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Solution Manual
to accompany
Introduction to Electric Circuits, 6e
By R. C. Dorf and J. A. Svoboda
1
Table of Contents
Chapter 1 Electric Circuit Variables
Chapter 2 Circuit Elements
Chapter 3 Resistive Circuits
Chapter 4 Methods of Analysis of Resistive Circuits
Chapter 5 Circuit Theorems
Chapter 6 The Operational Amplifier
Chapter 7 Energy Storage Elements
Chapter 8 The Complete Response of RL and RC Circuits
Chapter 9 The Complete Response of Circuits with Two Energy Storage Elements
Chapter 10 Sinusoidal Steady-State Analysis
Chapter 11 AC Steady-State Power
Chapter 12 Three-Phase Circuits
Chapter 13 Frequency Response
Chapter 14 The Laplace Transform
Chapter 15 Fourier Series and Fourier Transform
Chapter 16 Filter Circuits
Chapter 17 Two-Port and Three-Port Networks
2
Errata for Introduction to Electric Circuits, 6th Edition
Errata for Introduction to Electric Circuits, 6th Edition
Page 18, voltage reference direction should be + on the right in part B:
Page 28, caption for Figure 2.3-1: "current" instead of "cuurent"
Page 41, line 2: "voltage or current" instead of "voltage or circuit"
Page 41, Figure 2.8-1 b: the short circuit is drawn as an open circuit.
Page 42, line 11: "Each dependent source ..." instead of "Each dependent sources..."
Page 164, Table 5.5-1: method 2, part c, one should insert the phrase "Zero all independent sources,
then" between the "(c)" and "Connect a 1-A source. . ." The edited phrase will read:
"Zero all independent sources, then connect a 1-A source from terminal b to terminal a. Determine Vab.
Then Rt = Vab/1."
.
Page 340, Problem P8.3-5: The answer should be
Page 340, Problem P8.3-6: The answer should be
Page 341, Problem P.8.4-1: The answer should be
Page 546, line 4: The angle is
instead of
.
Page 554, Problem 12.4.1 Missing parenthesis:
Page 687, Equation 15.5-2: Partial t in exponent:
http://www.clarkson.edu/~svoboda/errata/6th.html (1 of 2)5/10/2004 7:41:43 PM
.
Errata for Introduction to Electric Circuits, 6th Edition
Page 757, Problem 16.5-7: Hb(s) = V2(s) / V1(s) and Hc(s) = V2(s) / Vs(s) instead of Hb(s) = V1(s) / V2
(s) and Hc(s) = V1(s) / Vs(s).
http://www.clarkson.edu/~svoboda/errata/6th.html (2 of 2)5/10/2004 7:41:43 PM
Chapter 1 – Electric Circuit Variables
Exercises
i (t ) = 8 t 2 − 4 t A
Ex. 1.3-1
q (t ) =
Ex. 1.3-3
∫
t
0
i dτ + q(0) =
∫
t
0
t
8
8
(8τ 2 − 4τ ) dτ + 0 = τ 3 −2τ 2 = t 3 − 2 t 2 C
0
3
3
q ( t ) = ∫ i (τ ) dτ + q ( 0 ) = ∫ 4sin 3τ dτ + 0 = −
t
t
0
0
4
4
4
t
cos 3τ 0 = − cos 3 t + C
3
3
3
Ex. 1.3-4
dq ( t )
i (t ) =
dt
Ex. 1.4-1
Ex. 1.4-2
Ex. 1.4-3
Ex. 1.4-4
0

i (t ) = 2

−2( t − 2 )
−2e
t <0
0< t < 2
t >2
i1 = 45 µA = 45 × 10-6 A < i2 = 0.03 mA = .03 × 10-3 A = 3 × 10-5 A < i3 = 25 × 10-4 A
∆ q = i∆ t =
i=
( 4000 A )( 0.001 s )
= 4 C
∆ q 45 × 10−9
=
= 9 × 10−6 = 9 µA
−3
∆t
5 × 10
electron  
C 

i = 10 billion
1.602 ×10−19
=


s
electron 


C 

9 electron  
1.602 × 10−19
10×10


s
electron 

electron
C
= 1010 × 1.602 ×10−19
electron
s
C
= 1.602 × 10−9 = 1.602 nA
s
1-1
Ex. 1.6-1
(a) The element voltage and current do not adhere to the passive convention in
Figures 1.6-1B and 1.6-1C so the product of the element voltage and current
is the power supplied by these elements.
(b) The element voltage and current adhere to the passive convention in Figures
1.6-1A and 1.6-1D so the product of the element voltage and current is the
power delivered to, or absorbed by these elements.
(c) The element voltage and current do not adhere to the passive convention in
Figure 1.6-1B, so the product of the element voltage and current is the power
delivered by this element: (2 V)(6 A) = 12 W. The power received by the
element is the negative of the power delivered by the element, -12 W.
(d) The element voltage and current do not adhere to the passive convention in
Figure 1.6-1B, so the product of the element voltage and current is the power
supplied by this element: (2 V)(6 A) = 12 W.
(e) The element voltage and current adhere to the passive convention in Figure
1.6-1D, so the product of the element voltage and current is the power
delivered to this element: (2 V)(6 A) = 12 W. The power supplied by the
element is the negative of the power delivered to the element, -12 W.
Problems
Section 1-3 Electric Circuits and Current Flow
i (t ) =
P1.3-1
(
)
d
4 1 − e −5t = 20 e −5t A
dt
(
)
t
t
t
t
4
4
q ( t ) = ∫ i (τ ) dτ + q ( 0 ) = ∫ 4 1 − e −5τ dτ + 0 = ∫ 4 dτ − ∫ 4 e−5τ dτ = 4 t + e−5t − C
0
0
0
0
5
5
P1.3-2
q ( t ) = ∫ i (τ ) dτ = ∫ 0 dτ = 0 C for t ≤ 2 so q(2) = 0.
P1.3-3
t
−∞
t
t
−∞
q ( t ) = ∫ i (τ ) dτ + q ( 2 ) = ∫ 2 dτ = 2 τ 2 = 2 t − 4 C for 2 ≤ t ≤ 4. In particular, q(4) = 4 C.
t
t
q ( t ) = ∫ i (τ ) dτ + q ( 4 ) = ∫ −1 dτ + 4 = − τ 4 + 4 = 8 − t C for 4 ≤ t ≤ 8. In particular, q(8) = 0 C.
2
t
2
t
t
q ( t ) = ∫ i (τ ) dτ + q ( 8 ) = ∫ 0 dτ + 0 = 0 C for 8 ≤ t .
4
t
4
t
8
8
1-2
P1.3-4
i = 600 A = 600
C
s
C
s
mg
Silver deposited = 600 ×20 min×60
= 8.05×105 mg=805 g
×1.118
s
min
C
Section 1-6 Power and Energy
P1.6-1
a.) q =
∫ i dt = i∆t = (10 A )( 2 hrs )( 3600s/hr ) = 7.2×10
b.) P = v i = (110 V )(10 A ) = 1100 W
c.) Cost =
P1.6-2
C
0.06$
× 1.1kW × 2 hrs = 0.132 $
kWhr
P = ( 6 V )(10 mA ) = 0.06 W
∆t =
P1.6-3
4
200 W⋅s
∆w
=
= 3.33×103 s
0.06 W
P
for 0 ≤ t ≤ 10 s:
v = 30 V and i =
for 10 ≤ t ≤ 15 s: v ( t ) = −
30
t = 2t A ∴ P = 30(2t ) = 60t W
15
25
t + b ⇒ v (10 ) = 30 V ⇒ b = 80 V
5
v(t ) = −5t + 80 and i (t ) = 2t A ⇒ P = ( 2t )( −5t +80 ) = −10t 2 +160t W
for 15 ≤ t ≤ 25 s: v = 5 V and i (t ) = −
i (25) = 0
30
t +b A
10
⇒ b = 75 ⇒ i (t ) = −3t + 75 A
∴ P = ( 5 )( −3t + 75 ) = −15t + 375 W
1-3
Energy = ∫ P dt =
= 30t 2
10
0
∫
10
0
60t dt + ∫10 (160t −10t 2 ) dt + ∫15 ( 375−15t ) dt
15
25
+ 80t 2 − 10 t 3 + 375t − 15 t 2 = 5833.3 J
3 10
2 15
15
25
P1.6-4
a.) Assuming no more energy is delivered to the battery after 5 hours (battery is fully
charged).
5( 3600 )
t
5 ( 3600 ) 
0.5 τ 
0.5 2
2 11 +
w = ∫ Pdt = ∫0 vi dτ = ∫0
 dτ = 22 t + 3600 τ
3600 

0
= 441× 103 J = 441 kJ
b.) Cost = 441kJ ×
P1.6-5
1 hr
10¢
×
= 1.23¢
3600s kWhr
p (t ) =
1
1
( cos 3 t )( sin 3 t ) = sin 6 t
3
6
1
p ( 0.5 ) = sin 3 = 0.0235 W
6
1
p (1) = sin 6 = −0.0466 W
6
1-4
Here is a MATLAB program to plot p(t):
clear
t0=0;
tf=2;
dt=0.02;
t=t0:dt:tf;
%
%
%
%
initial time
final time
time increment
time
v=4*cos(3*t);
i=(1/12)*sin(3*t);
% device voltage
% device current
for k=1:length(t)
p(k)=v(k)*i(k);
end
% power
plot(t,p)
xlabel('time, s');
ylabel('power, W')
P1.6-6
p ( t ) = 16 ( sin 3 t )( sin 3 t ) = 8 ( cos 0 − cos 6 t ) = 8 − 8cos 6 t W
Here is a MATLAB program to plot p(t):
clear
t0=0;
tf=2;
dt=0.02;
t=t0:dt:tf;
%
%
%
%
initial time
final time
time increment
time
v=8*sin(3*t);
i=2*sin(3*t);
% device voltage
% device current
for k=1:length(t)
p(k)=v(k)*i(k);
end
% power
plot(t,p)
xlabel('time, s');
ylabel('power, W')
1-5
(
)
(
)
p ( t ) = 4 1 − e −2 t × 2 e −2 t = 8 1 − e−2 t e −2t
P1.6-7
Here is a MATLAB program to plot p(t):
clear
t0=0;
tf=2;
dt=0.02;
t=t0:dt:tf;
%
%
%
%
initial time
final time
time increment
time
v=4*(1-exp(-2*t));
i=2*exp(-2*t);
% device voltage
% device current
for k=1:length(t)
p(k)=v(k)*i(k);
end
% power
plot(t,p)
xlabel('time, s');
ylabel('power, W')
P1.6-8
P = V I =3 × 0.2=0.6 W
w = P ⋅ t = 0.6 × 5 × 60=180 J
1-6
Verification Problems
VP 1-1
Notice that the element voltage and current of each branch adhere to the passive convention. The
sum of the powers absorbed by each branch are:
(-2 V)(2 A)+(5 V)(2 A)+(3 V)(3 A)+(4 V)(-5 A)+(1 V)(5 A) = -4 W + 10 W + 9 W -20 W + 5 W
=0W
The element voltages and currents satisfy conservation of energy and may be correct.
VP 1-2
Notice that the element voltage and current of some branches do not adhere to the passive
convention. The sum of the powers absorbed by each branch are:
-(3 V)(3 A)+(3 V)(2 A)+ (3 V)(2 A)+(4 V)(3 A)+(-3 V)(-3 A)+(4 V)(-3 A)
= -9 W + 6 W + 6 W + 12 W + 9 W -12 W
≠0W
The element voltages and currents do not satisfy conservation of energy and cannot be correct.
Design Problems
DP 1-1
The voltage may be as large as 20(1.25) = 25 V and the current may be as large as (0.008)(1.25)
= 0.01 A. The element needs to be able to absorb (25 V)(0.01 A) = 0.25 W continuously. A
Grade B element is adequate, but without margin for error. Specify a Grade B device if you trust
the estimates of the maximum voltage and current and a Grade A device otherwise.
1-7
DP1-2
(
)
(
)
p ( t ) = 20 1 − e −8 t × 0.03 e −8 t = 0.6 1 − e−8t e−8t
Here is a MATLAB program to plot p(t):
clear
t0=0;
tf=1;
dt=0.02;
t=t0:dt:tf;
v=20*(1-exp(-8*t));
i=.030*exp(-8*t);
for k=1:length(t)
p(k)=v(k)*i(k);
end
%
%
%
%
initial time
final time
time increment
time
% device voltage
% device current
% power
plot(t,p)
xlabel('time, s');
ylabel('power, W')
Here is the plot:
The circuit element must be able to absorb 0.15 W.
1-8
Chapter 2 - Circuit Elements
Exercises
Ex. 2.3-1
m ( i1 + i 2 ) = mi1 + mi 2
⇒ superposition is satisfied
m ( a i1 ) = a ( mi1 ) ⇒ homogeneity is satisfied
Therefore the element is linear.
Ex. 2.3-2
m ( i1 + i 2 ) + b = mi1 + mi 2 + b ≠ ( mi1 + b ) + ( mi 2 + b ) ⇒ superposition is not satisfied
Therefore the element is not linear.
v 2 (10 )
P= =
=1 W
R 100
Ex. 2.5-1
2
Ex. 2.5-2
P=
v 2 (10 cos t ) 2
=
= 10 cos 2 t W
R
10
Ex. 2.8-1
ic = − 1.2
A, v d = 24
id = 4 ( − 1.2) = − 4.8
V
A
id and vd adhere to the passive convention so
P = vd id = (24) (−4.8) = −115.2
W
is the power received by the dependent source
2-1
Ex. 2.8-2
vc = −2 V, id = 4 vc = −8 A and vd = 2.2 V
id and vd adhere to the passive convention so
P = vd id = (2.2) (−8) = −17.6 W
is the power received by the dependent source. The power supplied by the
dependent source is 17.6 W.
Ex. 2.8-3
ic = 1.25 A, vd = 2 ic = 2.5 V and id = 1.75 A
id and vd adhere to the passive convention so
P = vd id = (2.5) (1.75) = 4.375 W
is the power received by the dependent source.
2-2
Ex. 2.9-1
θ = 45° , I = 2 mA, R p = 20 kΩ
a=
θ
45
(20 kΩ) = 2.5 kΩ
⇒ aR =
p 360
360
vm = (2 ×10−3 )(2.5 ×103 ) = 5 V
Ex. 2.9-2
v = 10 V, i = 280 µA, k = 1
µA
°K
for AD590
 °K 
i
 = 280° K
i = kT ⇒ T = = (280µA)1
 µA 
k


Ex. 2.10-1
At t = 4 s both switches are open, so i = 0 A.
Ex. 2.10.2
At t = 4 s the switch is in the up position, so v = i R = (2 mA)(3 kΩ) = 6V .
At t = 6 s the switch is in the down position, so v = 0 V.
Problems
Section 2-3 Engineering and Linear Models
P2.3-1
The element is not linear. For example, doubling the current from 2 A to 4 A does not double the
voltage. Hence, the property of homogeneity is not satisfied.
P2.3-2
(a) The data points do indeed lie on a straight line. The slope of the line is 0.12 V/A and
the line passes through the origin so the equation of the line is v = 0.12 i . The element is indeed
linear.
(b) When i = 40 mA, v = (0.12 V/A)×(40 mA) = (0.12 V/A)×(0.04 A) = 4.8 mV
4
= 33 A = 33 A.
(c) When v = 4 V, i =
0.12
2-3
P2.3-3
(a) The data points do indeed lie on a straight line. The slope of the line is 256.5 V/A and
the line passes through the origin so the equation of the line is v = 256.5i . The element is indeed
linear.
(b) When i = 4 mA, v = (256.5 V/A)×(4 mA) = (256.5 V/A)×(0.004 A) = 1.026 V
12
(c) When v = 12 V, i =
= 0.04678 A = 46.78 mA.
256.5
Let i = 1 A , then v = 3i + 5 = 8 V. Next 2i = 2A but 16 = 2v ≠ 3(2i) + 5 = 11.. Hence,
the property of homogeneity is not satisfied. The element is not linear.
P2.3-4
Section 2-5 Resistors
P2.5-1
i = is = 3 A and v = Ri = 7 × 3 = 21 V
v and i adhere to the passive convention
∴ P = v i = 21 × 3 = 63 W
is the power absorbed by the resistor.
P2.5-2
i = is = 3 mA and v = 24 V
R =
24
v
=
= 8000 = 8 k Ω
.003
i
P = (3×10 −3 )× 24 = 72×10 −3 = 72 mW
P2.5-3
v = vs =10 V and R = 5 Ω
i =
v
10
=
=2 A
R
5
v and i adhere to the passive convention
∴ p = v i = 2⋅10 = 20 W
is the power absorbed by the resistor
2-4
v = vs = 24 V and i = 2 A
P2.5-4
R=
v 24
=
= 12 Ω
2
i
p = vi = 24⋅2 = 48 W
v1 = v 2 = vs = 150 V;
P2.5-5
R1 = 50 Ω; R2 = 25 Ω
v 1 and i1 adhere to the passive convention so
i1 =
v 1 150
=
=3 A
R 1 50
v
150
v2 and i 2 do not adhere to the passive convention so i 2 = − 2 = −
= −6 A
R2
25
The power absorbed by R1 is P1 = v1 i1 = 150 ⋅ 3 = 450 W
The power absorbed by R 2 is P 2 = − v 2i 2 = −150(−6) = 900 W
i1 = i 2 = is = 2 A ;
P2.5-6
R1 =4 Ω and R2 = 8 Ω
v 1 and i 1 do not adhere to the passive convention so
v 1 =− R 1 i 1 =−4⋅2=−8 V.
The power absorbed by R 1 is
P1 =−v 1i 1 =−(−8)(2) = 16 W.
v2 and i 2 do adhere to the passive convention so v2 = R 2 i 2 = 8 ⋅ 2 = 16 V .
The power absorbed by R 2 is P 2 = v 2i 2 = 16 ⋅ 2 = 32 W.
P2.5-7
Model the heater as a resistor, then
with a 250 V source: P =
(250) 2
v2
v2
⇒ R =
=
= 62.5 Ω
1000
R
P
v 2 (210) 2
= 705.6 W
with a 210 V source: P = =
R
62.5
2-5
P2.5-8
The current required by the mine lights is: i =
P 5000 125
A
=
=
3
v 120
Power loss in the wire is : i 2 R
Thus the maximum resistance of the copper wire allowed is
0.05P 0.05×5000
=
= 0.144 Ω
(125/3) 2
i2
now since the length of the wire is L = 2×100 = 200 m = 20,000 cm
R=
thus R = ρ L / A
with ρ = 1.7×10−6 Ω⋅ cm from Table 2.5−1
A=
ρL
R
=
1.7×10−6 ×20,000
= 0.236 cm 2
0.144
Section 2-6 Independent Sources
P2.6-1
v s 15
2
=
= 3 A and P = R i 2 = 5 ( 3 ) = 45 W
R
5
(b) i and P do not depend on is .
(a) i =
The values of i and P are 3 A and 45 W, both when i s = 3 A and when i s = 5 A.
P2.6-2
v 2 102
(a) v = R i s = 5 ⋅ 2 = 10 V and P = =
= 20 W
R
5
(b) v and P do not depend on v s .
The values of v and P are 10V and 20 W both when v s = 10 V and when v s = 5 V
2-6
P2.6-3
Consider the current source:
i s and v s do not adhere to the passive convention,
so Pcs =i s v s =3⋅12 = 36 W
is the power supplied by the current source.
Consider the voltage source:
i s and v s do adhere to the passive convention,
so Pvs = i s vs =3 ⋅12 = 36 W
is the power absorbed by the voltage source.
∴ The voltage source supplies −36 W.
P2.6-4
Consider the current source:
i s and vs adhere to the passive convention
so Pcs = i s vs =3 ⋅12 = 36 W
is the power absorbed by the current source.
Current source supplies − 36 W.
Consider the voltage source:
i s and vs do not adhere to the passive convention
so Pvs = i s vs = 3 ⋅12 =36 W
is the power supplied by the voltage source.
P2.6-5
(a) P = v i = (2 cos t ) (10 cos t ) = 20 cos 2 t mW
1 1

(b) w = ∫0 P dt = ∫0 20 cos t dt = 20 t + sin 2t  = 10 + 5 sin 2 mJ
2 4
0
1
1
1
2
2-7
Section 2-7 Voltmeters and Ammeters
P2.7-1
(a) R =
5
v
=
= 10 Ω
i 0.5
(b) The voltage, 12 V, and the
current, 0.5 A, of the voltage
source adhere to the passive
convention so the power
P = 12 (0.5) = 6 W
is the power received by the
source. The voltage source
delivers -6 W.
P2.7-2
The voltmeter current is zero
so the ammeter current is
equal to the current source
current except for the
reference direction:
i = -2 A
The voltage v is the voltage of
the current source. The power
supplied by the current source
is 40 W so
40 = 2 v ⇒ v = 20 V
2-8
Section 2-8 Dependent Sources
P2.8-1
r =
vb 8
= =4 Ω
ia 2
P2.8-2
vb = 8 V ; g v b = i a = 2 A ; g =
ia 2
A
= = 0.25
vb 8
V
P2.8-3
i b = 8 A ; d i b = i a = 32A ; d =
i a 32
A
=
=4
ib
8
A
P2.8-4
va = 2 V ; b va = vb = 8 V ; b =
vb 8
V
= =4
va 2
V
Section 2-9 Transducers
P2.9-1
a=
θ =
θ
360
,
θ =
360 vm
Rp I
(360)(23V)
= 75.27°
(100 kΩ)(1.1 mA)
P2.9-2
AD590 : k =1
µA
°
,
K
v =20 V (voltage condition satisfied)
4 µ A < i < 13 µ A 


i
T =

k
⇒
4 ° K< T <13° K
2-9
Section 2-10 Switches
P2.10-1
At t = 1 s the left switch is open and the
right switch is closed so the voltage
across the resistor is 10 V.
i=
v
10
= 2 mA
=
R
5×103
At t = 4 s the left switch is closed and the right switch is open so the voltage across the resistor is
15 V.
i=
v
15
= 3 mA
=
R
5×103
P2.10-2
At t = 1 s the current in the resistor
is 3 mA so v = 15 V.
At t = 4 s the current in the resistor
is 0 A so v = 0 V.
Verification Problems
VP2-1
vo =40 V and i s = − (−2) = 2 A. (Notice that the ammeter measures − i s rather than i s .)
So
vo
40
V
=
= 20
2
A
is
Your lab partner is wrong.
VP2-2
We expect the resistor current to be i =
vs 12
=
= 0.48 A. The power absorbed by
R 25
this resistor will be P = i vs = (0.48) (12) = 5.76 W.
A half watt resistor can't absorb this much power. You should not try another resistor.
2-10
Design Problems
DP2-1
1.)
10
10
> 0.04 ⇒ R <
= 250 Ω
R
0.04
2.)
102 1
<
⇒ R > 200 Ω
R
2
Therefore 200 < R < 250 Ω. For example, R = 225 Ω.
DP2-2
1.) 2 R > 40 ⇒ R > 20 Ω
15
2.) 2 2 R < 15 ⇒ R < = 3.75 Ω
4
Therefore 20 < R < 3.75 Ω. These conditions cannot satisfied simultaneously.
DP2-3
P1 = ( 30 mA ) ⋅ (1000 Ω ) = (.03) (1000 ) = 0.9 W < 1 W
2
2
P2 = ( 30 mA ) ⋅ ( 2000 Ω ) = (.03) ( 2000 ) = 1.8 W < 2 W
2
2
2
2
P3 = ( 30 mA ) ⋅ ( 4000 Ω ) = (.03) ( 4000 ) = 3.6 W < 4 W
2-11
Chapter 3 – Resistive Circuits
Exercises
Ex 3.3-1
Apply KCL at node a to get
Apply KCL at node c to get
Apply KCL at node b to get
2 + 1 + i3 = 0 ⇒ i3 = -3 A
2 + 1 = i4 ⇒ i4 = 3 A
i3 + i6 = 1 ⇒ -3 + i6 = 1 ⇒ i6 = 4 A
Apply KVL to the loop consisting of elements A and B to get
-v2 – 3 = 0 ⇒ v2 = -3 V
Apply KVL to the loop consisting of elements C, E, D, and A to get
3 + 6 + v4 – 3 = 0 ⇒ v4 = -6 V
Apply KVL to the loop consisting of elements E and F to get
v6 – 6 = 0 ⇒ v6 = 6 V
Check: The sum of the power supplied by all branches is
-(3)(2) + (-3)(1) – (3)(-3) + (-6)(3) – (6)(1) + (6)(4) = -6 - 3 + 9 - 18 - 6 + 24 = 0
3-1
Ex 3.3-2
Apply KCL at node a to
determine the current in the
horizontal resistor as shown.
Apply KVL to the loop
consisting of the voltages source
and the two resistors to get
-4(2-i) + 4(i) - 24 = 0 ⇒ i = 4 A
Ex 3.3-3
−18 + 0 − 12 − va = 0 ⇒ va = −30 V and im =
Ex 3.3-4
−va − 10 + 4va − 8 = 0 ⇒ va =
2
va + 3 ⇒ im = 9 A
5
18
= 6 V and vm = 4 va = 24 V
3
Ex 3.4-1
From voltage division

3 
 = 3V
 3+9 
v3 = 12 
then
v
i = 3 = 1A
3
The power absorbed by the resistors is: (12 ) ( 6 ) + (12 ) ( 3) + (12 ) ( 3) = 12 W
The power supplied by the source is (12)(1) = 12 W.
3-2
Ex 3.4-2
P = 6 W and R1 = 6 Ω
i2 =
6
P
=
= 1 or i =1 A
6
R1
v0 = i R1 =(1) (6)=6V
from KVL: − v
+ i (2 + 4 + 6 + 2) = 0
s
⇒ v = 14 i = 14 V
s
Ex 3.4-3
Ex 3.4-4
25
From voltage division ⇒ v =
(8) = 2 V
m 25+75
25
From voltage division ⇒ v =
( −8 ) = −2 V
m 25+75
Ex. 3.5-1
1
R
eq
=
1
1
1
1
4
+ 3+ 3+ 3= 3
3
10 10 10 10 10
⇒ R
eq
=
By current division, the current in each resistor =
Ex 3.5-2
103
1
kΩ
=
4
4
1 -3
1
(10 ) =
mA
4
4
10
From current division ⇒ i =
( −5 ) = − 1 A
m 10+40
3-3
Problems
Section 3-3 Kirchoff’s Laws
P3.3-1
Apply KCL at node a to get
2 + 1 = i + 4 ⇒ i = -1 A
The current and voltage of element B adhere to the passive convention so (12)(-1) = -12 W is
power received by element B. The power supplied by element B is 12 W.
Apply KVL to the loop consisting of elements D, F, E, and C to get
4 + v + (-5) – 12 = 0 ⇒ v = 13 V
The current and voltage of element F do not adhere to the passive convention so (13)(1) = 13 W
is the power supplied by element F.
Check: The sum of the power supplied by all branches is
-(2)(-12) + 12 – (4)(12) + (1)(4) + 13 – (-1)(-5) = 24 +12 – 48 + 4 +13 –5 = 0
3-4
P3.3-2
Apply KCL at node a to get
Apply KCL at node b to get
2 = i2 + 6 = 0 ⇒ i2 = -4 A
3 = i4 + 6 ⇒ i4 = -3 A
Apply KVL to the loop consisting of elements A and B to get
-v2 – 6 = 0 ⇒ v2 = -6 V
Apply KVL to the loop consisting of elements C, D, and A to get
-v3 – (-2) – 6 = 0 ⇒ v4 = -4 V
Apply KVL to the loop consisting of elements E, F and D to get
4 – v6 + (-2) = 0 ⇒ v6 = 2 V
Check: The sum of the power supplied by all branches is
-(6)(2) – (-6)(-4) – (-4)(6) + (-2)(-3) + (4)(3) + (2)(-3) = -12 - 24 + 24 + 6 + 12 – 6 = 0
3-5
KVL : −12 − R 2 (3) + v = 0 (outside loop)
P3.3-3
v = 12 + 3R 2 or R 2 =
KCL
i+
v − 12
3
12
− 3 = 0 (top node)
R1
i = 3−
12
12
or R1 =
3−i
R1
v = 12 + 3 ( 3) = 21 V
(a)
i = 3−
(b)
R2 =
12
=1 A
6
2 − 12
10
12
= − Ω ; R1 =
=8Ω
3
3
3 − 1.5
(checked using LNAP 8/16/02)
(c)
24 = − 12 i, because 12 and i adhere to the passive convention.
∴ i = − 2 A and R1 =
9 = 3v,
∴ v = 3V
12
= 2.4 Ω
3+ 2
because 3 and v do not adhere to the passive convention
and R 2 =
3 − 12
= −3 Ω
3
The situations described in (b) and (c) cannot occur if R1 and R2 are required to be nonnegative.
3-6
P3.3-4
12
i = =2A
1 6
20
= 5A
i =
2
4
i = 3−i = − 2 A
3
2
i = i +i = 3A
4 2 3
Power absorbed by the 4 Ω resistor = 4 ⋅ i 2 = 100 W
2
Power absorbed by the 6 Ω resistor = 6 ⋅ i 2 = 24 W
1
Power absorbed by the 8 Ω resistor = 8 ⋅ i 2 = 72 W
4
(checked using LNAP 8/16/02)
v1 = 8 V
P3.3-5
v2 = −8 + 8 + 12 = 12 V
(checked using LNAP 8/16/02)
P3.3-6
v3 = 2⋅ 4 = 8 V
v2
4Ω : P = 3 = 16 W
4
v22
= 24 W
6Ω : P =
6
v2
8Ω : P = 1 = 8 W
8
P2 mA = − 3 × ( 2 ×10−3 )  = −6 × 10−3 = −6 mW
P1 mA = −  −7 × (1×10−3 )  = 7 × 10−3 = 7 mW
(checked using LNAP 8/16/02)
3-7
P3.3-7
P2 V = +  2 × (1× 10−3 )  = 2 × 10−3 = 2 mW
P3 V = + 3 × ( −2 × 10−3 )  = −6 ×10−3 = −6 mW
(checked using LNAP 8/16/02)
P3.3-8
KCL: iR = 2 + 1 ⇒ iR = 3 A
KVL: vR + 0 − 12 = 0 ⇒ vR = 12 V
∴ R=
vR 12
=
=4Ω
3
iR
(checked using LNAP 8/16/02)
P3.3-9
KVL: vR + 56 + 24 = 0 ⇒ vR = −80 V
KCL: iR + 8 = 0 ⇒ iR = −8 A
∴ R=
vR −80
=
= 10 Ω
iR
−8
(checked using LNAP 8/16/02)
3-8
P3.3-10
KCL at node b:
−1.9
5.61 3.71 − 5.61 12 − 5.61
=
+
⇒ 0.801 =
+ 1.278
7
5
R1
R1
⇒ R1 =
KCL at node a:
1.9
= 3.983 ≈ 4 Ω
1.278 − 0.801
3.71 3.71 − 5.61 3.71 − 12
−8.29
+
+
= 0 ⇒ 1.855 + ( −0.475 ) +
=0
2
4
R2
R2
⇒ R2 =
8.29
= 6.007 ≈ 6 Ω
1.855 − 0.475
(checked using LNAP 8/16/02)
3-9
Section 3-4 A Single-Loop Circuit – The Voltage Divider
P3.4-1
6
6
12 =
12 = 4 V
v =
1 6+3+5+ 4
18
3
5
10
12 = 2 V ; v =
12 =
V
v =
2 18
3 18
3
4
8
12 = V
v =
4 18
3
(checked using LNAP 8/16/02)
P3.4-2
(a) R = 6 + 3 + 2 + 4 = 15 Ω
28 28
=
= 1.867 A
(b) i =
R 15
( c ) p = 28 ⋅ i =28(1.867)=52.27 W
(28 V and i do not adhere
to the passive convention.)
(checked using LNAP 8/16/02)
3-10
P3.4-3
i R2 = v = 8 V
12 = i R1 + v = i R1 + 8
⇒ 4 = i R1
8
8
4 4 ⋅ 100
; R1 = =
=
= 50 Ω
8
R 2 100
i
4
4
8 8 ⋅ 100
; R2 = =
= 200 Ω
(b) i = =
4
R1 100
i
4
8
( c ) 1.2 = 12 i ⇒ i = 0.1 A ; R1 = = 40 Ω; R2 = = 80 Ω
i
i
(a)
i=
(checked using LNAP 8/16/02)
P3.4-4
Voltage division
16
12 = 8 V
v1 =
16 + 8
4
v3 =
12 = 4 V
4+8
KVL: v3 − v − v1 = 0
v = −4 V
(checked using LNAP 8/16/02)
P3.4-5
v

⇒ R = 50  s − 1
v

 o 
with v = 20 V and v > 9 V, R < 61.1 Ω 
s
0

 R = 60 Ω
with v = 28 V and v < 13 V, R > 57.7 Ω 
s
0

 100 
v
using voltage divider: v = 
0  100 + 2 R  s
3-11
P3.4-6
 240 
a.) 
 18 = 12 V
 120 + 240 
18


b.) 18 
 = 0.9 W
 120 + 240 
 R 
c.) 
 18 = 2 ⇒ 18 R = 2 R + 2 (120 ) ⇒ R = 15 Ω
 R + 120 
R
d.) 0.2 =
⇒ ( 0.2 )(120 ) = 0.8 R ⇒ R = 30 Ω
R + 120
(checked using LNAP 8/16/02)
3-12
Section 3-5 Parallel Resistors and Current Division
P3.5-1
i =
1
i =
2
i =
3
i =
4
1
1
1
6
4=
4= A
1 + 1 + 1 +1
1+ 2 + 3 + 6
3
6 3 2 1
1
2
3
4 = A;
1 + 1 + 1 +1
3
6 3 2 1
1
2
4 =1 A
1 + 1 + 1 +1
6 3 2 1
1
4=2 A
1 + 1 + 1 +1
6 3 2
P3.5-2
(a)
(b)
(c)
1 1 1 1 1
= + + = ⇒ R = 2Ω
R 6 12 4 2
v = 6 ⋅ 2 = 12 V
p = 6 ⋅12 = 72 W
P3.5-3
i=
8
8
or R1 =
R1
i
8 = R 2 (2 − i ) ⇒ i = 2 −
(a)
(b)
8
8
or R 2 =
2−i
R2
i = 2−
8 4
8
= A ; R1 =
=6Ω
4
12 3
3
8 2
8
i = = A ; R2 =
=6Ω
2
12 3
2−
3
3-13
( c ) R1 = R 2
2 ⋅
R1 R 2
R1 + R 2
1
will cause i= 2 = 1 A. The current in both R1 and R 2 will be 1 A.
2
1
= 8 ; R1 = R 2 ⇒ 2 ⋅ R1 = 8 ⇒ R1 = 8 ∴ R1 = R 2 = 8 Ω
2
P3.5-4
Current division:
8
i =
−6 = −2 A
1 16 + 8 ( )
8
i =
−6 = −3 A
2 8+8( )
i = i −i
1 2
P3.5-5
= +1 A


R
1  i and
current division: i = 
2 R + R  s
 1
2
Ohm's Law: v = i R yields
o
2 2
 v  R + R 
2
i =  o  1
s
 R  R

 2 
1 
plugging in R = 4Ω, v > 9 V
o
1
and R = 6Ω, v < 13 V gives
o
1
So any 3.15 A < i < 3.47 A
s
gives i > 3.15 A
s
i < 3.47 A
s
keeps 9 V < v < 13 V.
o
3-14
P3.5-6
 24 
a) 
 1.8 = 1.2 A
 12 + 24 
 R 
b) 
 2 = 1.6 ⇒ 2 R = 1.6 R + 1.6 (12 ) ⇒ R = 48 Ω
 R + 12 
R
c) 0.4 =
⇒ ( 0.4 )(12 ) = 0.6 R ⇒ R = 8 Ω
R + 12
Section 3-7 Circuit Analysis
P3.7-1
(a)
(b)
(c)
48 ⋅ 24
= 32 Ω
48 + 24
32 ⋅ 32
v = 32 + 32 24 = 16 V ;
32 ⋅ 32
8+
32 + 32
16 1
= A
i=
32 2
48
1
1
⋅ = A
i2 =
48 + 24 2
3
R = 16 +
3-15
P3.7-2
3⋅ 6
=6Ω
3+ 6
1 1 1
=
+ + ⇒ R p = 2.4 Ω then
12 6 6
(a) R1 = 4 +
(b)
1
Rp
R 2 = 8 + R p = 10.4 Ω
(c) KCL: i2 + 2 = i1 and − 24 + 6 i2 + R 2i1 = 0
⇒ −24+6 (i1 −2)+10.4i1 = 0
36
=2.2 A ⇒ v1 =i1 R 2 =2.2 (10.4)=22.88 V
⇒ i1 =
16.4
1
6
(d ) i2 =
( 2.2 ) = 0.878 A,
1 1 1
+ +
6 6 12
v2 = ( 0.878 ) (6) = 5.3 V
(e) i3 =
6
2
i2 = 0.585 A ⇒ P = 3 i3 = 1.03 W
3+ 6
3-16
P3.7-3
Reduce the circuit from the right side by repeatedly replacing series 1 Ω resistors in parallel with
a 2 Ω resistor by the equivalent 1 Ω resistor
This circuit has become small enough to be easily analyzed. The vertical 1 Ω resistor is
equivalent to a 2 Ω resistor connected in parallel with series 1 Ω resistors:
i1 =
1+1
(1.5 ) = 0.75 A
2 + (1 + 1)
3-17
P3.7-4
(a)
1
1
1 1
=
+ +
⇒ R2 = 4 Ω
R2 24 12 8
and
R1 =
(10 + 8) ⋅ 9
= 6Ω
10 + 8 + 9
b
g
(b)
First, apply KVL to the left mesh to get −27 + 6 ia + 3 ia = 0 ⇒ ia = 3 A . Next,
apply KVL to the left mesh to get 4 ib − 3 ia = 0 ⇒ ib = 2.25 A .
(c)
1
8
2.25 = 1125
.
A
i2 =
1 1 1
+ +
24 8 12
b gLM b10 +98g + 9 3OP = −10 V
Q
N
and v1 = − 10
3-18
P3.7-5
30
v1 = 6 ⇒ v1 = 8 V
10 + 30
R2
12 = 8 ⇒ R2 = 20 Ω
R2 + 10
20 =
R1 10 + 30
R1 + 10 + 30
b
b
g
g
⇒ R1 = 40 Ω
Alternate values that can be used to change the numbers in this problem:
meter reading, V
6
4
4
4.8
Right-most resistor, Ω
30
30
20
20
R1, Ω
40
10
15
30
3-19
P3.7-6
P3.7-7
1× 10−3 =
24
12 ×103 + R p
12 × 10 = R p
3
⇒ R p = 12 ×103 = 12 kΩ
( 21×10 ) R
=
( 21×10 ) + R
3
3
⇒ R = 28 kΩ
P3.7-8


130 500
= 15.963 V
Voltage division ⇒ v = 50 
 130 500 + 200 + 20 


 100 
 10 
∴v = v 
 = (15.963)   = 12.279 V
h
 100 + 30 
 13 
v
∴ i = h = .12279 A
h 100
3-20
P3.7-9
3-21
P3.7-10
Req =
ia = −
15 ( 20 + 10 )
= 10 Ω
15 + ( 20 + 10 )
60
 30   60 
 20 
= −6 A, ib = 
 = 4 A, vc = 
 
 ( −60 ) = −40 V
Req
 30 + 15   Req 
 20 + 10 
P3.7-11
a)
Req = 24 12 =
b)
(24)(12)
=8Ω
24 + 12
from voltage division:
100
5
 20  100
v = 40 
V∴ i = 3 = A
=
x
x
3
20 3
 20 + 4 
from current division: i = i
5
 8 
=
A
x  8 + 8 
6
3-22
9 + 10 + 17 = 36 Ω
P3.7-12
a.)
b.)
36 (18 )
= 12 Ω
36+18
36 R
= 18 ⇒ 18 R = (18 )( 36 ) ⇒ R = 36 Ω
36+R
2 R( R ) 2
= R
2R + R 3
v 2 240
=
=1920 W
Pdeliv. =
Req 2 R
to ckt
3
Thus R =45 Ω
P3.7-13
Req =
P3.7-14
R = 2 + 1 + ( 6 12 ) + ( 2 2 ) = 3 + 4 + 1 = 8 Ω
eq
∴i =
40 40
=
=5 A
Req
8
( )
 6 
= ( 5) 1 = 5 A
i1 = i 

3
3
 6 + 12 
 2 
5
1
i2 = i 
 = ( 5) 2 = 2 A
 2+2
( )
from current division
3-23
Verification Problems
VP3-1
KCL at node a: i = i + i
3 1 2
− 1.167 = − 0.833 + ( −0.333)
− 1.167= − 1.166 OK
KVL loop consisting of the vertical
6 Ω resistor, the 3 Ω and4Ω resistors,
and the voltage source:
6i + 3i + v + 12 = 0
3
2
yields v = −4.0 V not v = −2.0 V
VP3-2
reduce circuit: 5+5=10 in parallel with 20 Ω gives 6.67Ω
 6.67 
by current division: i = 
 5 = 1.25 A
 20 + 6.67 
∴Reported value was correct.
VP3-3
320


v =
( 24 ) = 6.4 V
o  320 + 650 + 230 
∴Reported value was incorrect.
3-24
VP3-4
KVL bottom loop: − 14 + 0.1iA + 1.2iH = 0
KVL right loop: − 12 + 0.05iB + 1.2iH = 0
KCL at left node: iA + iB = iH
This alone shows the reported results were incorrect.
Solving the three above equations yields:
iA = 16.8 A
iH = 10.3 A
iB = −6.49 A
∴ Reported values were incorrect.
VP3-5
1


Top mesh: 0 = 4 i a + 4 i a + 2  i a + − i b  = 10 ( −0.5 ) + 1 − 2 ( −2 )
2


Lower left mesh: vs = 10 + 2 ( i a + 0.5 − i b ) = 10 + 2 ( 2 ) = 14 V
Lower right mesh: vs + 4 i a = 12 ⇒ vs = 12 − 4 (−0.5) = 14 V
The KVL equations are satisfied so the analysis is correct.
3-25
VP3-6
Apply KCL at nodes b and c to get:
KCL equations:
Node e: −1 + 6 = 0.5 + 4.5
Node a:
0.5 + i c = −1 ⇒ i c = −1.5 mA
Node d:
i c + 4 = 4.5 ⇒ i c = 0.5 mA
That's a contradiction. The given values of ia
and ib are not correct.
Design Problems
DP3-1
Using voltage division:
vm =
R 2 + aR p
R1 + (1 − a ) R p + R 2 + aR p
vm = 8 V when a = 0 ⇒
R2
R1 + R 2 + R p
vm = 12 V when a = 1 ⇒
R2 + R p
R1 + R 2 + R p
The specification on the power of the voltage source indicates
242
1
≤
⇒ R1 + R 2 + R p ≥ 1152 Ω
R1 + R 2 + R p 2
24 =
=
=
R 2 + aR p
R1 + R 2 + R p
24
1
3
1
2
Try Rp = 2000 Ω. Substituting into the equations obtained above using voltage division gives
3R 2 = R1 + R 2 + 2000 and 2 ( R 2 + 2000 ) = R1 + R 2 + 2000 . Solving these equations gives
R1 = 6000 Ω and R 2 = 4000 Ω .
With these resistance values, the voltage source supplies 48 mW while R1, R2 and Rp dissipate
24 mW, 16 mW and 8 mW respectively. Therefore the design is complete.
3-26
DP3-2
Try R1 = ∞. That is, R1 is an open circuit. From KVL, 8 V will appear across R2. Using voltage
200
division,
12 = 4 ⇒ R 2 = 400 Ω . The power required to be dissipated by R2
R 2 + 200
82
1
is
= 0.16 W < W . To reduce the voltage across any one resistor, let’s implement R2 as the
400
8
series combination of two 200 Ω resistors. The power required to be dissipated by each of these
42
1
= 0.08 W < W .
resistors is
200
8
Now let’s check the voltage:
190
210
11.88
< v < 12.12
0
190 + 420
210 + 380
3.700 < v0 < 4.314
4 − 7.5% < v0 < 4 + 7.85%
Hence, vo = 4 V ± 8% and the design is complete.
DP3-3
Vab ≅ 200 mV
v=
10
10
120 Vab =
(120) (0.2)
10 + R
10 + R
240
⇒ R=5Ω
let v = 16 =
10 + R
162
= 25.6W
∴P=
10
DP3-4
N 1
N
1
= N 
i = G v = v where G = ∑
T
T
R
 R
n = 1 Rn
∴N=
iR ( 9 )(12 )
=
= 18 bulbs
v
6
3-27
28
Chapter 4 – Methods of Analysis of Resistive Circuits
Exercises
Ex. 4.3-1
v −v
a
a
b
+
+ 3 = 0 ⇒ 5 v − 3 v = −18
a
b
3
2
v
KCL at a:
KCL at b:
v −v
b
a
− 3 −1 = 0 ⇒ v − v = 8
b
a
2
Solving these equations gives:
va = 3 V and vb = 11 V
Ex. 4.3-2
KCL at a:
v −v
a
a
b
+
+ 3 = 0 ⇒ 3 v − 2 v = −12
a
b
4
2
v
v −v
b
a
b
−
−4=0
3
2
⇒ − 3 v + 5 v = 24
a
b
v
KCL at a:
Solving:
va = −4/3 V and vb = 4 V
Ex. 4.4-1
Apply KCL to the supernode to get
v + 10 v
2+ b
+ b =5
20
30
Solving:
v = 30 V and v = v + 10 = 40 V
b
a b
4-1
Ex. 4.4-2
( vb + 8) − ( −12) + vb = 3
10
40
⇒ v = 8 V and v = 16 V
b
a
Ex. 4.5-1
Apply KCL at node a to express ia as a function of the node voltages. Substitute the result into
vb = 4 ia and solve for vb .
6 vb
+
=i
8 12 a
 9 + vb 
⇒ v = 4i = 4 
 ⇒ v = 4.5 V
b
a
b
 12 


Ex. 4.5-2
The controlling voltage of the dependent source is a node voltage so it is already expressed as a
function of the node voltages. Apply KCL at node a.
v −6 v −4v
a
a = 0 ⇒ v = −2 V
+ a
a
20
15
Ex. 4.6-1
Mesh equations:
−12 + 6 i + 3  i − i  − 8 = 0 ⇒ 9 i − 3 i = 20
1
1
2
 1 2
8 − 3  i − i  + 6 i = 0 ⇒ − 3 i + 9 i = −8
2
1
2
 1 2
Solving these equations gives:
13
1
i =
A and i = − A
1 6
2
6
The voltage measured by the meter is 6 i2 = −1 V.
4-2
Ex. 4.7-1
 3
Mesh equation: 9 + 3 i + 2 i + 4  i +  = 0 ⇒
 4
The voltmeter measures 3 i = −4 V
( 3 + 2 + 4 ) i = −9 − 3
⇒ i=
−12
A
9
Ex. 4.7-2
Mesh equation: 15 + 3 i + 6 ( i + 3) = 0 ⇒
( 3 + 6 ) i = −15 − 6 ( 3)
⇒ i=
−33
2
= −3 A
9
3
Ex. 4.7-3
3
3
= i1 − i 2 ⇒ i1 = + i 2 .
4
4
3

Apply KVL to the supermesh: −9 + 4i1 + 3 i 2 + 2 i 2 = 0 ⇒ 4  + i 2  + 5 i 2 = 9 ⇒ 9 i 2 = 6
4

2
4
so i 2 = A and the voltmeter reading is 2 i 2 = V
3
3
Express the current source current in terms of the mesh currents:
4-3
Ex. 4.7-4
Express the current source current in terms of the mesh currents: 3 = i1 − i 2
⇒ i1 = 3 + i 2 .
Apply KVL to the supermesh: −15 + 6 i1 + 3 i 2 = 0 ⇒ 6 ( 3 + i 2 ) + 3 i 2 = 15 ⇒ 9 i 2 = −3
Finally, i 2 = −
1
A is the current measured by the ammeter.
3
Problems
Section 4-3 Node Voltage Analysis of Circuits with Current Sources
P4.3-1
KCL at node 1:
0=
v −v
−4 − 4 − 2
1
1 2
+
+i =
+
+ i = −1.5 + i ⇒ i = 1.5 A
8
6
8
6
v
(checked using LNAP 8/13/02)
4-4
P4.3-2
KCL at node 1:
v −v
v
1 2
1
+
+ 1 = 0 ⇒ 5 v − v = −20
1 2
20
5
KCL at node 2:
v −v
v −v
1 2
2
3
+2=
⇒ − v + 3 v − 2 v = 40
1
2
3
20
10
KCL at node 3:
v −v
v
2
3
3
+1 =
⇒ − 3 v + 5 v = 30
2
3
10
15
Solving gives v1 = 2 V, v2 = 30 V and v3 = 24 V.
(checked using LNAP 8/13/02)
P4.3-3
KCL at node 1:
v −v
v
4 − 15 4
1 2
1
+
=i ⇒ i =
+
= −2 A
1
5
20 1
5
20
KCL at node 2:
v −v
v −v
1 2
2
3
+i =
2
5
15
 4 − 15  15 − 18
⇒ i = −
=2A
+
2
15
 5 
(checked using LNAP 8/13/02)
4-5
P4.3-4
Node equations:
−.003 +
−
v1 v1 − v2
+
=0
R1
500
v1 − v2 v2
+
− .005 = 0
500
R2
1
1
−1
+
= 0 ⇒ R1 =
= 200 Ω
1
R1 500
.003 +
500
2
−1 2
−
+
− .005 = 0 ⇒ R2 =
= 667 Ω
1
500 R2
.005 −
500
When v1 = 1 V, v2 = 2 V
−.003 +
(checked using LNAP 8/13/02)
P4.3-5
v1
v − v 2 v1 − v3
+ 1
+
=0
500
125
250
v − v3
v − v2
− 1
− .001 + 2
=0
125
250
v − v3 v1 − v3 v3
− 2
−
+
=0
250
250 500
Solving gives:
Node equations:
v1 = 0.261 V, v2 = 0.337 V, v3 = 0.239 V
Finally, v = v1 − v3 = 0.022 V
(checked using LNAP 8/13/02)
4-6
Section 4-4 Node Voltage Analysis of Circuits with Current and Voltage Sources
P4.4-1
Express the branch voltage of the voltage source in terms of its node voltages:
0 − va = 6 ⇒ va = −6 V
KCL at node b:
va − vb
v −v
+2= b c
6
10
KCL at node c:
Finally:
⇒
−6 − vb
v −v
+2= b c
6
10
vb − vc vc
=
10
8
⇒ −1−
⇒ 4 vb − 4 vc = 5 vc
9 
30 = 8  vc  − 3 vc
4 
vb
v −v
+2= b c
6
10
⇒ vb =
⇒ 30 = 8 vb − 3 vc
9
vc
4
⇒ vc = 2 V
(checked using LNAP 8/13/02)
P4.4-2
Express the branch voltage of each voltage source in terms of its node voltages to get:
va = −12 V, vb = vc = vd + 8
4-7
KCL at node b:
vb − va
= 0.002 + i ⇒
4000
vb − ( −12 )
= 0.002 + i ⇒ vb + 12 = 8 + 4000 i
4000
KCL at the supernode corresponding to the 8 V source:
v
0.001 = d + i ⇒ 4 = vd + 4000 i
4000
vb + 4 = 4 − vd ⇒ ( vd + 8 ) + 4 = 4 − vd ⇒ vd = −4 V
so
Consequently vb = vc = vd + 8 = 4 V and i =
4 − vd
= 2 mA
4000
(checked using LNAP 8/13/02)
P4.4-3
Apply KCL to the supernode:
va − 10 va va − 8
+
+
− .03 = 0 ⇒ va = 7 V
100
100 100
(checked using LNAP 8/13/02)
P4.4-4
Apply KCL to the supernode:
va + 8 ( va + 8 ) − 12 va − 12 va
+
+
+
=0
500
125
250
500
Solving yields
va = 4 V
(checked using LNAP 8/13/02)
4-8
P4.4-5
The power supplied by the voltage source is
v −v v −v 
 12 − 9.882 12 − 5.294 
va ( i1 + i 2 ) = va  a b + a c  = 12 
+

