G + G f Go + GiG f
sRx Cx + 1 s
H( s) =
G f Go
A0
+ Go Gs + G f + Gi + G f (Gs + Gi )
sRx Cx + 1
(
)
(
)
A0
R R + Rs + Ro Rs
sRx Cx + 1 i f
H( s) =
A0
(
)
RsRi
+ R f Ri + RsRi + RsR f + Ro ( Rs + Ri )
sRx Cx + 1
Substituting in numbers,
15 × 1013
15 × 1013
4
+ 5 × 10 4
+
×
5
10
−3
−3
×
+
8
10
1
×
+
8
10
1
s
s
≅
H( s) =
5 × 1013
5 × 1013
7
4
+ 20 × 10 7
+ 20 × 10 + 15 × 10
−3
−3
8 × 10 s + 1
8 × 10 s + 1
10-34
M. J. Roberts - 8/17/04
H( s) ≅
11
400 s + 15 × 1013
−4 s + 3.75 × 10
≅
×
2
5
10
.
s + 3.125 × 10 7
1.6 × 10 6 s + 5 × 1013
The low-frequency gain is
H(0) ≅ 2.5 × 10 −4
R f + Rs
3.75 × 1011
7 = 3=
3.125 × 10
Rs
as it should be. The closed-loop gain has a pole at s = −3.125 × 10 7 which sets a corner
frequency of approximately 5 MHz. The open-loop corner frequency was approximately
20 Hz. So the bandwidth has been increased by a factor of approximately 250,000.
43. Plot the unit step and ramp responses of unity-gain feedback systems with these
forward-path transfer functions.
(a)
H1 ( s) =
20
s( s + 2)( s + 6)
(b)
H1 ( s) =
20
s ( s + 2)( s + 6)
(c)
H1 ( s) =
100
s + 10 s + 34
(d)
H1 ( s) =
100
s( s + 10 s + 34 )
(e)
H1 ( s) =
100
s ( s + 10 s + 34 )
2
2
2
2
2
44. Draw pole-zero diagrams of these transfer functions.
(a)
H( s) =
(s + 3)(s − 1)
s( s + 2)( s + 6)
(b)
H( s) =
s
s + s +1
2
ω
1
(c)
H( s) =
s( s + 10)
s + 11s + 10
2
σ
-10
-1
(Pole-zero cancellation.)
10-35
M. J. Roberts - 8/17/04
(d)
H( s) =
1
(s + 1)(s + 1.618s + 1)(s2 + 0.618s + 1)
2
45. A second-order system is excited by a unit step and the response is as illustrated in
Figure E45 . Write an expression for the transfer function of the system.
Step Response
0.2
0.18
0.16
Amplitude
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0
10
20
30
40
50
60
Time (sec.)
Figure E45 Step response of a second-order system
From the graph it is apparent that system is highly underdamped and that the final
value of the step response is 0.1. So this second-order system has no zeros at zero.
Therefore the general form of the transfer function of this second-order system is
Aω 02
H( s) = 2
s + 2ζω o s + ω 02
and A = 0.1. From the graph there are 10 ringing cycles of response between 0 and 10
seconds. Therefore the resonant frequency is approximately 1Hz or 2π radians per second.
The impulse response is of the form,
(
)
h( t) = Ke −ζω 0 t cos ω 0 1 − ζ 2 t + θ .
1
. From the graph, the
ζω 0
time constant is approximately the time at which the ringing is at 36.8% of its maximum
value. That is at about 10 seconds. Therefore
So the characteristic exponential decay has a time constant of τ =
ζ=
1
1
=
= 0.0159 .
ω 0τ 20π
So the transfer function is
H( s) =
3.948
.
s + 0.2 s + 39.48
2
10-36
M. J. Roberts - 8/17/04
46. For each of the pole-zero plots below determine whether the frequency response is that
of a practical lowpass, bandpass, highpass or bandstop filter.
ω
ω
[s]
[s]
σ
(a)
σ
(b)
ω
ω
[s]
σ
(c)
(a)
[s]
σ
(d)
Highpass
47. A system has a transfer function,
H( s) =
(a)
A
.
s + 2ζω 0 s + ω 02
2
Let ω 0 = 1. Then let ζ vary continuously from 0.1 to 10 and plot in the s-plane the
paths that the two poles take while ζ is varying between those limits.
ω
ζ = 0.1
ζ = 10
-19.9
ζ = 10
-1
-0.1
σ
ζ = 0.1
(b)
Find the real-valued functional form of the impulse response for the case, ω 0 = 1 and
ζ = 0.5 .
10-37
M. J. Roberts - 8/17/04
4 −0.5 t  3 
Ae sin 
t
3
 4 
h (t ) =
(c)
Sketch the phase frequency response for the case, ω 0 = 1 and ζ = 0.1.
Phase of H( jω )
π
-4
4
ω
-π
(d)
Find the -3 dB bandwidth for for the case, ω 0 = 1 and ζ = 0.1.
H( jω ) =
( jω )
A
2
+ 0.2 jω + 1
=
A
1 − ω + 0.2 jω
2
The maximum value of the transfer function occurs at resonance, ω = ω 0 = 1. There
H( jω 0 ) =
A
= 5A
0.2 jω 02
At the -3 dB points the magnitude of the square of the transfer function is one-half of
the square of this value.
ω −3 dB = ±1.086, ±0.883
So the bandwidth is 0.0323 Hz.
(e)
The Q of a system is a measure of how “sharp” its frequency response is near a
resonance. It is defined as
1
Q=
.
2ζ
For very high-Q systems what is the relationship between Q, ω 0 and -3 dB
bandwidth?
If the Q is very high that means that the damping factor is very low. The peak of the
frequency response occurs at resonance and at that frequency,
H( jω 0 ) =
A
A
2 = Q
2ζω 0
ω 02
The -3 dB bandwidth is found by solving
10-38
M. J. Roberts - 8/17/04
2
H( jω −3 dB ) =
A
2
ω 02 − ω −23 dB + j
ω0
ω
Q −3 dB
 A
Q 2 
 ω0 
=
2
2
Solving,
ω −23 dB
1 1 1
1
1
4