6 
4
6


 4
= 12(0.5295 + 1.118) = 12(1.648) = 19.76 W
(checked using LNAP 8/13/02)
P4.4-6
Label the voltage measured by the meter. Notice that this is a node voltage.
Write a node equation at the node at which
the node voltage is measured.
 12 − v m  v m
v −8
−
+ 0.002 + m
=0
+
3000
 6000  R
That is
6000 
6000

3 +
 v m = 16 ⇒ R = 16
R 

−3
vm
(a) The voltage measured by the meter will be 4 volts when R = 6 kΩ.
(b) The voltage measured by the meter will be 2 volts when R = 1.2 kΩ.
4-9
Section 4-5 Node Voltage Analysis with Dependent Sources
P4.5-1
Express the resistor currents in terms of the
node voltages:
va − vc
= 8.667 − 10 = −1.333 A and
1
2 − 10
v −v
i 2= b c =
= −4 A
2
2
i 1=
Apply KCL at node c:
i1 + i 2 = A i1 ⇒ − 1.333 + ( −4 ) = A (−1.333)
⇒
A=
−5.333
=4
−1.333
(checked using LNAP 8/13/02)
P4.5-2
Write and solve a node equation:
va − 6
v
v − 4va
+ a + a
= 0 ⇒ va = 12 V
1000 2000 3000
ib =
va − 4va
= −12 mA
3000
(checked using LNAP 8/13/02)
P4.5-3
First express the controlling current in terms of
the node voltages:
2 − vb
i =
a
4000
Write and solve a node equation:
−
2 − vb
v
 2 − vb 
+ b − 5
 = 0 ⇒ vb = 1.5 V
4000 2000  4000 
(checked using LNAP 8/14/02)
4-10
P4.5-4
Apply KCL to the supernode of the CCVS to get
12 − 10 14 − 10 1
+
− + i b = 0 ⇒ i b = −2 A
4
2
2
Next
10 − 12
1
=− 
−2
V
=4
4
2 ⇒ r =
1
A
−
r i a = 12 − 14 
2
ia =
(checked using LNAP 8/14/02)
P4.5-5
First, express the controlling current of the CCVS in
v2
terms of the node voltages: i x =
2
Next, express the controlled voltage in terms of the
node voltages:
v2
24
⇒ v2 =
12 − v 2 = 3 i x = 3
V
2
5
so ix = 12/5 A = 2.4 A.
(checked using ELab 9/5/02)
4-11
Section 4-6 Mesh Current Analysis with Independent Voltage Sources
2 i1 + 9 (i1 − i 3 ) + 3(i1 − i 2 ) = 0
P 4.6-1
15 − 3 (i1 − i 2 ) + 6 (i 2 − i 3 ) = 0
−6 (i 2 − i 3 ) − 9 (i1 − i 3 ) − 21 = 0
or
14 i1 − 3 i 2 − 9 i 3 = 0
−3 i1 + 9 i 2 − 6 i 3 = −15
−9 i1 − 6 i 2 + 15 i 3 = 21
so
i1 = 3 A, i2 = 2 A and i3 = 4 A.
(checked using LNAP 8/14/02)
P 4.6-2
Top mesh:
4 (2 − 3) + R(2) + 10 (2 − 4) = 0
so R = 12 Ω.
Bottom, right mesh:
8 (4 − 3) + 10 (4 − 2) + v 2 = 0
so v2 = −28 V.
Bottom left mesh
−v1 + 4 (3 − 2) + 8 (3 − 4) = 0
so v1 = −4 V.
(checked using LNAP 8/14/02)
4-12
P 4.6-3
Ohm’s Law: i 2 =
−6
= −0.75 A
8
KVL for loop 1:
R i1 + 4 ( i1 − i 2 ) + 3 + 18 = 0
KVL for loop 2
+ (−6) − 3 − 4 ( i1 − i 2 ) = 0
⇒ − 9 − 4 ( i1 − ( −0.75 ) ) = 0
⇒ i 1 = −3 A
R ( −3) + 4 ( −3 − ( −0.75 ) ) + 21 = 0 ⇒ R = 4 Ω
(checked using LNAP 8/14/02)
P4.6-4
KVL loop 1:
25 ia − 2 + 250 ia + 75 ia + 4 + 100 (ia − ib ) = 0
450 ia −100 ib = −2
KVL loop 2:
−100(ia − ib ) − 4 + 100 ib + 100 ib + 8 + 200 ib = 0
−100 ia + 500 ib = − 4
⇒ ia = − 6.5 mA , ib = − 9.3 mA
(checked using LNAP 8/14/02)
P4.6-5
Mesh Equations:
mesh 1 : 2i1 + 2 (i1 − i2 ) + 10 = 0
mesh 2 : 2(i2 − i1 ) + 4 (i2 − i3 ) = 0
mesh 3 : − 10 + 4 (i3 − i2 ) + 6 i3 = 0
Solving:
5
i = i2 ⇒ i = − = −0.294 A
17
(checked using LNAP 8/14/02)
4-13
Section 4-7 Mesh Current Analysis with Voltage and Current Sources
P4.7-1
mesh 1: i1 =
1
A
2
mesh 2: 75 i2 + 10 + 25 i2 = 0
⇒ i2 = − 0.1 A
ib = i1 − i2 = 0.6 A
(checked using LNAP 8/14/02)
P4.7-2
mesh a: ia = − 0.25 A
mesh b: ib = − 0.4 A
vc = 100(ia − ib ) = 100(0.15) =15 V
(checked using LNAP 8/14/02)
P4.7-3
Express the current source current as a function of the mesh currents:
i1 − i2 = − 0.5 ⇒ i1 = i2 − 0.5
Apply KVL to the supermesh:
30 i1 + 20 i2 + 10 = 0 ⇒ 30 (i2 − 0.5) + 20i2 = − 10
50 i2 − 15 = − 10 ⇒ i2 =
i1 =−.4 A and v2 = 20 i2 = 2 V
5
= .1 A
50
(checked using LNAP 8/14/02)
4-14
P4.7-4
Express the current source current in terms
of the mesh currents:
ib = ia − 0.02
Apply KVL to the supermesh:
250 ia + 100 (ia − 0.02) + 9 = 0
∴ ia = − .02 A = − 20 mA
vc = 100(ia − 0.02) = −4 V
(checked using LNAP 8/14/02)
P4.7-5
Express the current source current in terms of the mesh currents:
i 3 − i 1 = 2 ⇒ i1 = i 3 − 2
Supermesh: 6 i1 + 3 i 3 − 5 ( i 2 − i 3 ) − 8 = 0 ⇒ 6 i1 − 5 i 2 + 8 i 3 = 8
Lower, left mesh: −12 + 8 + 5 ( i 2 − i 3 ) = 0 ⇒ 5 i 2 = 4 + 5 i 3
Eliminating i1 and i2 from the supermesh equation:
6 ( i 3 − 2 ) − ( 4 + 5 i 3 ) + 8 i 3 = 8 ⇒ 9 i 3 = 24
 24 
The voltage measured by the meter is: 3 i 3 = 3   = 8 V
 9 
(checked using LNAP 8/14/02)
4-15
P4.7-6
4 ( i − 2 ) + 2 i + 6 ( i + 3) = 0 ⇒ 12 i − 8 + 18 = 0 ⇒ i = −
Mesh equation for right mesh:
10
5
A=− A
12
6
(checked using LNAP 8/14/02)
P 4.7-7
i2 = −3 A
i1 − i2 = 5 ⇒ i1 − ( −3) = 5
⇒ i1 = 2 A
2 ( i3 − i1 ) + 4 i3 + R ( i3 − i2 ) = 0
⇒ 2 ( −1 − 2 ) + 4 ( −1) + R ( −1 − ( −3) ) = 0
⇒ R=5 Ω
(checked using LNAP 8/14/02)
4-16
P 4.7-8
Express the controlling voltage of the
dependent source as a function of the
mesh current
v2 = 50 i1
Apply KVL to the right mesh:
−100 (0.04(50i1 ) − i1 ) + 50i1 + 10 = 0 ⇒ i1 = 0.2 A
v2 = 50 i1 = 10 V
(checked using LNAP 8/14/02)
P 4.7-9
ib = 4ib − ia ⇒ ib =
1
ia
3
1 
−100  ia  + 200ia + 8 = 0
3 
⇒ ia = − 0.048 A
(checked using LNAP 8/14/02)
P4.7-10
Express the controlling current of
the dependent source as a function
of the mesh current:
ib = .06 − ia
Apply KVL to the right mesh:
−100 (0.06 − i a ) + 50 (0.06 − i a ) + 250 i a = 0 ⇒
Finally:
ia = 10 mA
vo = 50 i b = 50 (0.06 − 0.01) = 2.5 V
(checked using LNAP 8/14/02)
4-17
P4.7-11
Express the controlling voltage of
the dependent source as a function
of the mesh current:
vb = 100 (.006 − ia )
Apply KVL to the right mesh:
−100 (.006 − ia ) + 3[100(.006 − ia )] + 250 ia = 0 ⇒ ia = −24 mA
(checked using LNAP 8/14/02)
P4.7-12
apply KVL to left mesh : − 3 + 10 × 103 i1 + 20 × 103 ( i1 − i2 ) = 0 ⇒ 30 × 103 i1 − 20 × 103 i2 = 3
apply KVL to right mesh : 5 ×103 i1 + 100 × 103 i2 + 20 × 103 ( i2 − i1 ) = 0 ⇒ i1 = 8i2
Solving (1) & ( 2 ) simultaneously
Power delevered to cathode =
⇒ i1 =
( 5 i1 ) ( i2 ) + 100 ( i2 )2
( 55)( 3 220) + 100 ( 3 220)
= 5 6
∴ Energy in 24 hr. = Pt =
6
3
mA, i2 =
mA
55
220
2
( 2)
= 0.026 mW
( 2.6 ×10−5 W ) ( 24 hr ) (3600 s hr )
= 2.25 J
4-18
(1)
P4.7-13
(a)
(b)
vo = − g R L v and v =
∴
vo
= −g
vi
R2
R1 + R 2
vi ⇒
(5 ×103 )(103 ) = −170
1.1×103
RL R2
vo
= −g
vi
R1 + R 2
⇒ g = 0.0374 S
PSpice Problems
SP 4-1
4-19
SP 4-2
From the PSpice output file:
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V1
V_V2
-3.000E+00
-2.250E+00
V_V3
-7.500E-01
The voltage source labeled V3 is a short circuit used to measure the mesh current. The mesh
currents are i1 = −3 A (the current in the voltage source labeled V1) and i2 = −0.75 A (the current
in the voltage source labeled V3).
SP 4-3
The PSpice schematic after running the simulation:
The PSpice output file:
**** INCLUDING sp4_2-SCHEMATIC1.net ****
* source SP4_2
V_V4
0 N01588 12Vdc
4-20
R_R4
V_V5
R_R5
V_V6
I_I1
I_I2
N01588 N01565 4k
N01542 N01565 0Vdc
0 N01516 4k
N01542 N01516 8Vdc
0 N01565 DC 2mAdc
0 N01542 DC 1mAdc
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V4
V_V5
V_V6
-4.000E-03
2.000E-03
-1.000E-03
From the PSpice schematic: va = −12 V, vb = vc = 4 V, vd = −4 V. From the output file: i = 2 mA.
SP 4-4
The PSpice schematic after running the simulation:
The PSpice output file:
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V7
V_V8
-5.613E-01
-6.008E-01
The current of the voltage source labeled V7 is also the current of the 2 Ω resistor at the top of
the circuit. However this current is directed from right to left in the 2 Ω resistor while the current
i is directed from left to right. Consequently, i = +5.613 A.
4-21
Verification Problems
VP 4-1
Apply KCL at node b:
vb − va
v −v
1
−
+ b c = 0
4
2
5
−4.8 − 5.2
1 − 4.8 − 3.0
− +
≠0
4
2
5
The given voltages do not satisfy the KCL
equation at node b. They are not correct.
VP 4-2
Apply KCL at node a:
v
v −v 
− b a  − 2 + a = 0
2
 4 
4
 20 − 4 
−
= −4≠ 0
−2+
2
 4 
The given voltages do not satisfy the KCL
equation at node a. They are not correct.
4-22
VP 4-3
Writing a node equation:
 12 − 7.5  7.5 7.5 − 6
−
+
=0
+
R2
 R1  R3
so
−
4.5 7.5 1.5
+
+
=0
R1 R3 R2
There are only three cases to consider. Suppose
R1 = 5 kΩ and R 2 = R 3 = 10 kΩ. Then
−
4.5 7.5 1.5 −0.9 + 0.75 + 0.15
+
+
=
= 0
R1 R3 R2
1000
This choice of resistance values corresponds to branch
currents that satisfy KCL. Therefore, it is indeed possible
that two of the resistances are 10 kW and the other
resistance is 5 kW. The 5 kW is R1.
VP 4-4
KCL at node 1:
0=
v1 − v 2
20
+
v1
5
+1 ⇒
KCL at node 2:
v1 − v 2
20
= 2+
v 2 − v3
KCL at node 3:
10
−8 − ( −20 ) −8
+
+1 = 0
20
5
−8 − ( −20 )
−20 − ( −6 )
= 2+
20
10
12 6
⇒
=
20 10
⇒
v2 − v3
10
+1 =
v3
15
⇒
−20 − ( −6 )
−6
−4 −6
+1 =
⇒
=
10
15
10 15
KCL is satisfied at all of the nodes so the computer analysis is correct.
4-23
VP 4-5
Top mesh: 10 (2 − 4) + 12(2) + 4 (2 − 3) = 0
Bottom right mesh 8 (3 − 4) + 4 (3 − 2) + 4 = 0
Bottom, left mesh: 28 + 10 (4 − 2) + 8 (4 − 3) ≠ 0 (Perhaps the polarity of the 28 V source was
entered incorrectly.)
KVL is not satified for the bottom, left mesh so the computer analysis is not correct.
4-24
Design Problems
DP 4-1
Model the circuit as:
a)
We need to keep v2 across R2 as 4.8 ≤ v2 ≤ 5.4
display is active
0.3 A
For I = 
 0.1 A display is not active
KCL at a:
v2 − 15 v2
+
+I =0
R1
R2
Assumed that maximum I results in minimum v2 and visa-versa.
Then
4.8 V
v2 = 
5.4 V
when I = 0.3 A
when I = 0.1 A
Substitute these corresponding values of v2 and I into the KCL equation and solve for the resistances
4.8 − 15 4.8
+
+ 0.3 = 0
R1
R2
5.4 − 15 5.4
+
+ 0.1 = 0
R1
R2
⇒ R1 = 7.89 Ω, R2 = 4.83 Ω
b)
15 − 4.8
= 1.292 A ⇒ PR
= (1.292)2 (7.89) = 13.17 W
1max
7.89
5.4 )2
(
5.4
IR
=
= 1.118 A
⇒ PR
=
= 6.03 W
2max
2 max
4.83
4.83
maximum supply current = I R
= 1.292 A
IR
1max
=
1max
c) No; if the supply voltage (15V) were to rise or drop, the voltage at the display would drop
below 4.8V or rise above 5.4V.
The power dissipated in the resistors is excessive. Most of the power from the supply is
dissipated in the resistors, not the display.
4-25
DP 4-2
Express the voltage of the 8 V source in terms of its node voltages to get vb − va = 8 . Apply KCL
to the supernode corresponding to the 8 V source:
va − v1
R
+
va vb vb − ( −v 2 )
+ +
= 0 ⇒ 2 va − v1 + 2 vb + v 2 = 0
R R
R
⇒ 2 va − v1 + 2 ( va + 8 ) + v 2 = 0
⇒ 4 va − v1 + v 2 + 16 = 0
⇒ va =
Next set va = 0 to get
0=
v1 − v 2
4
v1 − v 2
4
−4
− 4 ⇒ v1 − v 2 = 16 V
For example, v1 = 18 V and v2 = 2 V.
4-26
DP 4-3
a)
pply KCL to left mesh:
−5 + 50 i1 + 300 (i1 − I ) = 0
Apply KCL to right mesh:
( R + 2) I + 300 ( I − i1 ) = 0
I=
150
1570 + 35 R
We desire 50 mA ≤ I ≤ 75 mA so if R = 100 Ω, then I = 29.59 mA fi l amp so the lamp will not
light.
Solving for I:
b) From the equation for I, we see that decreasing R increases I:
try R = 50 Ω ⇒ I = 45 mA (won't light)
try R = 25Ω ⇒ I = 61 mA ⇒ will light
Now check R±10% to see if the lamp will light and not burn out:
−10% → 22.5Ω → I = 63.63 mA  lamp will

+10% → 27.5Ω → I = 59.23 mA  stay on
DP 4-4
Equivalent resistance:
R = R1 || R 2 || ( R 3 + R 4 )
R
( 25 )
10 + R
We require vab = 10 V. Apply the voltage division principle in the left circuit to get:
Voltage division in the equivalent circuit: v1 =
4-27
10 =
(
)
R1 R 2 ( R3 + R4 )
R4
R4
×
× 25
v1 =
R3 + R4
R3 + R4 10 + R1 R 2 ( R3 + R4 )
(
)
This equation does not have a unique solution. Here’s one solution:
choose R1 = R2 = 25 Ω and R3 + R4 = 20 Ω
then 10 =
(12.5 20 ) × 25 ⇒ R = 18.4Ω
R4
×
4
20 10 + (12.5 20 )
and R3 + R4 = 20 ⇒ R3 = 1.6 Ω
DP 4-5
Apply KCL to the left mesh:
(R
1
+ R 3 ) i1 − R 3 i2 − v1 = 0
Apply KCL to the left mesh:
(R
− R 3 i1 +
2
+ R 3 ) i2 + v2 = 0
Solving for the mesh currents using Cramer’s rule:
− R3
 v1

( R1 + R 3 )


− R
 −v2 ( R 2 + R 3 ) 
3


i1 =
and i2 =
∆
∆
2
where ∆ = ( R1 + R 3 ) ( R 2 + R 3 ) − R 3
v1 
− v2 
Try R1 = R2 = R3 = 1 kΩ = 1000 Ω. Then ∆ = 3 MΩ. The mesh currents will be given by
=
[ 2v1 − v2 ] 1000
and i 2 =
[ −2v2 + v1 ] 1000
3 × 10
3 × 10
Now check the extreme values of the source voltages:
i
1
6
6
if v1 = v2 = 1 V ⇒ i = 2
mA
3
if v1 = v2 = 2 V ⇒ i = 4 mA
3
⇒ i = i1 − i2 =
v1 + v2
3000
okay
okay
4-28
Chapter 5 Circuit Theorems
Exercises
Ex 5.3-1
R = 10 Ω and is = 1.2 A.
Ex 5.3-2
R = 10 Ω and is = −1.2 A.
Ex 5.3-3
R = 8 Ω and vs = 24 V.
Ex 5.3-4
R = 8 Ω and vs = −24 V.
Ex 5.4-1


20
10
2
(15) + 20  −
( 2 ) = 6 + 20(− ) = −2 V
10 + 20 + 20
5
 10 + (20 + 20)

Ex 5.4-2
25
3
−
( 5) = 5 − 3 = 2 A
3+ 2 2+3
vm =
im =
Ex 5.4-3


3
3
vm = 3 
( 5) −
(18 ) = 5 − 6 = −1 A
 3 + (3 + 3)  3 + (3 + 3)
5-1
Ex 5.5-1
5-2
Ex 5.5-2
ia =
2 i a − 12
⇒ i a = −3 A
6
voc = 2 i a = −6 V
12 + 6 i a = 2 i a
3 i sc = 2 i a
Rt =
⇒ i a = −3 A
⇒ i sc =
2
( −3 ) = − 2 A
3
−6
=3Ω
−2
Ex 5.6-1
5-3
Ex 5.6-2
ia =
2 i a − 12
⇒ i a = −3 A
6
voc = 2 i a = −6 V
12 + 6 i a = 2 i a
3 i sc = 2 i a
Rt =
⇒ i a = −3 A
⇒ i sc =
2
( −3 ) = − 2 A
3
−6
=3Ω
−2
5-4
Ex 5.6-3
12 × 24 12 × 24
=
= 8Ω
12 + 24
36
24
voc =
( 30 ) = 20 V
12 + 24
Rt =
i=
20
8+ R
Ex 5.7-1
voc =
6
(18) = 12 V
6+3
Rt = 2 +
( 3)( 6 ) = 4 Ω
3+ 6
For maximum power, we require
R L = Rt = 4 Ω
Then
pmax =
voc
122
=
=9 W
4 Rt 4 ( 4 )
2
5-5
Ex 5.7-2
1
50
3
i sc =
5.6 ) =
(
( 5.6 ) = 5 A
1 1
1
50
1
5
+
+
+
+
3 150 30
150 ( 30 )
Rt = 3 +
= 3 + 25 = 28 Ω
150 + 30
2
R t i sc
28 ) 52
(
pmax =
=
= 175 W
4
4
Ex 5.7-3
 10   R L

100 R L
10 )  =
p=iv=
(


2
 Rt + R L   Rt + R L


 ( R t + R L )
The power increases as Rt decreases so choose Rt = 1 Ω. Then
pmax = i v =
100 ( 5 )
(1 + 5)
2
= 13.9 W
Ex 5.7-4
From the plot, the maximum power is 5 W when R = 20 Ω. Therefore:
Rt = 20 Ω
and
v
= oc
4 Rt
2
pmax
⇒ voc =
pmax 4 Rt = 5 ( 4 ) 20 = 20 V
5-6
Problems
Section 5-3: Source Transformations
P5.3-1
(a)
5-7
∴ Rt = 2 Ω
(b)
(c)
−9 − 4i − 2i + (−0.5) = 0
−9 + (−0.5)
i =
= −1.58 A
4+2
v = 9 + 4 i = 9 + 4(−1.58) = 2.67 V
ia = i = − 1.58 A
vt = − 0.5 V
(checked using LNAP 8/15/02)
P5.3-2
Finally, apply KVL:
−10 + 3 ia + 4 ia −
16
=0
3
∴ ia = 2.19 A
(checked using LNAP 8/15/02)
5-8
P5.3-3
Source transformation at left; equivalent resistor for parallel 6 and 3 Ω resistors:
Equivalents for series resistors, series voltage source at left; series resistors, then source
transformation at top:
Source transformation at left; series resistors at right:
Parallel resistors, then source transformation at left:
5-9
Finally, apply KVL to loop
− 6 + i (9 + 19) − 36 − vo = 0
i = 5 / 2 ⇒ vo = −42 + 28 (5 / 2) = 28 V
(checked using LNAP 8/15/02)
P5.3-4
− 4 − 2000 ia − 4000 ia + 10 − 2000 ia − 3 = 0
∴ ia = 375 µ A
(checked using LNAP 8/15/02)
5-10
P5.3-5
−12 − 6 ia + 24 − 3 ia − 3 = 0 ⇒ ia = 1 A
(checked using LNAP 8/15/02)
P5.3-6
A source transformation on the right side of the circuit, followed by replacing series resistors
with an equivalent resistor:
Source transformations on both the right side and the left side of the circuit:
5-11
Replacing parallel resistors with an equivalent resistor and also replacing parallel current sources
with an equivalent current source:
Finally,
va =
50 (100 )
100
( 0.21) = ( 0.21) = 7 V
50 + 100
3
(checked using LNAP 8/15/02)
5-12
Section 5-4 Superposition
P5.4–1
Consider 6 A source only (open 9 A source)
Use current division:
v1
 15 
= 6 
⇒ v1 = 40 V
20
15 + 30 
Consider 9 A source only (open 6 A source)
Use current division:
v2
 10 
= 9 
⇒ v2 = 40 V
20
10 + 35 
∴ v = v1 + v2 = 40 + 40 = 80 V
(checked using LNAP 8/15/02)
P5.4-2
Consider 12 V source only (open both current sources)
KVL:
20 i1 + 12 + 4 i1 + 12 i1 = 0
⇒ i1 = −1/ 3 mA
Consider 3 mA source only (short 12 V and open 9
mA sources)
Current Division:
4
 16 
i2 = 3 
=
mA

3
16 + 20 
5-13
Consider 9 mA source only (short 12 V and open 3
mA sources)
Current Division:
 12 
i3 = −9 
= −3 mA
 24 + 12 
∴ i = i1 + i2 + i3 = − 1/ 3 + 4 / 3 − 3 = − 2 mA
(checked using LNAP 8/15/02)
P5.4–3
Consider 30 mA source only (open 15 mA and short 15 V sources). Let i1 be the part of i due to
the 30 mA current source.
 2 
ia = 30 
 = 6 mA ⇒
 2+8
 6 
i1 = ia 
 = 2 mA
 6 + 12 
Consider 15 mA source only (open 30 mA source and short 15 V source) Let i2 be the part of i
due to the 15 mA current source.
 4 
ib = 15 
 = 6 mA ⇒
 4+6
 6 
i2 = ib 
 = 2 mA
 6 + 12 
5-14
Consider 15 V source only (open both current sources). Let i3 be the part of i due to the 15 V
voltage source.
 6 || 6 
 3 
i3 = − 2.5 
 = − 10 
 = −0.5 mA
 3 + 12 
 ( 6 || 6 ) + 12 
Finally,
i = i1 + i2 + i3 = 2 + 2 − 0.5 = 3.5 mA
(checked using LNAP 8/15/02)
P5.4–4
Consider 10 V source only (open 30 mA source and
short the 8 V source)
Let v1 be the part of va due to the
10 V voltage source.
v1 =
=
Consider 8 V source only (open 30 mA source and
short the 10 V source)
100 ||100
(10 )
(100 ||100 ) + 100
50
10
(10 ) = V
150
3
Let v2 be the part of va due to the
8 V voltage source.
v1 =
=
100 ||100
(8)
(100 ||100 ) + 100
50
8
(8) = V
150
3
5-15
Consider 30 mA source only (short both the 10 V
source and the 8 V source)
Let v2 be the part of va due to the
30 mA current source.
v3 = (100 ||100 ||100)(0.03)
=
va = v1 + v2 + v3 =
Finally,
100
(0.03) = 1 V
3
10 8
+ +1 = 7 V
3 3
(checked using LNAP 8/15/02)
P5.4-5
Consider 8 V source only (open the 2 A source)
Let i1 be the part of ix due to the 8 V
voltage source.
Apply KVL to the supermesh:
6 ( i1 ) + 3 ( i 1 ) + 3 ( i 1 ) − 8 = 0
i1 =
Consider 2 A source only (short the 8 V source)
8 2
= A
12 3
Let i2 be the part of ix due to the 2 A
current source.
Apply KVL to the supermesh:
6 ( i 2 ) + 3 ( i 2 + 2 ) + 3 i2 = 0
i2 =
Finally,
i x = i1 + i 2 =
1
−6
=− A
12
2
2 1 1
− = A
3 2 6
5-16
Section 5-5: Thèvenin’s Theorem
P5.5-1
(checked using LNAP 8/15/02)
5-17
P5.5-2
The circuit from Figure P5.5-2a can be reduced to its Thevenin equivalent circuit in four steps:
(a)
(b)
(c)
(d)
Comparing (d) to Figure P5.5-2b shows that the Thevenin resistance is Rt = 16 Ω and the open
circuit voltage, voc = −12 V.
5-18
P5.5-3
The circuit from Figure P5.5-3a can be reduced to its Thevenin equivalent circuit in five steps:
(a)
(b)
(c)
(d)
(e)
Comparing (e) to Figure P5.5-3b shows that the Thevenin resistance is Rt = 4 Ω and the open
circuit voltage, voc = 2 V.
(checked using LNAP 8/15/02)
5-19
P5.5-4
Find Rt:
Rt =
12 (10 + 2 )
=6Ω
12 + (10 + 2 )
Write mesh equations to find voc:
Mesh equations:
12 i1 + 10 i1 − 6 ( i2 − i1 ) = 0
6 ( i2 − i1 ) + 3 i 2 − 18 = 0
28 i1 = 6 i 2
9 i 2 − 6 i 1 = 18
36 i1 = 18 ⇒ i1 =
i2 =
Finally,
14  1  7
 = A
3 2 3
1
A
2
7
1
voc = 3 i 2 + 10 i1 = 3   + 10   = 12 V
 3
 2
(checked using LNAP 8/15/02)
5-20
P5.5-5
Find voc:
Notice that voc is the node voltage at node a. Express
the controlling voltage of the dependent source as a
function of the node voltage:
va = −voc
Apply KCL at node a:
 6 − voc  voc  3 
−
+  − voc  = 0
+
 8  4  4 
−6 + voc + 2 voc − 6 voc = 0 ⇒ voc = −2 V
Find Rt:
We’ll find isc and use it to calculate Rt. Notice that
the short circuit forces
va = 0
Apply KCL at node a:
6−0 0  3 
−
 + +  − 0  + i sc = 0
 8  4  4 
i sc =
Rt =
6 3
= A
8 4
voc −2
8
=
=− Ω
i sc 3 4
3
(checked using LNAP 8/15/02)
5-21
P5.5-6
Find voc:
2 va − va va
= + 3 + 0 ⇒ va = 18 V
3
6
The voltage across the right-hand 3 Ω resistor is zero so: va = voc = 18 V
Apply KCL at the top, middle node:
Find isc:
v
2 va − va va
= + 3 + a ⇒ va = −18 V
3
6
3
va −18
i sc = =
= −6 V
Apply Ohm’s law to the right-hand 3 Ω resistor :
3
3
v
18
Finally:
R t = oc =
= −3 Ω
i sc −6
Apply KCL at the top, middle node:
(checked using LNAP 8/15/02)
5-22
P5.5-7
(a)
−vs + R1 ia + ( d + 1) R 2 ia = 0
ia =
v oc =
vs
R1 + ( d + 1) R 2
( d + 1) R 2vs
R1 + ( d + 1) R 2
ia =
vs
R1
i sc = ( d + 1) ia =
−ia − d ia +
iT = ( d + 1)
625 Ω = R t =
and
5 = voc =
R1
vT
− iT = 0
R2
R1 ia = −vT
vT vT R 2 ( d + 1) + R1
+
=
× vT
R1 R 2
R1 R 2
Rt =
(b) Let R1 = R2 = 1 kΩ. Then
( d + 1) vs
R1 R 2
vT
=
iT R1 + ( d + 1) R 2
1000
1000
⇒ d=
− 2 = −0.4 A/A
625
d +2
( d + 1) vs
d +2
⇒ vs =
−0.4 + 2
5 = 13.33 V
−0.4 + 1
(checked using LNAP 8/15/02)
5-23
P5.5-8
From the given data:
2000

voc 
R t + 2000 
 voc = 1.2 V
 ⇒ 
4000
 R t = −1600 Ω
2=
voc 
R t + 4000 
6=
v=
When R = 8000 Ω,
R
voc
Rt + R
v=
8000
(1.2 ) = 1.5 V
−1600 + 8000
P5.5-9
From the given data:
voc

R t + 2000 
 voc = 24 V
 ⇒ 
voc
 R t = 4000 Ω

0.003 =
R t + 4000 
0.004 =
i=
voc
Rt + R
(a) When i = 0.002 A:
24
0.002 =
⇒ R = 8000 Ω
4000 + R
(b) Maximum i occurs when R = 0:
24
= 0.006 = 6 mA ⇒ i ≤ 6 mA
4000
P5.5-10
The current at the point on the plot where v = 0 is the short circuit current, so isc = 20 mA.
The voltage at the point on the plot where i = 0 is the open circuit voltage, so voc = −3 V.
The slope of the plot is equal to the negative reciprocal of the Thevenin resistance, so
1 0 − 0.002
−
=
⇒ R t = −150 Ω
Rt
−3 − 0
5-24
P5.5-11
−12 + 6000 ia + 2000 ia + 1000 ia = 0
ia = 4 3000 A
voc = 1000 ia =
4
V
3
ia = 0 due to the short circuit
−12 + 6000 isc = 0 ⇒ isc = 2 mA
4
voc
= 3 = 667 Ω
Rt =
isc .002
4
3
ib =
667 + R
ib = 0.002 A requires
4
3 − 667 = 0
R =
0.002
(checked using LNAP 8/15/02)
5-25
P5.5-12
10 = i + 0 ⇒ i = 10 A
voc + 4 i − 2 i = 0
⇒ voc = −2 i = −20 V
i + i sc = 10 ⇒ i = 10 − i sc
4 i + 0 − 2 i = 0 ⇒ i = 0 ⇒ i sc = 10 A
Rt =
−2 = iL =
voc −20
=
= −2 Ω
isc
10
−20
⇒ RL = 12 Ω
RL − 2
(checked using LNAP 8/15/02)
5-26
Section 5-6: Norton’s Theorem
P5.6-1
When the terminals of the boxes are open-circuited, no current flows in Box A, but the resistor in
Box B dissipates 1 watt. Box B is therefore warmer than Box A. If you short the terminals of
each box, the resistor in Box A will draw 1 amp and dissipate 1 watt. The resistor in Box B will
be shorted, draw no current, and dissipate no power. Then Box A will warm up and Box B will
cool off.
P5.6-2
(checked using LNAP 8/16/02)
5-27
P5.6-3
P5.6-4
To determine the value of the short circuit current, isc, we connect a short circuit across the
terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a)
shows the circuit from Figure 5.6-4a after adding the short circuit and labeling the short circuit
current. Also, the meshes have been identified and labeled in anticipation of writing mesh
equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.
In Figure (a), mesh current i2 is equal to the current in the short circuit. Consequently,
i2 = isc . The controlling current of the CCVS is expressed in terms of the mesh currents as
i a = i1 − i 2 = i1 − isc
Apply KVL to mesh 1 to get
3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) − 10 = 0 ⇒ 7 i1 − 4 i 2 = 10
(1)
Apply KVL to mesh 2 to get
5 i 2 − 6 ( i1 − i 2 ) = 0 ⇒ − 6 i1 + 11 i 2 = 0 ⇒ i1 =
11
i2
6
Substituting into equation 1 gives
 11 
7  i 2  − 4 i 2 = 10 ⇒ i 2 = 1.13 A ⇒ i sc = 1.13 A
6 
5-28
Figure (a) Calculating the short circuit current, isc, using mesh equations.
To determine the value of the Thevenin resistance, Rt, first replace the 10 V voltage
source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source across the
terminals of the circuit and then label the voltage across that current source as shown in Figure
(b). The Thevenin resistance will be calculated from the current and voltage of the current source
as
v
Rt = T
iT
In Figure (b), the meshes have been identified and labeled in anticipation of writing mesh
equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.
In Figure (b), mesh current i2 is equal to the negative of the current source current.
Consequently, i2 = i T . The controlling current of the CCVS is expressed in terms of the mesh
i a = i1 − i 2 = i1 + i T
currents as
Apply KVL to mesh 1 to get
3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) = 0 ⇒ 7 i1 − 4 i 2 = 0 ⇒ i1 =
4
i2
7
(2)
Apply KVL to mesh 2 to get
5 i 2 + vT − 6 ( i1 − i 2 ) = 0 ⇒ − 6 i1 + 11 i 2 = −vT
Substituting for i1 using equation 2 gives
4 
−6  i 2  + 11 i 2 = −vT
7 
Finally,
Rt =
⇒ 7.57 i 2 = −vT
vT −vT −vT
=
=
= 7.57 Ω
−iT
iT
i2
5-29
Figure (b) Calculating the Thevenin resistance, R t =
vT
, using mesh equations.
iT
To determine the value of the open circuit voltage, voc, we connect an open circuit across
the terminals of the circuit and then calculate the value of the voltage across that open circuit.
Figure (c) shows the circuit from Figure 4.6-4a after adding the open circuit and labeling the
open circuit voltage. Also, the meshes have been identified and labeled in anticipation of writing
mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.
In Figure (c), mesh current i2 is equal to the current in the open circuit. Consequently,
i2 = 0 A . The controlling current of the CCVS is expressed in terms of the mesh currents as
i a = i1 − i 2 = i1 − 0 = i1
Apply KVL to mesh 1 to get
3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) − 10 = 0 ⇒ 3 i1 − 2 ( i1 − 0 ) + 6 ( i1 − 0 ) − 10 = 0
⇒ i1 =
10
= 1.43 A
7
Apply KVL to mesh 2 to get
5 i 2 + voc − 6 ( i1 − i 2 ) = 0 ⇒ voc = 6 ( i1 ) = 6 (1.43) = 8.58 V
Figure (c) Calculating the open circuit voltage, voc, using mesh equations.
As a check, notice that R t isc = ( 7.57 )(1.13) = 8.55 ≈ voc
(checked using LNAP 8/16/02)
5-30
P5.6-5
To determine the value of the short circuit current, Isc, we connect a short circuit across the
terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a)
shows the circuit from Figure 4.6-5a after adding the short circuit and labeling the short circuit
current. Also, the nodes have been identified and labeled in anticipation of writing node
equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively.
In Figure (a), node voltage v1 is equal to the negative of the voltage source voltage.
Consequently, v1 = −24 V . The voltage at node 3 is equal to the voltage across a short, v3 = 0 .
The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The
voltage at node 3 is equal to the voltage across a short, i.e. v3 = 0 .
Apply KCL at node 2 to get
v1 − v 2
3
=
v 2 − v3
6
⇒ 2 v1 + v 3 = 3 v 2
⇒ − 48 = 3 v a
⇒ v a = −16 V
Apply KCL at node 3 to get
v2 − v3
6
+
4
v 2 = isc
3
⇒
9
v a = isc
6
⇒ isc =
9
( −16 ) = −24 A
6
Figure (a) Calculating the short circuit current, Isc, using mesh equations.
To determine the value of the Thevenin resistance, Rth, first replace the 24 V voltage
source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source circuit across
the terminals of the circuit and then label the voltage across that current source as shown in
Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current
source as
v
R th = T
iT
Also, the nodes have been identified and labeled in anticipation of writing node equations. Let
v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively.
5-31
In Figure (b), node voltage v1 is equal to the across a short circuit, i.e. v1 = 0 . The
controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The
voltage at node 3 is equal to the voltage across the current source, i.e. v3 = vT .
Apply KCL at node 2 to get
v1 − v 2
3
=
v 2 − v3
6
⇒ 2 v1 + v 3 = 3 v 2
⇒ vT = 3 v a
Apply KCL at node 3 to get
v2 − v3
6
+
4
v 2 + iT = 0 ⇒ 9 v 2 − v3 + 6 iT = 0
3
⇒ 9 v a − vT + 6 iT = 0
⇒ 3 v T − vT + 6 iT = 0 ⇒ 2 vT = −6 iT
Finally,
Rt =
vT
= −3 Ω
iT
Figure (b) Calculating the Thevenin resistance, R th =
vT
, using mesh equations.
iT
To determine the value of the open circuit voltage, voc, we connect an open circuit across
the terminals of the circuit and then calculate the value of the voltage across that open circuit.
Figure (c) shows the circuit from Figure P 4.6-5a after adding the open circuit and labeling the
open circuit voltage. Also, the nodes have been identified and labeled in anticipation of writing
node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively.
In Figure (c), node voltage v1 is equal to the negative of the voltage source voltage.
Consequently, v1 = −24 V . The controlling voltage of the VCCS, va, is equal to the node voltage
at node 2, i.e. va = v2 . The voltage at node 3 is equal to the open circuit voltage, i.e. v3 = voc .
Apply KCL at node 2 to get
5-32
v1 − v 2
3
=
v 2 − v3
6
⇒ 2 v1 + v 3 = 3 v 2
⇒ − 48 + v oc = 3 v a
Apply KCL at node 3 to get
v2 − v3
6
+
4
v 2 = 0 ⇒ 9 v 2 − v 3 = 0 ⇒ 9 v a = v oc
3
Combining these equations gives
3 ( −48 + voc ) = 9 v a = voc
⇒ voc = 72 V
Figure (c) Calculating the open circuit voltage, voc, using node equations.
As a check, notice that
R th I sc = ( −3)( −24 ) = 72 = Voc
(checked using LNAP 8/16/02)
Section 5-7: Maximum Power Transfer
P5.7-1
a) For maximum power transfer, set RL equal
to the Thevenin resistance:
R L = R t = 100 + 1 = 101 Ω
b) To calculate the maximum power, first replace the circuit connected to RL be its Thevenin
equivalent circuit:
5-33
The voltage across RL is
vL =
Then
pmax
101
(100 ) = 50 V
101 + 101
2
502
v
= L =
= 24.75 W
R L 101
P5.7-2
Reduce the circuit using source transformations:
Then (a) maximum power will be dissipated in resistor R when: R = Rt = 60 Ω and (b) the value
of that maximum power is
P
= i 2 ( R) = (0.03)2 (60) = 54 mW
max R
5-34
P5.7-3
 RL 
v L = vS 

 R S + R L 
v L2
v S2 R L
∴ pL =
=
R L ( RS + R L )2
By inspection, pL is max when you reduce RS to get the
smallest denominator.
∴ set RS = 0
P5.7-4
Find Rt by finding isc and voc:
The current in the 3 Ω resistor is zero because of the short circuit. Consequently, isc = 10 ix.
Apply KCL at the top-left node to get
ix + 0.9 = 10 ix
⇒ ix =
0.9
= 0.1 A
9
so
isc = 10 ix = 1A
Next
Apply KCL at the top-left node to get
5-35
ix + 0.9 = 10 ix
⇒ ix =
0.9
= 0.1 A
9
Apply Ohm’s law to the 3 Ω resistor to get
voc = 3 (10 ix ) = 30 ( 0.1) = 3 V
For maximum power transfer to RL:
R L = Rt =
voc 3
= =3Ω
isc 1
The maximum power delivered to RL is given by
pmax =
voc
32
3
=
= W
4 R t 4 ( 3) 4
2
P5.7-5
The required value of R is
R = Rt = 8 +
( 20 + 120 ) (10 + 50 ) = 50 Ω
( 20 + 120 ) + (10 + 50 )
30
 170
voc = 
( 20 )  10 − 
( 20 ) 50
170 + 30

170 + 30

170(20)(10) − 30(20)(50) 4000
=
=
= 20 V
200
200
The maximum power is given by
2
v
202
pmax = oc =
=2W
4 R t 4 ( 50 )
5-36
PSpice Problems
SP5-1
a = 0.3333
b = 0.3333
c =33.33 V/A
(a)
(b)
vo = 0.3333 v1 + 0.3333 v2 + 33.33 i 3
7 = 0.3333 (10 ) + 0.3333 ( 8 ) + 33.33 i 3
7−
18
3 = 3 = 30 mA
⇒ i3 =
100
100
3
5-37
SP5-2
Before the source transformation:
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V1
V_V2
-3.000E-02
-4.000E-02
After the source transformation:
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V2
-4.000E-02
5-38
SP5-3
voc = −2 V
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V3
V_V4
-7.500E-01
7.500E-01
isc = 0.75 A
Rt = −2.66 Ω
5-39
SP5-4
voc = 8.571 V
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V5
-2.075E+00
V_V6
1.132E+00
X_H1.VH_H1 9.434E-01
isc = 1.132 A
Rt = 7.571 Ω
5-40
Verification Problems
VP5-1
Use the data in the first two lines of the table to determine voc
and Rt:
voc 
R t + 0 
voc = 39.9 V
 ⇒ 
voc 
 R t = 410 Ω
0.0438 =
R t + 500 
0.0972 =
Now check the third line of the table. When R= 5000 Ω:
v
39.9
i = oc =
= 7.37 mA
R t + R 410 + 5000
which disagree with the data in the table.
The data is not consistent.
VP5-2
Use the data in the table to determine voc and isc:
voc = 12 V
(line 1 of the table)
isc = 3 mA
so
Rt =
voc
= 4 kΩ
isc
(line 3 of the table)
Next, check line 2 of the table. When R = 10 kΩ:
v
12
i = oc =
= 0.857 mA
3
R t + R 10 (10 ) + 5 (103 )
To cause i = 1 mA requires
which agrees with the data in the table.
v
12
0.001 = i = oc =
⇒ R = 8000 Ω
R t + R 10 (103 ) + R
I agree with my lab partner’s claim that R = 8000 causes i = 1 mA.
5-41
VP5-3
60
60
voc
11
i=
=
= 11 = 54.55 mA
R t + R 6 110 + 40 60 + 40
( )
11
The measurement supports the prelab calculation.
Design Problems
DP5-1
The equation of representing the straight line in Figure DP 5-1b is v = − R t i + voc . That is, the
slope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the
0−5
open circuit voltage. Therefore: R t = −
= 625 Ω and voc = 5 V.
0.008 − 0
Try R1 = R 2 = 1 kΩ . (R1 || R2 must be smaller than Rt = 625 Ω.) Then
5=
and
1
vs = vs
2
R1 + R 2
R2
⇒ vs = 10 V
625 = R 3 +
R1 R2
= R3 + 500 ⇒ R3 = 125 Ω
R1 + R2
Now vs, R1, R2 and R3 have all been specified so the design is complete.
DP5-2
The equation of representing the straight line in Figure DP 5-2b is v = − R t i + voc . That is, the
slope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the
0 − ( −3 )
= 500 Ω and voc = −3 V.
open circuit voltage. Therefore: R t = −
−0.006 − 0
From the circuit we calculate
R 3 ( R1 + R 2 )
R1 R 3
and voc = −
is
Rt =
R1 + R 2 + R 3
R1 + R 2 + R 3
so
R 3 ( R1 + R 2 )
R1 R 3
500 Ω =
is
and −3 V = −
R1 + R 2 + R 3
R1 + R 2 + R 3
5-42
Try R 3 = 1kΩ and R1 + R 2 = 1kΩ . Then R t = 500 Ω and
−3 = −
i s ⇒ 6 = R1 i s
2000
2
This equation can be satisfied by taking R1 = 600 Ω and is = 10 mA. Finally, R2 = 1 kΩ - 400 Ω =
400 Ω. Now is, R1, R2 and R3 have all been specified so the design is complete.
1000 R1
is = −
R1
DP5-3
The slope of the graph is positive so the Thevenin resistance is negative. This would require
R1 R 2
R3 +
< 0 , which is not possible since R1, R2 and R3 will all be non-negative.
R1 + R 2
Is it not possible to specify values of vs, R1, R2 and R3 that cause the current i and the
voltage v in Figure DP 5-3a to satisfy the relationship described by the graph in Figure DP 5-3b.
DP5-4
The equation of representing the straight line in Figure DP 5-4b is v = − R t i + voc . That is, the
slope of the line is equal to the Thevenin impedance and the "v - intercept" is equal to the open
−5 − 0
circuit voltage. Therefore: R t = −
= −625 Ω and voc = −5 V.
0 − 0.008
The open circuit voltage, voc, the short circuit current, isc, and the Thevenin resistance, Rt,
of this circuit are given by
R 2 ( d + 1)
voc =
vs
R1 + ( d + 1) R 2
,
( d + 1) v
isc =
s
R1
and
R1 R 2
Rt =
R1 + ( d + 1) R 2
Let R1 = R2 = 1 kΩ. Then
−625 Ω = R t =
and
−5 =
( d + 1) vs
d +2
1000
1000
⇒ d=
− 2 = −3.6 A/A
d +2
−625
⇒ vs =
−3.6 + 2
( − 5 ) = −3.077 V
−3.6 + 1
Now vs, R1, R2 and d have all been specified so the design is complete.
5-43
Chapter 6: The Operational Amplifier
Exercises
Ex. 6.4-1
vs vs − vo
+
+0 = 0
R1
R2
vo
R
= 1+ 2
vs
R1
Ex. 6.4-2
a)
va =
R2
vs
R1 + R2
va va − v0
+
+0 =0
R3
R4
vo
R
= 1+ 4
va
R3
b)
When R2 >> R1 then
⇒
vo  R2   R4 
=
 1 + 
vs  R1 + R2  R3 
R2
R
vo
R
− 2 = 1 and
− 1 + 4
R1 + R2
R2
vs
R3
6-1
Ex. 6.5-1
vo vo − vs
+
+0= 0
R2
R1
vo
R2
=
vs R1 + R2
Ex. 6.6–1
 Rf 
vin − vout vin
+ + 0 = 0 ⇒ vout = 1 +
 vin