+4
2 − 2  ±
2 − 2  ±
4 +
2
Q 
Q
Q
Q  Q Q2
2 
2 
= ω0
= ω0
2
2
For very large Q,
ω −3 dB ≅ ω 0 1 ±
1
1

≅ ω 0 1 ±

 2Q
Q
which means that the bandwidth is
∆ω −3dB ≅
ω0
.
Q
That is, for very-high-Q systems the -3 dB bandwidth is approximately the center frequency
divided by the Q.
48. Draw canonical system diagrams of the systems with these transfer functions.
(a)
H( s) = 10
s2 + 8
s3 + 3s2 + 7 s + 22
s3 Y1 ( s) = X( s) − 3s2 Y1 ( s) − 7 s Y1 ( s) − 22 Y1 ( s)
Y( s) = 10 s2 Y1 ( s) + 80 Y1 ( s)
10
+
80
X(s)
1
s
+
+
1
s
1
s
3
+
+
7
+
22
(b)
H( s) = 10
s + 20
(s + 4)(s + 8)(s + 14)
10-39
Y(s)
1
+
Y(s)
M. J. Roberts - 8/17/04
49. Draw cascade system diagrams of the systems with these transfer functions.
(a)
s2
H( s) = −50 3
s + 8 s2 + 13s + 40
Factoring the numerator and denominator,
H( s) = −50
s
s
2
s + 6.958 s + 1.042 s + 5.749
1
s
X(s)
1
s
+
0
6.958
+
1
s
0
+
-50
Y(s)
1.042
+
+
5.749
(b)
s3
H (s) = 3
s + 18 s 2 + 92 s + 120
50. Draw parallel system diagrams of the systems with these transfer functions.
(a)
H( s) = 10
(b)
H( s) =
s3
39.9 s + 73.84
0.09869
−
= 10 − 2
3
2
s + 4 s + 9s + 3
s + 3.604 s + 7.572 s + 0.3962
5
0.01667
0.15
0.1333
=
−
+
s+9
s + 2.333 s + 1.5
6 s + 77 s + 228 s + 189
3
2
51. Write state equations and output equations for the circuit of Figure E51 with the two
capacitor voltages, vC1 ( t) and vC 2 ( t) , as the state variables and the voltage at the input,
v i ( t) , as the excitation and the voltage, v R1 ( t) , as the response. Then, assuming the
capacitors are initially uncharged, find the unit step response of the circuit.
C1 = 1 µF
+
vi (t)
+
vC1(t)
R 2 = 10 kΩ
+
vR1(t)
R 1 = 10 kΩ
+
vC2(t)
-
Figure E51 A second-order RC circuit
10-40
C2 = 1 µF
M. J. Roberts - 8/17/04
 1
 1 
 1
1 
1 
v′C1 ( t) =  −
−
+
 vC1 ( t) +  −
 vC 2 ( t) + 
 v i ( t)
 R1C1 R2C1 
 R2C1 
 R1C1 R2C1 
v′C 2 ( t) = −
vC1 ( t) vC 2 ( t) v i ( t)
−
+
R2C2
R2C2 R2C2
and an output equation,
v R1 ( t) = − vC1 ( t) + v i ( t) .
Writing them in standard matrix form, (with numbers substituted)
and
The transfer function is
 vC′ 1 ( t)  −200 −100  vC1 ( t)  200 

= 
+


v′C 2 ( t)  −100 −100 vC 2 ( t)  100 
 v ( t) 
v R1 ( t) = [−1 0] C1  + v i ( t) .
 vC 2 ( t ) 
H( s) = C[ sI − A ] B + D
−1
which, in this case, is
H( s) =
s( s + 100)
s + 300 s + 10, 000
2
The step response is
h −1 ( t) = (0.7326e −261.8 t + 0.2764 e −38.2 t ) u( t)
52. Write state equations and output equations for the circuit of Figure E52 with the two
capacitor voltages, vC1 ( t) and vC 2 ( t) , as the state variables and the voltage at the input,
v i ( t) , as the excitation and the voltage at the output, v o ( t) , as the response. Then, find
and plot the response voltage for a unit step excitation assuming that the initial conditions
are
 vC1 (0)   2 

=   .
vC 2 (0)  −1
10-41
M. J. Roberts - 8/17/04
C1
+
R1
R2
K
+
vi (t)
vC1(t)
vx(t)
-
C2
+
vC2(t)
-
+
vo(t)
-
R1 = 6.8 kΩ , R2 = 12 kΩ
C1 = 6.8 nF , C2 = 6.8 nF
K = 3
Figure E52 A constant-K lowpass filter
v i ( t) − vC 2 ( t)

R1C1 R1C2 + R2C2  v′C1 ( t)  
=



R C

R1C2
 1 1
v′C 2 ( t)  v i ( t) − K vC 2 ( t) − vC1 ( t) 
 R1C2 + R2C2
−
v
 C′ 1 ( t)   R1R2C1C2


=
1
v′C 2 ( t)  

R2C2
R1C2 − K ( R1C2 + R2C2 ) 
  v ( t)   1 
R1R2C1C2
 C1  +  R1C1  v i ( t)
K −1
  vC 2 ( t )   0 



R2C2
 v ( t) 
v o ( t) = [0 K ] C1 
 vC 2 ( t ) 
Substituting numbers,
 s − 24510 −89389 
 12254
s + 33881

Φ(s) = 2
s + 9371s + 2.65 × 10 8
 2 s2 + 61995 s − 5.3 × 10 8 


s( s2 + 9371s + 2.65 × 10 8 ) 

Q( s) =
 s2 + 9373s − 2.65 × 10 8 
− 2 9371 2 65 10 8 
s+ . ×
)
 s( s +
10-42
M. J. Roberts - 8/17/04
Y( s) =

3 
4686
1.559 × 10 4
s + 4686
+
− 6
2
2
4

s
 ( s + 4686) + 2.43 × 10 8 1.559 × 10 ( s + 4686) + 2.43 × 10 8 
[
)]
(
y( t) = 3 − 6e −4686 t cos(1.559 × 10 4 t) + 0.3 sin(1.559 × 10 4 t) u( t)
10-43