Rf
R1
R1 

when R f = 100 kΩ and R1 = 25 kΩ then
 100 ⋅103 
vout
= 1 +
 = 5
25 ⋅103 
vin

6-2
Ex. 6.7-1

R2
R1 
 10 × 103  
 10 ×103  
v
+
+
+
v3 =  −
v
1
1






2
3  1
3
3
3
 10 × 10   R 2 + 10 × 10 
 10 × 10   10 × 10 


R2
R1 

= −
v2 + 2 1 +
v

3
3  1


 10 × 10 
 R 2 + 10 × 10 
1
We require v3 = ( 4 ) v1 −   v2 , so
5
R1 

⇒ R1 = 10 × 103 = 10 kΩ
4 = 2 1 +
3 
×
10
10


and
R2
1
=
5 R 2 + 10 × 103
⇒ R 2 + 10 × 103 = 5 R 2
⇒
R 2 = 2.5 kΩ
6-3
Ex. 6.7-2
As in Ex 6.7-1


R2
R1 

v3 = − 
v2 + 2 1 +
v

3
3  1
 R 2 + 10 ×10 
10
10
×




 4
We require v3 = ( 6 ) v1 −   v2 , so
5
R1 

⇒ R1 = 20 ×103 = 20 kΩ
6 = 2 1 +
3 
10
10
×


and
R2
4
=
5 R 2 + 10 × 103
⇒ 4 R 2 + 40 × 103 = 5 R 2
⇒ R 2 = 40 kΩ
Ex. 6.8-1
Analysis of the circuit in Section 6.7 showed that output offset voltage = 6 vos + (50 × 103 ) ib1
For a µ A741 op amp, vos ≤ 1 mV and ib1 ≤ 80 nA so
output offset voltage = 6 vos + (50 × 103 )ib1 ≤ 6 (10−3 )+(50.103 )(80×10−9 ) = 10 mV
Ex. 6.8-2
vo = −
 R 
R2
vin + 1 + 2  vos + R2ib1
R1
 R1 
When R2 = 10 kΩ, R1 = 2 kΩ, vos ≤ 5 mV and
(
) (
output offset voltage ≤ 6 5 × 10−3 + 10 × 103
ib1 ≤ 500 nA then
) ( 500.10 ) ≤ 35×10
−9
−3
= 35 mV
6-4
Analysis of this circuit in Section 6.7 showed that output offset voltage = 6 vos + ( 50 ×103 ) ib1
Ex. 6.8-3
For a typical OPA1O1AM, vos = 0.1 mV and ib = 0.012 nA so
output offset voltage ≤ 6 0.1×10−3  + ( 50 ×103 ) 0.012 ×10−9 
≤ 0.6 ×10−3 + 0.6 ×10−6 − 0.6 ×10−3 = 0.6 mV
Ex. 6.8-4
Writing node equations
v− − vs v− − vo
v−
+
+
=0
Ra
Rb
Ri + Rs


R
i v 
vo −  − A
−


R +R
i
s  + vo − v− = 0

R0
Rb
After some algebra
Av =
R0 ( Ri + Rs ) + ARi R f
vo
=
vs ( R f + R0 ) ( Ri + Rs ) + Ra ( R f + R0 + Ri + Rs ) − ARi Ra
For the given values, Av = −2.00006 V/V.
6-5
Problems
Section 6-4: The Ideal Operational Amplifier
P6.4-1
(checked using LNAP 8/16/02)
P6.4-2
Apply KVL to loop 1:
− 12 + 3000 i1 + 0 + 2000 i1 = 0
⇒ i1 =
12
= 2.4 mA
5000
The currents into the inputs of an ideal op
amp are zero so
io = i1 = 2.4 mA
i2 = − i1 = − 2.4 mA
va = i2 (1000 ) + 0 = −2.4 V
Apply Ohm’s law to the 4 kΩ resistor
vo = va − io ( 4000 )
= −2.4 − ( 2.4 ×10−3 ) ( 4000 ) = −12 V
(checked using LNAP 8/16/02)
6-6
P6.4-3
The voltages at the input nodes of an ideal op amp
are equal so va = −2 V .
Apply KCL at node a:
vo − ( −2 ) 12 − ( −2 )
+
= 0 ⇒ vo = −30 V
8000
4000
Apply Ohm’s law to the 8 kΩ resistor
io =
−2 − vo
= 3.5 mA
8000
(checked using LNAP 8/16/02)
P6.4-4
The voltages at the input nodes of an ideal
op amp are equal so v = 5 V .
Apply KCL at the inverting input node of
the op amp:
 v −5 
−3
− a
 − 0.1× 10 − 0 = 0 ⇒ va = 4 V
 10000 
Apply Ohm’s law to the 20 kΩ resistor
1
va
i =
= mA
20000
5
(checked using LNAP 8/16/02)
P6.4-5
The voltages at the input nodes of
an ideal op amp are equal so
va = 0 V . Apply KCL at node a:
 v − 0   12 − 0 
−3
− o
−
 − 2 ⋅10 = 0
 3000   4000 
⇒ vo = − 15 V
Apply KCL at the output node of
the op amp:
v
v
io + o + o = 0 ⇒ io = 7.5 mA
6000 3000
(checked using LNAP 8/16/02)
6-7
P6.4-6
The currents into the inputs of an ideal op amp
are zero and the voltages at the input nodes of
an ideal op amp are equal so va = 2.5 V .
Apply Ohm’s law to the 4 kΩ resistor:
2.5
v
ia = a =
= 0.625 mA
4000 4000
Apply KCL at node a:
ib = ia = 0.625 mA
Apply KVL:
vo = 8000 ib + 4000 ia
= (12 × 103 )( 0.625 × 10−3 ) = 7.5 V
(checked using LNAP 8/16/02)
P6.4-7
R2
 v − 0   va − 0 
vs
− s
 + 0 = 0 ⇒ va = −
 − 
R1
 R1   R 2 
io =
 R 2 + R3 
R 2 + R3
0 − va 0 − va
+
=−
v
va = 
 R1 R 3  s
R2
R3
R 2 R3


 v − 0   va − 0 
R4
R2 R4
− o
−
+ 0 = 0 ⇒ vo = −
vs
va =


 R 4   R3 
R3
R1 R 3

 

6-8
P6.4-8
The node voltages have been
labeled using:
1. The currents into the
inputs of an ideal op amp are
zero and the voltages at the
input nodes of an ideal op
amp are equal.
2. KCL
3. Ohm’s law
Then
v0 = 11.8 − 1.8 = 10 V
and
io =
10
= 2.5 mA
4000
(checked using LNAP 8/16/02)
P6.4-9
KCL at node a:
va − ( −18 )
v
+ a + 0 = 0 ⇒ va = −12 V
4000
8000
The node voltages at the input nodes of
ideal op amps are equal so vb = va .
Voltage division:
vo =
8000
vb = −8 V
4000 + 8000
(check using LNAP 8/16/02)
6-9
Section 6-5: Nodal Analysis of Circuits Containing Ideal Operational Amplifiers
P6.5-1
KCL at node b:
vb − 2
vb
v +5
1
+
+ b
= 0 ⇒ vb = − V
20000 40000 40000
4
The node voltages at the input nodes of an ideal op amp are equal so ve = vb = −
KCL at node e:
10
ve
v −v
+ e d = 0 ⇒ vd = 10 ve = −
V
1000 9000
4
1
V.
4
(checked using LNAP 8/16/02)
P6.5-2
Apply KCL at node a:
0=
va − 12
v
v −0
+ a + a
⇒ va = 4 V
6000 6000 6000
Apply KCL at the inverting input of the
op amp:
v −0
 0 − vo 
− a
+0+
 = 0
 6000 
 6000 
⇒ vo = −va = −4 V
Apply KCL at the output of the op amp:
vo
 0 − vo 
= 0
io − 
+
 6000  6000
v
⇒ io = − o = 1.33 mA
3000
(checked using LNAP 8/16/02)
6-10
P6.5-3
Apply KCL at the inverting input of
the op amp:
 v − 0   vs − 0 
− a
−
 = 0
 R2   R1 
R
⇒ va = − 2 vs
R1
Apply KCL at node a:
 1 1 1 
va −v0 va va −0
R R +R R +R R
+ +
= 0 ⇒ v0 = R4  + +  va = 2 3 2 4 3 4 va
R4
R3
R2
R2 R3
 R4 R3 R2 
Plug in values ⇒ yields
vo
30 + 900 + 30
=−
= −200 V/V
4.8
vs
=−
R2 R3 + R2 R4 + R3 R4
vs
R1 R3
P6.5-4
Ohm’s law:
i=
v1 − v2
R2
KVL:
v0 = ( R1 + R2 + R3 ) i =
R1 + R2 + R3
( v1 −v2 )
R2
6-11
P6.5-5
 R 
v1 −va v1 −v2
R
+
+ 0 = 0 ⇒ va = 1+ 1  v1 − 1 v2
R1
R7
R7
 R7 
 R 
v2 − vb
v −v
R
− 1 2 + 0 = 0 ⇒ vb =  1+ 2  v2 − 2 v1
R2
R7
R7
 R7 
 v −v  v −0
R6
vb
−  b c  + c + 0 = 0 ⇒ vc =
R4 + R6
 R4  R6
 v −v   v −v 
R
R
− a c  + c 0  + 0 = 0 ⇒ v0 = − 5 va + (1+ 5 )vc
R3
R3
 R3   R5 
 R R R ( R +R )
R
R (R +R ) R 
R 
R
v0 =  5 1 + 6 3 5 (1+ 2 )  v2 −  5 (1+ 1 ) + 6 3 5 2  v1
R7 
R7 R3 ( R4 + R6 ) R7 
 R3 R7 R3 ( R4 + R6 )
 R3
v −v
i0 = c 0 = "
R5
6-12
P6.5-6
KCL at node b:
KCL at node a:
So
5
va
vc
+
= 0 ⇒ vc = − va
3
3
20 × 10 25 × 10
4
 5 
va −  − va 
va − ( −12 )
va
va + 0
 4  = 0 ⇒ v = − 12 V
+
+
+
a
3
3
3
40 ×10
10 × 10 20 ×10
10 ×103
13
5
15
vc = − va = − .
4
13
6-13
P6.5-7
Apply KCL at the inverting input
node of the op amp
 ( va + 6 ) − 0 
 v −0 
− a
 = 0
+0−
 10000 
 30000 
⇒va = −1.5 V
Apply KCL to the super node
corresponding the voltage source:
va −0 va + 6− 0
+
10000 30000
( v + 6 )−vb = 0
v −v
+ a b + a
30000
10000
⇒
3va + va + 6 + va − vb
⇒
vb = 2va + 6 = 3 V
+ 3 ( va + 6 )− vb  = 0
Apply KCL at node b:
vb
v −v  v − v   ( v + 6 ) − vb 
+ b 0 − a b − a
 = 0
10000 30000  30000   10000 
⇒ 3vb +( vb − v0 )−( va − vb )−3( va + 6 ) − vb  = 0
⇒ v0 = 8vb − 4va −18 = 12 V
Apply KCL at the output node of the op amp:
i0 +
v0
v −v
+ 0 b = 0 ⇒ i0 = − 0.7 mA
30000 30000
6-14
P6.5-8
Apply KVL to the bottom mesh:
−i0 (10000) − i0 (20000) + 5 = 0
⇒ i0 =
1
mA
6
The node voltages at the input nodes of an
ideal op amp are equal. Consequently
va = 10000 i0 =
10
V
6
Apply KCL at node a:
va
v −v
+ a 0 = 0 ⇒
10000 20000
v0 = 3va = 5 V
P6.5-9
vb + 12
vb
+
= 0 ⇒ vb = −4 V
40000 20000
The node voltages at the input nodes of an ideal op amp are equal, so vc = vb = −4 V .
KCL at node b:
The node voltages at the input nodes of an ideal op amp are equal, so vd = vc + 0 × 10 4 = −4 V .
KCL at node g:
vg
 v f − vg 
2
−
+
= 0 ⇒ vg = v f
3 
3
3
 20 × 10  40 ×10
6-15
The node voltages at the input nodes of an ideal op amp are equal, so ve = vg =
2
vd − v f vd − 3 v f
vd − ve
+
=
+
KCL at node d: 0 =
20 ×103 20 ×103 20 ×103 20 ×103
vd − v f
Finally, ve = vg =
2
vf .
3
6
24
V
⇒ v f = vd = −
5
5
2
16
vf = −
V.
3
5
P6.5-10
By voltage division (or by applying KCL at
node a)
va =
R0
vs
R1 + R0
Applying KCL at node b:
vb − vs vb −v0
+
= 0
R1
R0 +∆R
⇒
R0 +∆R
( vb −vs )+ vb = v0
R1
The node voltages at the input nodes of an
ideal op amp are equal so vb = va .
 R +∆R  R0

R +∆R 
R0  ∆R
∆R
v0 =  0
+1
− 0
vs =  −vs
 vs = −

R1 
R1 + R0
R1 + R0  R0
 R1 + R0

 R1
6-16
Section 6-6: Design Using Operational Amplifier
P6.6-1
Use the current-to-voltage converter, entry (g) in Figure 6.6-1.
P6.6-2
Use the voltage –controlled current source, entry (i) in Figure 6.6-1.
P6.6-3
Use the noninverting summing amplifier, entry (e) in Figure 6.6-1.
6-17
P6.6-4
Use the difference amplifier, entry (f) in Figure 6.6-1.
P6.6-5
Use the inverting amplifier and the summing amplifier, entries (a) and (d) in Figure 6.6-1.
P6.6-6
Use the negative resistance converter, entry (h) in Figure 6.6-1.
6-18
P6.6-7
Use the noninverting amplifier, entry (b) in Figure 6.6-1. Notice that the ideal op amp forces the
current iin to be zero.
P6.6-8
Summing Amplifier: va = − ( 6 v1 + 2 v2 ) 
 ⇒ vo = 6 v1 + 2 v2
Inverting Amplifier:
vo = −va

P6.6-9
6-19
Using superposition, vo = v1 + v2 + v3 = −9 − 16 + 32 = 7 V
P6.6-10
R1
6
12
24
6||12
6||24
R2
-vo/vs
12||12||24
0.8
6||12||24
0.286
6||12||12
0.125
12||24
2
12||12
1.25
12||24
6||12||12
6||12||24
12||12||24
6||12
0.5
24
8
12
3.5
6
1.25
R1
R2
-vo/vs
12||12
6||24
0.8
6-20
Section 6-7: Operational Amplifier Circuits and Linear Algebraic Equations
P6.7-1
6-21
P6.7-2
6-22
Section 6-8: Characteristics of the Practical Operational Amplifier
P6.8-1
The node equation at node a is:
Solving for vout:
vout − vos
vos
=
+ ib1
3
100×10
10×103
 100×103 
vout = 1+
v + (100 ×103 ) ib1 = 11vos + (100 × 103 ) ib1
3  os
10
10
×


(
)
= 11 ( 0.03×10−3 )+(100×103 ) 1.2×10−9 = 0.45 mV
P6.8-2
The node equation at node a is:
vos
v −v
+ ib1 = 0 os
10000
90000
Solving for vo:
 90×103 
vo = 1+
v + ( 90 × 103 ) ib1 = 10 vos + ( 90 × 103 ) i b1
3  os
 10×10 
= 10(5 ×10−3 )+ ( 90 ×103 ) (.05 × 10−9 ) = 50.0045 × 10−3 − 50 mV
6-23
P6.8-3
v1 −vin v1 v1 −v0

+ +
= 0
R1
Rin
R2
v0
Rin ( R0 − AR2 )

=
 ⇒
v0 + Av1 v0 − v1
vin
( R1 + Rin )( R0 + R2 ) + R1 Rin (1+ A)
+
=0 

R0
R2
P6.8-4
a)
v0
49×103
R
= − 2 = −
= −9.6078
vin
R1
5.1×103
(
)
b)
2×106 ) 75−( 200,000 )( 50×103 )
(
v0
=
= −9.9957
vin
(5×103 + 2×106 )(75+50×103 ) + (5×103 )(2×106 )(1+ 200,000)
c)
v0
2×106 (75− (200,000)(49×103 ))
=
= −9.6037
vin
(5.1×103 +2×106 )(75+49×103 )+(5.1×103 )(2×106 )(1+200,000)
6-24
P6.8-5
Apply KCL at node b:
R3
(vcm − v p )
vb =
R2 + R3
Apply KCL at node a:
va −v0 va −(vcm + vn )
+
= 0
R4
R1
The voltages at the input nodes of
an ideal op amp are equal so
va = vb .
R
R +R
v0 = − 4 (vcm + vn ) + 4 1 va
R1
R1
R
v0 = − 4 (vcm + vn ) +
R1
when
so
R4
R1
R4
+1
( R4 + R1 ) R3
R3
R
R
R1
then
=
=
× 3 = 4
R3
R2
R1 ( R2 + R3 )
R1
+1 R2
R2
v0 = −
( R4 + R1 ) R3
(vcm − v p )
R1 ( R2 + R3 )
R4
R
R
(vcm + vn ) + 4 (vcm − v p ) = − 4 (vn + v p )
R1
R1
R1
6-25
PSpice Problems
SP6-1:
(a)
v z = a vw + b v x + c v y
The following three PSpice simulations show
1 V = vz = a when vw= 1 V, vx = 0 V and vy = 0 V
4 V = vz = b when vw= 0 V, vx = 1 V and vy = 0 V
-5 V = vz = c when vw= 0 V, vx = 0 V and vy = 1 V
Therefore
v z = vw + 4 v x − 5 v y
(b) When vw= 2 V:
vz = 4 vx − 5 v y + 2
1 V = vz = a when vw= 1 V, vx = 0 V and vy = 0 V:
6-26
4 V = vz = b when vw= 0 V, vx = 1 V and vy = 0 V:
-5 V = vz = c when vw= 0 V, vx = 0 V and vy = 1 V:
6-27
SP6-2
a) Using superposition and recognizing the inverting
and noninverting amplifiers:
vo = −
80
 80 
vs +  1 +  ( −2 ) = −3.2 vs − 8.4
25
 25 
b) Using the DC Sweep feature of PSpice produces the
plot show below. Two points have been labeled in
anticipation of c).
c) Notice that the equation predicts
( −3.2 ) ( −5 ) − 8.4 = 7.6
and
( −3.2 ) ( 0 ) − 8.4 = −8.4
Both agree with labeled points on the plot.
6-28
SP6-3
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V1
V_V2
-3.000E-04
-7.000E-04
v34 = −1.5 − −12 × 10−6 ≅ −1.5 V
v23 = 4.5 − ( −1.5 ) = 6 V
v50 = 12 − 0 = 12 V
io = −7 ×10−4 = −0.7 mA
6-29
SP6-4
V4 is a short circuit used to measure io.
The input of the VCCS is the voltage of the
left-hand voltage source. (The nominal value
of the input is 20 mV.) The output of the
VCCS is io.
A plot of the output of the VCCS versus the
input is shown below.
The gain of the VCVS is
gain =
50 × 10−6 − ( −50 × 10−6 )
1
A
= ×10−3
V
100 ×10 − ( −100 × 10 ) 2
−3
−3
(The op amp saturates for the inputs larger than
0.1 V, limiting the operating range of this
VCCS.)
6-30
Verification Problems
VP6-1
Apply KCL at the output node of the op
amp
io =
v − ( −5 )
vo
+ o
=0
10000
4000
Try the given values: io =−1 mA and
vo = 7 V
−1×10−3 ≠ 3.7 × 10−3 =
7 − ( −5 )
7
+
10000
4000
KCL is not satisfied. These cannot be the
correct values of io and vo.
VP6-2
va = ( 4 ×103 )( 2 ×10−3 ) = 8 V
12 ×103
va = −1.2 ( 8 ) = −9.6 V
10 ×103
So vo = −9.6 V instead of 9.6 V.
vo = −
6-31
VP6-3
First, redraw the circuit as:
Then using superposition, and recognizing of the inverting and noninverting amplifiers:
 6  4 
 4
vo =  −  −  ( −3) + 1 +  ( 2 ) = −18 + 6 = −12 V
 2  2 
 2
The given answer is correct.
VP6-4
First notice that ve = v f = vc is required by the ideal op amp. (There is zero current into the input
lead of an ideal op amp so there is zero current in the 10 kΩ connected between nodes e and f,
hence zero volts across this resistor. Also, the node voltages at the input nodes of an ideal op
amp are equal.)
The given voltages satisfy all the node equations at nodes b, c and d:
node b:
node c:
node d:
0− (−5)
0
0− 2
+
+
=0
10000 40000 4000
0− 2
2 −5
=
+0
4000 6000
2 −5
5
5−11
=
+
6000 5000 4000
Therefore, the analysis is correct.
6-32
VP6-5
The given voltages satisfy the node equations at nodes b and e:
node b:
node e:
−.25− 2 −.25 −.25−( −5 )
+
+
=0
20000 40000
40000
−2.5−( −0.25 ) −0.25
≠
+0
9000
1000
Therefore, the analysis is not correct.
Notice that
−2.5−( +0.25 ) +0.25
=
+0
9000
1000
So it appears that ve = +0.25 V instead of ve = −0.25 V.
Also, the circuit is an noninverting summer with Ra = 10 kΩ and Rb = 1 kΩ, K1 =1/ 2, K2
= 1/ 4 and K4 = 9. The given node voltages satisfy the equation
−2.5 = vd = K
4
( K v + K v ) = 10  12 ( 2 )+ 14 ( −5) 
1 a
2 c
None-the-less, the analysis is not correct.
6-33
Design Problems
DP6-1
From Figure 6.6-1g, this circuit
is described by vo = R f i in . Since i out =
Notice that i oa = i in +
i in <
(1250 ) i in
5000
=
Rf
vo
1 i out
, we require =
=
, or Rf = 1250 Ω
5000
4 i in 5000
5
5
i in . To avoid current saturation requires i in < i sat or
4
4
4
i sat . For example, if isat = 2 mA, then iin < 1.6 mA is required to avoid current saturation.
5
DP6-2
3
3
3
 3   12 
 3  12 
vo = − vi + 3 = − vi + 1 +    5 − vi +  1 + 
5
4
4
4
 4   35 
 4  12 + 23 
6-34
DP6-3
1
1
1
1

(a) 12 v i + 6 = 24  v i + ( 5 )  ⇒ K 4 = 24, K 1 = , and K 2 = . Take Ra = 18 kΩ and
20
2
20
2

Rb = 1 kΩ to get
(b)
(c)
(d)
6-35
DP6-4
6-36
DP6-5
We require a gain of
4
= 200 . Using an inverting amplifier:
20 ×10−3
Here we have 200 = −
R2
10 ×103 + R1
. For example, let R1 = 0 and R2 = 1 MΩ. Next, using the
noninverting amplifier:
Here we have 200 = 1 +
R2
R1
. For example, let R1 = 1 kΩ and R2 = 199 kΩ.
The gain of the inverting amplifier circuit does not depend on the resistance of the microphone.
Consequently, the gain does not change when the microphone resistance changes.
6-37
Chapter 7 Energy Storage Elements
Exercises
Ex. 7.3-1
2 2<t <4
d

and
i C ( t ) = 1 v s ( t ) = −1 4 < t < 8
dt
 0 otherwise

2 t − 2 2 < t < 4

so i (t ) = i C ( t ) + i R ( t ) =  7 − t
4<t <8
 0
otherwise

2 t − 4 2 < t < 4

i R (t ) = 1 v s (t ) =  8 − t
4<t <8
 0
otherwise

Ex. 7.3-2
1 t
1 t
v(t ) = ∫ i s (τ ) dτ + v(t 0 ) = ∫ i s (τ ) dτ − 12
1 0
C t0
3
v(t ) = 3∫ 4 dτ − 12 = 12 t − 12 for 0 < t < 4
t
0
v(t ) = 3∫ ( −2 ) dτ + 36 = 60 − 6 t for 4 < t < 10
t
4
In particular, v(4) = 36 V.
In particular, v(10) = 0 V.
v(t ) = 3∫ 0 dτ + 0 = 0 for 10 < t
t
10
Ex. 7.4-1
Cv 2
1
2
= ( 2×10 −4 ) (100 ) = 1 J
2
2
+
−
vc ( 0 ) = vc ( 0 ) = 100 V
W =
7-1
Ex. 7.4-2
a)
W ( t ) = W ( 0 ) + ∫ 0 vi dt
t
First, W ( 0 ) = 0 since v ( 0 ) = 0
Next, v( t ) = v( 0 ) +
∴ W (t ) =
b)
Ex. 7.4-3
b)
∫ ( 2×10 ) t ( 2 )dt
t
4
W (1s ) = 2 ×104 J = 20 kJ
0
W (100s ) = 2 × 104 (100 )
2
= 2 × 104 t 2
= 2 × 108 J = 200 MJ
We have v (0+ ) = v (0− ) = 3 V
vc ( t ) =
a)
1 t
4 t
10
2 dt = 2×104 t
i
dt
=
∫
∫
0
0
C
t
1 t
i (t ) dt + vc (0) = 5 ∫ 0 3 e5t dt + 3 = 3 ( e5t −1)+ 3 = 3 e5t V, 0<t <1
∫
0
C
v(t ) = vR ( t ) + vc ( t ) = 5 i ( t ) + vc ( t ) = 15 e5t + 3 e5t = 18 e5t V, 0 < t < 1
 W (t ) t =0.2 s = 6.65 J
2
W ( t ) = 1 Cvc2 ( t ) = 1 × 0.2 ( 3e5t ) = 0.9e10t J ⇒ 
2
2
W ( t ) t =0.8 s = 2.68 kJ
Ex. 7.5-1
7-2
Ex. 7.5-2
v1 = v2 ⇒
dv1 dv2
i
i
C
=
⇒ 1 = 2 ⇒ i1 = 1 i2
dt
dt
C1 C2
C2
C

KCL: i = i1 + i2 =  1 + 1 i2
 C2

⇒
i2 =
C2
i
C1 +C2
Ex. 7.5-3
(a) to (b) :
(c) to (d) :
1
1
= mF ,
1 1 1 9
+ +
1 1 1
3 3 3
1
1
1
1
= + +
2
2 10
Ceq
9
(b) to (c) : 1+
⇒ Ceq =
1 10
=
mF ,
9
9
10
mF
19
7-3
Ex. 7.6-1
 2 2<t <4
d

and
v L ( t ) = 1 i s ( t ) = −1 4 < t < 8
dt
 0 otherwise

2 t − 2 2 < t < 4

so v(t ) = v L ( t ) + v R ( t ) =  7 − t
4<t <8
 0
otherwise

Ex. 7.6-2
i (t ) =
2 t − 4 2 < t < 4

4<t <8
v R (t ) = 1 is (t ) =  8 − t
 0
otherwise

1 t
1 t
v s (τ ) dτ + i (t 0 ) = ∫ v s (τ ) dτ − 12
∫
1 0
L t0
3
i (t ) = 3∫ 4 dτ − 12 = 12 t − 12 for 0 < t < 4
t
i (t ) = 3∫ ( −2 ) dτ + 36 = 60 − 6 t for 4 < t < 10
0
t
4
In particular, i(4) = 36 A.
In particular, i(10) = 0 A.
i (t ) = 3∫ 0 dτ + 0 = 0 for 10 < t
t
10
Ex. 7.7-1
v = L
di
1 d
=  
4 t e − t ) = (1−t ) e− t V
(
dt
 4  dt
P = vi =  (1−t ) e− t  ( 4 t e− t ) = 4 t (1−t ) e −2t W
2
1
11
W = Li 2 =   ( 4 t e− t ) = 2 t 2 e − 2t J
2
24
7-4
v (t ) = L
 0
 2t

and i ( t ) = 
−2( t − 2 )
 0

di 1 di
=
dt 2 dt
p (t ) = v (t ) i (t )
Ex. 7.7-2
t <0
 0

0<t <1
 1
⇒ v( t ) = 
1<t < 2
 −1
 0
t >2
t <0
0<t <1
1<t < 2
t >2
t <0
 0
2t
0<t <1

= 
2( t − 2 ) 1<t < 2
 0
t >2

W ( t ) = W ( t0 ) + ∫ tt p( t ) dt
i (t ) = 0 for t < 0 ⇒ p ( t ) = 0 for t < 0 ⇒ W ( t0 ) = 0
0
∫ 2 t dt = t
1<t < 2 : W ( t ) = W (1)+ ∫ 2 ( t − 2 ) dt = t
0 < t < 1: W ( t ) =
t
2
0
t
t >2 : W ( t ) = W ( 2) = 0
1
2
− 4t + 4
Ex. 7.8-1
7-5
Ex. 7.8-2
Ex. 7.8-3
i1 =
1 t
∫ v dt + i1 ( t0 ) ,
L1 t 0
i2 =
1 t
∫ v dt + i 2 ( t0 )
L 2 t0
but i1 ( t0 ) = 0 and i 2 ( t0 ) = 0
1
1 t
1  t
1 t
t
∫ t v dt + ∫ t v dt =  +
∫ v dt
 ∫ t v dt =
L1 0
LP t 0
0
 L1 L2  0
1 t
1
∫ t v dt
i
L2
L 0
L1
∴1 = 1
=
=
1 t
1 1
i
L1 + L2
+
∫ t v dt
LP 0
L1 L2
i = i1 + i2 =
7-6
Problems
Section 7-3: Capacitors
P7.3-1
v (t ) = v (0) +
1 t
i (τ ) dτ
C ∫0
∴ Cv ( t ) = Cv ( 0 ) + i t
and q = Cv
In our case, the current is constant so
∫ i (τ ) dτ .
t
0
−6
−6
q −Cv( 0 ) 150×10 −(15×10 )( 5 )
∴ t=
=
= 3 ms
i
25×10−3
i (t ) = C
P7.3-2
P7.3-3
so
1d
1
d
12 cos ( 2t + 30° ) = (12 )( −2 ) sin ( 2t + 30° ) = 3cos ( 2t + 120° ) A
v (t ) =
8 dt
8
dt
( 3×10 ) cos ( 500t + 45 ) = C dtd
−3
°
12 cos ( 500t − 45° ) = C (12 )( −500 ) sin ( 500t − 45° )
= C ( 6000 ) cos ( 500t + 45° )
3×10−3 1
1
= ×10−6 = µ F
C=
3
6×10
2
2
7-7
v (t ) =
P7.3-4
is ( t ) = 0 ⇒ v ( t ) =
0 < t < 2 ×10−9
2 × 10−9 < t < 3 × 10−9
3 × 10−9 < t < 5 ×10−9
5 ×10−9 < t
P7.3-5
(b)
(a)
1 t
1
i (τ ) dτ + v ( 0 ) =
∫
0
C
2 × 10−12
is ( t ) = 4 ×10−6 A
1
2 × 10−12
∫ i (τ ) dτ − 10
t
−3
0
∫ 0 dτ − 10
t
0
−3
= −10−3
t
1
4 ×10−6 ) dτ − 10−3 = −5 × 10−3 + ( 2 × 106 ) t
−12 ∫2ns (
2 × 10
In particular, v ( 3 × 10−9 ) = −5 ×10−3 + ( 2 ×106 ) ( 3 × 10−9 ) = 10−3
⇒ v (t ) =
is ( t ) = −2 × 10−6 A
t
1
−2 × 10−6 ) dτ + 10−3 = 4 × 10−3 − (106 ) t
−12 ∫3ns (
2 × 10
In particular, v ( 5 × 10−9 ) = 4 ×10−3 − (106 ) ( 5 × 10−9 ) = −10−3 V
⇒ v (t ) =
is ( t ) = 0 ⇒ v ( t ) =
1
2 × 10−12
∫
t
5ns
0 dτ − 10−3 = −10−3 V
 0 0 < t <1

d
 4 1< t < 2
i (t ) = C v(t ) = 
dt
 −4 2 < t < 3
 0
3<t
t
1 t
v ( t ) = ∫ i (τ ) dτ + v ( 0 ) = ∫ i (τ ) dτ
0
C 0
v ( t ) = ∫ 0 dτ + 0 = 0 V
t
For 0 < t < 1, i(t) = 0 A so
For 1 < t < 2, i(t) = (4t − 4) A so
(
v ( t ) = ∫ ( 4τ − 4 ) dτ + 0 = 2τ 2 − 4τ
( )
t
0
) 1=2 t
t
2
− 4t + 2 V
v(2) = 2 22 − 4 ( 2 ) + 2 = 2 V . For 2 < t < 3, i(t) = (−4t + 12) A so
1
(
v ( t ) = ∫ ( −4τ + 12 ) dτ + 2 = −2τ 2 + 12τ
t
( )
v(3) = −2 32 + 12 ( 3) − 14 = 4 V
2
) 1 +2= ( −2 t
t
2
)
+ 12 t − 14 V
For 3 < t, i(t) = 0 A so v ( t ) = ∫ 0 dτ + 4 = 4 V
t
0
7-8
P7.3-6
(a)
(b)
 0 0<t <2
d

i (t ) = C v(t ) = 0.1 2 < t < 6
dt
0
6<t

t
1 t
v ( t ) = ∫ i (τ ) dτ + v ( 0 ) = 2 ∫ i (τ ) dτ
0
C 0
For 0 < t < 2, i(t) = 0 A so v ( t ) = 2 ∫ 0 dτ + 0 = 0 V
t
0
For 2 < t < 6, i(t) = 0.2 t − 0.4 V so
(
v ( t ) = 2 ∫ ( 0.2τ − 0.4 ) dτ + 0 = 0.2τ 2 − 0.8τ
t
( )
) 2 =0.2 t
t
2
− 0.8 t + 0.8 V
v(6) = 0.2 62 − 0.8 ( 6 ) + 0.8 = 3.2 V . For 6 < t, i(t) = 0.8 A so
1
v ( t ) = 2 ∫ 0.8 dτ + 3.2 = 1.6 t − 6.4 V
t
6
P7.3-7
v (t ) = v ( 0) +
t
1 t
i (τ ) dτ = 25 + 2.5 ×104 ∫ 0 ( 6×10−3 ) e −6τ dτ
∫
0
C
= 25 + 150∫ 0 e −6τ dτ
t
 1

= 2 5 + 150  − e −6τ  = 50 − 25e−6t V
 6
0
t
P7.3-8
v
1
= (1− 2e−2t ) × 10−3 = 25 (1− 2e −2t ) µ A
3
200×10
40
dv
iC = C
= 10×10−6 ( −2 ) ( −10 e−2t ) = 200 e−2 t µ A
dt
i = iR + i C = 200 e−2t + 25 − 50 e −2t
iR =
(
)
= 25 + 150e −2t
µA
7-9
Section 7-4: Energy Storage in a Capacitor
P7.4-1
Given
t<2
0


i ( t ) = 0.2 ( t − 2 ) 2 < t < 6
 0.8
t >6

The capacitor voltage is given by
v (t ) =
For t < 2
t
1 t
τ
τ
+
=
0
2
i
d
v
(
)
(
)
∫0 i (τ ) dτ + v ( 0 )
0.5 ∫ 0
v ( t ) = 2 ∫ 0 dτ + 0 = 0
t
In particular, v ( 2 ) = 0. For 2 < t < 6
0
v ( t ) = 2 ∫ 2 (τ − 2 ) dτ + 0 = ( 0.2τ 2 − 0.8τ ) = ( 0.2 t 2 − 0.8 t + 0.8 ) V = 0.2 ( t 2 − 4 t + 4 ) V
t
t
2
2
In particular, v ( 6 ) = 3.2 V. For 6 < t
v ( t ) = 2 ∫ 0.8 dτ + 3.2 = 1.6τ 2 + 3.2 = (1.6 t − 6.4 ) V = 1.6 ( t − 4 ) V
t
t
6
Now the power and energy are calculated as
0


2
p ( t ) = v ( t ) i ( t ) = 0.04 ( t − 2 )
 1.28 ( t − 4 )

and
W (t ) = ∫
t
0
t<2
2<t <6
6<t
0

t<2

4
2<t<6
p (τ ) dτ =  0.01( t − 2 )

2
6<t
0.8 ( t − 4 ) − 0.64
7-10
7-11
These plots were produced using three MATLAB scripts:
capvol.m
function v = CapVol(t)
if t<2
v = 0;
elseif t<6
v = 0.2*t*t - .8*t +.8;
else
v = 1.6*t - 6.4;
end
capcur.m
function i = CapCur(t)
if t<2
i=0;
elseif t<6
i=.2*t - .4;
else
i =.8;
end
c7s4p1.m
t=0:1:8;
for k=1:1:length(t)
i(k)=CapCur(k-1);
v(k)=CapVol(k-1);
p(k)=i(k)*v(k);
w(k)=0.5*v(k)*v(k);
end
plot(t,i,t,v,t,p)
text(5,3.6,'v(t), V')
text(6,1.2,'i(t), A')
text(6.9,3.4,'p(t), W')
title('Capacitor Current, Voltage and Power')
xlabel('time, s')
% plot(t,w)
% title('Energy Stored in the Capacitor, J')
% xlabel('time, s')
7-12
P7.4-2
ic ( 0 ) = 0.2 A

dv
= (10×10−6 ) ( −5 )( −4000 ) e −4000t = 0.2e−4000t A ⇒ 
−19
dt
 ic (10ms ) = 8.5×10 A
1
W ( t ) = Cv 2 ( t ) and v ( 0 ) = 5 − 5e0 = 0 ⇒ W ( 0 ) = 0
2
−3
v (10×10 ) = 5 − 5 e −40 = 5 − 21.2 × 10−18 ≅ 5 ⇒ W (10 ) = 1.25×10−4 J
ic = C
P7.4–3
i (t ) = C
dvc
so read off slope of vc (t ) to get i (t )
dt
p (t ) = vc (t ) i (t ) so multiply vc (t ) & i(t ) curves to get p (t )
π
1 t
1 t

i dτ = vc ( 0 ) + ∫0 50 cos10t +  dτ =
∫
0
2
6
C

π
5
Now since vc ( t )ave = 0 ⇒ vc ( 0 ) − sin = 0 ⇒ vc ( t )
2
6
vc ( t ) = vc ( 0 ) +
P7.4-4
∴ Wmax
( 2×10 ) ( 2.5) = 6.25 µ J
1
= C v2 =
c max
2
2
−6
π
5 π 5


 vc ( 0 ) − 2 sin 6  + 2 s in 10t + 6 
π
5

= sin 10t +  V
2
6

2
First non-negative t for max energy occurs when: 10t +
π
6
=
π
2
⇒t =
π
30
= 0.1047 s
7-13
P7.4-5
Max. charge on capacitor = C v = (10×10−6 ) ( 6 ) = 60 µ C
∆q
60×10−6
=
= 6 sec to charge
10×10−6
i
1
1
2
stored energy = W = C v 2 = (10×10−6 ) ( 6 ) =180 µ J
2
2
∆t =
Section 7-5: Series and Parallel capacitors
P7.5-1
2µ F 4 µ F = 6µ F
6µ F in series with 3µ F =
i (t ) = 2 µ F
P7.5-2
6µ F⋅3µ F
= 2µ F
6µ F+3µ F
d
(6 cos100t ) = (2×10−6 ) (6) (100) (− sin100t ) A = −1.2 sin100t mA
dt
4 µ F in series with 4 µ F =
2 µF 2 µF = 4 µF
4 µ F×4 µ F
= 2 µF
4 µ F+4 µ F
4 µ F in series with 4 µ F = 2 µ F
i (t ) =(2×10−6 )
P7.5-3
d
(5+3 e −250t ) = (2×10−6 ) (0+ 3(−250) e −250t ) A = −1.5 e −250t mA
dt
C in series with C =
C C
C
5
= C
2
2
C ⋅C C
=
C +C 2
C⋅ 5 C
5
5
2
C in series with C =
= C
5
2
7
C+ C
2
5  d
5 
(25×10−3 ) cos 250t =  C  (14sin 250t ) =  C (14)(250) cos 250t
 7  dt
7 
so 25×10−3 = 2500 C ⇒ C = 10×10−6 = 10 µF
7-14
Section 7-6: Inductors
P7.6-1
di
= 2 0 0 [1 0 0 ( 4 0 0 ) c o s 4 0 0 t ] V
dt
8 ×1 0 6 V
∴ v m ax = 8 ×1 0 6 V th u s h a ve a fie ld o f
= 4 ×1 0 6 V
m
m
2
6
w h ic h e x ce e d s d ie le c tric stre n g th in a ir o f 3 ×1 0 V /m
∴ W e g e t a d isc h a rg e a s th e a ir is io n iz e d .
F in d m a x . v o ltag e a c ro ss c o il:
v (t ) = L
P7.6-2
v=L
P7.6-3
(a)
(b)
di
+ R i = (.1) (4e− t − 4te− t ) + 10(4te− t ) = 0.4 e − t + 39.6t e − t V
dt
 0 0 < t <1
 4 1< t < 2
d

v(t ) = L i (t ) = 
dt
−4 2 < t < 3
3<t
 0
i (t ) =
t
1 t
v
d
i
0
τ
τ
+
=
(
)
(
)
∫0 v (τ ) dτ
L ∫0
For 0 < t < 1, v(t) = 0 V so
i ( t ) = ∫ 0 dτ + 0 = 0 A
t
For 1 < t < 2, v(t) = (4t − 4) V so
t
t
i ( t ) = ∫ ( 4τ − 4 ) dτ + 0 = ( 2τ 2 − 4τ ) =2 t 2 − 4 t + 2 A
0
1
0
i (2) = 4 ( 22 ) − 4 ( 2 ) + 2 = 2 A
For 2 < t < 3, v(t) = −4t + 12 V so
t
t
i ( t ) = ∫ ( −4τ + 12 ) dτ + 2 = ( −2τ 2 + 12τ ) +2= ( −2 t 2 + 12 t − 14 ) A
2
2
i (3) = −2 ( 32 ) + 12 ( 3) − 14 = 4 A
For 3 < t, v(t) = 0 V so i ( t ) = ∫ 0 dτ + 4 = 4 A
t
3
7-15
P7.6-4
P7.6-5
v (t ) = (250 × 10 −3 )
iL (t ) =
for
1
5 ×10−3
0< t < 1 µ s
iL (t ) =
1
5×10−3
d
(120 × 10 −3 ) sin(500t − 30° ) = (0.25)(0.12)(500) cos(500 t − 30° )
dt
= 15 cos(500t − 30° )
∫ v (τ ) dτ
t
0
s
− 2 ×10−6
vs (t ) = 4 mV
 4×10−3 
−3
−6
τ
d
t − 2×10−6 = 0.8− 2×10−6 A
×
−
×
=
4
10
2
10

−3 
∫0
×
5
10


t
 4×10−3

6
iL (1µs) = 
1×10−6 )  − 2×10−6 = − ×10−6 A =−1.2 A
−3 (
5
 5×10

for 1µ s <t < 3 µ s vs (t ) = −1 mV
t
1
6
1×10−3
6
−3
−6
τ
d
−
×
−
×
=−
1
10
10
(t −1×10−6 ) − ×10−6 =( −0.2 t −10−6 ) A
)
−3 ∫1µ s (
−3
5×10
5
5×10
5
 1×10−3

iL ( 3µ s ) =  −
+ 3×10−6  − 1×10−6 = −1.6 µA
−3
 5×10

iL (t ) =
for 3µ s < t vs (t ) = 0 so iL (t ) remains − 1.6 µA
7-16
P7.6-6
v(t ) =
( 2 ×10 ) i
3
s
for 0<t <1 µ s
(t ) +
( 4 ×10 ) dtd
−3
is (t ) (in general)
 1×10−3 
d
is (t ) = (1) 
t = 103 t ⇒ is (t ) = 1×103
−6 
dt
 1×10 
v(t ) = (2×103 )(1×103 ) t + 4×10−3 (1×103 ) = ( 2×106 t + 4 ) V
for 1µ s <t < 3µ s is (t ) = 1 mA ⇒
d
is (t ) = 0
dt
v(t ) = (2×103 )(1×10−3 ) + ( 4×10−3 )×0 = 2 V
 1×10−3 
t ⇒
for 3µ s< t < 5µ s is (t ) = 4×10−3 −
−6 
 1×10 
d
1×10−3
=−103
is (t ) = −
−6
dt
1×10
v(t ) = ( 2×103 )( 4×10−3 −103 t )+ 4×10−3 ( −103 ) = 4 − ( 2×106 ) t
when 5µ s <t< 7µ s
is (t ) = −1×10−3 and
v(t ) = ( 2×103 )(10−3 ) = − 2 V
d
is (t ) = 0
dt
 1×10−3 
d
when 7µ s< t < 8µ s is (t ) = 
t − 8×10−3 ⇒ is (t ) = 1×103
−6 
dt
 1×10 
v(t ) = ( 2 × 103 )(103 t − 8 × 10−3 ) + ( 4 × 10−3 )(103 ) = −12 + ( 2 × 106 ) t
when 8µ s < t , then is (t ) = 0 ⇒
d
is (t ) = 0
dt
v(t )= 0
P7.6-8
(a)
(b)
 0 0<t <2
d

v(t ) = L i (t ) = 0.1 2 < t < 6
dt
0
6<t

t
1 t
i ( t ) = ∫ v (τ ) dτ + i ( 0 ) = 2 ∫ v (τ ) dτ
0
L 0
For 0 < t < 2, v(t) = 0 V so i ( t ) = 2 ∫ 0 dτ + 0 = 0 A
t
0
For 2 < t < 6, v(t) = 0.2 t − 0.4 V so
t
t
i ( t ) = 2∫ ( 0.2τ − 0.4 ) dτ + 0 = ( 0.2τ 2 − 0.8τ ) =0.2 t 2 − 0.8 t + 0.8 A
2
2
( )
i (6) = 0.2 62 − 0.8 ( 6 ) + 0.8 = 3.2 A .
For 6 < t, v(t) = 0.8 V so
7-17
i ( t ) = 2 ∫ 0.8 dτ + 3.2 = (1.6 t − 6.4 ) A
t
6
Section 7-7: Energy Storage in an Inductor
P7.7-1
t<0
0
d

v( t ) =100×10
i ( t ) = 0.4 0≤ t ≤1
dt
0
t>1

t <0
0

p( t ) =v( t ) i( t ) = 1.6t 0≤t ≤1
0
t >1

−3
W (t ) = ∫
t
0
 0

p (τ ) dτ = 0.8t 2
 0.8

t <0
0<t <1
t >1
7-18
P7.7-2
P7.7-3
 d

p (t ) = v (t ) i (t ) = 5 (4sin 2t )  (4sin 2t )
 dt

= 5 (8cos 2t ) (4sin 2t )
= 80 [2 cos 2t sin 2t ]
= 80 [sin(2t + 2t ) + sin(2t − 2t )] = 80 sin 4t W
t
t
80
W(t ) = ∫ p(τ ) dτ = 80∫ sin4τ dτ = −
[cos 4τ |t0 ] = 20 (1 − cos 4t )
0
0
4
t
1
6 cos 100τ dτ + 0
25×10−3 ∫ 0
6
[sin 100τ | 0t ] = 2.4sin100 t
=
−3
(25×10 )(100)
p(t ) = v(t ) i(t ) = (6 cos100 t )(2.4 sin100t )
i(t ) =
= 7.2 [ 2(cos100 t )(sin100 t ) ]
= 7.2 [sin 200 t + sin 0] = 7.2 sin 200 t
W (t ) = ∫0 p (τ ) dτ = 7.2 ∫0 sin 200τ dτ
t
t
7.2
cos 200τ |t0 
200
= 0.036[1 − cos 200t ] J = 36 [1 − cos 200t ] mJ
=−
Section 7-8: Series and Parallel Inductors
P7.8-1
6×3
= 2 H and 2 H + 2 H = 4 H
6+3
1 t
6
sin100τ | t0  = 0.015sin100 t A = 15sin100 t mA
i (t ) = ∫ 0 6 cos100τ dτ =
4
4×100 
6H
3H =
P7.8-2
(8×10 )×(8×10 )
−3
4 mH + 4 mH = 8 mH , 8mH 8mH =
−3
8×10−3 +8×10−3
= 4 mH
and 4 mH + 4 mH = 8 mH
d
v(t ) = (8×10−3 )
(5+ 3e −250t ) = (8×10−3 ) (0+ 3(−250) e −250t ) =−6 e −250t V
dt
7-19
P7.8-3
L⋅ L
L
L
5
L
=
=
and L + L +
L+ L
2
2
2
5  d
5 
25cos 250 t =  L 
14×10−3 ) sin 250 t =  L (14×10−3 )(250) cos 250 t
(
 2  dt
2 
25
so L =
= 2.86 H
5
−3
(14×10 ) (250)
2
L
L =
(
)
Section 7-9: Initial Conditions of Switched Circuits
P7.9-1
Then
i L ( 0+ ) = i L ( 0− ) = 0 and v C ( 0+ ) = v C ( 0− ) = 12 V
Next
7-20
P7.9-2
Then
i L ( 0+ ) = i L ( 0− ) = 1 mA and v C ( 0+ ) = v C ( 0− ) = 6 V
Next
P7.9-3
Then
i L ( 0+ ) = i L ( 0− ) = 0 and v C ( 0+ ) = v C ( 0− ) = 0 V
Next
7-21
P7.9-4
at t = 0−
KVL: − vc (0− ) + 32 − 15 = 0 ⇒ vc (0− ) = vc (0+ ) =17 V
at t = 0+
Apply KCL to supernode shown above:
−15 − 9 15
ic ( 0+ ) +
−
+ 0.003 = 0 ⇒ ic ( 0+ ) = 6 mA
4000 5000
Then
ic ( 0+ ) 6 ×10−3
dvc
V
=
=
= 3000
−6
2 ×10
s
dt t =0+
C
7-22
Section 7-10: Operational amplifier Circuits and Linear Differential Equations
P7.10-1
P7.10-2
7-23
P7.10-3
P7.10-4
7-24
Verification Problems
VP7-1
We need to check the values of the inductor current at the ends of the intervals.
? − 1 + 0.065 = 0.025 ( Yes!)
at t = 1 0.025 =
25
at t = 3
−
at t = 9
3
3
+ 0.065 =?
− 0.115
25
50
−0.055 = −0.055
9
− 0.115 =? 0.065
50
( Yes!)
0.065 = 0.065 ( Yes!)
The given equations for the inductor current describe a current that is continuous, as
must be the case since the given inductor voltage is bounded.
VP7-2
We need to check the values of the inductor current at the ends of the intervals.
1
? − 1 + 0.03 ( Yes!)
+ 0.025 =
at t = 1 −
200
100
4
? 4 − 0.03 ( No!)
+ 0.03 =
100
100
The equation for the inductor current indicates that this current changes
instantaneously at t = 4s. This equation cannot be correct.
at t = 4
−
Design Problems
i (t )
d
= 0.5 F .
v ( t ) = −6 e −3t is proportional to i(t) so the element is a capacitor. C =
d
dt
v (t )
dt
v (t )
d
= 0.5 H .
b) i ( t ) = −6 e −3t is proportional to v(t) so the element is an inductor. L =
d
dt
i (t )
dt
v (t )
c) v(t) is proportional to i(t) so the element is a resistor. R =
= 2 Ω.
i (t )
DP7-1
a)
7-25
1.131cos ( 2t + 45° ) = 1.131  cos ( 45° ) cos ( 2t ) − sin ( 45° ) sin ( 2t ) 
DP7-2
= 0.8 cos 2 t − 0.8 sin 2 t
The first term is proportional to the voltage. Associate it with the
resistor. The noticing that
∫
t
−∞
v (τ ) dτ = ∫ 4 cos 2 t dτ = 2 sin 2t
t
−∞
d
d
v ( t ) = 4 cos 2 t = −8 sin 2 t
dt
dt
associate the second term with a capacitor to get the minus sign.
Then
4 cos 2 t
4 cos 2 t
R=
=
= 5 Ω and
0.8 cos 2 t
i1 (t )
−0.8 sin 2 t
i2 (t )
=
= 0.1 F
C=
d
t
−
8
sin
2
4 cos 2 t
dt
1.131cos ( 2t − 45° ) = 1.131 cos ( −45° ) cos ( 2t ) − sin ( −45° ) sin ( 2t ) 
= 0.8 cos 2 t + 0.8 sin 2 t
The first term is proportional to the voltage. Associate it with the
resistor. Then noticing that
∫ v (τ ) dτ = ∫
t
−∞
t
−∞
4 cos 2 t dτ = 2 sin 2t
d
d
v ( t ) = 4 cos 2 t = −8 sin 2 t
dt
dt
associate the second term with an inductor to get the plus sign. Then
R=
L=
∫
4 cos 2 t
4 cos 2 t
=
= 5 Ω and
0.8 cos 2 t
i1 (t )
t
−∞
4 cos 2 t dτ
i2 (t )
=
2 sin 2 t
= 2.5 H
0.8 sin 2 t
7-26
DP7-3
a)
11.31cos ( 2t + 45° ) = 11.31  cos ( 45° ) cos ( 2t ) − sin ( 45° ) sin ( 2t ) 
= 8 cos 2 t − 8 sin 2 t
The first term is proportional to the voltage. Associate it with the resistor. The noticing that
∫ i (τ ) dτ = ∫
t
−∞
t
−∞
4 cos 2 t dτ = 2 sin 2t
d
d
i ( t ) = 4 cos 2 t = −8 sin 2 t
dt
dt
associate the second term with an inductor to get the minus sign. Then
v (t )
8 cos 2 t
v2 (t )
−8 sin 2 t
R= 1
=
= 2 Ω and L =
=
=1H
d
4 cos 2 t 4 cos 2 t
4 cos 2 t −8 sin 2 t
dt
b)
11.31cos ( 2t + 45° ) = 11.31  cos ( −45° ) cos ( 2t ) − sin ( −45° ) sin ( 2t ) 
= 8 cos 2 t + 8 sin 2 t
The first term is proportional to the voltage. Associate it with the resistor. The noticing that
∫
t
−∞
i (τ ) dτ = ∫ 4 cos 2 t dτ = 2 sin 2t
t
−∞
d
d
i ( t ) = 4 cos 2 t = −8 sin 2 t
dt
dt
associate the second term with a capacitor to get the minus sign. Then
v (t )
8 cos 2 t
R= 1
=
= 2 Ω and C =
4 cos 2 t 4 cos 2 t
∫
t
−∞
4 cos 2 t dτ
v2 (t )
=
2 sin 2 t
= 0.25 F
8 sin 2 t
7-27
DP7-4
iL ( 0− ) = 0
at t=0−
By voltage division: vC ( 0− ) =
We require vC ( 0− ) = 3 V so
VB
4
VB = 12 V
at t=0+
Now we will check
First:
and
dvC
dt
t = 0+
iL ( 0+ ) = iL ( 0− ) = 0
vC ( 0+ ) = vC ( 0− ) = 3 V
Apply KCL at node a:
VB − vC ( 0+ )
+
+
iL ( 0 ) + iC ( 0 ) =
3
0 + iC ( 0+ ) =
12 − 3
⇒ iC ( 0+ ) = 3 A
3
iC ( 0
dvC
=
dt t =0+
C
as required.
Finally
+
)=
3
V
= 24
0.125
s
DP7-5
1
1
L i L2 = C v C2 where iL and vC are the steady state inductor current and capacitor
2
2
v
voltage. At steady state, i L = C . Then
R
We require
v 
L  C  = C vC2
R
2
⇒
C=
L
R2
⇒
R =
L
=
C
10−2
=
10−6
104 = 10 2 Ω
so R = 100 Ω .
7-28
Chapter 8 – The Complete Response of RL and RC Circuit
Exercises
Ex. 8.3-1
Before the switch closes:
After the switch closes:
2
= 8 Ω so τ = 8 ( 0.05 ) = 0.4 s .
0.25
Finally, v (t ) = voc + (v (0) − voc ) e− t τ = 2 + e−2.5 t V for t > 0
Therefore R t =
Ex. 8.3-2
Before the switch closes:
After the switch closes:
8-1
2
6
= 8 Ω so τ = = 0.75 s .
0.25
8
1
1
Finally, i (t ) = isc + (i (0) − isc ) e−t τ = + e−1.33 t A for t > 0
4 12
Therefore R t =
Ex. 8.3-3
At steady-state, immediately before t = 0:

12
 10  
i ( 0) = 
 
 = 0.1 A
 10 + 40   16+ 40||10 
After t = 0, the Norton equivalent of the circuit connected to the inductor is found to be
so I sc = 0.3 A, Rt = 40 Ω, τ =
Finally:
20
1
L
s
=
=
40
2
Rt
i (t ) = (0.1 − 0.3)e −2t + 0.3 = 0.3 − 0.2e −2t A
8-2
Ex. 8.3-4
At steady-state, immediately before t = 0
After t = 0, we have:
so Voc = 12 V, Rt = 200 Ω, τ = Rt C = (200)(20 ⋅10 −6 ) = 4 ms
Finally:
v(t ) = (12 − 12) e −t 4 + 12 = 12 V
Ex. 8.3-5
Immediately before t = 0, i (0) = 0.
After t = 0, replace the circuit connected to the inductor by its Norton equivalent to calculate the
inductor current:
I sc = 0.2 A, Rt = 45 Ω, τ =
L 25 5
=
=
Rth 45 9
So i (t ) = 0.2 (1 − e −1.8t ) A
Now that we have the inductor current, we can calculate v(t):
v(t ) = 40 i (t ) + 25
d
i (t )
dt
= 8(1 − e −1.8t ) + 5(1.8)e −1.8t
= 8 + e −1.8t V for t > 0
8-3
Ex. 8.3-6
At steady-state, immediately before t = 0
so i(0) = 0.5 A.
After t > 0: Replace the circuit connected to the inductor by its Norton equivalent to get
I sc = 93.75 mA, Rt = 640 Ω,
τ =
L
.1
1
s
=
=
Rt 640 6400
i (t ) = 406.25 e −6400t + 93.75 mA
Finally:
v (t ) = 400 i (t ) + 0.1
Ex. 8.4-1
d
i (t ) = 400 (.40625e −6400t + .09375) + 0.1 (−6400) (0.40625e −6400t )
dt
= 37.5 − 97.5e −6400t V
τ = ( 2×103 )(1×10−6 ) = 2 ×10−3 s
vc (t ) = 5 + (1.5−5 ) e −500 t
V
vc (0.001) = 5 − 3.5e − 0.5 = 2.88 V
So vc (t ) will be equal to vT at t = 1 ms if v = 2.88 V.
T
8-4
Ex. 8.4-2
iL (0) = 1 mA, I sc = 10 mA 

 ⇒ iL ( t ) = 10 − 9 e
L
Rt =500 Ω, τ =

500
vR (t ) = 300 iL ( t ) = 3 − 2.7 e
−
500
t
L
−500
t
L
mA
V
We require that vR ( t ) = 1.5 V at t = 10 ms = 0.01 s . That is
1.5 = 3 − 2.7e
Ex. 8.6-1
0 < t < t1
−
500
(0.01)
L
⇒ L=
5
= 8.5 H
0.588
v(t ) = v(∞)+Ae− t / RC where v(∞) = (1 A)(1 Ω ) = 1 V
v(t ) = 1 + A e− t /(1)(.1) = 1 + A e−10 t
Now v(0− ) = v(0+ ) = 0 = 1 + A ⇒ A = −1
∴ v(t ) =1−e−10t V
t > t1
t1 = 0.5 s,
v(t ) = v(t1 ) e
−
t −.5
(1)(.1)
= v(0.5) e−10(t −.5)
Now v(0.5) 1 − e −10(0.5) = 0.993 V
∴ v(t ) = 0.993 e−10(t −0.5) V
8-5
Ex. 8 6-2
t < 0 no sources ∴ v(0− ) = v(0+ ) = 0
0 < t <t1
v (t ) = v (∞ ) + A e
− t / RC
= v (∞ ) + A e
− t
2 × 1 0 5 (1 0 − 7 )
where for t = ∞ (steady-state)
∴capacitor becomes an open ⇒ v(∞) = 10 V
v(t ) = 10 + Ae −50t
Now v(0) = 0 = 10 + A ⇒ A = −10
t > t1 , t1 = .1 s
∴ v(t ) = 10(1− e −50t ) V
v(t ) = v(.1) e−50(t −.1)
where v(.1) = 10(1− e −50(.1) ) = 9.93 V
∴ v(t ) = 9.93e−50( t −.1) V
Ex. 8.6-3
For t < 0 i = 0.
For 0 < t < 0.2 s
KCL: −5 + v / 2 + i = 0 
 di
 + 10i = 50
di
also: v = 0.2
 dt
dt
∴ i (t ) = 5+ Ae−10t
i (0) = 0 = 5+ A ⇒ A = −5
so we have i (t ) = 5 (1−e−10t ) A
For t > 0.2 s
i (0.2) = 4.32 A ∴ i(t ) = 4.32e−10(t −.2) A
8-6
vs ( t ) = 10sin 20t V
Ex. 8.7-1
 d v( t ) 
KVL: −10sin 20t +10 .01
 + v( t ) = 0
dt 

d v( t )
⇒
+10 v( t ) = 100sin 20t
dt
Natural response: vn (t ) = Ae−t τ where τ = Rt C ∴ vn (t ) = Ae−10t
Forced response: try v f (t ) = B1 cos 20t + B2 sin 20 t
Plugging v f (t ) into the differential equation and
equating like terms yields: B1 = − 4 & B2 = 2
Complete response: v(t ) = vn (t ) + v f (t )
v(t ) = Ae −10t − 4 cos 20t + 2 sin 20t
Now v(0− ) = v(0+ ) = 0 = A − 4 ∴ A = 4
∴ v(t ) = 4 e −10t − 4 cos 20t + 2sin 20 t V
is ( t ) = 10 e −5t
Ex. 8.7-2
for t > 0
KCL at top node: −10e −5t + i ( t ) + v( t ) /10 = 0
Now v ( t ) = 0.1
Natural response: in (t ) = Ae −t τ
di ( t )
di ( t )
⇒
+100 i ( t ) = 1000 e −5t
dt
dt
where τ = L R t ∴ in (t ) = Ae −100t
Forced response: try i f (t ) = Be −5t & plug into the differential equation
−5 Be −5t + 100 Be −5t = 1000e −5t
Complete response: i (t ) = Ae
−100 t
+ 10.53e
⇒ B = 10.53
−5t
Now i (0 ) = i (0 ) = 0 = A + 10.53 ⇒ A = −10.53
−
+
∴ i (t ) = 10.53 (e −5t − e −100t ) A
Ex. 8.7-3
When the switch is closed, the inductor current is iL = vs / R = vs . When the switch opens, the
inductor current is forced to change instantaneously. The energy stored in the inductor
instantaneously dissipates in the spark. To prevent the spark, add a resistor (say 1 kΩ) across the
switch terminals.
8-7
Problems
Section 8.3: The Response of a First Order Circuit to a Constant Input
P8.3-1
Here is the circuit before t = 0, when the switch is
open and the circuit is at steady state. The open
switch is modeled as an open circuit.
A capacitor in a steady-state dc circuit acts like an
open circuit, so an open circuit replaces the
capacitor. The voltage across that open circuit is the
initial capacitor voltage, v (0).
v (0) =
By voltage division
6
(12 ) = 4 V
6+6+6
Next, consider the circuit after the switch closes. The
closed switch is modeled as a short circuit.
We need to find the Thevenin equivalent of the part
of the circuit connected to the capacitor. Here’s the
circuit used to calculate the open circuit voltage, Voc.
Voc =
6
(12 ) = 6 V
6+6
Here is the circuit that is used to determine Rt.
A short circuit has replaced the closed switch.
Independent sources are set to zero when
calculating Rt, so the voltage source has been
replaced by a short circuit.
( 6 )( 6 ) = 3 Ω
Rt =
6+6
Then
τ = R t C = 3 ( 0.25 ) = 0.75 s
Finally,
v ( t ) = Voc + ( v ( 0 ) − Voc ) e−t / τ = 6 − 2 e−1.33 t V for t > 0
8-8
P8.3-2
Here is the circuit before t = 0, when the switch is
closed and the circuit is at steady state. The closed
switch is modeled as a short circuit.
An inductor in a steady-state dc circuit acts like an
short circuit, so a short circuit replaces the
inductor. The current in that short circuit is the
initial inductor current, i(0).
i ( 0) =
12
=2 A
6
Next, consider the circuit after the switch opens.
The open switch is modeled as an open circuit.
We need to find the Norton equivalent of the part
of the circuit connected to the inductor. Here’s the
circuit used to calculate the short circuit current,
Isc.
I sc =
12
=1 A
6+6
Here is the circuit that is used to determine Rt.
An open circuit has replaced the open switch.
Independent sources are set to zero when
calculating Rt, so the voltage source has been
replaced by an short circuit.
( 6 + 6 )( 6 ) = 4 Ω
Rt =
( 6 + 6) + 6
Then
L 8
τ=
= =2 s
Rt 4
Finally,
i ( t ) = I sc + ( i ( 0 ) − I sc ) e −t / τ = 1 + e −0.5 t A for t > 0
8-9
P8.3-3
Before the switch closes:
After the switch closes:
−6
= 3 Ω so τ = 3 ( 0.05 ) = 0.15 s .
−2
Finally, v (t ) = voc + (v (0) − voc ) e− t τ = −6 + 18 e−6.67 t V for t > 0
Therefore R t =
P8.3-4
Before the switch closes:
After the switch closes:
8-10
−6
6
= 3 Ω so τ = = 2 s .
−2
3
t
−
10
Finally, i (t ) = isc + (i (0) − isc ) e τ = −2 + e−0.5 t A for t > 0
3
Therefore R t =
P8.3-5
Before the switch opens, v o ( t ) = 5 V ⇒ v o ( 0 ) = 5 V . After the switch opens the part of the
circuit connected to the capacitor can be replaced by it's Thevenin equivalent circuit to get:
Therefore τ = ( 20 × 103 )( 4 × 10−6 ) = 0.08 s .
Next, v C (t ) = voc + (v (0) − voc ) e
−
τ
t
= 10 − 5 e−12.5 t V for t > 0
Finally, v 0 (t ) = vC (t ) = 10 − 5 e −12.5 t V for t > 0
8-11
P8.3-6
Before the switch opens, v o ( t ) = 5 V ⇒ v o ( 0 ) = 5 V . After the switch opens the part of the
circuit connected to the capacitor can be replaced by it's Norton equivalent circuit to get:
Therefore τ =
5
= 0.25 ms .
20 × 103
Next, i L (t ) = isc + (i L (0) − isc ) e
Finally, vo ( t ) = 5
−
τ
t
= 0.5 − 0.25 e−4000 t mA for t > 0
d
i L ( t ) = 5 e −4000 t V
dt
for t > 0
P8.3-7
At t = 0− (steady-state)
Since the input to this circuit is constant, the
inductor will act like a short circuit when the
circuit is at steady-state:
for t > 0
iL ( t ) = iL ( 0 ) e − ( R L ) t = 6 e−20t A
8-12
P8.3-8
Before the switch opens, the circuit will be at steady state. Because the only input to this circuit
is the constant voltage of the voltage source, all of the element currents and voltages, including
the capacitor voltage, will have constant values. Opening the switch disturbs the circuit.
Eventually the disturbance dies out and the circuit is again at steady state. All the element
currents and voltages will again have constant values, but probably different constant values than
they had before the switch opened.
Here is the circuit before t = 0, when
the switch is closed and the circuit is at
steady state. The closed switch is modeled as
a short circuit. The combination of resistor
and a short circuit connected is equivalent to
a short circuit. Consequently, a short circuit
replaces the switch and the resistor R. A
capacitor in a steady-state dc circuit acts like
an open circuit, so an open circuit replaces
the capacitor. The voltage across that open
circuit is the capacitor voltage, vo(t).
Because the circuit is at steady state, the value of the capacitor voltage will be constant.
This constant is the value of the capacitor voltage just before the switch opens. In the absence of
unbounded currents, the voltage of a capacitor must be continuous. The value of the capacitor
voltage immediately after the switch opens is equal to the value immediately before the switch
opens. This value is called the initial condition of the capacitor and has been labeled as vo(0).
There is no current in the horizontal resistor due to the open circuit. Consequently, vo(0) is equal
to the voltage across the vertical resistor, which is equal to the voltage source voltage. Therefore
vo ( 0 ) = Vs
The value of vo(0) can also be obtained by setting t = 0 in the equation for vo(t). Doing so gives
vo ( 0 ) = 2 + 8 e0 = 10 V
Consequently,
Vs = 10 V
8-13
Next, consider the circuit after the switch
opens. Eventually (certainly as t →∞) the
circuit will again be at steady state. Here is
the circuit at t = ∞, when the switch is open
and the circuit is at steady state. The open
switch is modeled as an open circuit. A
capacitor in a steady-state dc circuit acts like
an open circuit, so an open circuit replaces
the capacitor. The voltage across that open
circuit is the steady-state capacitor voltage,
vo(∞). There is no current in the horizontal
resistor and vo(∞) is equal to the voltage
across the vertical resistor. Using voltage
division,
10
vo ( ∞ ) =
(10 )
R + 10
The value of vo(∞) can also be obtained by setting t = ∞ in the equation for vo(t). Doing so gives
vo ( ∞ ) = 2 + 8 e −∞ = 2 V
10
(10 ) ⇒ 2 R + 20 = 100 ⇒ R = 40 Ω
R + 10
−t τ
Finally, the exponential part of vo(t) is known to be of the form e
where τ = R t C and
Consequently,
2=
Rt is the Thevenin resistance of the part of the circuit connected to the capacitor.
Here is the circuit that is used to determine Rt.
An open circuit has replaced the open switch.
Independent sources are set to zero when
calculating Rt, so the voltage source has been
replaced by a short circuit.
R t = 10 +
( 40 )(10 ) = 18
40 + 10
Ω
τ = R t C = 18 C
so
From the equation for vo(t)
t
−0.5 t = −
⇒ τ =2s
Consequently,
τ
2 = 18 C ⇒ C = 0.111 = 111 mF
8-14
P8.3-9:
Before the switch closes, the circuit will be at steady state. Because the only input to this circuit
is the constant voltage of the voltage source, all of the element currents and voltages, including
the inductor current, will have constant values. Closing the switch disturbs the circuit by shorting
out the resistor R1. Eventually the disturbance dies out and the circuit is again at steady state. All
the element currents and voltages will again have constant values, but probably different constant
values than they had before the switch closed.
The inductor current is equal to the current in the 3 Ω resistor. Consequently,
− 0.35 t
vo (t ) 6 − 3 e
− 0.35 t
=
= 2− e
A when t > 0
i (t ) =
3
3
In the absence of unbounded voltages, the current in any inductor is continuous. Consequently,
the value of the inductor current immediately before t = 0 is equal to the value immediately after
t = 0.
Here is the circuit before t = 0, when the
switch is open and the circuit is at steady
state. The open switch is modeled as an open
circuit. An inductor in a steady-state dc
circuit acts like a short circuit, so a short
circuit replaces the inductor. The current in
that short circuit is the steady state inductor
current, i(0). Apply KVL to the loop to get
R1 i ( 0 ) + R 2 i ( 0 ) + 3 i ( 0 ) − 24 = 0
⇒ i (0) =
24
R1 + R 2 + 3
The value of i(0) can also be obtained by setting t = 0 in the equation for i(t). Do so gives
i ( 0 ) = 2 − e0 = 1 A
Consequently,
1=
24
⇒ R1 + R 2 = 21
R1 + R 2 + 3
Next, consider the circuit after the switch
closes. Here is the circuit at t = ∞, when the
switch is closed and the circuit is at steady
state. The closed switch is modeled as a
short circuit. The combination of resistor
and a short circuit connected is equivalent to
a short circuit. Consequently, a short circuit
replaces the switch and the resistor R1.
8-15
An inductor in a steady-state dc circuit acts like a short circuit, so a short circuit replaces the
inductor. The current in that short circuit is the steady state inductor current, i(∞). Apply KVL to
the loop to get
24
R 2 i ( ∞ ) + 3 i ( ∞ ) − 24 = 0 ⇒ i ( ∞ ) =
R2 + 3
The value of i(∞) can also be obtained by setting t = ∞ in the equation for i(t). Doing so gives
i ( ∞ ) = 2 − e −∞ = 2 A
Consequently
2=
24
⇒ R2 = 9 Ω
R2 + 3
R1 = 12 Ω
Then
Finally, the exponential part of i(t) is known to be of the form e
−t τ
where τ =
L
and
Rt
Rt is the Thevenin resistance of the part of the circuit that is connected to the inductor.
Here is shows the circuit that is used to determine Rt.
A short circuit has replaced combination of resistor
R1 and the closed switch. Independent sources are set
to zero when calculating Rt, so the voltage source
has been replaced by an short circuit.
R t = R 2 + 3 = 9 + 3 = 12 Ω
so
From the equation for i(t)
Consequently,
−0.35 t = −
2.857 =
τ
t
τ=
L
L
=
R t 12
⇒ τ = 2.857 s
L
⇒ L = 34.28 H
12
8-16
P8.3-10
First, use source transformations to obtain the equivalent circuit
for t > 0:
for t < 0:
So iL ( 0 ) = 2 A, I sc
and iL ( t ) = 2e−24t
1
L
1
s
= 0, Rt = 3 + 9 = 12 Ω, τ =
= 2 =
24
Rt 12
t >0
Finally v ( t ) = 9 iL ( t ) = 18 e−24t
t >0
8-17
Section 8-4: Sequential Switching
P8.4-1
Replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit to
get:
Before the switch closes at t = 0 the circuit is at steady state so v(0) = 10 V. For 0 < t < 1.5s, voc =
5 V and Rt = 4 Ω so τ = 4 × 0.05 = 0.2 s . Therefore
v (t ) = voc + (v (0) − voc ) e−t τ = 5 + 5e−5 t V for 0 < t < 1.5 s
At t =1.5 s, v (1.5) = 5 + 5e (
τ = 8 × 0.05 = 0.4 s . Therefore
−0.05 1.5)
= 5 V . For 1.5 s < t, voc = 10 V and Rt = 8 Ω so
v (t ) = voc + (v (1.5) − voc ) e
Finally
−( t −1.5) τ
= 10 − 5 e
−2.5 ( t −1.5 )
V for 1.5 s < t
−5 t
for 0 < t < 1.5 s
 5 + 5 e V
v (t ) = 
−2.5 ( t −1.5)
for 1.5 s < t
V
10 − 5 e
P8.4-2
Replace the part of the circuit connected to the inductor by its Norton equivalent circuit to get:
Before the switch closes at t = 0 the circuit is at steady state so i(0) = 3 A. For 0 < t < 1.5s, isc = 2
12
A and Rt = 6 Ω so τ = = 2 s . Therefore
6
i (t ) = isc + (i (0) − isc ) e−t τ = 2 + e−0.5 t A for 0 < t < 1.5 s
8-18
At t =1.5 s, i (1.5) = 2 + e
Therefore
−0.5 (1.5)
= 2.47 A . For 1.5 s < t, isc = 3 A and Rt = 8 Ω so τ =
i (t ) = isc + (i (1.5) − isc ) e
Finally
P8.4-3
At t = 0−:
−( t −1.5 ) τ
= 3 − 0.53 e
−0.667 ( t −1.5 )
12
= 1.5 s .
8
V for 1.5 s < t

2 + e −0.5 t A
for 0 < t < 1.5 s
i (t ) = 
−0.667 ( t −1.5 )
for 1.5 s < t
A
3 − 0.53 e
KVL : − 52 + 18 i + (12 8) i = 0
⇒
i (0− ) =104 39 A
 6 
+
∴ iL = i 
 = 2 A = iL (0 )
 6+ 2 
For 0 < t < 0.051 s
iL (t ) = iL (0) e −t τ
where τ = L R t
R t = 6 12 + 2 = 6 Ω
iL (t ) = 2 e −6t A
 6 +12 
−6 t
∴ i (t ) = iL (t ) 
 =6e A
 6 
For t > 0.051 s
i L ( t ) = i L (0.051) e − ( R
L ) ( t − 0.051)
i L (0.051) = 2 e − 6 (.051) = 1.473 A
i L ( t ) = 1.473 e − 14 ( t − 0.051) A
i ( t ) = i L ( t ) = 1.473 e − 14 ( t − 0.051) A
8-19
P8.4-4
At t = 0-: Assume that the circuit has reached steady state so that the voltage across the 100 µF
capacitor is 3 V. The charge stored by the capacitor is
q ( 0− ) = (100 × 10−6 ) ( 3) = 300 × 10−6 C
0 < t < 10ms: With R negligibly small, the circuit reaches steady state almost immediately (i.e. at
t = 0+). The voltage across the parallel capacitors is determined by considering charge
conservation:
q ( 0+ ) = (100 µ F) v ( 0+ ) + (400 µ F) v ( 0+ )
v (0
) = 100 ×10
( )
+
v 0+ = 0.6 V
q ( 0+ )
−6
+ 400 ×10−6
=
q ( 0− )
500 × 10−6
=
300 × 10−6
500 × 10−6
10 ms < t < l s: Combine 100 µF & 400 µF in parallel to obtain
v(t ) = v ( 0+ ) e − (t −.01) RC
= 0.6e −(t −.01) (10 ) (5 x10
v(t ) = 0.6 e−2(t −.01) V
3
−4 )
P8.4-5
For t < 0:
Find the Thevenin equivalent of the
circuit connected to the inductor (after
t >0). First, the open circuit voltage:
i1 =
40
=1 A
20 + 20
voc = 20 i1 − 5 i1 = 15 V
8-20
Next, the short circuit current:
20 i 1 = 5 i 1 ⇒ i 1 = 0
i sc + 0 =
Then
40
⇒ i sc = 2 A
20
Rt =
voc 15
= = 7.5 Ω
i sc
2
Replace the circuit connected to the
inductor by its Norton equivalent
circuit. First
1
L 15 × 10−3
τ=
=
=
7.5
500
Rt
Next
i ( t ) = 2 − 2 e − 500 t A
t >0
After t = 0, the steady state inductor
current is 2 A. 99% of 2 is 1.98.
1.98 = 2 − 2 e − 500 t
P8.4-6
⇒ t = 9.2 ms
v ( 0 ) = 5 V , v ( ∞ ) = 0 and τ = 105 ×10−6 = 0.1 s
∴ v ( t ) = 5 e −10 t V for t > 0
2.5 = 5 e −10 t1
i (t1 ) =
v (t1 )
100 ×10
3
t 1 = 0.0693 s
=
2.5
= 25 µ A
100 × 103
8-21
Section 8-5: Stability of First Order Circuits
P8.5-1
This circuit will be stable if the Thèvenin equivalent resistance of the circuit connected to the
inductor is positive. The Thèvenin equivalent resistance of the circuit connected to the inductor
is calculated as
100

v T (400− R) 100
iT 
=
100+ 400
 ⇒ Rt =
iT
100+ 400

vT = 400 i (t ) − R i (t ) 
i (t ) =
The circuit is stable when R < 400 Ω.
P8.5-2
The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as
Ohm’s law: v ( t ) = 1000 iT ( t )
KVL: A v ( t ) + vT ( t ) − v ( t ) − 4000 iT ( t ) = 0
∴ vT ( t ) = (1 − A ) 1000 iT ( t ) + 4000 iT ( t )
vT ( t )
= ( 5 − A ) ×1000
iT ( t )
The circuit is stable when A < 5 V/V.
Rt =
P8.5-3
The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as
v (t )
Ohm’s law: i ( t ) = − T
6000
v (t )
KCL: i ( t ) + B i ( t ) + i T ( t ) = T
3000
 v ( t )  vT ( t )
∴ i T ( t ) = − ( B + 1)  − T
+
 6000  3000
( B + 3) vT ( t )
=
6000
vT ( t ) 6000
Rt =
=
iT ( t ) B + 3
The circuit is stable when B > −3 A/A.
8-22
P8.5-4
The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as
(1000 )( 4000 ) i (t ) = 800 i (t )
v(t ) =
T
T
1000 + 4000
vT (t ) = v(t ) − Av(t ) = (1 − A ) v(t )
Rt =
vT (t )
= 800 (1 − A)
iT (t )
The circuit is stable when A < 1 V/V.
8-23
Section 8-6: The Unit Step Response
P8.6-1
The value of the input is one constant, 8 V, before time t = 0 and a different constant, −7 V, after
time t = 0. The response of the first order circuit to the change in the value of the input will be
vo ( t ) = A + B e − a t
for t > 0
where the values of the three constants A, B and a are to be determined.
The values of A and B are determined from the steady state responses of this circuit
before and after the input changes value.
Capacitors act like open circuits when the input is constant
and the circuit is at steady state. Consequently, the
capacitor is replaced by an open circuit.
The value of the capacitor voltage at time t = 0,
will be equal to the steady state capacitor voltage before
the input changes. At time t = 0 the output voltage is
−a 0
vo ( 0 ) = A + B e ( ) = A + B
The steady-state circuit for t < 0.
Consequently, the capacitor voltage is labeled as A + B.
Analysis of the circuit gives
A+ B = 8 V
Capacitors act like open circuits when the input is constant
and the circuit is at steady state. Consequently, the
capacitor is replaced by an open circuit.
The value of the capacitor voltage at time t = ∞, will be
equal to the steady state capacitor voltage after the input
changes. At time t = ∞ the output voltage is
vo ( ∞ ) = A + B e
The steady-state circuit for t > 0.
− a (∞)
=A
Consequently, the capacitor voltage is labeled as A.
Analysis of the circuit gives
A = -7 V
Therefore
B = 15 V
8-24
The value of the constant a is determined from the time constant, τ, which is in turn
calculated from the values of the capacitance C and of the Thevenin resistance, Rt, of the circuit
connected to the capacitor.
1
=τ = Rt C
a
Here is the circuit used to calculate Rt.
Rt = 6 Ω
Therefore
a=
1
1
= 2.5
−3
s
( 6 ) ( 66.7 ×10 )
(The time constant is τ = ( 6 ) ( 66.7 × 10−3 ) = 0.4 s .)
Putting it all together:
8 V
for t ≤ 0

vo ( t ) = 
− 2.5 t
V for t ≥ 0
−7 + 15 e
P8.6-2
The value of the input is one constant, 3 V, before time t = 0 and a different constant, 6 V, after
time t = 0. The response of the first order circuit to the change in the value of the input will be
vo ( t ) = A + B e − a t
for t > 0
where the values of the three constants A, B and a are to be determined.
The values of A and B are determined from the steady state responses of this circuit
before and after the input changes value.
Capacitors act like open circuits when the input is
constant and the circuit is at steady state. Consequently,
the capacitor is replaced by an open circuit.
The steady-state circuit for t < 0.
The value of the capacitor voltage at time t = 0,
will be equal to the steady state capacitor voltage
before the input changes. At time t = 0 the output
voltage is
−a 0
vo ( 0 ) = A + B e ( ) = A + B
Consequently, the capacitor voltage is labeled as A + B.
Analysis of the circuit gives
8-25
A+ B =
6
( 3) = 2 V
3+ 6
Capacitors act like open circuits when the input is
constant and the circuit is at steady state. Consequently,
the capacitor is replaced by an open circuit.
The value of the capacitor voltage at time t = ∞, will be
equal to the steady state capacitor voltage after the
input changes. At time t = ∞ the output voltage is
vo ( ∞ ) = A + B e
The steady-state circuit for t > 0.
− a (∞)
=A
Consequently, the capacitor voltage is labeled as A.
Analysis of the circuit gives
A=
Therefore
6
(6) = 4 V
3+ 6
B = −2 V
The value of the constant a is determined from the time constant, τ, which is in turn
calculated from the values of the capacitance C and of the Thevenin resistance, Rt, of the circuit
connected to the capacitor.
1
=τ = Rt C
a
Here is the circuit used to calculate Rt.
Rt =
Therefore
a=
( 3)( 6 ) = 2
3+ 6
( 2 )(.5 )
1
=1
Ω
1
s
(The time constant is τ = ( 2 )( 0.5 ) = 1 s .)
Putting it all together:
2 V
for t ≤ 0

vo ( t ) = 
− t
 4 − 2 e V for t ≥ 0
8-26
P8.6-3
The value of the input is one constant, −7 V, before time t = 0 and a different constant, 6 V, after
time t = 0. The response of the first order circuit to the change in the value of the input will be
vo ( t ) = A + B e − a t
for t > 0
where the values of the three constants A, B and a are to be determined.
The values of A and B are determined from the steady state responses of this circuit
before and after the input changes value.
Inductors act like short circuits when the input is
constant and the circuit is at steady state.
Consequently, the inductor is replaced by a short
circuit.
The value of the inductor current at time t = 0,
will be equal to the steady state inductor current before
the input changes. At time t = 0 the output current is
io ( 0 ) = A + B e
The steady-state circuit for t < 0.
− a ( 0)
= A+ B
Consequently, the inductor current is labeled as A + B.
Analysis of the circuit gives
A+ B =
−7
= −1.4 A
5
Inductors act like short circuits when the input is
constant and the circuit is at steady state.
Consequently, the inductor is replaced by a short
circuit.
The value of the inductor current at time t = ∞, will be
equal to the steady state inductor current after the
input changes. At time t = ∞ the output current is
io ( ∞ ) = A + B e
The steady-state circuit for t > 0.
− a (∞)
=A
Consequently, the inductor current is labeled as A.
Analysis of the circuit gives
A=
Therefore
6
= 1.2 A
5
B = −2.6 V
8-27
The value of the constant a is determined from the time constant, τ, which is in turn
calculated from the values of the inductance L and of the Thevenin resistance, Rt, of the circuit
connected to the inductor.
1
L
=τ =
a
Rt
Here is the circuit used to calculate Rt.
Rt =
Therefore
( 5 ) ( 4 ) = 2.22
a=
5+ 4
2.22
1
= 1.85
1.2
s
(The time constant is τ =
1.2
= 0.54 s .)
2.22
for t ≤ 0
−1.4 A

io ( t ) = 
− 1.85 t
A for t ≥ 0
1.2 − 2.6 e
Putting it all together:
P8.6-4
Ω
v (t ) = 4u (t ) − u (t − 1) − u (t − 2) + u (t − 4) − u (t − 6)
P8.6-5
0

vs ( t ) = 4
0

t <1
1<t < 2
t >2
∴
Assume that the circuit is at steady
state at t = 1−. Then
v ( t ) = 4 − 4 e − (t −1) V for 1 ≤ t ≤ 2
τ = R C = ( 5 ×105 )( 2 ×10−6 ) = 1 s
0

v(t ) = 4− 4e − ( t −1)
2.53e − (t − 2)

so v ( 2 ) = 4 − 4 e − (2−1) = 2.53 V
and
v ( t ) = 2.53 e − (t − 2) V for t ≥ 2
t ≤1
1≤t ≤ 2
t ≥2
8-28
P8.6-6
The capacitor voltage is v(0−) = 10 V immediately before the switch opens at t = 0.
For 0 < t < 0.5 s the switch is open:
1 1
v ( 0 ) = 10 V, v ( ∞ ) = 0 V, τ = 3 × =
s
6 2
so v ( t ) = 10 e − 2 t V
In particular, v ( 0.5 ) = 10 e
For t > 0.5 s the switch is closed:
− 2 ( 0.5)
= 3.679 V
v ( 0 ) = 3.679 V, v ( ∞ ) = 10 V, Rt = 6 || 3 = 2 Ω,
τ = 2× =
1
6
so
1
s
3
v ( t ) = 10 + ( 3.679 − 10 ) e
= 10 − 6.321 e
− 3 ( t − 0.5 )
− 3 ( t − 0.5)
V
V
P8.6-7
Assume that the circuit is at steady state before t = 0. Then the initial inductor current is
i(0−) = 0 A.
For 0 < t < 1 ms:
The steady state inductor current will be
30
i ( ∞ ) = lim i ( t ) =
( 40 ) = 24 A
t →∞
30 + 20
The time constant will be
50 × 10−3
1
τ=
= 10−3 =
s
30 + 20
1000
The inductor current is i ( t ) = 24 (1 − e −1000 t ) A
In particular, i ( 0.001) = 24 (1 − e −1 ) = 15.2 A
For t > 1 ms
Now the initial current is i(0.001) = 15.2 A and
the steady state current is 0 A. As before, the time
constant is 1 ms. The inductor current is
i ( t ) = 15.2 e
−1000 ( t − 0.001)
A
8-29
The output voltage is
480 (1 − e −1000 t ) V t < 1 ms
v ( t ) = 20 i ( t ) = 
−1000 ( t − 0.001)
V t > 1 ms
303 e
P8.6-8
For t < 0, the circuit is:
After t = 0, replace the part of the circuit connected to the capacitor by its Thevenin equivalent
circuit to get:
vc ( t ) = −15 + ( −6 − ( −15 ) ) e
= −15 + 9 e − 5 t V
− t / ( 4000×0.00005 )
8-30
Section 8-7 The Response of an RL or RC Circuit to a Nonconstant Source
P8.7-1
Assume that the circuit is at steady state before t = 0:
KVL : 12ix + 3(3 ix ) + 38.5 = 0 ⇒ ix = −1.83 A
Then vc (0− ) = −12 ix = 22 V = vc (0+ )
After t = 0:
KVL : 12i (t ) − 8e −5t + v ( t ) = 0
x
c
dv ( t )
1 dvc ( t )
= 0 ⇒ ix ( t ) =
KCL : −ix ( t )−2ix ( t ) + (1 36) c
dt
108 dt
 1 dvc ( t ) 
−5t
∴ 12 
+ v (t ) = 0
 − 8e
c
108 dt 
dv (t )
c + 9v (t ) = 72e−5t ⇒ v (t ) = Ae −9t
c
cn
dt
Try v (t ) = Be −5t & substitute into the differential equation ⇒ B = 18
cf
∴ v (t ) = Ae −9t + 18 e−5t
c
v (0) = 22 = A + 18 ⇒ A = 4
c
∴ v (t ) = 4e−9t + 18e−5t V
c
8-31
P8.7-2
Assume that the circuit is at steady state before t = 0:
iL (0+ ) = iL (0− ) =
12
= 3A
4
After t = 0:
v( t ) −12
v( t )
+ iL ( t ) +
= 6 e −2t
4
2
di ( t )
also : v ( t ) = (2 / 5) L
dt
di ( t ) 
3
iL ( t ) + (2 / 5) L  = 3 + 6 e −2t
4
dt 
KCL :
diL ( t ) 10
+ iL ( t ) = 10 + 20 e −2t
3
dt
∴ in (t ) = Ae
− (10 / 3)t
, try i f (t ) = B + Ce−2t , substitute into the differential equation,
and then equating like terms ⇒ B =3, C =15 ⇒ i f (t ) =3+15 e−2t
∴iL (t ) =in (t ) + i f (t ) = Ae −(10 / 3)t + 3+15e−2t , iL (0) = 3 = A + 3 + 15 ⇒ A = −15
Finally, v(t ) =( 2 / 5 )
∴ iL (t ) = −15e − (10 / 3) t + 3 + 15e−2t
diL
= 20 e − (10 / 3) t −12 e −2t V
dt
P8.7-3
Assume that the circuit is at steady state before t = 0:
 6 
Current division: iL (0− ) = −5 
 = −1 mA
 6 + 24 
8-32
After t = 0:
KVL: − 25sin 4000 t + 24iL ( t ) + .008
di L ( t )
25
+3000i L ( t ) =
sin4000t
dt
.008
diL ( t )
=0
dt
in (t ) = Ae −3000t , try i f (t ) = B cos 4000t + C sin 4000t , substitute into the differential equation
and equate like terms ⇒ B = −1/ 2, C = 3 / 8 ⇒ i f (t ) = −0.5cos 4000 t + 0.375 sin 4000 t
iL (t ) = in (t ) + i f (t ) = Ae −3000t − 0.5cos 4000 t + 0.375 sin 4000 t
iL (0+ ) = iL (0− ) =−10−3 = A− 0.5 ⇒ A ≅ 0.5
∴ iL (t ) = 0.5 e −3000t − 0.5cos 4000 t + 0.375 sin 4000 t mA
but v(t ) = 24iL (t ) = 12 e −3000t − 12 cos 4000t + 9sin 4000t V
P8.7-4
Assume that the circuit is at steady state before t = 0:
Replace the circuit connected to the capacitor by its Thevenin equivalent (after t=0) to get:
dvc ( t ) 

+v t =0 ⇒
KVL: − 10 cos 2t + 15  1
30 dt  c ( )


dvc ( t )
+ 2vc ( t ) = 20 cos 2t
dt
vn (t ) = Ae−2t , Try v f (t ) = B cos 2t + C sin 2t & substitute into the differential equation to get
B = C = 5 ⇒ v f (t ) = 5cos 2t + 5sin 2t. ∴ vc (t ) = vn (t ) + v f (t ) = Ae−2t + 5cos 2t + 5sin 2t
Now vc (0) = 0 = A + 5 ⇒ A = −5 ⇒ vc (t ) = −5e−2t + 5cos 2t + 5sin 2t V
8-33
P8.7-5
Assume that the circuit is at steady state before t = 0. There are no sources in the circuit so i(0) =
0 A. After t = 0, we have:
di ( t )
KVL : − 10sin100t + i ( t ) + 5
+ v (t ) = 0
dt
v( t )
Ohm's law : i ( t ) =
8
dv( t )
∴
+18 v( t ) = 160sin100t
dt
∴ vn (t ) = Ae−18t , try v f (t ) = B cos100t + C sin100t , substitute into the differential equation
and equate like terms ⇒ B = −1.55 & C = 0.279 ⇒ v f (t ) = −1.55cos100t + 0.279sin100t
∴ v(t ) = vn (t ) + v f (t ) = Ae−18t −1.55 cos100 t + 0.279 sin100 t
v(0) = 8 i (0) = 0 ⇒ v (0) = 0 = A−1.55 ⇒ A = 1.55
so v(t ) = 1.55e−18t −1.55cos100t + 0.279 sin100t V
P8.7-6
Assume that the circuit is at steady state
before t = 0.
vo ( t ) = −vc ( t )
vC (0+ ) = vC (0− ) = −10 V
After t = 0, we have
v (t )
8 e−5 t
i (t ) = s
=
= 0.533 e−5 t mA
15000
15000
The circuit is represented by the differential
dv ( t ) vC ( t )
equation: i ( t ) = C C
+
. Then
dt
R
( 0.533 ×10 ) e
−3
−5 t
= ( 0.25 ×10−6 )
dvc ( t )
+ (10−3 ) vc ( t ) ⇒
dt
dvc ( t )
+ 4000 vc ( t ) = 4000 e−5t
dt
Then vn ( t ) = Ae−4000t . Try v f ( t ) = Be−5t . Substitute into the differential equation to get
(
d B e−5t
dt
) + 4000 ( B e ) = 4000 e
−5t
−5t
⇒ B=
4000
= −1.00125 ≅ −1
−3995
8-34
vC (t ) = v f ( t ) + vn ( t ) = e −5t + Ae −4000t
vC (0) = −10 = 1 + A ⇒ A = −11 ⇒ vC (t ) = 1 e−2t − 11 e−4000t V
vo (t ) = − vC (t ) = 11e−4000t −1e −5t V , t ≥ 0
Finally
P8.7-7
From the graph iL (t ) = 1 t mA . Use KVL to get
4
(1) iL (t ) + 0.4
Then
diL (t )
= v1 (t ) ⇒
dt
diL (t )
+ 2.5 iL (t ) = 2.5 v1 (t )
dt
di  1 
t + 2.5  1 t  = 2.5 v1 (t ) ⇒
4 
dt  4 
v1 = 0.1+ 0.25t V
P8.7-8
Assume that the circuit is at steady state before
t = 0.
v (0+ ) = v (0− ) =
2
30 = 10 V
4+ 2
After t = 0 we have
KVL :
 1 d v( t ) 
5 d v( t )
+ v (t ) + 4 
−i  = 30
2 dt
 2 dt


1 d v( t ) 
−3t
2 i ( t ) + 4 i ( t ) −
 + 30 = e
2
dt


The circuit is represented by the differential equation
Take vn ( t ) = Ae
d v( t )
6
6
2
v (t ) =
(10 + e −3t )
+
dt
19
19
3
−( 6 /19 ) t
. Try , v f ( t ) = B + Ce −3t , substitute into the differential equation to get
8-35
−3Ce −3t +
Equate coefficients to get
B = 10 , C = −
4
4
⇒ v f ( t ) = e −3t + Ae− (6 /19) t
51
51
v ( t ) = vn ( t ) + v f ( t ) = 10 −
Then
Finally
6
60 4 −3t
( B + Ce −3t ) =
+ e
19
19 19
4 −3t
e + Ae− (6 /19) t
51
vc (0+ ) = 10 V, ⇒ 10 = 10 −
∴ vc (t ) = 10 +
4
+A ⇒
51
A=
4 − (6 /19) t −3t
(e
−e ) V
51
4
51
P8.7-9
We are given v(0) = 0. From part b of the
figure:
5t 0 ≤ t ≤ 2 s
vs ( t ) = 
t > 2s
10
Find the Thevenin equivalent of the part of the
circuit that is connected to the capacitor:
The open circuit voltage:
The short circuit current:
(ix=0 because of the short across the right 2 Ω
resistor)
Replace the part of the circuit connected to the
capacitor by its Thevenin equivalent:
KVL:
dv( t )
+ v ( t ) − vs ( t ) = 0
dt
dv( t ) v ( t ) vs ( t )
+
=
dt
2
2
2
vn ( t ) = Ae−0.5 t
8-36
For 0 < t < 2 s, vs ( t ) = 5 t . Try v f ( t ) = B + C t . Substituting into the differential equation and
equating coefficients gives B = −10 and C =5. Therefore v ( t ) = 5t − 10 + A e − t / 2 . Using v(0) = 0,
we determine that A =10. Consequently, v ( t ) = 5t + 10(e−t / 2 − 1) .
At t = 2 s, v( 2 ) = 10e−1 = 3.68 .
Next, for t > 2 s, vs ( t ) = 10 V . Try v f ( t ) = B . Substituting into the differential equation and
equating coefficients gives B = 10. Therefore v ( t ) = 10 + Ae
determine that A = −6.32. Consequently, v ( t ) = 10 − 6.32 e
P8.7-10
− (t −2) / 2
− (t − 2) / 2
. Using v ( 2 ) = 3.68 , we
.
 d v (t ) 
KVL: − kt + Rs C C  + vC ( t ) = 0
dt 

d vC ( t )
k
1
vC ( t ) =
t
⇒
+
Rs C
Rs C
dt
vc ( t ) = vn ( t ) + v f ( t ) , where vc ( t ) = Ae− t / Rs C . Try v f ( t ) = B0 + B1 t
& plug into D.E. ⇒ B1 +
1
k
t thus B0 = − kRs C , B1 = k .
[ B0 + B1t ] =
Rs C
Rs C
Now we have vc (t ) = Ae − t / Rs C + k (t − Rs C ). Use vc (0) = 0 to get 0 = A − kRs C ⇒ A = kRs C.
∴ vc (t ) = k[t − Rs C (1− e −t / Rs C )]. Plugging in k =1000 , Rs = 625 kΩ & C = 2000 pF get
vc (t ) = 1000[t − 1.25 × 10−3 (1 − e −800 t )]
v(t) and vC(t) track well on a millisecond time scale.
8-37
Spice Problems
SP 8-1
8-38
SP 8-2
8-39
SP 8-3
v(t ) = A + B e −t / τ
for t > 0
⇒ 7.2 = A + B 
 ⇒ B = −0.8 V
8.0 = v(∞) = A + B e −∞ ⇒ A = 8.0 V 
0.05
 8 − 7.7728 
7.7728 = v(0.05) = 8 − 0.8 e −0.05 / τ ⇒ −
= ln 
 = −1.25878
0.8
τ


0.05
⇒ τ=
= 39.72 ms
1.25878
7.2 = v(0) = A + B e 0
Therefore
v(t ) = 8 − 0.8 e −t / 0.03972 V for t > 0
8-40
SP 8-4
i (t ) = A + B e −t / τ
0 = i (0) = A + B e
0
4 × 10−3 = i (∞) = A + B e −∞
⇒ 0 = A+ B
⇒
A = 4 × 10−3
2.4514 ×10 = v(5 ×10 ) = ( 4 ×10
−3
⇒ −
−6
5 × 10−6
τ
for t > 0
−3

−3
 ⇒ B = −4 ×10 A
A 
) − ( 4 ×10 ) e
−3
(
)
− 5×10−6 / τ
 ( 4 − 2.4514 ) × 10−3 
= ln 
 = −0.94894
4 × 10−3


5 × 10−6
⇒ τ=
= 5.269 µ s
0.94894
Therefore
i (t ) = 4 − 4 e −t / 5.269×10
−6
mA for t > 0
8-41
Verification Problems
VP 8-1
First look at the circuit. The initial capacitor voltage is vc(0) = 8 V. The steady-state capacitor
voltage is vc = 4 V.
We expect an exponential transition from 8 volts to 4 volts. That’s consistent with the plot.
Next, let’s check the shape of the exponential transition. The Thevenin resistance of the part of
( 2000 )( 4000 ) = 4 kΩ so the time constant is
the circuit connected to the capacitor is R t =
2000 + 4000
3
2
4


τ = R t C =  × 103  ( 0.5 × 10−6 ) = ms . Thus the capacitor voltage is
3
3

vc (t ) = 4 e− t
0.67
+4 V
where t has units of ms. To check the point labeled on the plot, let t1 = 1.33 ms. Then
vc (t1 ) =
 1.33 
−

4 e  .67 
+ 4 = 4.541 ~ 4.5398 V
So the plot is correct.
VP 8-2
The initial and steady-state inductor currents shown on the plot agree with the values obtained
from the circuit.
Next, let’s check the shape of the exponential transition. The Thevenin resistance of the part of
( 2000 )( 4000 ) = 4 kΩ so the time constant is
the circuit connected to the inductor is R t =
2000 + 4000
3
L
5
15
ms . Thus inductor current is
τ=
=
=
R t 4 ×103
4
3
iL (t ) − 2 e− t 3.75 + 5 mA
where t has units of ms. To check the point labeled on the plot, let t1 = 3.75 ms. Then
iL (t1 ) =
 3.75 
−

−2 e  3.75 
+ 5 = 4.264 mA ≠ 4.7294 mA
so the plot does not correspond to this circuit.
8-42
VP 8-3
Notice that the steady-state inductor current does not depend on the inductance, L. The initial
and steady-state inductor currents shown on the plot agree with the values obtained from the
circuit.
After t = 0
So I sc = 5 mA and τ =
The inductor current is given by iL (t ) = −2e −1333t
has units of Henries. Let t 1 = 3.75 ms, then
L
L
1333
+ 5 mA , where t has units of seconds and L
4.836 = iL (t1 ) = −2 e −(1333)⋅(0.00375) L + 5 = −2e−5 L + 5
4.836−5
= e −5 L
−2
so
and
L=
−5
=2 H
 4.836−5 
ln 

 −2 
is the required inductance.
VP 8-4
First consider the circuit. When t < 0 and the circuit is at steady-state:
For t > 0
So
Voc =
R2
R1 R2
RRC
( A + B) , Rt =
and τ = 1 2
R1 + R2
R1 + R2
R1 + R2
8-43
Next, consider the plot. The initial capacitor voltage is (vc (0)=) –2 and the steady-state capacitor
voltage is (Voc =) 4 V, so
vC (t ) = − 6e −t τ + 4
At t 1 = 1.333 ms
3.1874 = vC (t1 ) = − 6 e −0.001333 τ + 4
so
τ =
−0.001333
= 0.67 ms
 −4+ 3.1874 
ln 

−6


Combining the information obtained from the circuit with the information obtained from the plot
gives
R2
R2
R1 R2C
A = −2,
( A + B ) = 4,
= 0.67 ms
R1 + R2
R1 + R2
R1 + R2
There are many ways that A, B, R , R , and C can be chosen to satisfy these equations. Here is
one convenient way. Pick R = 3000 and R = 6000. Then
1
1
2
2
2A
= −2 ⇒ A = −3
3
2( A+ B)
= 4 ⇒ B −3 = 6 ⇒ B = 9
3
2
1
µF = C
2000 ⋅ C = ms ⇒
3
3
Design Problems
DP 8-1
R3
6
12 ⇒ R1 + R 2 = R 3 .
=
Steady-state response when the switch is open:
R1 + R 2 + R 3
Steady-state response when the switch is open: 10 =
10 ms = 5 τ = ( R 1 || R 3 ) C =
R3
6
R3
R1 + R 3
12
⇒ R1 =
R3
5
.
C
Let C = 1 µF. Then R 3 = 60 kΩ, R 1 = 30 kΩ and R 2 = 30 kΩ.
8-44
DP 8-2
12
steady state response when the switch is open : 0.001 = R + R
1
2
steady state response when the switch is open: 0.004 =
12
R1
⇒ R + R = 12 kΩ .
1
2
⇒ R1 = 3 kΩ .
Therefore, R 2 = 9 kΩ.
 L 
L
10 ms = 5 τ = 5 
⇒ L = 240 H
=
 R1 + R 2  2400


DP 8-3
Rt = 50 kΩ when the switch is open and Rt = 49 kΩ ≈ 50 kΩ when the switch is closed so use
Rt = 50 kΩ.
10−6
(a) ∆t = 5 Rt C ⇒ C =
= 4 pF
5 50×103
(b) ∆ t = 5 ( 50×103 )( 2×10−6
(
)
) = 0.5 s
DP 8-4
Rt = 50 kΩ when the switch is open and Rt = 49 kΩ ≈ 50 kΩ when the switch is closed so use
Rt = 50 kΩ.
∆t
−∆t τ
When the switch is open: 5 e
= (1 − k ) 5 ⇒ ln (1− k ) = −
⇒ ∆ t = −τ ln (1− k )
When the switch is open: 5 − 5 e
(a) C =
−∆t τ
10−6
= 6.67 pF
− ln (1−.95 ) ( 50×103 )
τ
= k 5 ⇒ ∆ t = −τ ln (1− k )
(b) ∆ t = − ln(1 − .95) ( 50×103 )( 2×10−6 ) = 0.3 s
DP 8-5
i (0) =
R1
20
×
40 R1 R1 + 40
40 +
40 + R1
8-45
For
t > 0:
i (t ) = i (0) e
−t
τ
where τ =
L
10−2
=
R t 40+ R 2
At t < 200 µ s we need i ( t ) > 60 mA and i ( t ) <180 mA
First let's find a value of R 2 to cause i (0) < 180 mA.
Try R 2 = 40 Ω . Then i (0) =
1
t
A = 166.7 mA so i (t ) = 0.1667 e− τ .
6
Next, we find a value of R 2 to cause i (0.0002) > 60 mA.
10−2
1
s.
= 0.2 ms =
50
5000
i (0.0002) = 166.7×10−3 e −5000×0.0002 = 166.7×10−3 e −1 = 61.3 mA
Try R 2 = 10Ω, then τ =
DP 8-6
The current waveform will look like this:
We only need to consider the rise time:
−t
Vs
A
τ
−t
τ
iL (t ) =
(1 − e ) =
(1 − e )
R+2
R+2
where
L 0.2 1
s
τ =
=
=
Rt
3 15
A
∴ iL (t ) = (1 − e −15t )
3
Now find A so that iL2 R fuse ≥ 10 W during 0.25 ≤ t ≤ 0.75 s
∴ we want [iL2 (0.25)]R fuse = 10 W ⇒
A2
(1−e −15(.25) ) 2 (1) =10 ⇒ A = 9.715 V
9
8-46
Chapter 9 - Complete Response of Circuits with Two Energy
Storage Elements
Exercises
Apply KVL to right mesh:
di ( t )
+ v ( t ) + 1( i ( t ) −is ( t ) ) = 0
dt
di ( t )
⇒ v ( t ) = −2
− ( i ( t )−is ( t ) )
dt
Ex. 9.3-1
2
The capacitor current and voltage are related by
i (t ) =
 1 dis ( t ) 1 di ( t ) d 2i ( t )
1 dv ( t ) 1 d  di ( t )
=
−
−
+
−
−
i
t
i
t
2
(
)
(
)

=
s
dt
dt 2
2 dt
2 dt 
2 dt
 2 dt
∴
d 2i ( t ) 1 di ( t )
1 dis ( t )
+
+ i( t ) =
2
dt
2 dt
2 dt
The inductor voltage is related to the inductor
current by
di ( t )
v (t ) = 1
dt
Ex. 9.3-2
Apply KCL at the top node:
is ( t ) =
Using the operator s =
v (t )
1 dv ( t )
+ i (t ) +
1
2 dt
d
we have
dt
v (t ) = s i (t )

v (t ) 1

+ sv ( t )
 ⇒ is ( t ) = v ( t ) +
1
s
2
is ( t ) = v ( t ) + i ( t ) + sv ( t ) 

2
Therefore
2s is ( t ) = 2 s v ( t ) + 2 v ( t ) + s 2 v ( t ) ⇒
d 2v ( t )
dv ( t )
d i (t )
+2
+ 2 v (t ) = 2 s
2
dt
dt
dt
9-1
Using the operator s =
Ex. 9.3-3
d
, apply KVL to the
dt
left mesh:
i1 ( t ) + s ( i1 ( t ) −i 2 ( t ) ) = vs ( t )
Apply KVL to the right mesh:
1
2 i 2 ( t ) + 2 i 2 ( t ) + s ( i 2 ( t ) −i1 ( t ) ) = 0
s
1
2
i1 ( t ) = 2 i 2 ( t ) + 2 i 2 ( t ) + i 2 ( t )
s
s
Combining these equations gives:
3s i2 ( t ) + 4si2 ( t ) + 2i2 ( t ) = s vs ( t )
2
2
or
Ex. 9.4-1
d 2i2 ( t )
di2 ( t )
d 2 vs ( t )
3
+4
+ 2i2 ( t ) =
dt 2
dt
dt 2
Using the operator s =
top node:
i s (t ) =
d
, apply KCL at the
dt
v (t )
1
+ i (t ) + s v (t )
4
4
Apply KVL to the right-most mesh:
v (t ) − ( s i (t ) + 6 i (t )) = 0
Combining these equations gives:
The characteristic equation is: s 2 + 7 s + 10 = 0 .
The natural frequencies are: s = −2 and s = −5 .
Ex. 9.4-2
s 2 i ( t ) + 7 s i ( t ) + 10 i ( t ) = 4 i s ( t )
Assume zero initial conditions . Write mesh
d
equations using the operator s =
:
dt
1
s i1 ( t ) − i 2 ( t )  + 7 + 10 i1 ( t ) − 10 = 0
2 
and
1
v ( t ) − 7 − s i1 ( t ) − i 2 ( t )  = 0
2
9-2
Now 0.005 s v ( t ) = i 2 ( t ) ⇒ v ( t ) = 200
200
i2 (t )
s
i2 (t )
s
so the second mesh equation becomes:
1
− 7 − s i1 ( t ) − i 2 ( t )  = 0
2
Writing the mesh equation in matrix form:
s

10 + 2

 −1 s
 2
1

s 
 i1 ( t )   3 
2

=
1
200  i 2 ( t )  7 
s+
2
s 
−
Obtain the characteristic equation by calculating a determinant:
10 +
−
−
1
s
2
= s 2 + 20s + 400 = 0 ⇒ s1,2 = −10 ± j 17.3
1
200
s+
2
s
s
2
1
s
2
Ex. 9.5-1
After t = 0 , we have a parallel RLC circuit with
1
1
7
1
1
=
=
=
=6
α =
and ω o2 =
2 RC
2(6)(1/ 42) 2
(7)(1/ 42)
LC
7
  − 6 = −1, − 6
 2
dv ( t )
∴ vn (t ) = A1e −t + A2 e−6t . We need vn (0) and n
dt
7
∴ s 1 , s 2 = −α ± α −ω = − ±
2
2
2
2
o
t =0
At t = 0+ we have:
to evaluate A1 & A2 .
iC ( 0+ ) = −10 A ⇒
dv ( t )
10
=
= 420 V s
1
dt t =0+
42


 A1 = −84 , A2 = 84
dvn
=−420= − A1 −6 A2 
dt t =0+

Then
vn (0+ ) = 0 = A1 + A2
9-3
Finally
∴ vn (t ) = −84e −t + 84e −6t V
s2 +
Ex. 9.5-2
vn ( t ) = A1 e −2.68t + A2 e−37.3t , v(0) = 0 = A1 + A2
1
1
= 0 ⇒ s 2 + 40s + 100 = 0
s+
RC
LC
Therefore
s 1 , s 2 = −2.68 , −37.3
v(0+ )
1 dv(0+ )
+
+ i (0 ) +
= 0 so
KCL at t = 0 yields
1
40 dt
dv(0+ )
= − 40 v(0+ ) − 40 i (0+ ) = − 40(0) − 40(1) = − 2.7 A1 − 37.3 A2
dt
Therefore: A1 = −1.16 , A2 = 1.16 ⇒ v(t ) = vn (t ) = −1.16e −2.68t + 1.16e−37.3t
+
Ex. 9.6-1
For parallel RLC circuits: α =
1
1
1
1
=
= 50, ω o2 =
=
= 2500
−3
2 RC 2(10)(10 )
LC (0.4)(10−3 )
The roots of the characteristic equations are:∴ s1,2 = −50 ±
(50) 2 − 2500 = −50, −50
The natural response is vn (t ) = A1 e−50t + A2 t e−50t .
At t = 0+ we have:
−v(0+ )
ic (0 ) =
= − .8V
10Ω
dv( t )
i (0+ )
∴
= c
= −800 V / s
dt
c
+
t =0
+
9-4
So vn (0+ ) = 8 = A1 ⇒ vn (t ) = 8e−50t + A2 t e−50t
dv(0+ )
= −800 = − 400 + A2 ⇒ A2 = − 400
dt
∴ vn (t ) = 8 e −50t − 400 t e−50t V
α=
1
1
=
= 8000
2 RC 2(62.5)(10−6 )
1
1
=
= 1 08
ω o2 =
−6
LC (.01)(10 )
Ex. 9.7-1
∴ s = − α ± α 2 −ω o2 = − 8000 ±
(8000) 2 −108 = − 8000 ± j 6000
∴ vn (t ) = e −8000t [ A1 cos 6000 t + A2 sin 6000 t ]
at t = 0+
0.08 +
∴
10
+ ic (0+ ) = 0
62.5
⇒ ic (0+ ) = −.24 A
dv(0+ ) ic (0+ )
=
= −2.4 ×10+5 V/s
dt
C
dvn (0+ )
= −2.4 × 105 = 6000 A2 − 8000 A1 ⇒ A2 = −26.7
vn (0 ) = 10 = A1 and
dt
+
∴ vn (t ) = e−8000t [10 cos 6000 t − 26.7 sin 6000 t ] V
d 2v ( t )
dv ( t )
+5
+ 6 v ( t ) = vs ( t ) so the characteristic equation is
2
dt
dt
s 2 + 5 s + 6 = 0 . The roots are s 1 , s 2 = −2, − 3 .
Ex. 9.8-1
The differential equation is
d 2v ( t )
dv ( t )
(a)
+5
+ 6 v ( t ) = 8 . Try v f ( t ) = B . Substituting into the differential equation gives
2
dt
dt
6 B = 8 ∴ v f (t ) = 8 / 6 V .
9-5
(b)
d 2v ( t )
2
+5
dv ( t )
+ 6 v ( t ) = 3 e − 4 t . Try v f ( t ) = B e− 4 t . Substituting into the differential
3
3
equation gives (−4) 2 B + 5(−4) B + 6 B = 3 ⇒ B = . ∴ v f ( t ) = e−4t .
2
2
2
d v (t )
dv ( t )
(c)
+5
+ 6 v ( t ) = 2 e − 2 t . Try v f ( t ) = B t e− 2 t because –2 is a natural frequency.
2
dt
dt
Substituting into the differential equation gives
dt
dt
(4t − 4) B + 5 B (1 − 2t ) + 6 Bt = 2 ⇒ B = 2. ∴ v f ( t ) = 2 t e −2t .
d 2i ( t )
Ex. 9.8-2
di ( t )
+ 20 i ( t ) = 36 + 12 t . Try i f ( t ) = A + B t . Substituting into the differential
dt 2
dt
equation gives 0 + 9 B + 20( A + Bt ) = 36 + 12t ⇒ B = 0.6 and A = 1.53.
+9
∴ i f ( t ) = 1.53 + 0.6 t A
Ex. 9.9-1
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
vC ( ∞ ) =
R2
R1 + R 2
1
Next, represent the circuit by a 2nd order differential equation:
KCL at the top node of R2 gives:
KVL around the outside loop gives:
vC ( t )
R2
vs ( t ) = L
+C
d
vC ( t ) = iL ( t )
dt
d
iL ( t ) + R1 iL ( t ) + vC ( t )
dt
Use the substitution method to get
9-6
vs ( t ) = L

 v (t )

d  vC ( t )
d
d
+ C vC ( t )  + R1  C
+ C vC ( t )  + vC ( t )


 R2

dt  R 2
dt
dt



= LC
 L
d

R1 
d2
+
+
+
+
v
t
R
C
v
t
v t
1
(
)
(
)



1
C
 R2
 dt C
 R 2  C ( )
dt 2




(a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω
Use the steady state response as the forced response:
R2
1
1= V
v f = vC ( ∞ ) =
2
R1 + R 2
The characteristic equation is
R 

1+ 1 

 1
R1 
R2  2
+ s+
= s + 6 s + 8 = ( s + 2 )( s + 4 ) = 0
s2 + 
 R 2 C L   LC 

 




so the natural response is
vn = A1 e −2 t + A2 e −4 t V
vc ( t ) =
The complete response is
iL ( t ) =
vC ( t )
1.309
+
d
vC ( t ) = −1.236 A1 e −2 t − 3.236 A2 e−4 t + 0.3819
dt
+
At t = 0
1
+ A1 e −2 t + A2 e−4 t V
2
( )
0 = vc 0+ = A1 + A2 + 0.5
( ) = −1.236 A − 3.236 A
0 = iL 0
+
1
2
+ 0.3819
Solving these equations gives A1 = −1 and A2 = 0.5, so
vc ( t ) =
1 −2 t 1 −4 t
−e + e V
2
2
(b) C = 1 F, L = 1 H, R1 = 3 Ω, R2 = 1 Ω
Use the steady state response as the forced response:
R2
1
v f = vC ( ∞ ) =
1= V
R1 + R 2
4
The characteristic equation is
9-7
R 

1+ 1 

 1
R1 
R2  2
2
+ s+
= s + 4s + 4 = ( s + 2 ) = 0
s2 + 
 R 2 C L   LC 

 




so the natural response is
vn = ( A1 + A2 t ) e −2 t V
vc ( t ) =
The complete response is
iL ( t ) = vC ( t ) +
At t = 0+
1
+ ( A1 + A2 t ) e −2 t V
4
d
1
vC ( t ) = +
dt
4
( )
(( A
0 = vc 0+ = A1 +
( )
0 = iL 0+ =
2
)
− A1 ) − A2 t e −2 t
1
4
1
+ A2 − A1
4
Solving these equations gives A1 = −0.25 and A2 = −0.5, so
vc ( t ) =
1  1 1  −2 t
−  + t e V
4 4 2 
(c) C = 0.125 F, L = 0.5 H, R1 = 1 Ω, R2 = 4 Ω
Use the steady state response as the forced response:
R2
4
v f = vC ( ∞ ) =
1= V
R1 + R 2
5
The characteristic equation is
R 

1+ 1 

 1
R1 
R2  2
+ s+
= s + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) = 0
s2 + 
 R 2 C L   LC 

 




so the natural response is
vn = e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
The complete response is
iL ( t ) =
vc ( t ) = 0.8 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
vC ( t )
4
+
A2 −2 t
A1
1d
vC ( t ) = 0.2 +
e cos 4 t − e−2 t sin 4 t
8 dt
2
2
9-8
( )
At t = 0+
0 = vc 0+ = 0.8 + A1
( )
0 = iL 0+ = 0.2 +
A2
2
Solving these equations gives A1 = -0.8 and A2 = -0.4, so
vc ( t ) = 0.8 − e −2 t ( 0.8cos 4 t + 0.4sin 4 t ) V
Ex 9.9-2
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
vC ( ∞ ) =
R2
R1 + R 2
1, iL ( ∞ ) =
1
R1 + R 2
and
vo ( ∞ ) =
R2
R1 + R 2
1
Next, represent the circuit by a 2nd order differential equation:
KVL around the right-hand mesh gives:
KCL at the top node of the capacitor gives:
vC ( t ) = L
d
iL ( t ) + R 2 iL ( t )
dt
vs ( t ) − vC ( t )
d
− C vC ( t ) = iL ( t )
R1
dt
Use the substitution method to get
vs ( t ) = R1 C
Using iL ( t ) =
d  d
  d

 L iL ( t ) + R 2 iL ( t )  +  L iL ( t ) + R 2 iL ( t )  + R1 iL ( t )
dt  dt
  dt

d2
d
= R1 LC 2 iL ( t ) + ( L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t )
dt
dt
vo ( t )
gives
R2
 L
d
 R1 + R 2 
d2
vs ( t ) =
LC 2 vo ( t ) + 
v t
+ R1 C  vo ( t ) + 
 R2
 dt
 R 2  o ( )
R2
dt




R1
9-9
(a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω
Use the steady state response as the forced response:
R2
1
1= V
v f = vo ( ∞ ) =
2
R1 + R 2
The characteristic equation is
R 

1+ 2 

 1
R2  
R1  2
+
= s + 6 s + 8 = ( s + 2 )( s + 4 ) = 0
s2 + 
s+

 R1 C L   LC 

 




so the natural response is
vn = A1 e −2 t + A2 e −4 t V
The complete response is
vo ( t ) =
1
+ A1 e −2 t + A2 e −4 t V
2
A1 −2 t
A2 −4 t
v (t )
1
iL ( t ) = o
e +
e V
=
+
1.309 2.618 1.309
1.309
vC ( t ) = 1.309 iL ( t ) +
At t = 0+
1 d
1
iL ( t ) = + 0.6180 A1 e −2 t + 0.2361 A2 e −4 t
4 dt
2
( )
0 = iL 0+ =
( )
0 = vC 0+ =
A1
A2
1
+
+
2.618 1.309 1.309
1
+ 0.6180 A1 + 0.2361 A2
2
Solving these equations gives A1 = −1 and A2 = 0.5, so
vo ( t ) =
1 −2 t 1 −4 t
−e + e V
2
2
(b) C = 1 F, L = 1 H, R1 = 1 Ω, R2 = 3 Ω
Use the steady-state response as the forced response:
R2
3
v f = vo ( ∞ ) =
1= V
R1 + R 2
4
The characteristic equation is
R 

1+ 2 

 1
R2  
R1  2
2
+
= s + 4s + 4 = ( s + 2 ) = 0
s2 + 
s+

 R1 C L   LC 

 




9-10
vn = ( A1 + A2 t ) e −2 t V
so the natural response is
vo ( t ) =
The complete response is
iL ( t ) =
3
+ ( A1 + A2 t ) e −2 t V
4
vo ( t )
3
vC ( t ) = 3 iL ( t ) +
=
1  A1 A2  −2 t
t e V
+ +
4  3
3 
3   A1 A2  A2  −2 t
d
iL ( t ) = +   +
t e
+
4  3
3  3 
dt
( )
At t = 0+
0 = iL 0+ =
( )
0 = vC 0+
A1
+
1
4
3
3 A1 A2
= + +
4 3
3
Solving these equations gives A1 = -0.75 and A2 = -1.5, so
vo ( t ) =
3  3 3  −2 t
−  + t e V
4 4 2 
(c) C = 0.125 F, L = 0.5 H, R1 = 4 Ω, R2 = 1 Ω
Use the steady state response as the forced response:
R2
1
v f = vo ( ∞ ) =
1= V
R1 + R 2
5
The characteristic equation is
R 

1+ 2 

 1
R2  
R1  2
+
= s + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) = 0
s2 + 
s+

 R1 C L   LC 

 




so the natural response is
vn = e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
The complete response is
vo ( t ) = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
iL ( t ) =
vo ( t )
1
= 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
9-11
vC ( t ) = iL ( t ) +
At t = 0+
1 d
iL ( t ) = 0.2 + 2 A2 e−2 t cos 4 t − 2 A1 e−2 t sin 4 t
2 dt
( )
( 0 ) = 0.2 + 2 A
0 = iL 0+ = 0.2 + A1
0 = vC
+
2
Solving these equations gives A1 = -0.8 and A2 = -0.4, so
vc ( t ) = 0.2 − e −2 t ( 0.2 cos 4 t + 0.1sin 4 t ) V
Ex. 9.10-1
At t = 0+ no initial stored energy ⇒ v1 (0+ ) = v2 (0+ ) = i (0+ ) = 0
3 di (0+ )
di (0+ )
+0= 0 ⇒
=0
10 dt
dt
0
dv1 (0+ )
+
⇒
= 0
KCL at A : + i1 (0 ) + 0 = 0
1
dt
5 dv2 (0+ )
= 10 ⇒
KCL at B : − 0 + i2 (0+ ) − 10 = 0 ⇒ i2 (0+ ) =
6
dt
KVL : − 0 +
For t > 0:
dv2 (0+ )
= 12 V s
dt
v1 1 d v1
+
+i = 0
1 12 dt
5 d v2
= 10
KCL at B : − i +
6 dt
3 di
− v1 +
+ v2 = 0
KVL:
10 dt
Eliminating i yields
1 d v1 5 d v2
v1 +
+
− 10 = 0
12 dt 6 dt
3  5 d 2 v2 
−v1 + 
 + v2 = 0
10  6 dt 2 
KCL at A :
Next
9-12
v1 = v2 +
Now, eliminating v1
1 d 2 v2
4 dt 2
⇒
d v1 d 2 v2 1 d 3 v2
=
+
4 dt 3
dt
dt 2
1 d 2 v2 1  d v2 1 d 3 v2  5 d v2
+ 
+
= 10
v2 +
+
4 dt 2 12  dt 4 dt 3  6 dt
Finally, the circuit is represented by the differential equation:
d 3 v2
d 2 v2
dv
+
12
+ 44 2 + 48v2 = 480
3
2
dt
dt
dt
The characteristic equation is s 3 + 12s 2 + 44s + 48 = 0 . It’s roots are s1,2,3 = −2, −4, −6 . The
vn = A1e −2t + A2 e −4t + A3e −6t
natural response is
Try v f = B as the forced response. Substitute into the differential equation and equate
coefficients to get B = 10. Then
v2 (t ) = vn (t ) + v f (t ) = A1e −2t + A2 e −4t + A3e−6t + 10
We have seen that v2 (0+ ) = 0 and
Then
d 2 v2 (0+ )
dv2 (0+ )
= 12 V/s . Also
= 4[v1 (0+ ) − v2 (0+ )] = 0 .
dt
dt 2
v2 (0+ ) = 0 = A1 + A2 + A3 + 10
dv2 (0+ )
= 12 = −2 A1 − 4 A2 − 6 A3
dt
d 2 v2 (0+ )
= 0 = 4 A1 + 16 A2 + 36 A3
dt 2
Solving these equations yields A1 = −15, A2 = 6, A3 = −1 so
v2 (t ) = ( −15 e −2t + 6 e −4t − e −6t + 10 ) V
Ex. 9.11-1
1
1
s2 +
s+
=0
RC
LC
In our case L = 0.1, C = 0.1 so we have s 2 +
10
s + 100 = 0
R
9-13
a)
R = 0.4 Ω ⇒ s 2 + 25s + 100 = 0
s1,2 = −5, − 20
b)
R = 1 Ω ⇒ s 2 + 10 s + 100 = 0
s1,2 = − 5 ± j 5 3
9-14
Problems
Section 9-3: Differential Equations for Circuits with Two Energy Storage Elements
P9.3-1
KCL: iL ( t ) =
v (t )
dv ( t )
+C
R2
dt
KVL: vs ( t ) = R1iL ( t ) + L
diL ( t )
+ v (t )
dt
dv ( t )  L dv ( t )
d 2v ( t )
 v (t )
+C
+ LC
+ v (t )
vs ( t ) = R1 
+
dt  R2 dt
dt 2
 R2
d 2v ( t )
 R1 

L  dv ( t )
=
+
+
+
+
]
1
[
vs ( t ) 
LC
 v ( t )  R1C

R2  dt
dt 2
 R2 

In this circuit R1 = 2 Ω, R 2 = 100 Ω, L = 1 mH, C = 10 µ F so
2
dv ( t )
−8 d v ( t )
+ 10
vs ( t ) = 1.02v ( t ) + .00003
dt
dt 2
dv ( t ) d 2 v ( t )
8
8
10 vs ( t ) = 1.02 × 10 v ( t ) + 3000
+
dt
dt 2
P9.3-2
Using the operator s =
KCL: is ( t ) =
d
we have
dt
v (t )
+ iL ( t ) + Csv ( t )
R1
KVL: v ( t ) = R2iL ( t ) + LsiL ( t )
Solving by usingCramer's rule for iL ( t ) : iL ( t ) =
is ( t )
R2 Ls
+
+ R2Cs + LCs 2 + 1
R1 R1
 R2 
L

2
1 +  iL ( t ) +  + R2C  siL ( t ) + [ LC ] s iL ( t ) = is ( t )
R
R
1 

 1

9-15
In this circuit R1 = 100 Ω, R 2 = 10 Ω, L = 1 mH, C = 10 µ F so
1.1iL ( t ) + .00011siL ( t ) + 10−8 s 2iL ( t ) = is ( t )
diL ( t ) d 2iL ( t )
+
= 108 is ( t )
1.1× 10 iL ( t ) + 11000
2
dt
dt
8
P9.3-3
After the switch closes, a source
transformation gives:
KCL:
iL ( t ) + C
dvc ( t ) vs ( t ) + vc ( t )
+
=0
dt
R2
diL ( t )
− vc ( t ) − vs ( t ) = 0
dt
di ( t )
vc ( t ) = R1is ( t ) + R1iL ( t ) + L L
− vs ( t )
dt
Differentiating
KVL:
R1is ( t ) + R1iL ( t ) + L
d vc ( t )
d is ( t )
d iL ( t )
d 2iL ( t ) d vs ( t )
= R1
+ R1
+L
−
dt
dt
dt
dt 2
dt
 d is ( t )
d iL ( t )
d 2iL ( t ) d vs ( t )  vs ( t )
+ R1
+L
−
iL ( t ) + C  R1
+
dt
dt
dt 2
dt  R2

di ( t )

1 
+  R1is ( t ) + R1iL ( t ) + L L
− vs ( t )  = 0
R2 
dt

Then
Solving for i L ( t ) :
− R1
d 2i L ( t )  R1
R1 di s ( t ) 1 dv s ( t )
1  di L ( t )  R1
1 
+ +
+
+
+
i s (t ) −

 i L (t ) =
2
dt
LCR 2
L dt
L dt
 L R 2C  dt
 LR 2C LC 
9-16
Section 9-4: Solution of the Second Order Differential Equation - The Natural Response
P9.4-1
From Problem P 9.3-2 the characteristic equation is:
1.1×108 +11000 s + s 2 = 0 ⇒ s1 , s2 =
−11000± (11000) 2 − 4(1.1×108 )
= −5500± j8930
2
KVL: 40 ( is ( t ) − iL ( t ) ) = (100 × 10−3 )
P9.4-2
diL ( t )
+ vc ( t )
dt
The current in the inductor is equal to the current in
the capacitor so
1
 dv ( t )
iL ( t ) =  ×10−3  c
3
 dt
2
 40
 dv ( t )  100
 d vC ( t )
vC ( t ) = 40 is ( t ) −  × 10−3  C − 
× 10−6 
2
 3
 dt
 3
 dt
d 2 vC ( t )
dv ( t )
+ 400 C + 30000 vC ( t ) = 40 is ( t )
2
dt
dt
2
s + 400s + 30000 = 0 ⇒ ( s + 100)( s + 300) = 0 ⇒ s1 = −100, s2 = −300
P9.4-3
v ( t ) − vs ( t )
dv ( t )
+ iL ( t ) + (10 × 10−6 )
=0
1
dt
di ( t )
KVL: v ( t ) = 2iL ( t ) + (1× 10−3 ) L
dt
KCL:
0 = 2 iL ( t ) + (1× 10−3 )
vs ( t ) = 3iL ( t ) + 0.00102
diL ( t )
dt
diL ( t )
dt
− vs ( t ) + iL ( t ) + (10 × 10−6 ) ( 2 )
+ 10−8
d 2iL ( t )
dt
2
⇒
d 2iL ( t )
dt
diL ( t )
dt
+ (10 × 10−6 )(10−3 )
+ 102000
diL ( t )
dt
d 2iL ( t )
dt
+ 3 × 108 iL ( t ) = 108 vs ( t )
s 2 + 102000s + 3 ×108 = 0 ⇒ s1 = 3031, s2 = −98969
9-17
Section 9.5: Natural Response of the Unforced Parallel RLC Circuit
dv( 0 )
d
= −3000 V/s . Using the operator s = , the node
dt
dt
−
v( t ) v( t ) vs ( t )
L


+
= 0 or  LCs 2 + s + 1 v ( t ) = vs ( t )
equation is Csv ( t ) +
R
sL
R


1
1
s+
= 0 ⇒ s 2 + 500 s + 40, 000 = 0
The characteristic equation is: s 2 +
RC
LC
P9.5-1
The initial conditions are v ( 0 ) = 6 V,
The natural frequencies are: s1,2 = −250 ± 2502 − 40, 000 = −100, −400
The natural response is of the form v ( t ) = Ae−100t + Be−400t . We will use the initial conditions
to evaluate the constants A and B.
v ( 0) = 6 = A + B


 ⇒
dv ( 0 )
= −3000 = −100 A − 400 B 
dt

A = −2 and B = 8
Therefore, the natural response is
v ( t ) = −2e−100t + 8e−400t
t >0
P9.5-2
The initial conditions are v ( 0 ) = 2 V, i ( 0 ) = 0 .
The characteristic equation is: s 2 +
1
1
s+
= 0 ⇒ s 2 + 4s + 3 = 0
RC
LC
The natural frequencies are: s1 , s2 = −1, − 3
The natural response is of the form v ( t ) = Ae − t + Be −3t . We will use the initial conditions to
evaluate the constants A and B.
dv ( t )
= − Ae − t − 3Be −3t . At t = 0 this becomes
Differentiating the natural response gives
dt
dv ( t ) v ( t )
dv ( t )
v (t ) i (t )
dv ( 0 )
−
.
+
+ i ( t ) = 0 or
= −
= − A − 3B . Applying KCL gives C
dt
R
dt
RC
C
dt
dv ( 0 )
v ( 0) i ( 0)
At t = 0 this becomes
= −
−
. Consequently
dt
RC
C
−1A − 3B = −
v ( 0) i ( 0)
2
−
=−
−0 = − 8
14
RC
C
9-18
Also, v ( 0 ) = 2 = A + B . Therefore A = −1 and B = 3. The natural response is
v ( t ) = −e − t + 3e −3t V
P9.5-3
di1 ( t )
di ( t )
−3 2
=0
dt
dt
di ( t )
di ( t )
KVL : − 3 1
+3 2
+ 2i2 ( t ) = 0
dt
dt
KVL : i1 + 5
(1)
( 2)
Using the operator s =
d
, the KVL equations are
dt
(1+5s )i1 + ( −3s )i2 = 0 
3s
2
i1 = 0 ⇒ (1 + 5s )( 3s + 2 ) − ( 3s )  i1 = 0
 ⇒ (1 + 5s ) i1 − ( 3s )


3s + 2
( −3s ) i1 + ( 3s + 2 ) i2 = 0
1
The characteristic equation is (1+5s )( 3s + 2 ) − 9 s 2 = 6 s 2 + 13s + 2 = 0 ⇒ s1,2 = − , −2
6
The currents are i1 ( t ) = Ae 6 + Be−2t and i2 ( t ) = Ce 6 + De −2t , where the constants A, B, C
and D must be evaluated using the initial conditions. Using the given initial values of the
currents gives
i1 ( 0 ) = 11 = A + B and i 2 ( 0 ) = 11 = C + D
−t
−t
Let t = 0 in the KCL equations (1) and (2) to get
di1 ( 0 )
dt
=−
di2 ( 0 )
A
C
33
143
= − − 2 B and
=−
=− −D
2
6
6
6
dt
So A = 3, B = 8, C = −1 and D = 12. Finally,
i1 (t ) = 3e− t / 6 + 8e−2t A
and
i 2 (t ) = −e −t / 6 + 12e −2t A
9-19
Section 9.6: Natural Response of the Critically Damped Unforced Parallel RLC Circuit
P9.6-1
After t = 0
Using KVL: 100 ic ( t ) + 0.025
dic ( t )
+ vc ( t ) = 0
dt
The capacitor current and voltage are related by:
∴
d 2 vc ( t )
dt
2
ic ( t ) = 10−5
+ 4000
dvc ( t )
dt
dvc ( t )
dt
+ 4 × 106 vc ( t ) = 0
The characteristic equation is: s 2 + 4000 s + 4 × 106 = 0
The natural frequencies are: s1,2 = − 2000, − 2000
The natural response is of the form: vc ( t ) = A1e−2000t + A2 t e−2000t
Before t = 0 the circuit is at steady state
(The capacitor current is continuous at t = 0
in this circuit because it is equal to the
inductor current.)
vc ( 0+ ) = 3 = A1
dvc ( 0+ )
= 0 = −2000 A1 + A2 ⇒ A2 = 6000
dt
∴ vc ( t ) = ( 3 + 6000t ) e−2000t V for t ≥ 0
P9.6-2
After t = 0
∫
Using KCL:
t
−∞
1 dvc ( t )
=0
4 dt
d 2 vc ( t )
dv ( t )
⇒
+4 c
+ 4vc ( t ) = 0
dt
dt
vc (τ ) dτ + vc ( t ) +
The characteristic equation is: s 2 + 4 s + 4 = 0
9-20
The natural frequencies are: s1,2 = −2, −2
The natural response is of the form: vc ( t ) = A1e−2t + A2 t e−2 t
i L ( 0+ ) = i L ( 0 − ) and vC ( 0+ ) = vC ( 0− )
Before t = 0 the circuit is at steady state
i C ( 0+ ) = −i L ( 0+ ) = −2 A
dv ( 0+ )
vc ( 0
+
)=0=
A1 and
dvc ( 0+ )
dt
dt
= −8 = A2
=
i C ( 0+ )
14
= −8 V
⇒ vc ( t ) = −8 t e−2 t V
P9.6-3
Assume that the circuit is at steady state before t = 0. The initial conditions are
vc ( 0− ) = 104 V & iL ( 0− ) = 0 A
After t = 0
− vc ( t ) + .01
KVL:
diL ( t )
+ 106 iL ( t ) = 0
dt
 d 2iL ( t )
dvC ( t )
di ( t ) 
iL ( t ) = −C
= −C .01
+ 106 L 
2
dt
dt
dt 

KCL:
∴ 0.01 C
(1)
( 2)
d 2iL ( t )
di ( t )
+ 106 C L
+ iL ( t ) = 0
2
dt
dt
The characteristic equation is: ( 0.01 C ) s 2 + (106 C ) s + 1 = 0
The natural frequencies are: s1,2 =
−106 C ±
(10 C )
2
− 4 ( 0.01C )
2 ( 0.01C )
6
For critically-damped response: 1012 C 2 − .04C = 0 ⇒ C = 0.04 pF so s1,2 = −5 ×107 , −5 ×107 .
9-21
The natural response is of the form: iL ( t ) = A1e −5×10 t + A2 t e −5×10
diL +
( 0 ) = 100
dt
di ( 0 )
= 106
So iL ( 0 ) = 0 = A1 and L
dt
Now from (1) ⇒
7
7
vc ( 0+ ) − 106 iL ( 0+ )  = 106 A


s
t
= A2 ∴ iL ( t ) = 106 t e−5×10 t A
7
Now v ( t ) = 106 iL ( t ) = 1012 t e−5×10 t V
7
P9.6-4
The characteristic equation can be shown to be: s 2 +
The natural frequencies are: s1,2 = −250, − 250
1
1
= s 2 + 500 s + 62.5 × 103 = 0
s+
RC
LC
The natural response is of the form: v ( t ) = Ae −250t + B t e−250t
dv ( 0 )
= −3000 = −250 A + B ⇒ B = −1500
dt
− 1500 t e−250t V
v ( 0 ) = 6 = A and
∴ v ( t ) = 6e−250t
P9.6-5
After t=0, using KVL yields:
di ( t )
t
+ Ri ( t ) + 2 + 4 ∫0 i (τ ) dτ = 6
(1)
dt
v( t )
Take the derivative with respect to t:
d 2i ( t ) di ( t )
+R
+ 4i ( t ) = 0
dt 2
dt
The characteristic equation is s 2 + Rs + 4 = 0
Let R = 4 for critical damping ⇒
( s + 2)
2
=0
So the natural response is i ( t ) = A t e −2t + B e −2t
i ( 0 ) = 0 ⇒ B = 0 and
∴ i ( t ) = 4 t e −2t A
di ( 0 )
= 4 − R ( i( 0 ) ) = 4 − R ( 0 ) = 4 = A
dt
9-22
Section 9-7: Natural Response of an Underdamped Unforced Parallel RLC Circuit
P9.7-1
After t = 0
KCL:
vc ( t )
dv ( t )
+ iL ( t ) + 5 ×10−6 c
= 0
dt
250
KVL:
vc ( t ) = 0.8
(1)
diL ( t )
dt
( 2)
d 2vc ( t )
dv ( t )
+ 800 c + 2.5 ×105 v ( t ) = 0 ⇒ s 2 + 800 s + 250, 000 = 0, s1,2 = −400 ± j 300
c
dt
dt 2
The natural response is of the form v
c
(t )
Before t = 0 the circuit is at steady state:
= e −400t  A1 cos 300t + A 2sin 300t 
iL ( 0+ ) = iL ( 0− ) =
−6
A
500
vc ( 0+ ) = vc ( 0− ) = 250 −6
(
500
)+6 = 3 V
From equation (1) :
dvc ( 0+ )
dt
= − 2 × 105 iL ( 0+ ) − 800vc ( 0+ ) = 0
vc ( 0+ ) = 3 = A1
dvc ( 0+ )
= 0 = −400 A1 + 300 A2 ⇒ A2 = 4
∴ vc ( t ) = e −400t [3cos 300t + 4sin 300t ] V
dt
9-23
P9.7-2
Before t = 0
v ( 0+ ) = v ( 0− ) = 0 V
i ( 0+ ) = i ( 0− ) = 2 A
After t = 0
KCL :
t
v (t ) 1 d v (t )
+
+ 2 ∫ v (τ ) dτ + i ( 0 ) = 0
0
1
4 dt
Using the operator s =
d
we have
dt
1
 2
v (t ) +   s v (t ) +   v (t ) + i (0) = 0 ⇒
 4
s
(s
2
+ 4s + 8) v ( t ) = 0
The characteristic equation and natural frequencies are: s 2 + 4s + 8 = 0
The natural response is of the form: v ( t ) = e
v ( 0 ) = 0 = B1 and
−2 t
 B1 cos 2t + B 2 sin 2t 
dv ( 0 )
= 4  −i ( 0 ) − v ( 0 )  = −4 [ 2] = −8 = 2 B2 or B2 = −4
dt
v ( t ) = −4e −2t sin 2t V
so
P9.7-3
After t = 0
1 dvc ( t ) vc ( t )
+
+ iL ( t ) = 0
4 dt
2
4 diL ( t )
KVL : vc ( t ) =
+ 8 iL ( t )
dt
KCL :
Characteristic Equation:
⇒ s = −2 ± j 2
d 2i L ( t )
dt 2
+4
di L ( t )
dt
(1)
( 2)
+ 5 i L ( t ) = 0 ⇒ s 2 + 4 s + 5 = 0 ⇒ s1,2 = −2 ± i
Natural Response: i L ( t ) = e −2t  A1 cos t + A2 sin t 
9-24
vc ( 0− )
 48 
+
−
= 7 
 ⇒ vc ( 0 ) = vc ( 0 ) = 8 V
+
2
4
8
2


8
iL ( 0+ ) = iL ( 0− ) = − = −4 A
2
Before t = 0
diL ( 0+ )
dt
=
vc ( 0+ )
4
iL ( 0+ ) = −4 = A1
diL ( 0+ )
∴ iL ( t ) = e −2t [ −4 cos t + 2sin t ] A
dt
− 2iL ( 0+ ) =
8
A
− 2 ( −4 ) = 10
s
4
= 10 = − 2 A1 + A2 ⇒ A2 = 2
P9.7-4
The plot shows an underdamped response, i.e. v ( t ) = e − α t [ k1 cos ω t + k2 sin ω t ] + k3 .
Examining the plot shows v ( ∞ ) = 0 ⇒ k3 = 0, v ( 0 ) = 0 ⇒ k1 = 0 .
Therefore, v ( t ) = k2 e − α t sin ω t .
Again examining the plot we see that the maximum voltage is approximately 260 mV the time is
approximately 5 ms and that the minimum voltage is approximately −200 mV the time is
approximately 7.5 ms. The time between adjacent maximums is approximately 5 ms so
2π
ω≈
= 1257 rad/s . Then
5 ×10−3
−α ( 0.005 )
0.26 = k2 e
sin (1257 (.005 ) )
(1)
−0.2 = k2 e
−α ( 0.0075 )
To find α we divide (1) by (2) to get
( 6.29 rad )  ⇒


 sin ( 9.43 rad ) 
α ( 0.0025 )  sin
− 1.3 = e
From (1) we get k2 = 544 . Then
sin (1257 (.0075 ) )
( 2)
e0.0025α = 1.95 ⇒ α = 267
v ( t ) = 544e −267t sin1257t V
9-25
P9.7-5
After t = 0
The characteristic equation is:
s2 +
1
1
s+
= 0 or s 2 + 2 s + 5 = 0
RC
LC
The natural frequencies are: s1,2 = −1 ± j 2
The natural response is of the form:
v(0 ) = 2 = B1 . From KCL, ic ( 0
+
v ( 0+ )
v(t ) = e − t  B1 cos 2t + B 2 sin 2t 
) = − 5 − i ( 0 ) = − 52 − 101 = − 12 Vs
dv ( 0 )
3
 1
= 10 − = − B + 2 B ⇒ B = −
dt
+
+


 2
3
Finally, v ( t ) = 2e − t cos 2t − e− t sin 2t V
2
+
1
2
2
so
2
t≥0
9-26
Section 9-8: Forced Response of an RLC Circuit
P9.8-1
KCL : is ( t ) =
After t = 0
v (t )
R
KVL : v ( t ) = L
is ( t ) =
dt
dt
2
diL ( t )
dv ( t )
dt
d 2iL ( t )
L diL ( t )
+ iL ( t ) + LC
R dt
dt 2
d 2iL ( t )
2
+ iL ( t ) + C
d 2iL ( t )
+
dt
1 diL ( t ) 1
1
+
iL ( t ) =
is ( t )
RC dt
LC
LC
+ ( 650 )
diL ( t )
dt
+ (105 ) iL ( t ) = (105 ) is ( t )
(a) Try a forced response of the form i f ( t ) = A . Substituting into the differential equations gives
0+0+ A
1
1
=
⇒
−3
(.01) (1×10 ) (.01) (1×10−3 )
A = 1 . Therefore i f ( t ) = 1 A .
(b) Try a forced response of the form i f ( t ) = A t + B . Substituting into the differential equations
gives 0 + A
1
65
+ ( A t + B)
= 0.5 t . Therefore A = 0.5 and
( 0.01)( 0.001)
(100 ) ( 0.001)
B = −3.25 × 10 −3 . Finally i f ( t ) = 5 t − 3.25 × 10−3 A .
(c) Try a forced response of the form i f ( t ) = A e−250t . It doesn’t work so try a forced response of
the form i f ( t ) = B t e−250t . Substituting into the differential equation gives
( −250 )2 B e−250t − 500 B e−250t  + 650 ( −250 ) B t e−250t + B e−250t  + 105 B t e−250t = 2 e−250t .




( 250 )
2
B + 650 ( −250 ) B + 105 B = 0 ⇒ ( 250 ) + 650 ( −250 ) + 105  B = 0 ⇒


Equating coefficients gives
and
2
−500 B + 650 B = 2 ⇒
Finally i f ( t ) = 0.0133 t e −250t A .
[ 0] B = 0
B = 0.0133
9-27
P9.8-2
After t = 0
d 2v ( t )
dt
d 2v ( t )
dt
+
v (t )
R dv ( t ) 1
+
v (t ) = s
L dt
LC
LC
+ 70
dv ( t )
dt
+ 12000v ( t ) = 12000 vs ( t )
(a) Try a forced response of the form v f ( t ) = A . Substituting into the differential equations gives
0 + 0 + 12000 A = 24000 ⇒ A = 2 . Therefore v f ( t ) = 2 V .
(b) Try a forced response of the form v f ( t ) = A + B t . Substituting into the differential equations
gives 70 A + 12000 A t + 12000 B = 2400 t . Therefore A = 0.2 and B =
Finally v f ( t ) = ( −1.167 × 10−3 ) t + 0.2 V .
−70 A
= −1.167 × 10 −3 .
12000
(c) Try a forced response of the form v f ( t ) = A e −30t . Substituting into the differential equations
gives 900 Ae −30t − 2100 Ae −30 t + 12000 Ae −30 t = 12000 e −30t . Therefore A =
v f ( t ) = 1.11e−250 t V .
12000
= 1.11 . Finally
10800
9-28
Section 9-9: Complete Response of an RLC Circuit
P9.9-1
First, find the steady state response for t < 0, when the switch is open. Both inputs are constant
so the capacitor will act like an open circuit at steady state, and the inductor will act like a short
circuit. After a source transformation at the left of the circuit:
i L ( 0) =
and
11 − 4
= 2.33 mA
3000
v C ( 0) = 4 V
After the switch closes
Apply KCL at node a:
vC
R
+C
d
vC + iL = 0
dt
Apply KVL to the right mesh:
L
d
d
i L + Vs − vC = 0 ⇒ vC = L i L + Vs
dt
dt
After some algebra:
V
1 d
1
d2
i +
iL +
iL = − s
2 L
dt
R C dt
LC
R LC
⇒
d2
d
i + ( 500 ) i L + (1.6 × 105 ) i L = −320
2 L
dt
dt
The characteristic equation is
s 2 + 500 s + 1.6 × 105 = 0 ⇒ s1,2 = −250 ± j 312 rad/s
9-29
After the switch closes the steady-state inductor current is iL(∞) = -2 mA so
i L ( t ) = −0.002 + e −250 t ( A1 cos 312 t + A2 sin 312 t )
v C ( t ) = 6.25
d
i L (t ) + 4
dt
= 6.25 e −250 t  −250 ( A1 cos 312 t − A2 sin 312 t ) − 312 ( A1 sin 312 t − A2 cos 312 t )  + 4
= 6.25 e −250 t ( 312 A2 − 250 A1 ) cos 312 t + ( 250 A2 + 312 A1 ) sin 312 t  + 4
Let t = 0 and use the initial conditions:
iL ( 0+ ) = 0.00233 = −0.002 + A1 ⇒ 0.00433 = A1
vC ( 0+ ) = 4 = 6.25 ( 312 A2 − 250 A1 ) + 4 ⇒
Then
A2 =
250
250
A2 =
( 0.00433) = 0.00347
312
312
i L ( t ) = −0.002 + e −250 t ( 0.00433cos 312 t + 0.00345sin 312 t )
= −0.002 + 0.00555 e −250 t cos ( 312 t − 36.68° ) A
v C ( t ) = 4 + 13.9 e −250 t sin ( 312 t ) V
i (t ) =
vC (t )
2000
= 2 + 6.95 e −250 t sin ( 312 t ) mA
(checked using LNAP on 7/22/03)
P9.9-2
First, find the steady state response for t < 0. The input is constant so the capacitor will act like
an open circuit at steady state, and the inductor will act like a short circuit.
i ( 0) =
and
v (0) =
−1
= −0.2 A
1+ 4
4
( −1) = −0.8 V
1+ 4
9-30
For t > 0
Apply KCL at node a:
v − Vs
d
+C v+i = 0
R1
dt
Apply KVL to the right mesh:
R2 i + L
d
d
i − v = 0 ⇒ v = R2 i + L iL
dt
dt
After some algebra:
L + R1 R 2C d
R1 + R 2
d2
Vs
i+
i+
i=
2
dt
R1 L C dt
R1 L C
R1 L C
The forced response will be a constant, if = B so 1 =
⇒
d2
d
i +5 i +5 i =1
2
dt
dt
d2
d
B + 5 B + 5 B ⇒ B = 0.2 A .
2
dt
dt
To find the natural response, consider the characteristic equation:
0 = s 2 + 5 s + 5 = ( s + 3.62 )( s + 1.38 )
The natural response is
so
Then
At t=0+
in = A1 e −3.62 t + A2 e−1.38 t
i ( t ) = A1 e −3.62 t + A2 e−1.38 t + 0.2
d


v ( t ) =  4 i ( t ) + 4 i ( t )  = −10.48 A1 e−3.62 t − 1.52 A2 e−1.38 t + 0.8
dt


−0.2 = i ( 0 + ) = A1 + A2 + 0.2
−0.8 = v ( 0 + ) = −10.48 A1 − 1.52 A2 + 0.8
so A1 = 0.246 and A2 = −0.646. Finally
i ( t ) = 0.2 + 0.246 e −3.62 t − 0.646 e−1.38 t A
9-31
P9.9-3
First, find the steady state response for t < 0. The input is constant so the capacitors will act like
an open circuits at steady state.
v1 ( 0 ) =
1000
(10 ) = 5 V
1000 + 1000
v2 ( 0 ) = 0 V
and
For t > 0,
Node equations:
v1 − 10  1
v −v
 d
+  × 10−6  v1 + 1 2 = 0
1000  6
1000
 dt
1
 d
⇒ 2 v1 +  × 10−3  v1 − 10 = v2
6
 dt
v1 − v2  1
 d
=  × 10−6  v2
1000  16
 dt
1
 d
⇒ v1 − v2 =  × 10−3  v2
 16
 dt
After some algebra:
d2
d
v + ( 2.8 × 104 ) v1 + ( 9.6 ×107 ) v1 = 9.6 ×108
2 1
dt
dt
The forced response will be a constant, vf = B so
d2
d
B + ( 2.8 × 104 ) B + ( 9.6 × 107 ) B = 9.6 × 108
2
dt
dt
⇒ B = 10 V .
To find the natural response, consider the characteristic equation:
s 2 + ( 2.8 × 104 ) s + ( 9.6 × 107 ) = 0 ⇒ s1,2 = −4 × 103 , −2.4 × 104
The natural response is
vn = A1 e −4×10 t + A2 e−2.4×10
3
so
v1 ( t ) = A1 e−4×10 t + A2 e−2.4×10
3
4t
4t
+ 10
At t = 0
9-32
5 = v1 ( 0 ) = A1 e
Next
At t = 0
+ A2 e
1
 d
2 v1 +  ×10−3  v1 − 10 = v2
6
 dt
⇒
−2.4×104 ( 0 )
+ 10 = A1 + A2 + 10
(1)
d
v1 = −12000v1 + 6000 v2 + 6 × 104
dt
d
v1 ( 0 ) = −12000v1 ( 0 ) + 6000 v2 ( 0 ) + 6 × 104 = −12000 ( 5 ) + 6000 ( 0 ) + 6 × 104 = 0
dt
(
)
(
)
3
4
d
v1 ( t ) = A1 −4 × 103 e −4×10 t + A2 −2.4 × 104 e−2.4×10 t
dt
so
At t = 0+
0=
−4×103 ( 0 )
(
)
(
)
(
)
(
d
−4×103 ( 0 )
−2.4×104 ( 0 )
+ A2 −2.4 × 104 e
= A1 −4 × 103 + A2 −2.4 × 104
v1 ( 0 ) = A1 −4 × 103 e
dt
)
so A1 = −6 and A2 = 1. Finally
v1 ( t ) = 10 + e −2.4 ×10
4t
P9.9-4
For t > 0
− 6 e−4 ×10
3t
V for t > 0
KCL at top node:
diL ( t )


1 dv ( t )
− 5 cos t  + iL ( t ) +
=0
 0.5
dt
12 dt


(1)
KVL for right mesh:
0.5
diL ( t ) 1 dv ( t )
=
+ v (t )
12 dt
dt
( 2)
Taking the derivative of these equations gives:
d 2iL ( t ) diL ( t ) 1 d 2 v ( t )
of (1) ⇒ 0.5
+
+
= −5 sin t
dt
12 dt 2
dt 2
dt
2
2
d of ( 2 ) ⇒ 0.5 d iL ( t ) = 1 d v ( t ) + dv ( t )
dt
12 dt 2
dt 2
dt
d
(3)
( 4)
9-33
d 2iL ( t )
di ( t )
Solving for
in ( 4 ) and L
in ( 2 ) & plugging into ( 3) gives
2
dt
dt
d 2v ( t )
dv ( t )
+
+ 12v ( t ) = −30sin t
7
dt 2
dt
The characteristic equation is: s 2 + 7s+12 = 0 .
The natural frequencies are s1,2 = −3, −4 .
The natural response is of the form vn (t ) = A1e −3t + A2 e −4t . Try a forced response of the form
v f ( t ) = B1 cos t + B 2 sin t . Substituting the forced response into the differential equation and
equating like terms gives B1 =
21
33
and B 2 = − .
17
17
v ( t ) = vn (t ) + v f ( t ) = A1e−3t + A2 e−4t +
21
33
cos t − sin t
17
17
We will use the initial conditions to evaluate A1 and A2. We are given iL ( 0 ) = 0 and v ( 0 ) = 1 V .
Apply KVL to the outside loop to get
1 iC ( t ) + iL ( t )  + 1( iC ( t ) ) + v ( t ) − 5cos t = 0
At t = 0+
iC ( 0 ) =
5cos ( 0 ) + iL ( 0 ) − v ( 0 ) 5 + 0 − 1
=
=2 A
2
2
dv ( 0 ) iC ( 0 )
2
=
=
= 24 V/s
dt
1 12 1 12


 ⇒
33 
dv(0+ )
= 24 = −3 A1 − 4 A2 −
17 
dt
v(0+ ) = 1 = A1 + A2 +
Finally,
∴ v(t ) = 25e
−3t
21
17
A1 = 25
429
A2 = −
17
429e −4t − 21cos t + 33sin t
−
V
17
9-34
P9.9-5
Use superposition. Find the response to inputs 2u(t) and –2u(t-2) and then add the two responses.
First, consider the input 2u(t):
Using the operator s =
For 0< t < 2 s
d
we have
dt
KVL:
vc ( t ) + siL ( t ) + 4 iL ( t ) − 2  = 0
(1)
iL ( t ) =
1
3
s vc ( t ) ⇒ vc ( t ) = iL ( t )
(2)
3
s
Plugging (2) into (1) yields the characteristic equation: ( s 2 + 4 s + 3) = 0 . The natural frequencies
are s1,2 = −1 , −3 . The inductor current can be expressed as
KCL:
iL (t ) = in (t ) + i f (t ) = ( A1 e −t + A2 e−3t ) + 0 = A1 e− t + A2 e −3t .
Assume that the circuit is at steady state before t = 0. Then vc (0+ ) = 0 and iL (0+ ) = 0 .
Using KVL we see that (1)
diL (0+ )
= 4  2 − iL (0+ )  − vC (0+ ) = 8 A/s . Then
dt
iL (0) = 0 = A1 + A2


 A1 = 4 , A2 = −4 .
diL (0)
= 8 = − A1 − 3 A2 
dt

Therefore iL (t ) = 4e − t − 4e−3t A . The response to 2u(t) is
0
t<0

v1 (t ) = 8 − 4 iL (t ) = 
−t
−3t
8 − 16 e + 16 e V t > 0 .
= 8 − 16 e − t + 16 e −3t  u ( t ) V
The response to –2u(t-2) can be obtained from the response to 2u(t) by first replacing t by t-2
everywhere is appears and the multiplying by –1. Therefore, the response to –2u(t-2) is
v2 (t ) =  −8 + 16e − (t − 2) − 16 e−3(t − 2)  u ( t - 2 ) V .
By superposition, v(t ) = v1 (t ) + v2 (t ) . Therefore
v(t ) = 8 − 16e −t + 16e−3t  u (t ) +  −8 + 16e− (t − 2) − 16 e −3(t − 2)  u (t − 2) V
9-35
P9.9-6
First, find the steady state response for t < 0, when the switch is closed. The input is constant so
the capacitor will act like an open circuit at steady state, and the inductor will act like a short
circuit.
i ( 0) = −
and
After the switch opens
5
= −1.25 A
4
v (0) = 5 V
Apply KCL at node a:
v
d
+ 0.125 v = i
2
dt
Apply KVL to the right mesh:
−10 cos t + v + 4
After some algebra:
The characteristic equation is
Try
d
i+4i =0
dt
d2
d
v + 5 v + 6 v = 20 cos t
2
dt
dt
s 2 + 5 s + 6 = 0 ⇒ s1,2 = −2, − 3
vf = A cos t + B sin t
d2
d
A cos t + B sin t ) + 5 ( A cos t + B sin t ) + 6 ( A cos t + B sin t ) = 20 cos t
2 (
dt
dt
9-36
( − A cos t − B sin t ) + 5 ( − A sin t + B cos t ) + 6 ( A cos t + B sin t ) = 20 cos t
( − A + 5 B + 6 A) cos t + ( − B − 5 A + 6 B ) sin t = 20 cos t
Equating the coefficients of the sine and cosine terms yields A =2 and B =2. Then
vf = 2 cos t + 2 sin t
v ( t ) = 2 cos t + 2 sin t + A1 e −2 t + A2 e −3 t
v (t )
d
+ 0.125 v ( t ) = i ( t ) ⇒
2
dt
Next
d
v (t ) = 8 i (t ) − 4 v (t )
dt
d
V
 5
v ( 0 ) = 8 i ( 0 ) − 4 v ( 0 ) = 8  −  − 4 ( 5 ) = −30
dt
s
 4
Let t = 0 and use the initial conditions:
5 = v ( 0 ) = 2 cos 0 + 2 sin 0 + A1 e −0 + A2 e −0 = 2 + A1 + A2
−30 =
d
v ( t ) = −2 sin t + 2 cos t − 2 A1 e −2 t − 3 A2 e−3 t
dt
d
v ( 0 ) = −2 sin 0 + 2 cos 0 − 2 A1 e −0 − 3 A2 e −0 = 2 − 2 A1 − 3 A2
dt
So A1 = −23 and A2 = 26 and
v ( t ) = 2 cos t + 2 sin t − 23 e −2 t + 26 e −3 t V for t > 0
9-37
P9.9-7
First, find the steady-state response for t < 0. The input is constant so the capacitor will act like
an open circuit at steady state, and the inductor will act like a short circuit.
i ( 0) = 0 A
and
After t = 0
Apply KCL at node a:
v (0) = 0 V
C
d
v=i
dt
Apply KVL to the right mesh:
d
8 i + v + 2 i + 4 (2 + i) = 0
dt
d
12 i + v + 2 i = −8
dt
2
 1 
4
d
d
After some algebra:
v + ( 6) v + 
v = −
2
dt
dt
C
2C
The forced response will be a constant, vf = B so
 1 
4
d2
d
B + ( 6) B + 
B = −
2
dt
dt
C
2C
(a)
⇒ B = −8 V
d2
d
v + ( 6 ) v + ( 9 ) v = −72 .
2
dt
dt
2
The characteristic equation is s + 6 s + 9 = 0 ⇒ s1,2 = −3, −3
When C = 1/18 F the differential equation is
Then v ( t ) = ( A1 + A2 t ) e −3t − 8 .
Using the initial conditions:
9-38
0 = i ( 0) = C
So
(b)
0 = v ( 0 ) = ( A1 + A2 ( 0 ) ) e0 − 8 ⇒
A1 = 8
d
v ( 0 ) = C  −3 ( A1 + A2 ( 0 ) ) e0 + A2 e0  ⇒
dt
A2 = 3 A1 = 24
v ( t ) = ( 8 + 24 t ) e −3t − 8 V for t > 0
d2
d
v + ( 6 ) v + ( 5 ) v = −40
2
dt
dt
2
The characteristic equation is s + 6 s + 5 = 0 ⇒ s1,2 = −1, −5
When C = 1/10 F the differential equation is
Then v ( t ) = A1 e − t + A2 e −5 t − 8 .
Using the initial conditions:
0 = v ( 0 ) = A1 e0 + A2 e0 − 8 ⇒ A1 + A2 = 8 

 ⇒ A1 = 10 and A2 = −2
d
0
0
0 = v ( 0 ) = − A1 e − 5 A2 e ⇒ − A1 − 5 A2 = 0 
dt

So
v ( t ) = 10 e − t − 2 e −5 t − 8 V for t > 0
(c)
d2
d
v + ( 6 ) v + (10 ) v = −80
2
dt
dt
2
The characteristic equation is s + 6 s + 10 = 0 ⇒ s1,2 = −3 ± j
When C = 1/20 F the differential equation is
Then v ( t ) = e −3 t ( A1 cos t + A2 sin t ) − 8 .
Using the initial conditions:
0 = v ( 0 ) = e0 ( A1 cos 0 + A2 sin 0 ) − 8 ⇒
0=
So
A1 = 8
d
v ( 0 ) = −3 e0 ( A1 cos 0 + A2 sin 0 ) + e0 ( − A1 sin 0 + A2 cos 0 ) ⇒
dt
A2 = 3 A1 = 24
v ( t ) = e −3 t ( 8cos t + 24 sin t ) − 8 V for t > 0
9-39
P9.9-8
The circuit will be at steady state for t<0:
so
iL(0+) = iL(0−) = 0.5 A
and
vC(0+) = vC(0−) = 2 V.
For t>0:
Apply KCL at node b to get:
1
1 d
1 1 d
= i (t ) +
v (t ) ⇒ i (t ) = −
v (t )
L
C
L
4
4 dt
4 4 dt C
Apply KVL at the right-most mesh to get:
4i
L(
t) + 2
d
1 d

i (t ) = 8 
vc ( t )  + v ( t )
L
dt
 4 dt
 c
Use the substitution method to get
d 1 1 d
1 1 d


1 d

vC ( t )  + 2  −
vC ( t )  = 8 
vc ( t )  + v ( t )
4 −
dt  4 4 dt
 4 4 dt


 4 dt
 c
or
d2
d
v t + 6 v (t ) + 2 v (t )
2 C ( )
C
dt
dt C
d2
d
The forced response will be a constant, vC = B so 2 =
B + 6 B + 2B ⇒ B = 1 V .
dt
dt 2
2=
To find the natural response, consider the characteristic equation:
0 = s 2 + 6 s + 2 = ( s + 5.65 )( s + 0.35 )
The natural response is
vC ( t ) = A1 e
so
Then
vn = A1 e
iL ( t ) =
−5.65 t
+ A2 e
−0.35 t
−5.65 t
+ A2 e
−0.35 t
+1
1 1 d
1
−5.65 t
−0.35 t
vC ( t ) = + 1.41A1 e
+
+ 0.0875 A2 e
4 4 dt
4
9-40
2 = vC ( 0 + ) = A1 + A + 1
At t=0+
1
1
= iL ( 0+ ) = + 1.41A1 + 0.0875 A
2
2
4
2
so A1 = 0.123 and A2 = 0.877. Finally
vC ( t ) = 0.123 e
−5.65 t
+ 0.877 e
−0.35 t
+1 V
P9.9-9
The inductor current and voltage are related be
After t = 0
v (t ) = L
di ( t )
dt
(1)
Apply KCL at the top node to get
C
dv( t )
v( t )
+ i (t ) +
=5
dt
2
(2)
d
, and substituting (1) into (2) yields ( s 2 + 4s + 29 ) i ( t ) = 5 .
dt
The characteristic equation is s 2 + 4 s + 29 = 0 . The characteristic roots are s1,2 = − 2 ± j 5 .
Using the operator s =
The natural response is of the form in ( t ) = e −2t [ A cos 5t + B sin 5t ] .
Try a forced response of the form i f ( t ) = A . Substituting into the differential equation gives
A = 5 . Therefore i f ( t ) = 5 A .
The complete response is i(t ) = 5 + e−2t [ A cos 5t + B sin 5t ] where the constants A and B are
yet to be evaluated using the initial condition:
i (0) = 0 = A + 5 ⇒ A = −5
di (0)
di (0)
2A
⇒
= 0 = −2 A + 5 B ⇒ B =
= −2
0 = v ( 0) = L
dt
dt
5
Finally, i (t ) = 5 + e −2t [ −5cos 5t − 2sin 5t ] A .
P9.9-10
9-41
Assume that the circuit is at steady before t = 0.
i (0+ ) = i (0− ) =
2
×9 = 6 A
2+1
1
v(0+ ) =v(0− ) =
× 9× 1.5 = 4.5 V
2+1
Apply KCL at the top node of the current source
to get
After t = 0:
i ( t ) + 0.5
dv( t ) v( t )
+
= is ( t )
dt
1.5
(1)
Apply KVL and KCL to get
dv( t ) v( t ) 
5di( t )

v ( t ) + 0.5
+
+ i (t )
 0.5 =
dt
dt
1.5 

(2)
Solving for i(t) in (1) and plugging into (2) yields
d 2 v( t ) 49 dv( t ) 4
di ( t )
2
+
+ v ( t ) = is ( t ) + 2 s
2
dt
30 dt
5
5
dt
where is ( t ) = 9 + 3e −2t A
Using the operator s =
d
49
4
, the characteristic equation is s 2 + s + = 0 and the characteristic
30
5
dt
roots are s1,2 =−.817 ± j.365 . The natural response has the form
vn (t ) = e−0.817t  A1 cos (0.365 t ) + A2 sin (0.365 t ) 
Try a forced response of the form v f (t ) = B0 + B1e
−2 t
. Substituting into the differential equations
gives B0 = 4.5 and B1 = −7.04 . The complete response has the form
v(t ) = e−.817 t  A1 cos(0.365 t ) + A2 sin (0.365 t )  + 4.5 − 7.04 e −2t
Next, consider the initial conditions:
v (0) = 4.5 = A1 + 4.5 − 7.04
⇒
A1 = 7.04
9-42
4
4
d v(0)
= 2 is (0) - 2 i (0) − v(0) = 2(9 + 3) − 2(6) − (4.5) = 6
3
3
dt
6=
d v( 0 )
= −0.817 A1 + 0.365 A2 + 14.08 ⇒ A2 = −6.38
dt
So the voltage is given by
v(t ) = e−0.817 t 7.04 cos(0.365 t ) + A2 sin (0.365 t )  + 4.5 − 7.04 e−2t
Next the current given by
Finally
i (t ) = is (t ) −
v(t )
d v(t )
− 0.5
1.5
dt
i (t ) = e −0.817 t [ 2.37 cos(0.365t ) + 7.14sin(0.365t ) ] + 6 + 0.65e −2t A
P9.9-11
First, find the steady state response for t < 0. The input is constant so the capacitor will act like
an open circuit at steady state, and the inductor will act like a short circuit.
va ( 0− ) = −4 i ( 0− )
(
)
i ( 0− ) = 2va ( 0− ) = 2 −4 i ( 0− ) ⇒ i ( 0− ) = 0 A
and
v ( 0− ) = 10 V
9-43
For t > 0
Apply KCL at node 2:
va
d
+ K va + C
v=0
R
dt
KCL at node 1 and Ohm’s Law:
va = − R i
d
1+ K R
v=
i
dt
C
so
Apply KVL to the outside loop:
L
d
i + R i + v − Vs = 0
dt
After some algebra:
d2
R d
1+ K R
1+ K R
v+
v+
v=
Vs
2
dt
L dt
LC
LC
⇒
d2
d
v + 40 v + 144 v = 2304
2
dt
dt
The forced response will be a constant, vf = B so
d2
d
B + ( 40 ) B + (144 ) B = 2304 ⇒ B = 16 V
2
dt
dt
The characteristic equation is s 2 + 40 s + 144 = 0 ⇒ s1,2 = −4, −36 .
v ( t ) = A1 e− 4 t + A2 e−36 t + 16 .
Then
Using the initial conditions:
0=
So
10 = v ( 0+ ) = A1 e0 + A2 e0 + 16 ⇒
d
v ( 0+ ) = −4 A1 e0 − 36 A2 e0
dt


 ⇒
⇒ − 4 A1 − 36 A2 = 0 

A1 + A2 = −6
A1 = −6.75 and A2 = 0.75
v ( t ) = 0.75 e −36 t − 6.75 e−4 t + 16 V for t > 0
(checked using LNAP on 7/22/03)
9-44
Section 9-10: State Variable Approach to Circuit Analysis
P9.10-1
At t = 0− the circuit is source free ∴ iL (0) = 0 and v(0) = 0.
Apply KCL at the top node to get
After t = 0
iL ( t ) +
1 dv( t )
=4
5 dt
(1)
Apply KVL to the right mesh to get
v( t ) −(1)
Solving for i1 ( t ) in (1) and plugging into (2) ⇒
diL ( t )
−6 iL ( t ) = 0
dt
(2)
d 2 v( t )
dv( t )
+ 6
+ 5v ( t ) = 120 .
2
dt
dt
The characteristic equation is s 2 + 6 s + 5= 0 . The natural frequencies are s1,2 =−1, −5 . The natural
response has the form vn (t ) = A1 e−t + A2 e−5t . Try v f ( t ) = B as the forced response.
Substituting into the differential equation gives B = 24 so v f ( t ) = 24 V. The complete response
has the form v (t ) = A1 e−t + A2 e−5t + 24 .
Now consider the initial conditions. From (1)
v(0) = 0 = A1 + A2 + 24


 ⇒
dv(0)
= 20 = − A1 −5 A2 
dt

dv(0)
= 20 − 5 iL (0) = 20 V . Then
s
dt
A1= −25, A2 = 1
Finally v (t ) = − 25e −t + e −5t + 24 V .
9-45
P9.10-2
Before t = 0 there are no sources in the circuit so iL (0) = 0 and v(0) = 0 .
After t = 0 we have:
Apply KCL at the top node to get
iL ( t ) = 4 −
1 dv ( t )
10 dt
(1)
Apply KVL to the left mesh to get
v( t ) −
diL ( t )
− 6iL ( t ) = 0
dt
(2)
Substituting iL ( t ) from (1) into (2) gives
d 2 v( t )
dv( t )
+6
+ 10v( t ) = 240
2
dt
dt
The characteristic equation is s 2 + 6 s + 10 = 0 . The natural frequencies are s1,2 = −3 ± j . The
natural response has the form vn (t ) = e −3t  A1 cos t + A2 sin t  . Try v f ( t ) = B as the forced
response. Substituting into the differential equation gives B = 24 so v f ( t ) = 24 V. The complete
response has the form v(t ) = e −3t  A1 cos t + A2 sin t  + 24 .
Now consider the initial conditions. From (1)
dv(0)
= 40 − 10 iL (0) = 40 V . Then
s
dt
v (0) = 0 = A1 + 24 ⇒ A1 = −24
dv(0)
= 40 = −3 A1 + A2 = 72 + A2 ⇒ A2 = −32
dt
Finally, v (t ) = e −3t [ −24 cos t −32sin t ] + 24 V
9-46
P9.10-3
Assume that the circuit is at steady state before t = 0 so iL (0) = −3 A and v(0) = 0 V .
After t = 0 we have
KCL: i ( t ) + C
KVL: v( t ) = L
i (t ) + C
d 2i ( t )
−6
1 di ( t )
1
+
+
i (t ) =
2
dt
R C dt
LC
LC
⇒
dv( t ) v( t )
+
+ 6=0
dt
R
di ( t )
dt
d  di( t )  1  di ( t ) 
L
 + L
+ 6=0
dt  dt  R  dt 
d 2i ( t )
di ( t )
+ 100
+ 250i ( t ) = −1500
2
dt
dt
The characteristic equation is s 2 + 100 s + 250 = 0 . The natural frequencies are
s1,2 = −2.57, − 97.4 . The natural response has the form in (t ) = A1 e−2.57 t + A2 e−97.4t . Try
( t ) = B as the forced response. Substituting into the differential equation gives B = −6 so
i f ( t ) = −6 A . The complete response has the form i(t ) = A1 e−2.57 t + A2 e−97.4t − 6 .
i
f
Now consider the initial conditions:
i (0) = A1 + A2 − 6 = −3

 A1 = 3 .081

di (0)
= 0 = − 2.57 A1 −97.4 A2  A2 =−0.081
dt

Finally:
i (t ) = 3. 081 e
v(t ) = .2
−2.57 t
−.081e
di ( t )
= −1.58e
dt
−97.4 t
−2.57 t
−6 A
+1.58e−97.4t V
9-47
P9.10-4
Apply KCL to the supernode corresponding to
the dependent voltage source to get
ix ( t ) − 2ix ( t ) − 0.01
dv ( t ) vx ( t )
+
=0
2
dt
Apply KCL at node 1 to get
i ( t ) − 2ix ( t ) +
vx ( t )
=0
2
(Encircled numbers are node numbers.)
Apply KVL to the top-right mesh to get
vx ( t ) + v ( t ) − 0.1
Apply KVL to the outside loop to get ix ( t ) = −2 vx ( t ) − v ( t ) .
Eliminate ix ( t ) to get
Then eliminate vx ( t ) to get
Using the operator s =
di ( t )
=0
dt
dv ( t )
5
=0
vx ( t ) + v ( t ) − 0.01
2
dt
9
i ( t ) + vx ( t ) + 2 v ( t ) = 0
2
d i (t )
vx ( t ) = −v ( t ) + 0.01
dt
dv ( t )
di ( t )
=0
+ 0.25
dt
dt
di ( t )
−2.5 v ( t ) + i ( t ) + 0.45
=0
dt
−1.5 v ( t ) − 0.01
d
we have
dt
(−1.5 − .01s )v ( t ) + (.25s ) i ( t ) = 0
(−2.5)v( t ) + (1+.45s ) i ( y ) = 0
The characteristic equation is s 2 +13.33 s + 333.33 = 0 . The natural frequencies
are s1 , s2 = −6.67 ± j 17 . The natural response has the form vn (t ) = [ A cos17 t + B sin17 t ] e−6.67 t .
v(t ) = [ A cos17 t
( t ) = 0 . The complete response has the form
+ B sin 17 t ] e−6.67 t .
The forced response is v
f
9-48
The given initial conditions are i (0) = 0 and v (0) = 10 V. Then
v(0) =10 = A
and
dv(0)
=−111= − 6.67 A +17 B ⇒ B =−2.6
dt
Finally i (t ) = [3.27 sin 17 t ] e−6.67 t A .
(Checked using LNAP on 7/22/03)
P9.10-5
Assume that the circuit is at steady state before t = 0 so v(0) = 10 V and iL (0) =
The switch is open when 0 < t < 0.5 s
v(0) 10
=
A.
3
3
For this series RLC circuit we have:
α=
1
R
= 3 and ω 02 =
= 12
2L
LC
−α ±
α 2 −ω02 = s1,2 = −3 ± j 3
The natural response has the form vn (t ) = e −3t ( A cos1.73 t + B sin1.72 t ) . There is no source so
v f ( t ) =0 . The complete response has the form v (t ) = e −3t ( A cos1.73 t + B sin1.72 t ) .
v(0) = 10 = A
Next

 A= 10

dv(0) i ( 0 ) 10 3
 ⇒ 
=−
=−
= − 20 = −3 A +1.73 B 
 B =5.77
dt
C
16

v(t ) = e−3t (10 cos1.73 t + 5.77 sin1.73 t ) V
so
i(t ) = e−3t ( 3.33 cos1.73 t − 5.77 sin1.73 t ) A
1.73
1.73 

v (0.5) = e −1.5  10 cos
+ 5.77 sin
 = 0.2231× ( 6.4864 + 4.3915 ) = 2.43 V
2
2 

In particular,
and
1.73
1.73 

i (0.5) = e −1.5  3.33cos
−5.77 sin
 = 0.2231× ( 2.1600 − 4.3915 ) = −0.50 A
2
2 

9-49
The switch is closed when t > 0.5 s
Apply KCL at the top node:
v( t ) −30
1 dv( t )
+ iL ( t ) +
=0
6
6 dt
dv( t ) 
1
⇒ iL ( t ) = 5 −  v( t ) +

dt 
6
⇒
2
d iL ( t )
1  dv( t ) d v( t ) 
=− 
+

dt
dt 2 
6  dt
Apply KVL to the right mesh:
1 diL ( t )
v( t ) = 3 iL ( t ) +
2 dt
The circuit is represented by the differential equation
d 2v ( t )
dv ( t )
+7
+ 18 v ( t ) = 180
2
dt
dt
The characteristic equation is 0 = s 2 + 7 s + 18 . The natural frequencies are s1,2 = −3.7 ± j 2.4 .
The natural response has the form vn (t ) = e −3.5 t ( A cos 2.4 t + B sin 2.4 t ) . The forced response is
v
f
( t ) = 10
V . The complete response has the form v (t ) = e −3.5 t ( A cos 2.4 t + B sin 2.4 t ) + 10 .
Next
v (0.5) = e −3.5×0.5 ( A cos1.2 + B sin1.2 ) = 0.063 A + 0.162 B
dv ( 0.5 )
= e−3.5×0.5 ( −3.5 A + 2.4 B ) cos1.2 − ( 3.5B + 2.4 A ) sin1.2 
dt
= e−3.5×0.5 ( −3.5cos1.2 − 2.4sin1.2 ) A + e−3.5×0.5 ( 2.4 cos1.2 − 3.5sin1.2 ) B
= −0.6091A − 0.4158B
Using the initial conditions yields
2.43=v(0.5) = 0.063 A+ 0.162 B

A=−20.65

dv(0.5) i ( 0.5 ) −1 2
 ⇒
B = 23.03
=−
=−
=3=−0.6091A− 0.4158B 
dt
C
16

Finally
In summary
v(t ) = e −3.5 t ( −20.65 cos 2.4 t + 23.03 sin 2.4 t ) + 10
e −3t (10 cos1.73 t + 5.77 sin1.73 t ) V
0<t <0.5

v(t ) =  −3.5 t
e ( −20.65 cos 2.4 t + 23.03 sin 2.4 t )+10 V 0.5<t
9-50
Section 9-11: Roots in the Complex Plane
P9.11-1
After t = 0
i1 + i 2 +
2 ×10−3
d i1
dt
2 ×10−3
d i1
dt
2000
−6
=0
= 3000 i 2 + 2 ×10−3
Using the operator s =
d i2
dt
d
yields
dt
 2000 + 2 × 10−3 s
  i1   6 
2000
= 

−3
−3  
2 × 10 s
3000 + 2 × 10 s  i 2   0 

s 2 + 3.5 ×106 s + 1.5 ×1012 = 0
⇒ s1,2 = −5 × 105 , −3 ×106
P9.11-2
From P9.7-1
s2 +
1
1
1
1
= 0 ⇒ s2 +
=0
s+
s+
−6
RC
LC
( 250 ) ( 5 ×10 ) ( 0.8) ( 5 ×10−6 )
⇒ s 2 + 800 s + 250000 = 0
s1,2 = 400 ± j 300
P9.11-3
dv( t ) v( t )
1
KCL: i L ( t ) = × 10−6
+
dt
4
4000
di ( t )
KVL: vs ( t ) = 4 L + v( t )
dt
2
dv ( t )
dv( t ) v( t ) 
d 1
−6 d v ( t )
10
v ( t ) = 4  ×10−6
v
t
+
+
=
+ 10−3
+ v (t )
(
)

2
s
dt  4
dt 4000 
dt
dt
9-51
Characteristic equation: s 2 + 103 s + 106 = 0
Characteristic roots: s1,2 = −500 ± j 866
P9.11-4
Before t = 0 the voltage source voltage is 0 V so vb (0+) = vb (0−) = 0 V and
i (0+) = i (0−) = 0 A . Apply KCL at node a to get
va (0+ ) −36
v (0+ ) − vb (0+ )
− i (0+) + a
= 0 ⇒ va (0+) + 2 va (0+) = 36 ⇒ va (0) = 12 V
12
6
After t = 0 the node equations are:
−
va ( t ) − vs ( t ) 1 t
v ( t ) − va ( t )
+ ∫ ( vb (τ ) − va (τ ) ) dτ + b
=0
0
12
6
L
C
Using the operator s =
d vb ( t ) vb ( t ) − va ( t ) 1 t
+
+ ∫ ( vb (τ ) − va (τ ) ) dτ = 0
6
dt
L 0
d
we have
dt
v (t )
1 1
1
 1 1
+ +  va ( t ) +  − −  vb ( t ) = s

6
12
s
 12
 6 s
1 1
 1 1
1
 − −  va ( t ) +  s + +  vb ( t ) = 0
 6 s
 18 6 s 
Using Cramer’s rule
9-52
( s 2 +5s + 6) vb ( t ) = ( s + 6) vs ( t ) =( s + 6) ( 36 )
The characteristic equation is s 2 + 5s + 6 = 0 . The natural
frequencies are s1,2 = −2, −3 . The natural response has the
form vn (t ) = A1 e −2 t + A2 e −3 t . Try v f ( t ) = B as the forced
response. Substituting into the differential equation gives
B = 36 so v f ( t ) = 36 V. The complete response has the form
vb (t ) = A1 e −2 t + A2 e −3t + 36 .
vb (0+ ) = 36 + A1 + A2
Next
dvb +
(0 ) = −2 A1 − 3 A2
dt
Apply KCL at node a to get
At t = 0+
So
Finally
1 dvb ( t ) vb ( t ) − va ( t )
+
+ i (t ) = 0
18 dt
6
1
1 d vb ( 0
(−2 A1 − 3 A2 ) =
18
18 dt
+
) = v ( 0 )−v ( 0 ) − i
+
b
6
0 = vb (0+ ) =36+ A1 + A 2 

 ⇒
1
( −2 A 1 − 3 A 2 ) = 2 
18

vb ( t ) = 36 − 72e
+
a
−2 t
(0+ ) = 126−0 − 0 = 2
A1 = −72, A2 = 36
+ 36e −3t V for t ≥ 0
9-53
PSpice Problems
SP 9-1
Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure
to edit the labels of the parts so, for example, there is only one R1.)
9-54
V(C1:2), V(C2:2) and V(C3:2) are the capacitor voltages, listed from top to bottom.
9-55
SP 9-2
Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure
to edit the labels of the parts so, for example, there is only one R1.)
9-56
V(R2:2), V(R4:2) and V(R6:2) are the output voltages, listed from top to bottom.
9-57
SP 9-3
9-58
SP 9-4
9-59
Verification Problems
VP 9-1
This problem is similar to the verification example in this chapter. First, check the steady-state
inductor current
v ( t ) 25
i (t ) = s
=
= 250 mA
100 100
This agrees with the value of 250.035 mA shown on the plot. Next, the plot shows an
underdamped response. That requires
12 ×10−3 = L < 4 R 2C = 4(100) 2 (2 × 10−6 ) = 8 × 10−2
This inequality is satisfied, which also agrees with the plot.
The damped resonant frequency is given by
ω =
d
1
LC
 1 
−

 2 RC 
2


1
1
=
−
= 5.95 × 103
−6 
−6
−3
×
2(100)(2
10
)
( 2 ×10 )(12 ×10 ) 

2
The plot indicates a maxima at 550.6µs and a minima at 1078.7µs. The period of the damped
oscillation is
T d = 2 (1078.7 µ s − 550.6 µ s) = 1056.2 µ s
Finally, check that
5.95 ×103 = ω d =
2π
2π
=
= 5.949 × 103
−6
Td
1056.2 ×10
The value of ω d determined from the plot agrees with the value obtained from the circuit.
The plot is correct.
VP 9-2
This problem is similar to the verification example in this chapter. First, check the steady-state
inductor current.
v ( t ) 15
i (t ) = s
=
= 150 mA
100 100
9-60
This agrees with the value of 149.952 mA shown on the plot.
Next, the plot shows an underdamped response. This requires
8 ×10−3 = L < 4 R 2C = 4 (100)2 (0.2 ×10−6 ) = 8 × 10−3
This inequality is not satisfied. The values in the circuit would produce a critically damped, not
underdamped, response.
This plot is not correct.
Design Problems
DP 9-1
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
The specifications require that vC ( ∞ ) =
vC ( ∞ ) =
1
so
2
R2
1
=
2 R1 + R 2
R2
R1 + R 2
1
⇒ R1 = R 2
Next, represent the circuit by a 2nd order differential equation:
KCL at the top node of R2 gives:
KVL around the outside loop gives:
vC ( t )
R2
vs ( t ) = L
+C
d
vC ( t ) = iL ( t )
dt
d
iL ( t ) + R1 iL ( t ) + vC ( t )
dt
Use the substitution method to get
9-61
vs ( t ) = L

 v (t )

d  vC ( t )
d
d
+ C vC ( t )  + R1  C
+ C vC ( t )  + vC ( t )


 R2

dt  R 2
dt
dt



= LC
 L
d

R1 
d2
v
t
R
C
v
t
v t
+
+
+
+
1
(
)
(
)



1
C
 R2
 dt C
 R 2  C ( )
dt 2




The characteristic equation is
R

1+ 1

 1
R1 
R2
+ s+
s2 + 
 R 2 C L   LC

 




 = s 2 + 6 s + 8 = ( s + 2 )( s + 4 ) = 0



Equating coefficients of like powers of s:
1
R2 C
Using R1 = R 2 = R gives
+
R1
L
= 6 and
1
R
+ =6 ⇒
RC L
1+
R1
=8
R2
LC
1
=4
LC
These equations do not have a unique solution. Try C = 1 F. Then L =
1
H and
4
1
3
1
+ 4 R = 6 ⇒ R 2 − R + = 0 ⇒ R = 1.309 Ω or R = 0.191 Ω
R
2
4
Pick R = 1.309 Ω. Then
iL ( t ) =
At t = 0+
vC ( t )
1.309
vc ( t ) =
+
1
+ A1 e −2 t + A2 e−4 t V
2
d
vC ( t ) = −1.236 A1 e−2 t − 3.236 A2 e−4 t + 0.3819
dt
( )
0 = vc 0+ = A1 + A2 + 0.5
( ) = −1.236 A − 3.236 A
0 = iL 0
+
1
2
+ 0.3819
Solving these equations gives A1 = -1 and A2 = 0.5, so
vc ( t ) =
1 −2 t 1 −4 t
−e + e V
2
2
9-62
DP 9-2
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
The specifications require that vC ( ∞ ) =
vC ( ∞ ) =
1
so
4
R2
1
=
4 R1 + R 2
R2
R1 + R 2
1
⇒ 3 R 2 = R1
Next, represent the circuit by a 2nd order differential equation:
KCL at the top node of R2 gives:
KVL around the outside loop gives:
vC ( t )
R2
vs ( t ) = L
+C
d
vC ( t ) = iL ( t )
dt
d
iL ( t ) + R1 iL ( t ) + vC ( t )
dt
Use the substitution method to get
vs ( t ) = L

 v (t )

d  vC ( t )
d
d
+ C vC ( t )  + R1  C
+ C vC ( t )  + vC ( t )


 R2

dt  R 2
dt
dt



 L
d

R1 
d2
+ R1 C  vC ( t ) +  1 +
v t +
v t
2 C ( ) 
 R2
 dt
 R 2  C ( )
dt




The characteristic equation is
R 

1+ 1 

 1
R1 
R2  2
2
+ s+
= s + 4s + 4 = ( s + 2 ) = 0
s2 + 
 R 2 C L   LC 

 




Equating coefficients of like powers of s:
R
1+ 1
R1
R2
1
+
= 4 and
=4
R2 C L
LC
= LC
9-63
Using R 2 = R and R1 = 3R gives
1
3R
+
=4 ⇒
RC L
1
=1
LC
These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and
1
4
1
1
+ 3 R = 4 ⇒ R 2 − R + = 0 ⇒ R = 1 Ω or R = Ω
R
3
3
3
Pick R = 1 Ω. Then R1 = 3 Ω and R 2 = 1 Ω .
vc ( t ) =
iL ( t ) = vC ( t ) +
At t = 0+
1
+ ( A1 + A2 t ) e −2 t V
4
d
1
vC ( t ) = +
dt
4
( )
(( A
0 = vc 0+ = A1 +
( )
0 = iL 0+ =
2
)
− A1 ) − A2 t e −2 t
1
4
1
+ A2 − A1
4
Solving these equations gives A1 = -0.25 and A2 = -0.5, so
vc ( t ) =
1  1 1  −2 t
−  + t e V
4 4 2 
DP 9-3
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
R2
vC ( ∞ ) =
1
R1 + R 2
4
The specifications require that vC ( ∞ ) = so
5
9-64
R2
4
=
5 R1 + R 2
⇒ 4 R1 = R 2
Next, represent the circuit by a 2nd order differential equation:
vC ( t )
KCL at the top node of R2 gives:
R2
vs ( t ) = L
KVL around the outside loop gives:
+C
d
vC ( t ) = iL ( t )
dt
d
iL ( t ) + R1 iL ( t ) + vC ( t )
dt
Use the substitution method to get
vs ( t ) = L

 v (t )

d  vC ( t )
d
d
+ C vC ( t )  + R1  C
+ C vC ( t )  + vC ( t )


 R2

dt  R 2
dt
dt



= LC
 L
d

R1 
d2
v t +
v t
+ R1 C  vC ( t ) +  1 +
2 C ( ) 
 R2
 dt
 R 2  C ( )
dt




The characteristic equation is
R

1+ 1

 1
R1 
R2
+ s+
s2 + 
 R 2 C L   LC

 




 = s 2 + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) = 0



Equating coefficients of like powers of s:
1
R2 C
Using R1 = R and R 2 = 4 R gives
+
R1
L
= 4 and
1+
R1
R2
LC
= 20
1
R
+ = 4 and
4R C L
1
= 16
LC
1
1
These equations do not have a unique solution. Try C = F . Then L = H and
8
2
2
+ 2 R = 4 ⇒ R2 − 2R + 1 = 0 ⇒ R = 1 Ω
R
Then R1 = 1 Ω and R 2 = 4 Ω . Next
vc ( t ) = 0.8 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
9-65
iL ( t ) =
vC ( t )
4
+
A 2 −2 t
A1
1d
vC ( t ) = 0.2 +
e cos 4 t − e−2 t sin 4 t
8 dt
2
2
( )
+
At t = 0
0 = vc 0+ = 0.8 + A1
( )
0 = iL 0+ = 0.2 +
A2
2
Solving these equations gives A1 = −0.8 and A2 = −0.4, so
vc ( t ) = 0.8 − e −2 t ( 0.8cos 4 t + 0.4sin 4 t ) V
DP 9-4
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
The specifications require that vC ( ∞ ) =
vC ( ∞ ) =
1
so
2
R2
1
=
2 R1 + R 2
R2
R1 + R 2
1
⇒ R1 = R 2
Next, represent the circuit by a 2nd order differential equation:
KCL at the top node of R2 gives:
KVL around the outside loop gives:
vC ( t )
R2
vs ( t ) = L
+C
d
vC ( t ) = iL ( t )
dt
d
iL ( t ) + R1 iL ( t ) + vC ( t )
dt
Use the substitution method to get
9-66

 v (t )

d  vC ( t )
d
d
+ C vC ( t )  + R1  C
+ C vC ( t )  + vC ( t )


 R2

dt  R 2
dt
dt



vs ( t ) = L
= LC
 L
d

R1 
d2
1
v
t
R
C
v
t
v t
+
+
+
+
(
)
(
)



1
C
 R2
 dt C
 R 2  C ( )
dt 2




The characteristic equation is
R

1+ 1

 1
R1 
R2
+ s+
s2 + 
 R 2 C L   LC

 




 = s 2 + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) = 0



Equating coefficients of like powers of s:
1
R2 C
Using R1 = R 2 = R gives
Substituting L =
+
R1
L
= 4 and
1
R
+ = 4 and
RC L
1+
R1
R2
LC
= 20
1
= 10
LC
1
into the first equation gives
10 C
( RC )
2
0.4 ± 0.42 − 4 ( 0.1)
4
1
− ( RC ) + = 0 ⇒ RC =
10
10
2
Since RC cannot have a complex value, the specification cannot be satisfied.
DP 9-5
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
9-67
vC ( ∞ ) =
R2
R1 + R 2
1, iL ( ∞ ) =
The specifications require that vo ( ∞ ) =
1
so
2
R2
1
R1 + R 2
1
=
2 R1 + R 2
and vo ( ∞ ) =
R2
R1 + R 2
1
⇒ R1 = R 2
Next, represent the circuit by a 2nd order differential equation:
KVL around the right-hand mesh gives:
KCL at the top node of the capacitor gives:
vC ( t ) = L
d
iL ( t ) + R 2 iL ( t )
dt
vs ( t ) − vC ( t )
d
− C vC ( t ) = iL ( t )
R1
dt
Use the substitution method to get
vs ( t ) = R1 C
Using iL ( t ) =
= R1 LC
vo ( t )
gives
R2
vs ( t ) =
d  d
  d

 L iL ( t ) + R 2 iL ( t )  +  L iL ( t ) + R 2 iL ( t )  + R1 iL ( t )
dt  dt
  dt

d2
d
i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t )
2 L( ) (
dt
dt
R1
R2
LC
The characteristic equation is
 L
d
 R1 + R 2 
d2
+ R1 C  vo ( t ) + 
v t +
v t
2 o( ) 
 R2
 dt
 R 2  o ( )
dt




R

1+ 2

 1
R2  
R1
+
s2 + 
s+

 R1 C L   LC

 


Equating coefficients of like powers of s:
Using R1 = R 2 = R gives


 = s 2 + 6 s + 8 = ( s + 2 )( s + 4 ) = 0



R2
1
+
= 6 and
R1 C L
1
R
+ =6 ⇒
RC L
1+
R2
R1
LC
=8
1
=4
LC
These equations do not have a unique solution. Try C = 1 F. Then L =
1
H and
4
9-68
1
3
1
+ 4 R = 6 ⇒ R 2 − R + = 0 ⇒ R = 1.309 Ω or R = 0.191 Ω
R
2
4
Pick R = 1.309 Ω. Then
vo ( t ) =
1
+ A1 e −2 t + A2 e −4 t V
2
A1 −2 t
A2 −4 t
v (t )
1
iL ( t ) = o
e +
e V
=
+
1.309 2.618 1.309
1.309
vC ( t ) = 1.309 iL ( t ) +
1 d
1
iL ( t ) = + 0.6167 A1 e−2 t + 0.2361 A2 e−4 t
4 dt
2
( )
At t = 0+
0 = iL 0+ =
( )
0 = vC 0+ =
A1
A2
1
+
+
2.618 1.309 1.309
1
+ 0.6167 A1 + 0.2361 A2
2
Solving these equations gives A1 = -1 and A2 = 0.5, so
vo ( t ) =
1 −2 t 1 −4 t
−e + e V
2
2
DP 9-6
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
vC ( ∞ ) =
R2
R1 + R 2
1, iL ( ∞ ) =
The specifications require that vo ( ∞ ) =
1
R1 + R 2
and
vo ( ∞ ) =
R2
R1 + R 2
1
3
so
4
R2
3
=
4 R1 + R 2
⇒ 3R1 = R 2
9-69
Next, represent the circuit by a 2nd order differential equation:
vC ( t ) = L
KVL around the right-hand mesh gives:
KCL at the top node of the capacitor gives:
d
iL ( t ) + R 2 iL ( t )
dt
vs ( t ) − vC ( t )
d
− C vC ( t ) = iL ( t )
R1
dt
Use the substitution method to get
vs ( t ) = R1 C
Using iL ( t ) =
= R1 LC
vo ( t )
gives
R2
vs ( t ) =
d  d
  d

 L iL ( t ) + R 2 iL ( t )  +  L iL ( t ) + R 2 iL ( t )  + R1 iL ( t )
dt  dt
  dt

d2
d
i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t )
2 L( ) (
dt
dt
R1
R2
LC
 L
d
 R1 + R 2 
d2
+
+
+
v
t
R
C
v
t
(
)
(
)



 vo ( t )
o
o
1
 R2
 dt
dt 2


 R2 
The characteristic equation is
R

1+ 2

 1
R2  
R1
+
s2 + 
s+

 R1 C L   LC

 


Equating coefficients of like powers of s:
Using R1 = R and R 2 = 3R gives


 = s 2 + 4s + 4 = ( s + 2 )2 = 0



R2
1
+
= 4 and
R1 C L
1
3R
+
= 4 and
RC L
1+
R2
R1
LC
=4
1
=1
LC
These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and
1
4
1
1
+ 3 R = 4 ⇒ R 2 − R + = 0 ⇒ R = 1 Ω or R = Ω
R
3
3
3
Pick R = 1 Ω. Then R1 = 1 Ω and R 2 = 3 Ω .
vo ( t ) =
3
+ ( A1 + A2 t ) e−2 t V
4
9-70
iL ( t ) =
vo ( t )
vC ( t ) = 3 iL ( t ) +
At t = 0+
=
3
1  A1 A2  −2 t
t e V
+ +
4  3
3 
3   A1 A2  A2  −2 t
d
iL ( t ) = +   +
t e
+
4  3
3  3 
dt
0 = iL ( 0 + ) =
A1
+
1
4
3
3 A1 A2
0 = vC ( 0 + ) = + +
4 3
3
Solving these equations gives A1 = −0.75 and A2 = −1.5, so
vo ( t ) =
3  3 3  −2 t
−  + t e V
4 4 2 
DP 9-7
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
vC ( ∞ ) =
R2
R1 + R 2
1, iL ( ∞ ) =
1
The specifications require that vo ( ∞ ) = so
5
R2
1
=
5 R1 + R 2
1
R1 + R 2
and
vo ( ∞ ) =
R2
R1 + R 2
1
⇒ R1 = 4 R 2
vC ( t ) = L
d
iL ( t ) + R 2 iL ( t )
dt
vs ( t ) − vC ( t )
d
− C vC ( t ) = iL ( t )
R1
dt
Next, represent the circuit by a 2nd order differential equation:
KVL around the right-hand mesh gives:
KCL at the top node of the capacitor gives:
9-71
Use the substitution method to get
vs ( t ) = R1 C
Using iL ( t ) =
= R1 LC
vo ( t )
gives
R2
vs ( t ) =
d  d
  d

 L iL ( t ) + R 2 iL ( t )  +  L iL ( t ) + R 2 iL ( t )  + R1 iL ( t )
dt  dt
  dt

d2
d
i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t )
2 L( ) (
dt
dt
R1
R2
LC
 L
d
 R1 + R 2 
d2
+ R1 C  vo ( t ) + 
v t +
v t
2 o( ) 
 R2
 dt
 R 2  o ( )
dt




The characteristic equation is
R

1+ 2

 1
R2  
R1
+
s2 + 
s+

 R1 C L   LC

 




 = s 2 + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) = 0



Equating coefficients of like powers of s:
R2
1
+
= 4 and
R1 C L
Using R 2 = R and R1 = 4 R gives
1+
1
R
+ = 4 and
4R C L
These equations do not have a unique solution. Try C =
R2
R1
LC
= 20
1
= 16
LC
1
1
F . Then L = H and
8
2
2
+ 2 R = 4 ⇒ R2 − 2R + 2 = 0 ⇒ R = 1 Ω
R
Then R1 = 4 Ω and R 2 = 1 Ω . Next
vo ( t ) = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
iL ( t ) =
vo ( t )
vC ( t ) = iL ( t ) +
At t = 0+
1
= 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
1 d
iL ( t ) = 0.2 + 2 A2 e−2 t cos 4 t − 2 A1 e−2 t sin 4 t
2 dt
9-72
( )
( 0 ) = 0.2 + 2 A
0 = iL 0+ = 0.2 + A1
0 = vC
+
2
Solving these equations gives A1 = −0.8 and A2 = −0.4, so
vc ( t ) = 0.2 − e −2 t ( 0.2 cos 4 t + 0.1sin 4 t ) V
DP 9-8
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
vC ( ∞ ) =
R2
R1 + R 2
1, iL ( ∞ ) =
The specifications require that vC ( ∞ ) =
1
R1 + R 2
and
vo ( ∞ ) =
R2
R1 + R 2
1
1
so
2
R2
1
=
⇒ R1 = R 2
2 R1 + R 2
Next, represent the circuit by a 2nd order differential equation:
KVL around the right-hand mesh gives:
KCL at the top node of the capacitor gives:
vC ( t ) = L
d
iL ( t ) + R 2 iL ( t )
dt
vs ( t ) − vC ( t )
d
− C vC ( t ) = iL ( t )
R1
dt
Use the substitution method to get
vs ( t ) = R1 C
Using iL ( t ) =
= R1 LC
vo ( t )
gives
R2
d  d
  d

 L iL ( t ) + R 2 iL ( t )  +  L iL ( t ) + R 2 iL ( t )  + R1 iL ( t )
dt  dt
  dt

d2
d
i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t )
2 L( ) (
dt
dt
9-73
vs ( t ) =
R1
R2
LC
The characteristic equation is
 L
d
 R1 + R 2 
d2
v t +
v t
+ R1 C  vo ( t ) + 
2 o( ) 
 R2
 dt
 R 2  o ( )
dt




R

1+ 2

 1
R2  
R1
+
s2 + 
s+

 R1 C L   LC

 




 = s 2 + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) = 0



Equating coefficients of like powers of s:
Using R1 = R 2 = R gives
Substituting L =
R2
1
+
= 4 and
R1 C L
1
R
+ =4 ⇒
RC L
1+
R2
R1
LC
= 20
1
= 10
LC
1
into the first equation gives
10 C
( RC )
2
0.4 ± 0.42 − 4 ( 0.1)
4
1
− ( RC ) + = 0 ⇒ RC =
10
10
2
Since RC cannot have a complex value, the specification cannot be satisfied.
9-74
DP 9-9
Let’s simulate the three copies of the circuit simultaneously. Each copy uses a different value of
the inductance.
The PSpice transient response shows that when L = 1 H the inductor current has its maximum at
approximately t=0.5 s.
Consequently, we choose L = 1 H.
9-75
Chapter 10 – Sinusoidal Steady-State Analysis
Exercises
Ex. 10.3-1
(a)
(b)
Ex. 10.3-2
T = 2π / ω = 2π / 4
v leads i by 30 − (−70) = 100°
v ( t ) = 3cos 4 t + 4sin 4 t =
( 3 )) = 5 cos(4 t −53° )
(
(3) 2 + (4) 2 cos 4 t − tan −1 4
Ex. 10.3-3


 12   
i ( t ) = −5 cos 5t + 12 sin 5t = (−5) 2 + (12) 2 cos  5 t − 180 + tan −1     = 13 cos (5 t −112.6°)
 −5   


Ex. 10.4-1
KCL: is ( t ) =
v( t )
d
I
= m cos ω t
v (t ) +
dt
RC
C
v( t )
d
+ C v (t ) ⇒
R
dt
Try v f ( t ) = A cos ω t + B sin ω t & plug into above differential equation to get
−ω A sin ω t + ω B cos ω t +
Equating sin ω t & cos ω t terms yields
1
I
( A cos ω t + B sin ω t ) = m cos ω t
RC
C
R Im
ω R2 C Im
and B =
A =
1+ω 2 R 2 C 2
1+ω 2 R 2 C 2
Therefore
v f (t ) =
ω R2 C Im
R Im
cos
sin ω t =
+
ω
t
1+ω 2 R 2 C 2
1+ω 2 R 2 C 2
R Im
1+ ω R C
2
2
2
cos ω t − tan −1 (ω RC ) 
10-1
Ex. 10.4-2
KVL : − 10 + j 3 I + 2 I = 0 ⇒ I =
Therefore
Ex. 10.5-1
i (t ) =
10
=
2+ j 3
10∠0D
10
=
∠ − 56.3D A
D
13 ∠56.3
13
10
cos (3 t − 56.3D ) A
13
10
= 4.24 e− j 45 = 3− j 3
2.36 e j 45
Ex. 10.5-2
32e j 90
32 j (90-111)
j 32
=
=
e
= 3.75 e − j 21
−3+ j 8 8.54 e j111 8.54
Ex. 10.6-1
(a)
i = 4 cos(ω t − 80D ) = Re{4 e j ω t e − j 80° } ⇒ I = 4 e − j 80° = 4∠ − 80° A
(b)
i = 10 cos(ω t + 20° ) = Re{10 e j ω t e j 20° } ⇒ I = 10e j 20° = 10∠20°
(c)
i = 8sin (ω t − 20D ) = 8cos (ω t − 110D ) = 8 Re{e j ω t e − j110° } ⇒ I = 8e − j110° = 8∠ − 110° A
Ex. 10.6-2
(a) V = 10∠ − 140° = 10 e − j140° V ⇒ v(t ) = Re{10 e − j140° e j ω t } = 10 cos (ω t − 140°) V
(b) V = 80 + j 75 = 109.7∠43.2° = 109.7 e j 43.2° ⇒
v (t ) = Re{109.7 e j 43.2° e jω t } = 109.7 cos(ω t + 43.2°) V
10-2
Ex. 10.6-3
d
v + v = 10 cos 100 t
dt
( 0.01)( j 100 )V + V =10
0.01
V=
10
=7.071 ∠− 45°
1+ j
v = 7.071 cos 100 t V
Ex. 10.6-4
{
vs = 40 cos100t = Re 4 e j100 t
KVL:
i (t ) + 10 × 10 −3
1
di (t )
+
5×10 −3
dt
∫
t
−∞
}
i (t ) dt = vS
Assume i (t ) = Ae j100 t where A is complex number to be determined. Plugging into
the differential equation yields
Ae j100 t + j Ae j100 t + (− j 2 A)e j100 t = 4 e j100 t
In the time domain:
{
}
{
⇒
}
A=
4
= 2 2 e j 45°
1− j
i (t ) = Re 2 2 e j100 t e j 45° = Re 2 2 e j (100 t -45°) = 2 2 cos (100 t + 45° ) A
Ex. 10.7-1
(a) v = R i = 10 (5 cos100 t ) = 50 cos 100 t
(b)
di
v = L = 0.01[5(−100) sin100 t ] = −5sin100 t = 5cos (100 t + 90° ) V
dt
(c)
1
v = ∫ i dt = 103 ∫ 5cos100 t dt = 50sin100 t = 50 cos (100 t − 90°) V
C
Ex. 10.7-2
i=C
dv
= 10 × 10−6 [100(−500) sin (500 t + 30°)]
dt
= − 0.5sin (500 t +30°) = 0.5sin (500 t + 210°) = 0.5cos(500t +120°) A
10-3
Ex. 10.7-3
From Figure E10.7-3 we get
i (t ) = I m sin ω t = I m cos(ω t − 90°) A ⇔ I = I m ∠ − 90°A
v(t ) = Vm cos ω t ⇔ V = Vm ∠0° V
The voltage leads the current by 90° so the element is an inductor:
Z eq =
Also
Ex. 10.8-1
V ∠0°
V
V
= m
= m ∠90° Ω
I m ∠−90°
Im
I
Z eq = j ω L = ω L ∠90° ⇒ ω L =
ZR = 8 Ω, ZC =
=
1
Vm
Im
⇒ L =
Vm
ω Im
2.4 j 2.4
=
= − j 2.4 Ω, ZL1 = j 5 (2) = j 10 Ω,
j
j× j
1
12
ZL2 = j 5 (4) = j 20 Ω and VS = 5 ∠-90° V.
j5
Ex. 10.8-2
ZR = 8 Ω, ZC =
1
=
j4
4
=
= − j 4 Ω, ZL1 = j 3 (2) = j 6 Ω,
j j× j
1
12
ZL2 = j 3 (4) = j 12 Ω and IS = 4 ∠15° V.
j3
10-4
Ex 10.9-1
V1 (ω ) =
j10
5 e − j 90 = 3.9 e − j 51
8 + j10
V 2 (ω ) =
j 20
5 e − j 90 = 5.68 e − j 90
j 20 − j 2.4
V (ω ) = V1 (ω ) − V 2 (ω ) = 3.9 e − j 51 − 5.68 e − j 90
= 3.58 e j 47
Ex 10.9-2
V1 (ω ) =
V 2 (ω ) =
8 ( j6)
4 e j15 = 19.2 e j 68
8 + j6
j12 ( − j 4 )
4 e j15 = 24 e − j 75
j12 − j 4
V (ω ) = V1 (ω ) + V 2 (ω ) = 14.4 e − j 22
10-5
Ex. 10.10-1
KCL at Va:
Va
V −V
+ a b =1
4− j 2 − j10
(4 − j12) Va + (−4 + j 2) Vb = −20 − j 40
KCL at Vb:
Vb − Va
V
+ b + 0.5∠ − 90° = 0 ⇒ (−2 − j 4) Va + (2 − j 6) Vb = 10 + j 20
− j10 2+ j 4
(−20− j 40) (−4+ j 2)
(10+ j 20)
(2 − j 6)
−200+ j100
=
= 5∠296.5° V
V =
a
(4 − j12) (−4+ j 2)
−80− j 60
(−2− j 4) (2- j 6)
Cramer’s rule yields:
Therefore
v (t ) = 5 cos (100 t + 296.5° ) = 5 cos (100 t − 63.5° ) V
a
Ex. 10.10-2
The mesh equations are:
j15 I1 + 10 (I1 − I 2 ) = 20 ⇒ (10 + j15) I1 − 10 I 2 = 20
− j 5 I 2 +10(I 2 − I1 ) = −30∠−90° ⇒ −10I1 + (10− j 5) I 2 = j 30
20
−10
j 30 10− j 5
200+ j 200
=
= 2.263∠ − 8.1° A
I1 =
10 + j15 −10
75+ j100
10 − j 5
−10
Cramer’s rule yields:
Next
Therefore
VL = ( j15) I1 = (15∠90°) (2.263∠−8.1°) = 24 2∠82° V
vL (t ) = 24 2 cos (ω t + 82°) V
10-6
Ex. 10.10-3
The mesh equations are:
(10+ j 50) I1 −10 I 2 = j 30
−10 I1 + (10− j 20) I 2 + j 20 I 3 = j 50
j 20 I 2 + (30− j10) I 3 = 0
Solving the mesh equations gives:
I1 = − 0.87 − j 0.09 A,
Then
I 2 = −1.32+ j 1.27 A, I 3 = 0.5+ j 1.05 A
Va = 10 (I1 − I 2 ) = 14.3∠ − 72° V and Vb = Va + j 50 = 36.6 ∠83° V
Ex 10.11-1
V1 =
V2 =
j10
5 e − j 90 = 3.9 e− j 51
8 + j10
j 20
5 e − j 90 = 5.68 e− j 90
j 20 − j 2.4
Vt = V1 − V 2 = 3.9 e − j 51 − 5.68 e− j 90
= 3.58 e j 47
10-7
Zt =
8 ( j10 ) − j 2.4 ( j 20 )
+
= 4.9 + j 1.2
8 + j10 − j 2.4 + j 20
Ex 10.11-2
V1 (ω ) =
V 2 (ω ) =
j10
5 e − j 90 = 3.9 e− j 51
8 + j10
j 20
5 e − j 90 = 5.68 e − j 90
j 20 − j 2.4
V (ω ) = V1 (ω ) − V 2 (ω ) = 3.9 e − j 51 − 5.68 e − j 90
= 3.58 e j 47
V1 (ω ) =
V 2 (ω ) =
8 ( j6)
4 e j15 = 19.2 e j 68
8 + j6
j12 ( − j 4 )
4 e j15 = 24 e − j 75
j12 − j 4
V (ω ) = V1 (ω ) + V 2 (ω ) = 14.4 e − j 22
Using superposition: v(t) = 3.58 cos ( 5t + 47° ) + 14.4 cos ( 3t - 22° )
10-8
Ex. 10.11-3
Use superposition. First, find the response to the voltage source acting alone:
Z eq = − j
10⋅10
= 5(1 − j ) Ω
10− j10
Replacing the parallel elements by the equivalent impedance. The write a mesh equation :
−10 + 5 I1 + j15 I1 + 5(1 − j) I1 = 0 ⇒ I1 =
10
= 0.707∠ − 45° A
10+ j10
i1 (t ) = 0.707 cos(10 t − 45° ) A
Therefore:
Next, find the response to the dc current source acting alone:
Current division:
Using superposition:
Ex. 10.12-1
ω2 =
I2 = −
10
× 3 = −2 A
15
i (t ) = 0.707 cos(10 t − 45°) − 2 A
1
1
=
= 106
−3
−3
LC (1×10 )(1×10 )
⇒
ω = 1000 rad sec
Ex. 10.12-2
Diagram drawn with relative magnitudes
arbitrarily chosen:
10-9
Ex. 10.12-3
Two possible phasor diagrams for
currents:
In both cases:
I CL = I LC =
( 25 )−(15 )
2
2
= 20 A
I LC = I L − I C ⇒ I C = 6 − 20 = −14 A
In the first case:
That isn’t possible. Turning to the second case:
I CL = I C − I L ⇒ I C = 20 + 6 = 26 A
Ex. 10.14-1
Z1 =
1
1
R1 X 1 ( X 1 − jR1 )
and R1 = 1 kΩ , X 1 =
=
= 1 kΩ
2
2
ω C1 (1000 )(10−6 )
R1 + X 1
Z1 =
(1)(1)(1− j1) 1
1
= − j kΩ and Z 2 = R 2 = 1 kΩ
1+1
2
2
Vo
Z
−1
=− 2 =
= −1− j
1
1
Vs
Z1
−j
2
2
10-10
Problems
Section 10-3: Sinusoidal Sources
P10.3-1
(a)
i (t ) = 2 cos(6t + 120° ) + 4 sin(6t − 60° )
= 2 (cos 6t cos120° −sin 6t sin120° ) + 4 (sin 6t cos 60° − cos 6t sin 60° )
= 2.46 cos 6t + 0.27 sin 6t = 2.47 cos(6t − 6.26° )
(b)
v(t ) = 5 2 cos8t + 10 sin(8t + 45° )
= 5 2 cos8t +10[sin 8t cos 45° + cos8t sin 45° ]
= 10 2 cos8t + 5 2 sin 8t
v(t ) =
P10.3-2
250 cos(8t − 26.56° ) = 5 10 sin(8t + 63.4° ) V
2π
2π
=
= 6283 rad sec
T 1×10−3
v(t ) = Vm sin(ω t + φ ) = 100 sin(6283 t + φ )
ω = 2π f =
v(0) = 10 = 100 sin φ ⇒ φ = sin −1 (0.1) = 6°
v(t ) = 100 sin(6283 t + 6°) V
P10.3-3
ω 1200π
=
= 600 Hz
2π
2π
i (2 × 10−3 ) = 300 cos(1200 π (2 × 10−3 ) + 55°) = 3cos(2.4π + 55°)
f =
 180° 
−3
2.4π × 
 = 432° ⇒ i (2 ×10 ) = 300 cos(432°+55°) = 300 cos(127°) = −180.5 mA
 π 
P10.3-4
10-11
P10.3-5
A = 18 V
T = 18 − 2 = 16 ms
ω=
2π
2π
=
= 393 rad/s
T
0.016
16 = 18 cos (θ ) ⇒ θ = 27°
v ( t ) = 18 cos ( 393 t + 27° ) V
P10.3-6
A = 15 V
T = 43 − 21 = 32 ms
ω=
2π
2π
=
= 196 rad/s
T
0.032
8 = 15 cos (θ ) ⇒ θ = 58°
v ( t ) = 15 cos (196 t + 58° ) V
10-12
Section 10-4: Steady-State Response of an RL Circuit for a Sinusoidal Forcing Function
P10.4-1
L
di
+ R i = − vs
dt
Try i f = A cos 300 t + B sin 300 t then
equating coefficients gives
⇒
di f
dt
di
+ 120 i = −400 cos 300 t
dt
= −300 A sin 300 t + 300 B cos 300 t . Substituting and
−300 A+120 B = 0  A = − 0.46
 B = −1.15
300 B +120 A =−400 
i (t ) = −0.46 cos 300 t − 1.15sin 300 t = 1.24 cos (300 t − 68°) A
Then
P10.4-2
−is +
v
dv
dv
+C
=0 ⇒
+ 500 v = 500 cos1000 t
dt
dt
2
dv f
Try v f = A cos1000 t + B sin1000 t then
= −1000 A cos1000 t + 1000 B cos1000 t .
dt
Substituting and equating coefficients gives
−1000 A+ 500 B =0 
 ⇒
1000 B +500 A=500 
A = 0.2
B = 0.4
v (t ) = 0.2 cos1000 t + 0.4 sin1000 t = 0.447 cos (1000t − 63°) V
Then
P10.4-3
( j 4) (.05) = j (0.2)
I (ω ) =
12 e j 45° ~ 12 e j 45°
j 45°
= (2 ⋅10−3 ) e
⇒ i (t ) = 2 cos (4 t + 45°) mA
6000 + j (0.2)
6000
10-13
Section 10.5: Complex Exponential Forcing Function
(5∠36.9° ) (10∠−53.1° ) 50∠−16.2°
10∠−16.2°
=
=
= 2 5∠10.36°
°
(4 + j3)(6− j8)
10− j5
5∠− 26.56
P10.5-1

3 2∠− 45° 
3
5∠ + 81.87°  4− j 3+
 = 5∠ + 81.87°[4 − j 3 + ∠ − 36.87°]
5
5 2∠−8.13° 

= 5∠+81.87° (4.48− j 3.36) = 5∠+81.87° (5.6∠−36.87°)
P10.5-2
= 28∠+ 45°= 14 2 + j14 2
P10.5-3
A*C* (3− j 7) 5e − j 2.3
=
= 0.65 − j 6.31
B
6 e j15
P10.5-4
(6∠120° ) (−4 + j 3 + 2e j15 ) = −12.1 − j 21.3 ⇒ a =−12.1 and b =−21.3
P10.5-5
 3− b 
(a)
j tan −1 

2
2
j120
 −4 
Ae
= −4 + j (3 − b) = 4 + (3−b) e
 3−b 
°
120 = tan −1 
 ⇒ b = 3 + 4 + tan (120 ) = −3.93
 −4 
A = 42 + (3−b) 2 =
(b)
(c)
42 + (3− (−3.93)) 2 = 8.00
−4 + 8 cos θ + j (b + 8 sin θ ) = 3e − j120 = − 1.5 − j 2.6
2.5
−4+8 cos θ = −1.5 ⇒ θ = cos −1
= 72°
8
b + 8 sin (72° ) = − 26 ⇒ b = −10.2
−10 + j 2a = Ae j 60 = A cos 60° − j A sin 60
A=
−10
−20sin 60°
= −20 and a =
= −8.66
cos 60°
2
10-14
d 

5  0.1 v  + v = cos 2 t ⇒
 dt 
P 10.5-6
d
v + 2 v = 2 cos 2 t
dt
Replace the real excitation by a complex exponential excitation to get
Let ve = A e j 2t so
d
v + 2 v = 2 e j 2t
dt
d
ve = j 2 A e j 2t and
dt
d
ve + 2 ve = 2 e j 2t
dt
⇒
( j 2 + 2 ) A e j 2t = 2 e j 2 t
so
Finally v ( t ) = Re ( ve ) =
P 10.5-7
j 2 A e j 2 t + 2 A e j 2t = 2 e j 2 t
⇒
A=
2
1
=
∠ − 45°
2 + j2
2
1 j ( 2t −45°)
 1 − j 45°  j 2t
ve = 
e
e
e =
2
 2

1
cos ( 2t − 45° ) V
2
d
d2
0.45 v + v + 0.15 2 v = 4 cos 5 t ⇒
dt
dt
20
80
d2
d
cos 5 t
v+3 v+
v=
2
3
3
dt
dt
Replace the real excitation by a complex exponential excitation to get
d2
dt
2
v+3
d
20
80 j 5t
v+
v=
e
dt
3
3
d
d2
ve = j 5 A e j 5 t , and 2 ve = −25 A e j 5 t
dt
dt
2
d
d
20
80 j 5t
20
80 j 5t
v+3 v+
v=
e
A e j 5t =
e
⇒ − 25 A e j 5t + 3 j 5 A e j 5t +
2
dt
3
3
3
3
dt
80
20 
80 j 5t
80

j 5t
3
⇒ A=
=
= 1.126∠ − 141
e
 −25 + j15 +  A e =
20
3 
3
55
45
−
+
j

−25 + j15 +
3
Let ve = A e j 5 t so
(
(
)
(
)
)
j 5t −141° )
ve = 1.126 e− j141° e j 5t = 1.126 e (
Finally v ( t ) = Re ( ve ) = 1.126 cos ( 2t − 141° ) V
so
10-15
Section 10-6: The Phasor Concept
P10.6-1
Apply KVL
6i+2
or
2
d
i − 15 cos 4 t = 0
dt
d
i + 6 i = 15 cos 4t
dt
j 4 t +θ )
Now use i = I m Re{e (
} and 15 cos 4 t = 15 Re{e 4 t } to write
2
(
) (
)
d
j 4 t +θ
j 4 t +θ
I m Re{e ( ) } + 6 I m Re{e ( ) } = 15 Re{e 4 t }
dt
 d

Re 2 ( I m e j 4 t e jθ ) + 6 ( I m e j 4 t e jθ )  = Re{15 e 4 t }
 dt

{
}
Re 2 ( j 4 I m e j 4 t e jθ ) + 6 ( I m e j 4 t e jθ ) = Re{15 e4 t }
j8 ( I m e jθ ) + 6 ( I m e jθ ) = 15
I m e jθ =
15
15
=
= 1.5∠ − 53°
6 + j8 10∠53°
i ( t ) = 1.5 cos ( 4 t − 53° ) A
Finally
v (t ) = 2
d
d
i ( t ) = 2 (1.5 cos ( 4 t − 53° ) ) = 3 ( −4sin ( 4 t − 53° ) )
dt
dt
= −12 ( cos ( 4 t − 143° ) )
= 12 cos ( 4 t + 37° ) V
10-16
P10.6-2
v − 4 cos 2 t
d
+ 0.25 v + i = 0
1
dt
Apply KCL at node a:
Apply KVL to the right mesh:
4i + 4
After some algebra:
d
d
i − v = 0 ⇒ v = 4 i + 4 iL
dt
dt
d2
d
i + 5 i + 5 i = 4 cos 2t
2
dt
dt
j 2 t +θ )
} and 4 cos 2 t = 4 Re{e 2 t } to write
Now use i = I m Re{e (
d2 
d
j 2 t +θ
j 2 t +θ
j 2 t +θ
I Re{e ( ) } + 5  I m Re{e ( ) } + 5  I m Re{e ( ) } = 4 Re{e 2 t }
2  m
dt
dt
 d2

d
j 2 t +θ
j 2 t +θ
j 2 t +θ
Re  2  I m e ( )  + 5  I m e ( )  + 5  I m e ( )   = Re{4 e 2 t }
dt
 dt

{
}
Re −4 e jθ I m e j 2 t + 5 ( j 2 e jθ I m e j 2 t ) + 5 e jθ I m e j 2 t = Re{4 e2t }
I m e jθ =
−4 e j θ I m + 5 ( j 2 e j θ I m ) + 5 e j θ I m = 4
4
4
4
=
=
= 0.398∠ − 84°
−4 + 5 ( j 2 ) + 5 1 + j 10 10.05∠84
i ( t ) = 0.398 cos ( 2 t − 85° ) A
10-17
P10.6-3
VS = 2∠ − 90° V
Z R = R; Z C =
−j
−j
= − j 16000 Ω
=
ω C (500)(0.125×10−6 )
 − j 16000 
(16000∠−90° )( 2∠−90° ) = 1.25∠ − 141° V
V (ω ) = 
 ( 2∠−90° ) =
25612∠−39°
 20000 − j 16000 
therefore
v(t) = 1.25 cos (500t −141° ) V
Section 10-7: Phasor Relationships for R, L, and C Elements
P10.7-1
P10.7-2
10-18
P10.7.3
P10.7-4
10-19
P10.7-5
(a)
(b)
v = 15cos (400 t + 30°) V
i = 3 sin(400 t+30°) = 3 cos (400 t − 60°) V
v leads i by 90° ⇒ element is an inductor
v
15
Z L = peak =
= 5 = ω L = 400 L ⇒ L = 0.0125 H = 12.5 mH
3
ipeak
i leads v by 90° ⇒ the element is a capacitor
8
1
1
=4=
=
⇒ C = 277.77 µ F
ω C 900 C
ipeak 2
v = 20 cos (250 t + 60°) V
Zc =
(c)
P10.7-6
vpeak
=
i = 5sin (250 t +150°) =5cos (250 t + 60°) A
Since v & i are in phase ⇒ element is a resistor
v
20
∴ R = peak =
=4Ω
5
ipeak
V1 = 150 cos(−30°) + j150sin(−30°) = 130 − j 75 V
V2 = 200 cos 60°+ j 200sin 60° = 100+ j173 V
V = V1 + V2 = 230+ j 98 = 250∠23.1° V
Thus v(t ) = v1 (t ) + v2 (t ) = 250 cos (377 t + 23.1°) V
10-20
Section 10-8: Impedance and Admittance
ω = 2π f = 2π (10 ×103 ) = 62830 rad sec
P10.8-1
Z R = R = 36 Ω ⇔ YR =
1
1
= 0.0278 S
=
Z R 36
Z L = jω L = j (62830)(160×10−6 ) = j10 Ω ⇔ YL =
ZC =
1
= − 0.1 j S
ZL
−j
−j
1
=
= − j 16 Ω ⇔ YC =
= 0.0625 j S
−6
ZC
ω C (62830)(1×10 )
Yeq = YR + YL + YC = 0.0278 − j0.00375 = 0.027 ∠9° S
Z eq =
P10.8-2
Z=
1
= 36.5∠ 9° = 36 − j5.86 Ω
Yeq
V
−10 ∠40°
=
= − 5000∠ − 155°Ω = 4532 + 2113 j = R + j ω L
−I 2×10−3 ∠195°
so R =4532 Ω and L =
P10.8-3
ω
2113
=
2113
= 1.06 m H
2×106
j
L
R
( R + jω L)
−j
C ωC
=
Z(ω ) = ω C
j
1 
−
+ ( R + jω L) R + j  ω L −
ωC
ω C 

−
1 
R 
L

 −j
 R − j  ω L −
ω C  
 C ω C 

=
2
1 

R 2 + ω L −
ω C 

=
Z(ω ) will be purely resistive when
1   R2 L 
1 
RL R 
−
+ ω L−
ω L−
− j 
ωC   ωC C 
ω C  
C ωC 
1 

R 2 + ω L −
ω C 

2
10-21
R2 L 
1 
1 R
2
+ ω L−
− 
= 0 ⇒ ω =
CL  L
ωC C
ωC
2
when R =6 Ω , C = 22 µ F, and L = 27 mH, then ω = 1278 rad/s.
P10.8-4
Zc R
R + j (ω L −ω R 2 C +ω 3 R 2 L C 2 )
jω C
= jω L +
=
1
1+ (ω R C ) 2
R +Z c
R+
jω C
Set real part equal to 100 Ω to get C
R
Z = ZL +
R
= 100 ⇒ C = 0.158 µF
1+ (ω R C ) 2
Set imaginary part of numerator equal to 0 to get L ( ω = 2π f = 6283 rad sec )
L − R 2C + ω 2 R 2 LC 2 = 0 ⇒ L = 0.1587 H
P10.8-5
Z L = j ω L = j (6.28×106 ) (47×10−6 ) = j 300 Ω
 1 

( 300 + j 300 )
jω C 

Z eq = Z c ||(Z R +Z L ) =
= 590.7 Ω
1
+ 300+ j 300
jω C
300+300 j
590.7 =
⇒ 590.7 −(590.7)(300 ω C ) + j (590.7)(300ω C ) = 300 + j 300
1+300 j ω C −300 ω C
(
Equating imaginary terms ω =2π f = 6.28×106 rad sec
)
(590.7) (300ω C ) = 300 ⇒ C = 0.27 nF
10-22
Section 10-9: Kirchhoff’s Laws Using Phasors
P10.9-1
(a)
(b)
(c)
P10.9-2
Z1 =3+ j 4 = 5∠53.1° Ω
and
Z 2 =8− j8 = 8 2 ∠− 45° Ω
Total impedance = Z1 + Z 2 = 3 + j 4 + 8 − j8 = 11 − j 4 = 11.7∠− 20.0° Ω
I=
100∠0°
100
100
=
=
∠20.0° ⇒
Z1 +Z 2 11.7 ∠− 20° 11.7
i(t ) = 8.55 cos (1250 t + 20.0°) A
V1 (ω ) = Vs (ω ) − V2 (ω ) = 7.68∠47° − 1.59∠125°
= ( 5.23 + j 5.62 ) − ( −0.91 + 1.30 )
= ( 5.23 + 0.91) + j ( 5.62 − 1.30 )
= 6.14 + j 4.32
= 7.51∠35°
v1 ( t ) = 7.51 cos ( 2 t + 35° ) V
P10.9-3
I = I 1 + I 2 = 0.744∠ − 118° + 0.5405∠100 = ( −0.349 − j 0.657 ) + ( −0.094 + j 0.532 )
= ( −0.349 − 0.094 ) + j ( −0.657 + 0.532 )
= −0.443 − j 0.125
= 0.460∠196°
i ( t ) = 460 cos (2 t + 196°) mA
P10.9-4
Vs = 2 ∠30° V and I =
2 ∠30°
= 0.185 ∠ − 26.3° A
6+ j12+ 3 / j
i (t ) = 0.185 cos (4 t − 26.3°) A
10-23
j 15 = j (2π ⋅ 796) (3 ⋅10−3 )
P10.9-5
I=
12
= 0.48 ∠ − 37° A
20 + j15
i (t ) = 0.48 cos (2π ⋅ 796 t − 37°) A
P10.9-6
Z1 = R = 8 Ω, Z 2 = j 3 L, I = B ∠ − 51.87° and I s = 2 ∠ − 15° A
8
I B ∠−51.87°
Z1
=
=
=
=
2 ∠−15°
Is
Z1 + Z 2 8+ j 3L
8 ∠0°
 3L 
82 + (3L) 2 ∠ tan −1  
 8 
Equate the magnitudes and the angle.
 3L 
angles: + 36.87 = + tan −1   ⇒ L = 2 H
 8 
8
B
magnitudes:
=
⇒ B =1.6
64+ 9 L2 2
P10.9-7
The voltage V can be calculated using Ohm's Law.
V = (1.72 ∠ - 69°) (4.24∠45°) = 7.29∠ - 24° V
The current I can be calculated using KCL.
10-24
I = (3.05 ∠ - 77°) - (1.72∠ - 69°) = 1.34∠ - 87° A
Using KVL to calculate the voltage across the inductor and then Ohm's Law gives:
j 2L =
24 - 4(1.34∠-87°)
⇒ L=4 H
3.05∠-77°
P10.9-8
 10 
10


V10 = Vs 
= 20∠0° 


 10 2∠− 45° 
 10 − j10 
= 0 2∠45°
v10 (t ) = 10 2 cos (100 t + 45°) V
P10.9-9
(a)
(b)
160 ∠0°
160 ∠0°
=
(−1326) (300 + j 37.7) 303 ∠−5.9°
− j1326 + 300 + j 37.7
= 0.53 ∠5.9° A
i(t ) = 0.53cos (120π t +5.9°) A
I=
I=
160∠0°
160∠0°
=
(− j199)(300 + j 251) 256∠−59.9°
− j199+ 300+ j 251
= 0.625∠59.9° A
i(t ) =0.625 cos (800π t +59.9°) A
10-25
Section 10-10:
Node Voltage and Mesh Current Analysis Using Phasors
P10.10-1
Draw frequency domain circuit and write node equations:
VA VA − VC
+
= 0 ⇒ (2 + j )VA − 2VC = j 20
10
j5
VC − VA VC
+
− (1+ j ) = 0 ⇒ 4VA + VC = 20− j 20
KCL at C:
j5
− j4
KCL at A − 2 +
Solve using Cramers rule:
(2 + j )
j 20
4
20 − j 20
60 − j100 116.6 ∠−59°
Vc =
=
=
= 11.6 ∠ − 64.7° V
(2 + j ) −2
10 + j
101 ∠5.7°
4
1
P10.10-2
KCL:
V
V
V
(V −100)
+
+
+
= 0 ⇒ V = 57.6 ∠22.9° V
− j125 j80 250
150
IS =
100− V
= 0.667 − 0.384 ∠22.9° = 0.347 ∠− 25.5° A
150
IC =
V
= 0.461 ∠112.9° A
125 ∠−90°
10-26
IL =
V
= 0.720 ∠− 67.1° A
80∠90°
IR =
V
= 0.230∠22.9° A
250
P10.10-3
KCL at node A:
Va Va − Vb
+
=0
200
j 100
(1)
KCL at node B:
Vb − Va
V
V −1.2
+ b + b
= 0
j 100 − j 50
j 80
1
3
⇒ Va = Vb −
4
2
(2)
Substitute Eqn (2) into Eqn (1) to get
Vb = 2.21 ∠ − 144° V
Then Eqn (2) gives
Finally
Va = ( 0.55∠−144° ) − 1.5 = 1.97∠ − 171° V
va (t ) = 1.97 cos (4000 t − 171°) V and vb (t ) = 2.21cos (4000 t − 144°) V
10-27
P10.10-4
ω = 104 rad s
I s = 20∠53° A
 1 1 j 
 1 
KCL at a:  + +  Va +  −  Vb = 20∠53.13°
 20 40 60 
 40 
 1 
 1 j j 
KCL at b:  −  Va + − +  Vb − j Vc = 0
80
 40 
 40 40 80 
−j
 1 j 
KCL at c:
Vb + +  Vc = 0
80
 40 80 
The node equations are:
Solving thes equation yields
Va = 2 ⋅ 240∠45° V ⇒
va (t ) = 339.4 cos (ω t + 45°) V
vs = sin (2π ⋅ 400 t ) V
P10.10-5
R = 100 Ω
LR = 40 mH
 40 mH
LS = 
 60 mH
door opened
door closed
With the door open VA − VB = 0 since the bridge circuit is balanced.
With the door closed Z LR = j (800π )(0.04) = j100.5 Ω and Z LS = j (800π )(0.06) = j150.8 Ω.
The node equations are:
KCL at node B:
VB − VC VB
j100.5
+
= 0 ⇒ VB =
VC
R
j100.5+100
Z LR
KCL at node A :
VA − VC VA
+
=0
Z LS
R
10-28
Since VC = Vs =1 V
VB =0.709∠44.86° V and VA =0.833∠33.55 V
Therefore
VA − VB = 0.833∠33.55° − 0.709∠44.86° = (0.694 + j.460) − (0.503 + j 0.500) = 0.191 − j 0.040
= 0.195∠ − 11.83° V
P10.10-6
The node equations are:
V1 −(−1+ j ) V1 V1 − V2
+ +
=0
−j2
j2
2
V2 − V1 V2
+
−I C = 0
− j2
− j2
Also, expressing the controlling signal of the dependent source in terms of the node voltages
yields
 −1+ j 
−1 + j
⇒ IC = 2 I x = 2 
Ix =
 = −1 − j A
-2 j
 -2 j 
Solving these equations yields
V2 =
−3− j
= 2 ∠ − 135° V ⇒ v(t ) = v2 (t ) = 2 cos (40 t − 135°) V
1+ j 2
10-29
P10.10-7
V2 = 0.7571∠66.7° V
V3 = 0.6064∠ − 69.8° V

I1 =I 2 + I 3 

 I 3 =0.3032 ∠20.2° A
V3 − V2 

I2 =
 yields  I 2 = 0.1267∠−184° A
j 10 
 I =0.195∠36° A
 1
V3 
I3 =

−j2 
i1 (t ) =0.195cos (2 t + 36°) A
therefore
P10.10-8
The mesh equations are
(4 + j 6) I1 − j 6 I 2 = 12 + j12 3
- j 6 I1 + (8 + j 2) I 2 = 0
Using Cramer’s rule yields
Then
I1 =
(12+ j 12 3) (8+ j 2)
= 2.5∠29° = 2.2 + j 1.2 A
(4+ j 6) (8+ j 2) − (− j 6) (− j 6)
I2 =
j6
6∠90°
(2.5∠29°) =
(2.5∠29°) = 1.82∠105° A
8+ j 2
68∠14°
and
VL = j 6(I1 − I 2 ) = (6∠90°) (2.5∠29° − 1.82∠105°) = (6∠90°) (2.71∠ − 11.3°) = 16.3∠78.7° V
Finally
V = − j 4I 2 = (4∠ − 90°)(1.82∠105°) = 7.28∠15° V
c
10-30
P10.10-9
The mesh equations are:
(10 − j ) I1 + ( j ) I 2 + 0 I 3 = 10
j I1 − j I 2 + j I 3 = 0
0 I1 + j I 2 + (1− j ) I 3 = j10
Solving these mesh equations using Cramer’s rule yields:
(10− j )
j
0
I2 =
(10− j )
j
0
10
0
j
0
j 10 (1− j ) 90 − j 20
=
= 8.38∠77.5° A ⇒ i (t ) = 8.38cos (103 t + 77.5° ) A
0
j
−11 j
j
−j
j (1− j )
(checked using LNAPAC on 7/3/03)
P10.10-10
The mesh equations are:
−1
− j 4   I1  10∠30°
 (2 + j 4)
 −1
(2 +1/ j 4)
−1  I 2  =  0 

 − j 4
(3+ j 4)   I 3   0 
−1
Using Cramer’s rule yields
I3 =
Then
2 + j8
(10∠30° ) = 3.225∠44° A
12+ j 22.5
V = 2 I 3 = 2 ( 3.225∠44° ) = 6.45∠44° V ⇒ v(t ) = 6.45cos (105 t + 44° ) V
10-31
P10.10-11
Mesh Equations:
j 75 I1 − j 100 I 2 = 375
− j 100 I1 + (100+ j 100) I 2 = 0
Solving for I 2 yields I 2 = 4.5 + j 1.5 A
⇒ i 2 (t ) = 4.74 ∠18.4° A
10-32
Section 10-11: Superposition, Thèvenin and Norton Equivalents and Source
Transformations
P10.11-1
Use superposition
I1 =
12∠45°
= 3.3∠11.3° mA
3000 + j 2000
I2 =
−5∠0°
= 1.5∠153° mA
3000+ j1500
i (t ) = 3.3cos (4000 t + 11.3°) + 1.5cos (3000 t + 153°) mA
P10.11-2
Use superposition
I1 =
I 2 (ω ) =
3
= 0.5 mA
6000
−1∠45°
= −0.166 ×10−3 ∠45° A
6000+ j 0.2
i (t ) = i 2 (t ) + i1 (t ) = − 0.166 cos (4 t + 45°) + 0.5 mA
= 0.166 cos (4 t − 135°) + 0.5 mA
P10.11-3
Use superposition
12∠ 45°
= 2∠45° mA
I1 (ω ) =
6000 + j 0.2
5∠−90°
= 0.833∠ − 90° mA
I 2 (ω ) =
6000 + j 0.15
i (t ) = i1 (t ) − i 2 (t ) = 2 cos (4 t + 45°) − 0.833cos (3 t − 90°) mA
10-33
P10.11-4
Find Voc :

80 + j80 
Voc = ( 5 ∠−30° ) 

 80 + j80 − j 20 
 80 2∠− 21.9° 
= ( 5 ∠−30° ) 

 100∠36.90° 
= 4 2∠ − 21.9° V
Find Z t :
Zt =
( − j 20 )( 80 + j80 ) = 23 ∠ − 81.9°
− j 20 + 80 + j80
Ω
The Thevenin equivalent is
10-34
P10.11-5
First, determine Voc :
The mesh equations are
600 I1 − j 300 (I1 − I 2 ) = 9 ⇒ (600 − j 300) I1 + j 300 I 2 = 9∠0°
−2 V + 300 I 2 − j 300 (I1 − I 2 ) = 0 and V = j 300 (I1 − I 2 ) ⇒
j 3 I1 + (1 − j 3) I 2 = 0
I 2 = 0.0124∠ − 16° A
Using Cramer’s rule:
Voc = 300 I 2 = 3.71∠ − 16° V
Then
Next, determine I sc :
−2 V − V = 0 ⇒ V = 0 ⇒ I sc =
The Thevenin impedance is
ZT =
9∠0°
= 0.015∠0° A
600
Voc 3.71∠−16°
=
= 247∠ − 16° Ω
0.015∠0°
I sc
The Thevenin equivalent is
10-35
P10.11-6
First, determine Voc :
The node equation is:
Voc Voc − (6 + j8) 3  Voc − (6+ j8) 
+
− 
=0
− j4
j2
j2
2

Voc =3+ j 4=5∠53.1° V
Vs = 10∠53° = 6 + j 8 V
Next, determine I sc :
The node equation is:
V
V
V − (6 + j8) 3  V − (6 + j8) 
+
+
− 
=0
2 − j4
j2
2
j2

V=
I sc =
3 + j4
1− j
V 3+ j 4
=
2 2− j 2
The Thevenin impedance is
ZT =
Vs = 10∠53° = 6 + j 8 V
 2− j 2 
Voc
= 3 + j4 
 = 2 − j2 Ω
I sc
 3+ j 4 
The Thevenin equivalent is
10-36
Y = G + YL + YC
1
Y = G when YL + YC = 0 or
+ jω C = 0
jω L
1
1
1
, fO =
ωO =
=
2π LC 2π 39.6×10−15
LC
P10.11-7
= 0.07998×107 Hz =800 KHz
(80 on the dial of the radio)
P10.11-8
In general:
I=
Voc
ZL
and V =
Voc
Z t +Z L
Z t +Z L
In the three given cases, we have
Z1 = 50 Ω ⇒
Z2 =
I1 =
V1 25
=
= 0.5 A
Z1 50
1
1
=
= − j 200 Ω ⇒
jω C j (2000)(2.5×10 −6 )
Z 3 = jω L = j (2000)(50 × 10−3 ) = j100 Ω ⇒
I2 =
I3 =
V2 100
=
= 0.5 A
Z 2 200
V3
50
=
= 0.5 A
Z 3 100
Since |I| is the same in all three cases, Z t +Z1 = Z t +Z 2 = Z t +Z3 . Let Z t = R + j X . Then
( R + 50) 2 + X 2 = R 2 + ( X − 200) 2 = R 2 + ( X + 100) 2
This requires
Then
( X − 200) 2 = ( X + 100) 2 ⇒ X = 50 Ω
( R + 50) 2 + (50) 2 = R 2 + (−150) 2 ⇒ R = 175 Ω
so
Z t =175+ j 50 Ω
and
Voc = I1 Z t + R1 =(0.5) (175+ 50) 2 + (50) 2 =115.25 V
10-37
P10.11-9
Z1 =
(− j 3)(4)
= 2.4∠ − 53.1° Ω
− j 3+ 4
=1.44 − j1.92 Ω
Z 2 = Z1 + j 4
= 1.44 + j 2.08
= 2.53∠55.3° Ω
Z3 = 3.51∠ − 37.9° Ω
= 2.77 − j 2.16 Ω
 3.51∠−37.9° 
( 3.51∠−37.9° ) = 1.9∠ − 92° A
I = ( 2.85∠− 78.4° ) 
 = ( 2.85∠− 78.4° )
( 5.24∠− 24.4° )
 2.77 − j 2.16 + 2 
10-38
P10.11-10
Z2 =
I=
(200)(− j 4)
= 4∠ − 88.8° Ω
200− j 4
0.4∠− 44°
= 4∠ − 44° mA
−4 j +100+ j 4
i (t ) = 4 cos (25000 t − 44°) mA
10-39
Section 10-12: Phasor Diagrams
P10-12-1
V = V1 − V2 + V3 = ( 3+ j 3) − ( 4 + j 2 ) + ( 3+ j 2 ) = −4 + j 3
*
*
P10.12-2
I=
10∠0°
= 0.74∠42° A
10+ j1− j10
VR = R I = 7.4∠42° V
VL = Z L I = (1∠90°)(0.74∠42°) = 0.74∠132° V
VC = Z C I = (10∠−90°)(0.74∠42°) = 7.4∠− 48° V
VS = 10∠0° V
P10.12-3
I = 72 3 + 36 3∠(140° − 90°) + 144∠210° + 25∠φ = 40.08 − j 24.23 + 25∠φ
= 46.83∠ − 31.15° + 25∠φ
To maximize I , require that the 2 terms on the right side have the same angle ⇒ φ = −31.15°.
10-40
Section 10-14: Phasor Circuits and the Operational Amplifier
Vo (ω )
 104 || − j104 
10 − j 225
−j
e
= −
=
 = −10
Vs (ω )
1− j
2
 1000 
 10 − j 225 
− j 225
Vs (ω ) = 2 ⇒ Vo (ω ) = 
e
 2 = 10e
 2

vo ( t ) = 10 cos (1000t − 225°) V
P10.14-1
H (ω ) =
P10.14-2
Node equations:
V1 − VS
VS
+ j ω C 1 V1 = 0 ⇒ V1 =
1 + j ω C1 R1
R1
 R3 
V1 V1 − V0
V
+
= 0 ⇒ V0 =  1 +
 R 2  1
R2
R3


Solving:
1+
R3
R2
V0
=
VS 1 + j ω C 1 R1
P10.14-3
Node equations:
j ω C 1 VS
V1
+ j ω C 1 ( V1 − VS ) = 0 ⇒ V1 =
R1
1 + j ω C 1 R1
 R3 
V1 V1 − V0
V
+
= 0 ⇒ V0 =  1 +
 R 2  1
R2
R3


Solving:
 R 
j ω C 1 1 + 3 
 R2 
V0


=
1 + j ω C 1 R1
VS
10-41
P10.14-4
Node equations:
V1 − VS
V
VS
+ 1 = 0 ⇒ V1 =
175
1 + j 109
− j1.6
V − V0
V1
+ 1
= 0 ⇒ V0 = 11 V1
1000 10000
Solving:
V0 =
11
11
VS =
( 0.005∠0° )
1 + j 109
110∠89.5°
= 0.5∠89.5° mV
Therefore
v0 (t ) = 0.5cos (ω t − 89.5°) mV
10-42
PSpice Problems
SP10-1
10-43
SP10-2
10-44
SP 10-3
10-45
SP 10-4 The following simulation shows that k1 = 0.4and k2 = -3 V/A. The required values of
Vm and Im are Vm = 12.5 V and Im = -1.667 A.
10-46
Verification Problems
VP 10-1
Generally, it is more convenient to divide complex numbers in polar form. Sometimes, as in this
case, it is more convenient to do the division in rectangular form.
Express V1 and V2 as: V1 = − j 20 and V 2 = 20 − j 40
KCL at node 1:
2−
V1
10
−
V1 − V 2
j 10
= 2−
− j 20 − j 20 − ( 20 − j 40 )
−
= 2+ j2−2− j2 = 0
10
j 10
KCL at node 2:
V1 − V 2
j 10
−
 V1  − j 20 − ( 20 − j 40 ) 20 − j 40
 − j 20 
+ 3  =
−
+ 3
 = ( 2 + j 2) − ( 2 − j4) − j 6 = 0
j 10
10
10
 10 
 10 
V2
The currents calculated from V1 and V2 satisfy KCL at both nodes, so it is very likely that the V1
and V2 are correct.
VP 10-2
I 1 = 0.390 ∠ 39° and I 2 = 0.284 ∠ 180°
Generally, it is more convenient to multiply complex
numbers in polar form. Sometimes, as in this case, it is more
convenient to do the multiplication in rectangular form.
Express I1 and I2 as: I 1 = 0.305 + j 0.244 and I 2 = −0.284
KVL for mesh 1:
8 ( 0.305 + j 0.244 ) + j 10 ( 0.305 + j 0.244 ) − (− j 5) = j 10
Since KVL is not satisfied for mesh 1, the mesh currents are
not correct.
Here is a MATLAB file for this problem:
10-47
% Impedance and phasors for Figure VP 10-2
Vs = -j*5;
Z1 = 8;
Z2 = j*10;
Z3 = -j*2.4;
Z4 = j*20;
% Mesh equations in matrix form
Z = [ Z1+Z2
0;
0
Z3+Z4 ];
V = [ Vs;
-Vs ];
I = Z\V
abs(I)
angle(I)*180/3.14159
% Verify solution by obtaining the algebraic sum of voltages for
% each mesh. KVL requires that both M1 and M2 be zero.
M1 = -Vs + Z1*I(1) +Z2*I(1)
M2 = Vs + Z3*I(2) + Z4*I(2)
VP 10-3
V1 = 19.2 ∠ 68° and V 2 = 24 ∠ 105° V
KCL at node 1 :
19.2 ∠ 68° 19.2 ∠ 68°
+
− 4∠15 = 0
2
j6
KCL at node 2:
24 ∠105° 24 ∠105°
+
+ 4∠15 = 0
j12
− j4
The currents calculated from V1 and V2 satisfy
KCL at both nodes, so it is very likely that the V1
and V2 are correct.
Here is a MATLAB file for this problem:
% Impedance and phasors for Figure VP 10-3
Is = 4*exp(j*15*3.14159/180);
Z1 = 8;
Z2 = j*6;
Z3 = -j*4;
10-48
Z4 = j*12;
% Mesh equations in matrix form
Y = [ 1/Z1 + 1/Z2
0;
0
1/Z3 + 1/Z4 ];
I = [ Is;
-Is ];
V = Y\I
abs(V)
angle(V)*180/3.14159
% Verify solution by obtaining the algebraic sum of currents for
% each node. KCL requires that both M1 and M2 be zero.
M1 = -Is + V(1)/Z1 + V(1)/Z2
M2 = Is + V(2)/Z3 + V(2)/Z4
VP 10-4
First, replace the parallel resistor and capacitor by an equivalent impedance
ZP =
The current is given by
(3000)(− j 1000)
= 949 ∠ − 72° = 300 − j 900 Ω
3000− j 1000
I=
VS
100 ∠0°
=
= 0.2∠53° A
j 500+ Z P j 500+ 300− j 900
 − j 1000 
I1 = 
 ( 0.2 ∠53° ) = 63.3 ∠ − 18.5° mA
 3000 − j 1000 
Current division yields


3000
I2 = 
 ( 0.2 ∠53° ) = 190∠71.4° mA
 3000 − j 1000 
The reported value of I1 is off by an order of magnitude.
10-49
Design Problems
DP 10-1
R2
jω C
1
=
R2
1 + jω CR 2
1 + jω CR 2
R1
Vo (ω )
=−
=−
Vi (ω )
R1
1 + jω CR 2
R2
R2
R1
Vo (ω )
j (180 − tan −1 ω CR 2 )
e
=
2
Vi (ω )
1 + (ω CR 2 )
R2
In this case the angle of
tan (180 − 76 )
Vo (ω )
= 0.004 and the
is specified to be 104° so CR 2 =
1000
Vi (ω )
Vo (ω )
8
is specified to be
so
2.5
Vi (ω )
R2
= 132 . One set of values
R1
1 + 16
that satisfies these two equations is C = 0.2 µ F, R1 = 1515 Ω, R 2 = 20 kΩ .
magnitude of
R1
=
8
⇒
2.5
R2
DP 10-2
R2
Vo (ω )
=
Vi (ω )
where K =
jω C
1
=
R2
1 + jω CR 2
R2
1 + jω CR 2
K
=
R2
1 + jω CR p
R1 +
1 + jω CR 2
R1
R1 + R 2
and R p =
R1 R 2
R1 + R 2
Vo (ω )
K
− j tan −1 ω CR p
=
e
2
Vi (ω )
1 + (ω CR p )
10-50
In this case the angle of
C Rp = C
R1 R 2
R1 + R 2
=−
Vo (ω )
is specified to be -76° so
Vi (ω )
tan ( −76 )
V (ω )
2.5
= 0.004 and the magnitude of o
so
is specified to be
12
1000
Vi (ω )
R2
2.5
K
=
⇒ 0.859 = K =
. One set of values that satisfies these two equations is
R1 + R 2
1 + 16 12
C = 0.2 µ F, R1 = 23.3 kΩ, R 2 = 142 kΩ .
jω L R 2
DP 10-3
jω
L
R 2 + jω L
R1
Vo (ω )
=
=−
jω L R 2
L
Vi (ω )
1 + jω
R1 +
Rp
R 2 + jω L
where R p =
R1 R 2
R1 + R 2
Vo (ω )
Vi (ω )
=
ω
L
R1
 L 
1 + ω
 R p 


2
e

L 
j  90 − tan −1 ω



R
p 

Vo (ω )
L L ( R1 + R 2 ) tan ( 90 − 14 )
is specified to be 14° so
=
=
= 0.1
Rp
R1 R 2
40
Vi (ω )
L
40
R1
V (ω )
2.5
2.5
L
=
⇒
= 0.0322 . One
is specified to be
so
and the magnitude of o
8
R1
Vi (ω )
8
1 + 16
In this case the angle of
set of values that satisfies these two equations is L = 1 H, R1 = 31 Ω, R 2 = 14.76 Ω .
10-51
jω L R 2
DP 10-4
jω
L
R 2 + jω L
R1
Vo (ω )
=
=−
jω L R 2
L
Vi (ω )
1 + jω
R1 +
Rp
R 2 + jω L
where R p =
R1 R 2
R1 + R 2
Vo (ω )
Vi (ω )
In this case the angle of
=
ω
L
R1
 L 
1 + ω
 R p 


2
e

L 
j  90 − tan −1 ω



R
p 

Vo (ω )
is specified to be -14°. This requires
Vi (ω )
L L ( R1 + R 2 ) tan ( 90 + 14 )
=
=
= −0.1
Rp
R1 R 2
40
This condition cannot be satisfied with positive
DP 10-5
Z1 =10 Ω
1
Z2 =
jω C
Z3 = R + jω L
Y1 =
1
S
10
Y2 = jω C
1
Y3 =
R + jω L
v(t ) = 80 cos (1000 t − θ ) V ⇒ V = 8∠ − θ V
iS (t ) = 10 cos 100 t A ⇒ I s =10∠0° A
so
1

1
+
+ jω C  = 10∠0° ⇒ R + 10 − 10 ω 2 LC + j (ω L + 10 ω RC ) = 1.25 R + j1.25 ω L
10 R + jω L

(80∠−θ ) 
Equate real part: 40 − 40ω 2 LC = R where ω = 1000 rad sec
Equate imaginary part: 40 RC = L
Solving yields R = 40(1− 4×107 RC 2 )
10-52
Now try R = 20 Ω ⇒ 1 − 2(1 − 4 × 107 (20)C 2 )
which yields C = 2.5×10−5 F= 25 µ F so L = 40 RC = 0.02 H=20 mH
Now check the angle of the voltage. First
Y1 = 1/10 = 0.1 S
Y2 = j 0.25 S
Y3 = 1/(20+ j 20) = .025− j.025 S
then
Y = Y1 + Y2 + Y3 = 0.125 , so V =YI s = (0.125∠0°)(10∠0°) = 1.25∠0° V
So the angle of the voltage is θ =0° , which satisfies the specifications.
10-53
Chapter 11: AC Steady State Power
Exercises:
Ex. 11.3-1
Z = j3 +
∴I =
4(− j 2)
= 0.8 + j1.4
4− j 2
= 1.6 ∠60.3° Ω
7∠0°
V
=
= 4.38 ∠ − 60.3° A
°
Z 1.6∠60.3
i (t ) = 4.38cos (10 t − 60.3° ) A
The instantaneous power delivered by the source is given by
(7)(4.38)
[cos (60.3°) + cos (20 t −60.3°)]
2
= 7.6 + 15.3cos (20 t − 60.3°) W
p (t ) = v(t ) ⋅ i (t ) = (7 cos 10 t )(4.38cos (10 t − 60.3°)) =
The inductor voltage is calculated as
VL = I ⋅ Z L = (4.38 ∠ − 60.3°)(j3) = 13.12 ∠29.69° V
vL (t ) = 13.12 cos (10 t + 29.69°) V
The instantaneous power delivered to the inductor is given by
pL (t ) = vL (t ) ⋅ i (t ) =  (13.12 cos (10t + 29.69°)(4.38cos (l0t − 60.3° ) 
57.47
[ cos (29.69°+ 60.3°)+ cos (20 t + 29.69°−60.3°)]
2
= 28.7 cos (20t − 30.6°) W
=
11-1
Ex. 11.3-2
(a) When the element is a resistor, the current has the same phase angle as the voltage:
i (t ) =
v(t ) Vm
=
cos (ω t + θ ) A
R
R
The instantaneous power delivered to the resistor is given by
2
2
2
V
V
Vm
V
pR (t ) = v(t ) ⋅ i (t ) = Vm cos (ω t + θ ) ⋅ cos (ω t + θ ) = m cos 2 (ω t + θ ) = m + m cos (2ω t + θ )
R
R
2R 2R
(b) When the element is an inductor, the current will lag the voltage by 90°.
Z L = jω L = ω L∠90° Ω
⇒
I=
V ∠θ
V
V
= m
= m ∠ (θ −90° )
Z ω L∠90° ω L
The instantaneous power delivered to the inductor is given by
V
V
pL (t ) = i (t ) ⋅ v(t ) = m cos (ω t + θ − 90° ) ⋅Vm cos (ω t + θ ) = m cos ( 2ω t + 2θ −90° ) W
ωL
2ω L
2
11-2
Ex. 11.3-3
(a) Psource
The equivalent impedance of the parallel resistor
(1)( j ) = 1 1+ j Ω . Then
and inductor is Z =
( )
1+ j
2
I V
=
cos θ =
2
10∠0°
20
I=
=
∠ − 18.4° A
1
10
1+ (1+ j )
2
 20 

 10  cos −18.4° = 30.0 W
(
)
2
(10 ) 
 20 
(1)
2
I max R1  10 
=
= 20 W
(b) PR 1 =
2
2
2
Ex. 11.4-1
I eff =
3
1 t 2
1 2
2
=
+
i
t
dt
dt
(
)
(10)
(5) 2 dt  = 8.66
∫
∫
∫
0
0
2


T
3
Ex. 11.4-2
(a)
I
2
i (t ) = 2 cos 3 t A ⇒ I eff max =
= 2A
2
2
(b)
(c)
i (t ) = cos (3 t − 90° ) + cos (3 t + 60° ) A
1
3
I = (1∠−90° ) + (1∠60° ) = − j + + j
= 0.518∠ − 15° A
2
2
0.518
i (t ) = 0.518 cos (3t − 15°) A ⇒ I eff =
= 0.366 A
2
 2   3 
2
I eff = 
 +

 2  2
2
2
⇒ I eff = 2.55 A
11-3
Ex. 11.4-3
Use superposition:
V1 = 5∠0° V
V2 = 2.5 V (dc)
V3 = 3∠−90° V
V1 and V2 are phasors having the same frequency, so we can add them:
V1 + V3 = ( 5∠0° ) + ( 3∠ − 90° ) = 5 − j3 = 5.83∠ − 31.0° V
Then
vR (t ) = v1 (t ) + ( v2 (t ) + v3 (t ) ) = 2.5 + 5.83cos (100 t − 31.0°) V
Finally
 5.83 
= (2.5) + 
 = 23.24 V ⇒ VR eff = 4.82 V
 2 
2
2
R eff
V
2
Ex. 11.5-1
Analysis using Mathcad (ex11_5_1.mcd):
Enter the parameters of the voltage source:
Enter the values of R and L
R := 10
L := 4
The impedance seen by the voltage source is:
The mesh current is: I :=
A := 12
ω := 2
Z := R + j ⋅ω ⋅L
A
Z
11-4

I ⋅( I ⋅Z)
The complex power delivered by the source is: Sv :=
Sv = 4.39 + 3.512i
2

I ⋅( I ⋅R)
The complex power delivered to the resistor is: Sr :=
Sr = 4.39
2

I ⋅( I ⋅j ⋅ω ⋅L)
The complex power delivered to the inductor is: Sl :=
Sl = 3.512i
2
Sr + Sl = 4.39 + 3.512i
Verify Sv = Sr + Sl :
Sv = 4.39 + 3.512i
Ex. 11.5-2
Analysis using Mathcad (ex11_5_2.mcd):
Enter the parameters of the voltage source: A := 12
R := 10
Enter the values of R, L an
dC
ω := 2
L := 4
C := 0.1
The impedance seen by the voltage source is: Z := R + j ⋅ω ⋅L +
The mesh current is:
I :=
A
Z
The complex power delivered by the sourceis:
The complex power delivered to the resistor is:
The complex power delivered to the inductor is:
Sv :=
Sr :=
Sl :=
1
j ⋅ω ⋅C

I ⋅( I ⋅Z)
Sv = 6.606 + 1.982i

I ⋅( I ⋅R)
2

I ⋅( I ⋅j ⋅ω ⋅L)
2

I ⋅ I ⋅

j ⋅ω ⋅C 
The complex power delivered to the capacitor is: Sc := 
Verify Sv = Sr + Sl + Sc :
Sr + Sl + Sc = 6.606 + 1.982i
2
1
2
Sr = 6.606
Sl = 5.284i
Sc = −3.303i
Sv = 6.606 + 1.982i
11-5
Ex. 11.5-3
Analysis using Mathcad (ex11_5_3.mcd):
Enter the parameters of the voltage source:
A := 12
ω := 2
Enter the Average and Reactive Power delivered to the RL circuit: P := 8
The complex power delivered to the RL circuit is:
The impedance seen by the voltage source is:
S := P + j ⋅Q
Z :=
Calculate the required values of R and L R := Re( Z)
The mesh current is: I :=
2

2 ⋅S
A
L :=
Im( Z)
A
ω

I ⋅( I ⋅Z)
The complex power delivered by the source is: Sv :=
2

I ⋅( I ⋅R)
The complex power delivered to the resistor is: Sr :=
2

I ⋅( I ⋅j ⋅ω ⋅L)
The complex power delivered to the inductor is: Sl :=
2
Verify Sv = Sr + Sl :
Z
Sr + Sl = 8 + 6i
Q := 6
R = 5.76
L = 2.16
Sv = 8 + 6i
Sr = 8
Sl = 6i
Sv = 8 + 6i
( )
Ex. 11.6-1
(377) (5) 

pf = cos (∠ Z ) = cos  tan −1 ω L  = cos  tan −1
= 0.053
R 
100 


Ex. 11.6-2
pf = cos (∠ Z) = cos  tan −1 X  = cos  tan −1 80  = 0.53 lagging
50 
R 


( )
XC =
( )
(50)2 + (80)2
= −111.25 Ω ⇒ ZC = − j 111.25 Ω
50 tan (cos −1 1) −80
11-6
Ex. 11.6-3
PT = 30 + 86 = 116 W and QT = 51 VAR
S T = PT + j QT = 116+ j 51 = 126.7 ∠23.7° VA
pf plant = cos 23.7° = 0.915
Ex. 11.6-4
P = V I cos θ
Z=
⇒ I =
P
4000
=
= 44.3 A
V cos θ
(110)(.82)
V
∠ cos −1 (0.82) = 2.48 ∠34.9° = 2.03+ j 1.42 = R + j X
I
To correct power factor to 0.95 requires
R2 + X 2
(2.03) 2 + (1.42) 2
=
= − 8.16 Ω
R tan (cos −1 pfc ) − X
(2.03) tan (18.19° ) −1.42
−1
C=
=325 µ F
ω X1
X1 =
Ex. 11.7-1
(a) I = I1 + I 2 = ( 0.4714∠135° ) + (1.414∠ − 45° ) = 0.9428∠ − 45° A
0.94282
( 6 ) = 2.66 W
2
(b)
1.22
I1 = 1.2∠53° A ⇒ p1 =
( 6 ) = 4.32 W
2
0.47142
I 2 = 0.4714∠135° A ⇒ p1 =
( 6 ) = 0.666 W
2
∴ p = p1 + p2 = 4.99 W
⇒
p=
Ex. 11.8-1
For maximum power, transfer
Z L = Z*t = 10 − j14 Ω
100
I=
=5 A
(10+ j14) (10− j14)
 5 
PL = 
 Re(10 − j14) = 125 W
 2
2
11-7
Ex. 11.8-2
If the station transmits a signal at 52 MHz
then
ω = 2π f = 104π ×106 rad/sec
so the received signal is
vs (t ) = 4 cos (104π × 106 t ) mV
(a)
If the receiver has an input impedance of Zin =300 Ω then
300
1  1   2.4×10−3 
Zin
Vin =
Vs =
× 4 ×10−3 = 2.4 mV ⇒ P = Vin2 
=
 = 9.6 nW
R + Zin
200+300
2  R L   2(300) 
2
(b)
If two receivers are connected in parallel then Zin = 300||300 = 150 Ω and
Vin =
total P =
(c)
Zin
150
VS =
(4 ×10−3 ) =1.71× 10−3 V
R + Zin
200+150
2
Vin  1  (1.71×10−3 ) 2
= 9.7 nW or 4.85 nW to each set

=
2  Zin 
2(150)
In this case, we need Zin = R || R = 200 Ω ⇒ R = 400 Ω , where R is the input
impedance of each television receiver. Then
Ptotal =
Vm 2 (2×10−3 ) 2
=
= 10 nW ⇒ 5 nW to each set
2 Zin
2(200)
11-8
Ex 11.9-1
Coil voltages:
V1 = j 24 I 1 + j 16 I 2 = j 40 I
V 2 = j 16 I 1 + j 40 I 2 = j 56 I
Mesh equation:
24 = V1 + V 2 = j 40 I + j 56 I = j 96 I
I=
24
1
=−j
4
j 96
 1
Vo = V 2 = ( j 56 )  − j  = 14
 4
vo = 14 cos 4t V
Ex 11.9-2
Coil voltages:
V1 = j 24 I 1 − j 16 I 2 = j 8 I
V 2 = − j 16 I 1 + j 40 I 2 = j 24 I
Mesh equation:
24 = V1 + V 2 = j 8 I + j 24 I = j 32 I
I=
24
3
=−j
4
j 32
 3
Vo = V 2 = ( j 24 )  − j  = 18
 4
vo = 18 cos 4t V
Ex 11.9-3
0 = V 2 = j 16 I 1 + j 40 I 2
⇒ I1 = −
40
I 2 = −2.5 I 2
16
V s = V1 = j 24 I 1 + j 16 I 2
= j (24(−2.5) + 16) I 2
= − j 44 I 2
I2 =
24
6
= j
− j 44
11
I o = I 1 − I 2 = (−2.5 − 1) I 2
= −3.5 I 2
 6
= −3.5  j  = − j 1.909
 11 
io = 1.909 cos ( 4t - 90° ) A
11-9
0 = V 2 = − j 16 I 1 + j 40 I 2
Ex 11.9-4
⇒ I1 =
40
I 2 = 2.5 I 2
16
V s = V1 = j 24 I 1 − j 16 I 2
= j (24(2.5) − 16) I 2
= j 44 I 2
I2 =
24
6
=−j
11
j 44
I o = I 1 − I 2 = (2.5 − 1) I 2
= 1.5 I 2
6

= 1.5  − j  = − j 0.818
 11 
io = 0.818 cos ( 4t - 90° ) A
Ex. 11.10-1
I1 =
5∠0°
5∠0°
=
= 2∠0° A
100 − j 75 (1 + j 3) + ( 4 − j 3)
(1 + j 3) +
52
V1 = ( 4 − j 3) 2∠0° = 10∠ − 36.9° V
11-10
Ex. 11.10-2
I1 =
5∠0°
5∠0°
5∠0°
=
=
= 0.68∠42° A
2+ j 0.2 5.5 + j 4.95 7.4∠− 42°
−
+
j
5
5
(
)
22
1
I 2 = I1 = 0.34∠42° A
2
V2 = ( 2 + j 0.2 ) I 2 = ( 2.01∠5.7° )( 0.34∠42° ) = 0.68∠47.7° V
v2 (t ) = 0.68cos (10 t + 47.7°) V and i 2 (t ) = 0.34 cos (10 t + 42°) A
so
Ex. 11.10-3
Z1 =
then
1 
Z
Z
1
Z Z

=
=
= 9  Z +  and Z 3 = 2 (Z + Z 2 )
,
Z
Z
+

2
2
2 
2 

n3
4
n1
n2 
n3 
4

1
Z 

Z ab = Zin = Z + Z3 = Z +  Z +9 Z+   = 4.0625 Z
4
4 

11-11
Problems
Section 11-3: Instantaneous Power and Average Power
P11.3-1
1∠0° =
V
V
V
+
+
⇒ V = 14.6∠ − 43° V
20 j 63 − j16
I=
V
= 0.23∠ − 133° A
j 63
p(t ) = i (t )v(t ) = 0.23(cos (2π ⋅103 t − 133° )) × 14.6 cos (2π ⋅103 t − 43° )
= 3.36 cos (2π ⋅103 t −133° ) cos (2π ⋅103 t − 43° )
=1.68 (cos (90° ) + cos (4π ⋅103 t −176° ))
=1.68 cos (4π ⋅103 t −176° )
P11.3-2
Current division:
 1800 − j 2400 
I=4 5

1800− j 2400+ 600 
=5
5
∠ − 8.1° mA
2
I 600
5
=
= 300(25)   = 1.875 ×104 µW = 18.75 mW
2
 2
2
P600Ω
Psource =
(
)
 5
V I cos θ 1
= (600)  5  4 5 cos(−8.1° ) = 2.1×104 µW = 21 mW
2
2
 2
P11.3-3
11-12
Node equations:
20I X −100
20I X − V
+ IX +
= 0 ⇒ I X (20 − j15) − V = − j 50
− j5
10
V − 20I X
V
− 3I X + = 0 ⇒ I X (−40 + j 30) + V (−2 − j ) = 0
− j5
10
Solving the node equations using Cramer’s rule yields
IX =
50 5∠63.4°
j 50(2 − j )
=
= 2 5∠10.3° A
(40 − j 30) − (20 − j15)(2 − j )
25∠53.1°
Then
2
PAVE
(
I
= X (20) = 10 2 5
2
)
2
= 200 W
P11.3-4
A node equation:
(V − 16) V
+ − (2 2∠45°) = 0
8
j4

2
⇒ V =  16
 ∠18.4° V
3 

Then
I=
16 − V
= 3.2 ∠ − 116.6° A
j4

2
16

5
1 
= ×
= 6.4 W absorbed
2
8
2
PAVE 8Ω
PAVE current source = −
(
1 V
= ×
2 8
)
2
(
)
1
1
2
V 2 2 cos θ = − 16
 2 2 cos ( 26.6° ) = −12.8 W absorbed
2
2
5
PAVE inductor = 0
1
1
PAVE voltage source = − (16) I cosθ = − (16)( 3.2)cos( − 116.6°) = 6.4 W absorbed
2
2
11-13
P11.3-6
A node equation:
−20 +
Then
V1 V1 + (3 / 2)V1
+
= 0 ⇒ V1 = 50 5 ∠ − 26.6° V
10
15 − j 20
I=
V1 + ( 3/2 ) V1
( 5 / 2 ) V1 = 5 5 ∠26.6° A
=
15 − j 20
25∠ − 53.1°
Now the various powers can be calculated:
1 V1
1 (50 5)
=
=
= 625 W absorbed
2 10
2 10
2
2
PAVE 10Ω
PAVE current source = −
1
1
V (20) cos θ = − (50 5)(20) cos (−26.6°) = −1000 W absorbed
2
2
PAVE 15Ω =
PAVE voltage source = −
I
2
2
(5 5 )
(15 ) = −
(
2
2
)(
(15 ) = 937.5 W absorbed
)
1 3
1
I V1 cos θ = − 5 5 75 5 cos ( −53.1° ) = −562.5 W absorbed
2 2
2
PAVE capacitor = 0 W
P11.3-7
Z=
I =
200 ( j 200 ) 200 ∠90° 200
=
=
∠45° Ω
200 (1 + j )
2 ∠45°
2


120∠0D
200
= 0.85 ∠ − 45° A, I R = 
 I = 0.6∠0° A
200
200
200
j
+
D


∠45
2
P = I R = ( 0.6 ) ( 200 ) = 72 W and w = ( 72 )(1) = 72 J
2
2
11-14
Section 11-4: Effective Value of a Periodic Waveform
P11.4-1
(a)
( Treat i as two sources of different frequencies.)
i = 2 − 4 cos 2t = i1 + i 2
2A source: I eff = lim
T →∞
and
4 cos 2t source:
1
T
∫
T
o
(2) dt = 2 A
2
Ieff =
4
A
2
The total is calculated as
 4 
= ( 2) + 
 = 12 A ⇒ I rms = I eff = 12 = 2 3 A
 2
2
I eff
2
2
i ( t ) = 3cos (π t − 90° ) + 2 cos π t ⇒ I = ( 3∠ − 90° ) +
(b)
(
2 ∠0°
)
= 2 − j 3 = 3.32∠ − 64.8° A
I rms =
3.32
= 2.35 A
2
i ( t ) = 2 cos 2t + 4 2 cos ( 2t + 45° ) + 12 cos ( 2 t − 90° )
(c)
(
)
I = ( 2∠0° ) + 4 2 ∠45° + (12∠ − 90° ) = ( 2 + 4 ) + ( j 4 − j12 ) =10∠ − 53.1° A
I rms =
P11.4-2
(a)
1
Vrms =
5
(b)
(c)
Vrms =
Vrms =
1
5
1
5
10
=5 2 A
2
( ∫ 6 dt +∫ 2 dt ) = 15 ( ∫ 36 dt +∫ 4 dt ) =
2
2
5
2
2
5
( ∫ 2 dt +∫ 6 dt ) = 15 ( ∫ 4 dt +∫ 36 dt ) =
0
2
2
2
5
0
2
2
2
5
( ∫ 2 dt +∫ 6 dt ) = 15 ( ∫ 4 dt +∫ 36 dt ) =
0
3
0
2
2
5
3
2
0
2
3
5
0
3
1
84
= 4.10 V
( 72 + 12 ) =
5
5
1
116
= 4.81 V
(8 + 108) =
5
5
1
84
= 4.10 V
(12 + 72 ) =
5
5
11-15
P11.4-3
2
(a)
1 4 4 2 
4 4
4 4 2
2
Vrms =
t +  dt =
2 t + 1) dt =
(
( 4 t + 4 t + 1) dt

∫
∫
3 1  3 3
27 1
27 ∫1
4  4 t3

=
27  3

=
=
(b)
4
4t 2
+
2
4

4
+t 1 


4
( (85.33 − 1.33) + ( 2 )(16 − 1) + 3)
27
1
1
4
(117 ) = 4.16 V
27
1 4  4 22 
4 4
4 4 2
2
t
dt
t
dt
=
−
+
=
−
+
=
2
11
(
)
( 4 t − 44 t + 121) dt


3 ∫1  3
3 
27 ∫1
27 ∫1
2
Vrms
4  4 t3

=
27  3

=
=
(c)
4
44t 2
−
2
4

4
+ 121 t 1 


4
(84 + ( −22 )15 + (121) 3)
27
1
1
4
(117 ) = 4.16 V
27
1 3 4
4 3
4 3 2
2

t + 2  dt =
=
2 t + 3) dt =
(
( 4 t + 12 t + 9 ) dt

∫
∫
3 03
27 0
27 ∫ 0

2
Vrms
4  4 t3

=
27  3

=
=
12t 2
+
2
0
3
4
( 36 + 54 + 27 )
27

3
+ 9t 0 

0

3
4
(117 ) = 4.16 V
27
11-16
 2π
v ( t ) = 1 + cos 
 T
P11.4-4
(a)
vdc eff
2
t
1 T 
=  ∫ 0 1dt  =
T
 T
T
0
T

2
=  − 0  = 1 V and vac eff =
T

veff = vdc eff + vac eff
2
2
ω=
(b)
I rms
2
1 T /2
A2
2
= ∫ 0 ( A sin ω t ) dt =
T
T
I rms =
P11.4-5
∫
2
=
2π
, I rms =
T
T /2
0
 1 
1 +
 = 1.225 V
 2
2
2
1 T 2
i ( t ) dt
T ∫0
A2
A
=
, where A = 10 mA ⇒
4
2
rms
I rms = 5 mA
0 ≤ t ≤ 0.1
0.1≤ t ≤ 0.2
0.2 ≤ t ≤ 0.3
2
0.2
2
1  0.1
2
 = 90
+

−

90
90
0.2
t
dt
t
dt
(
)
(
)
  .3
∫0.1 
.3  ∫ 0
902
=
.3
V
1
V
2
T /2
1
A2  T / 2
A2

(1−cos 2ω t ) dt =  ∫0 dt − ∫0 cos 2ω t dt  = A
2
2T
4
90 t

v ( t ) = 90 ( 0.2−t )
0

2
=
Vrms

t  = vdc + vac

 0.1 t 2 dt + 0.2 0.2−t 2 dt 
) 
∫0.1 (
 ∫ 0

 .001 .001 
 3 + 3  = 18 V
= 18 = 4.24 V
11-17
Section 11-5: Complex Power
P11.5-1
I* =
R+ j4 L =
P 11.5-2
2 ( 3.6 + j 7.2 )
2S
=
= 0.6 + j 1.2 = 1.342∠63.43° A
12∠0°
12∠0°
12∠0°
= 8.94∠63.43 = 4 + j 8 ⇒ R = 4 Ω and L = 2 H
1.342∠ − 63.43
I* =
2 (18 + j 9 )
2S
=
= 3 + j 1.5 = 3.35∠26.56° A
12∠0°
12∠0°
1
1
1
1
3.35∠ − 26.56°
+
= −j
=
= 0.2791∠ − 26.56 = 0.250 + j 0.125
R j4 L R
4L
12∠0°
⇒ R = 4 Ω and L = 2 H
P11.5-3
Let
Zp =
8 ( j 8)
j 8 1− j 8 + j 8
=
×
=
= 4+ j4 V
8 + j 8 1+ j 1− j
2
Next
12∠0°
12∠0°
I=
=
= 1.342∠ − 26.6° A
4 + Z p 4 + (8 + j 8)
Finally
(12∠0° )(1.342∠ − 26.6° )
S=
*
2
= 7.2 + j 3.6 VA
11-18
P11.5-4
Before writing node equations, we can simplify the circuit using a source transformation:
The node equations are:
V1 V1 − V2
+
= 0 ⇒ V1 (1 + j ) − V2 = 10
2
j2
V2 − V1 1
V2
+ V1 +
= 0 ⇒ V1 ( −4 + j ) + V2 ( j 8 ) = 0
j2
8
0.8 + j 0.4
−5 +
Using Cramers’s rule
then
V1 =
80
= (16 / 3) ∠126.9° V
(4 − j ) − j8(1+ j )
1
1
I = − V1 − V2 = − V1 − V1 (1+ j )+ j 10= 2.66∠126.9° A
8
8
(
Now the complex power can be calculated as
)(
2.66∠−126.9° −(2 / 3)∠36.9°
I* ( −(1/ 8)V1 )
S=
=
2
2
Finally
S = P + jQ = j
)
=−j
8
VA
9
8
8
⇒ P = 0, Q = VAR
9
9
11-19
P11.5-5
I=
S = VI* =
50∠120°
50∠120°
=
= 2.5∠83.13° A
16+ j12 20∠36.87°
( 50∠120° )( 2.5∠−83.13° ) =
125 ∠36.87° = 100 + j 75 VA
P11.5-6
KVL:
(10+ j 20 ) I1 = 5∠0° − j 2 I 2
⇒ (10 + j 20 ) I1 + j 2 I 2 = 5∠0°
KCL:
I1 + I 2 = 6∠0°
Solving these equations using Cramer’s rule:
∆=
I1 =
10 + j 20
1
j2
= 10 + j18
1
1 5 j2
5 − j12
=
= 0.63∠232° A = −0.39 − j 0.5 A
10 + j18
∆ 6 1
I 2 = 6 − I1 = 6 + 3.9 + j.5 = 6.39 + j.5 = 6.41∠4.47° A
Now we are ready to calculate the powers. First, the powers delivered:
(
)
1
*
5∠0° ) −I 2 = 2.5 ( 6.41∠(180− 4.47) ) = −16.0 + j1.1 VA
(
2
1
S 6∠ 0° = [5− j 2 I 2 ]( 6∠0° ) = 5− j 2( 6.39+ j.5 )  3 = 18.0− j 38.3 VA
2
S Total = S5∠ 0° + S 6∠ 0° = 2.0− j 37.2 VA
S5∠ 0° =
delivered
11-20
Next, the powers absorbed:
1
10
2
2
10 I1 = (.63) = 2.0 VA
2
2
j 20 2
S j20Ω =
I1 = j 4.0 VA
2
1
2
2
S − j2Ω = ( − j 2 ) I 2 = − j ( 6.41) = − j 41.1 VA
2
S Total = 2.0 − j 37.1 VA
S10Ω =
absorbed
To our numerical accuracy, the total complex power delivered is equal to the total complex
power absorbed.
P11.5-7
(a)
(b)
Z=
P=
V 100∠20°
=
= 4∠30° Ω
I 25∠ − 10°
I V cos θ (100 )( 25 ) cos 30°
=
= 1082.5 W
2
2
1
= 0.25∠ − 30° = 0.2165 − j 0.125 S . To cancel the phase angle we add a capacitor
Z
having an admittance of YC = j 0.125 S . That requires ω C =0.125 ⇒ C = 1.25 mF .
(c) Y =
P11.5-8
Apply KCL at the top node to get
10 − V1
V1 3
− V1 +
= 0 ⇒ V1 = 4∠36.9° V
3
3 4
1− j
2
Then
V
I1 = 1 = 1∠36.9° A
4
The complex power delivered by the source is calculated as
−
S=
Finally
(1∠36.9° )* (10∠0° ) =
2
5∠ − 36.9° VA
pf = cos ( −36.9° ) = .8 leading
11-21
Section 11.6: Power Factor
P11.6-1
Heating:
P = 30 kW
Motor:
θ = cos −1 ( 0.6 ) = 53.1° 
Total (plant):
P = 30 + 90 = 120 kW 
 ⇒ S = 120 + j120 = 170∠45° VA
Q = 0 + 120 = 120 kVAR 
S = 150 kVA
 P = 150 cos 53.1° = 90 kW
 ⇒ 

Q = 150sin 53.1° = 120 kVAR
The power factor is pf = cos 45° = 0.707 lagging.
The current required by the plant is I =
P11.6-2
Load 1:
Load 2:
Total:
S 170 kVA
=
= 42.5 A .
V
4 kV
P1 = S cosθ = (12 kVA )( 0.7 ) = 8.4 kW
Q1 = S sin ( cos −1 (.7 ) ) = (12 kVA ) sin ( 45.6° ) = 8.57 kVAR
P2 = (10 kVA )( 0.8 ) = 8 kW
Q2 = 10sin ( cos −1 ( 0.8 ) ) = 10sin ( 36.9° ) = 6.0 kVAR
S = P + jQ = 8.4 + 8 + j ( 8.57 + 6.0 ) = 16.4 + j14.57 = 21.9∠41.6° kVA
The power factor is pf = cos ( 41.6° ) = 0.75 . The average power is P = 16.4 kW. The
apparent power is |S| = 21.9 kVA.
11-22
P11.6-3
The source current can be calculated from the apparent
power:
2( 50∠ cos -1 0.8 )
Vs I s∗
2S
∗
S=
⇒I =
=
= 5∠36.9° A
s
Vs
20∠0°
2
I s = 5∠ − 36.9° = 4 − j3 A
Next
I1 =
Vs
20∠0°
=
= 2∠ − 53.1° = 1.2 − j1.6 A
6 + j8 10∠53.1°
I 2 = I s − I1 = 4 − j 3 − 1.2 + j1.6 = 2.8 − j1.4
= 3.13∠ − 26.6° A
Finally,
Z=
Vs
20∠0°
=
= 6.39∠26.6° Ω
I z 3.13∠ − 26.6°
P11.6-4
(Using all rms values.)
2
(a)
V
2
2
P = I R=
⇒ V = P ⋅ R = ( 500 )( 20 ) ⇒
R
(b)
Is = I + I L =
V = 100 Vrms
V
V 100∠0° 100∠0°
+
=
+
= 5 − j 5 = 5 2∠ − 45° A
20 j 20
20
j 20
(c)
Zs = − j 20 +
( 20 )( j 20 ) = 10
20 + j 20
pf = cos ( −45° ) =
2∠ − 45° Ω
1
leading
2
11-23
(d)
No average power is dissipated in the capacitor or inductor. Therefore,
PAVE = PAVE = 500 W ⇒
20Ω
source
Vs I s cos θ = 500 ⇒ Vs =
500
=
I s cos θ
(
)
500
= 100 V
 1 
5 2 

 2
V = 100∠160° V
I = 2∠190° A = −1.97 − j 0.348 A
P1 = 23.2 W, Q1 = 50 VAR
P11.6-5
Load 1:
S1 = P1 + jQ1 = 23.2 + j 50 = 55.12∠65.1° VA
pf1 = cos 65.1° = 0.422 lagging
I1 =
∗
S1 55.12∠65.1°
=
= 0.551∠ − 94.9°, so I1 = 0.551∠94.9° A
100∠160°
Vs
I 2 = I - I1 = −1.97 − j 0.348 + 0.047 − j.549 = 2.12∠ − 155° A
S 2 = VI 2 = (100∠160° )( 2.12∠155° ) = 212∠ − 45° = 150 − j150 VA
Load 2:
∗
pf 2 = cos ( −45° ) = 0.707 leading
S = S1 + S 2 = ( 23.2 + j50 ) + (150 − j150 ) = 173.2 − j100 = 200∠ − 30° VA
pf = cos ( −30° ) = 0.866 leading
Total:
P11.6-6
Z refrig =
120
= 14.12 Ω
8.5
Z refrig = 14.12∠45° = 10 + j10 Ω
V 2 (120 )
=
=
= 144 Ω
P
100
2
R lamp
R range
Ι refrig =
(a)
( 240 )
=
2
12, 000
= 4.8 Ω
120∠0°
120∠0°
= 8.5∠ − 45° Arms , I lamp =
= 0.83∠0° Arms
10 + j10
144
and
11-24
I range =
240∠0°
= 50∠0° A
4.8
I1 = I refrig + I range = 56 − j 6 = 56.3∠ − 6.1° A
From KCL:
I 2 = −I lamp − I range = 50.83∠180° A
I N = −I1 − I 2 = 7.92∠ − 49° A
(b)
Prefrig = I refrig
2
Rrefrig = 722.5 W and Qrefrig = I refrig X refrig = 722.5 VAR
2
Plamp = 100 W and Q lamp = 0
Ptotal = 722 + 100 + 12, 000 = 12.82 kW 
 ⇒ S = 12,822 + j 722 = 12.84∠3.2° kVA
Qtotal = 722 + 0 + 0 = 722 VAR

The overall power factor is pf = cos ( 3.2° ) = 0.998 ,
(c)
Mesh equations:
Solve to get:
− 20
 30 + j10
 − 20
164

 −10 − j10 − 144
− 10 − j10  I A  120∠0°
− 144   I B  = 120∠0°

158.8 + j10   I C   0
I A = 54.3 − j1.57 = 54.3∠ − 1.7° Arms
I B = 51.3 − j 0.19 = 51.3∠ − 0.5°Arms
I C = 50 + j 0 = 50∠0° Arms
The voltage across the lamp is
Vlamp = Rlamp I B − IC = 144 1.27∠ − 8.6° = 183.2 V
11-25
VI=220 ( 7.6 ) = 1672 VA
P11.6-7
(a)
pf =
P 1317
=
= .788
VI 1672
θ =cos −1pf = 38.0° ⇒ Q=VIsinθ =1030VAR
To restore the pf to 1.0, a capacitor is required to eliminate Q by introducing –Q, then
V2
(220)2
⇒ X c = 47Ω
1030 =
=
Xc
Xc
(b)
1
ω X = (377)(47) = 56.5µ F
∴ C= 1
P = VI cosθ
(c)
where θ = 0°
then 1317 = 220I
∴ I = 6.0A for corrected pf
*
Note I = 7.6A for uncorrected pf
P11.6-8
First load:
S1 = P + jQ = P (1 + j tan (cos −1 (.6))) = 500(1 + j tan 53.1°) = 500 + j 677 kVA
S 2 = 400 + j 600 kVA
Second load:
S = S1 + S 2 = 900 + j1277 kVA
Total:
S desired = P + jP tan (cos −1 (.90)) = 900 + j 436 VA
From the vector diagram: S desired = S + Q . Therefore
900 + j 436 = 900 + j1277 + Q ⇒ Q = − j841 VAR
V
V
j
(1000) 2
*
=
−
⇒
=
=
= j1189 ⇒ Z = − j1189 = −
j841
Z
*
377 C
− j841 − j841
Z
2
Finally,
2
C=
1
= 2.20 µ F
(1189)(377)
11-26
P11.6-9
(a)
S = P + jQ = P + jP tan (cos −1 pf ) = 1000 + j1000 tan (cos −1 0.8) = 1000 + j 750 VA
S 1000 + j750
=
= 10 + j7.5 ⇒ I = 10 − j 7.5 A
Let VL = 100∠0° Vrms. Then I* =
VL
100∠0°
V
100∠0°
ZL = L =
= 8∠36.9° = 6.4 + j 4.8 V
I 12.5∠−36.9°
(b)
VL =[6.4 + j (200)(.024) + Z L )(I ) = (12.8 + j 9.6)(10 − j 7.5) = 200∠0° V
1
For maximum power transfer, we require ( 6.4 + j 4.8 )* = Z L || Z new =
.
YL +Ynew
1
1
1
= YL + Ynew ⇒ Ynew =
−
= j 0.15 S
(6.4 − j4.8)
6.4 − j 4.8 6.4 + j 4.8
Then Z new = − j 6.67 Ω so we need a capacitor given by
1
1
= 6.67 ⇒ C =
= 0.075 µ F
(6.67)(200)
ωC
Section 11-7: The Power Superposition Principle
P11.7-1
Use superposition since we have two different
frequency sources. First consider the dc source
(ω = 0):
 12 
I1 = 14 
 = 12 A
 12 + 2 
P1 = I12 R = (12) 2 (2) = 288 W
Next, consider the ac source (ω = 20 rad/s):
11-27
After a source transformation, current division gives
− j 60




25
(12− j 5)
∠116.6° A
I 2 = −9.166 
 =
60
−
j
5


+ 2+ j 4
 (12 − j 5)

Then
I
(125)(2)
P2 = 2 (2) =
=125 W
2
2
2
Now using power superposition
P = P1 + P2 = 288 + 125 = 413 W
P11.7-2
Use superposition since we have two different frequency sources. First consider ω = 2000 rad/s
source:
Current division yields
 8 
 −j2  5
I1 = 5 
∠63.4° A
=
5
 8 +8 
 − j 2 
Then
I 8
P1 = 1
= 20 W
2
2
Next consider ω = 8000 rad/s source.
Current division yields
11-28
 8 
 j7 
5
I 2 = − j5 
∠ − 171.9° A
=
8
50
 +8 
 j 7 
Then
I 8
P2 = 2
=2W
2
2
Now using power superposition
P = P1 + P2 = 22 W
P11.7-3
Use superposition since we have two different frequency. First consider the dc source (ω = 0):
1
i 2 (t ) = 0 and i1 (t ) = 10   = 1 A
 10 
PR1 = i12 R1 = 12 (10) = 10 W
PR 2 = 0 W
Next consider ω = 5 rad/s sources.
Apply KCL at the top node to get
−6 I 2 + I1 + I 2 − ( 4 ∠−30° ) +
(10 I1 −10∠40°)
=0
j10
−10 I1 + (5− j 2) I 2 = 0
Apply KVL to get
Solving these equations gives
I1 = −0.56 ∠ − 64.3° A and I 2 = −1.04∠ − 42.5° A
Then
11-29
PR1 =
I12 R1
2
( 0.56 )
=
2
(10)
2
= 1.57 W and PR 2 =
I 22 R 2
2
=
(1.04 )
2
(5)
2
= 2.7 W
Now using power superposition
PR = 10 + 1.57 = 11.57 W and PR 2 = 0 + 2.7 = 2.7 W
1
P11.7-4
Use superposition since we have two different frequency. First consider the ω = 10 rad/s source:
V1 4∠0°
=
= 0.28 + j 0.7 A
Z 2− j5
V = 2 I1 = 2(0.28 + j.7) = 0.56 + j1.4 = 1.51 ∠68.2° V
I1 =
V = − j 5 I1 = 3.77 ∠ − 21.8° V
R1
C1
Next consider ω = 5 rad/s source.
V2 6∠−90°
=
= 0.577 − j 0.12 A
Z
2 − j10
VR = 2 I 2 = 2(.577 − j 0.12) =1.15− j 0.24
I2 =
2
V
C2
=1.17∠−11.8° V
= − j10I 2 =5.9∠258.3° V
Now using superposition
vR (t) =1.51cos (10 t + 68.2°) + 1.17 cos (5 t − 11.8°) V
vC (t) = 3.77 cos (10 t − 21.8°) +5.9 cos (5 t − 258.3°) V
Then
 1.51   1.17 
=
 +
 = 1.82 ⇒ VReff =1.35 V
 2  2 
2
2
Reff
V
 3.77   5.9 
=
 +
 = 24.52 ⇒ VCeff = 4.95 V
 2   2
2
2
Ceff
V
2
2
11-30
11-31
Section 11-8: Maximum Power Transfer Theorem
Z t = 4000 || − j 2000 = 800 − j1600 Ω
P11.8-1
Z L = Z*t =800 + j1600 Ω
 R =800 Ω
R + j1000 L = 800 + j1600 ⇒ 
 L =1.6 H
P11.8-2
Z t = 25, 000 || − j 50, 000 = 20, 000 − j10, 000 Ω
Z L =Z*t = 20,000+ j10,000 Ω
 R = 20 kΩ

R + jω L = 20, 000 + j10, 000 ⇒ 100 L =10,000

L =100 H

After selecting these values of R and L,
I = 1.4 mA and Pmax
 0.14×10−2 
3
= 
 ( 20×10 ) = 19.5 mW
2


2
Since Pmax > 12 mW , yes, we can deliver 12 mW to the load.
 −j 
R
ω C  R − jω R 2C

Z t = 800 + j1600 Ω and Z L =
=
j
1 + (ω RC ) 2
R−
ωC
 −j 
R

2
 ω C  = R − jω R C = 800 − j1600 Ω
Z L = Z*t ⇒
j
1+ (ω RC ) 2
R−
ωC
Equating the real parts gives
P11.8-3
800 =
R
4000
=
2
1+ (ω RC ) 1+[(5000)(4000)C ]2
⇒ C = 0.1 µ F
11-32
P11.8-4
Z t = 400 + j 800 Ω and Z L = 2000 || − j1000 = 400 − j 800 Ω
Since Z L = Z*t the average power delivered to the load is maximum and cannot be increased by
adjusting the value of the capacitance. The voltage across the 2000 Ω resistor is
VR = 5
So
ZL
= 2.5 − j 5 = 5.59e− j 63.4 V
Zt + ZL
 5.59  1
P=
= 7.8 mW

 2  2000
2
is the average power delivered to the 2000 Ω resistor.
P11.8-5
Notice that Zt,not ZL, is being adjusted .When Zt is fixed, then the average power delivered to
the load is maximized by choosing ZL = Zt*. In contrast, when ZL is fixed, then the average
power delivered to the load is maximized by minimizing the real part of Zt. In this case, choose
R = 0. Since no average power is dissipated by capacitors or inductors, all of the average power
provided by source is delivered to the load.
Section 11-9: Mutual Inductance
P11-9-1
Vs + I jω L1 + I jω M + I jω L2 − I jω M = 0
⇒ jω ( L1 + L2 − 2 M ) =
Therefore
Vs
I
Lab = L1 + L2 − 2M
11-33
P11.9-2
I1 + I 2 = I s
KCL:
The coil voltages are given by:
V = I1 jω L1 + I 2 jω M
Then
I2 =
V = I 2 jω L2 + ( I s − I 2 ) jω M
and
Then
V − jω L1I s
jω ( M − L1 )
V = I 2 jω L2 + I1 jω M
V=
(V − jω L1I s )  jω ( L2 − M ) 
jω ( M − L1 )
Finally
Lab =
+ jω MI s
⇒
 L L −M 2 
V
= jω  1 2

Is
 L1 + L2 − 2M 
L1 L2 − M 2
L1 + L2 − 2M
P11.9-3
Mesh equations:
−141.4∠0° + 2 I1 + j 40 I1 − j 60 I 2 = 0
l200 I 2 + j 60 I 2 − j 60 I1 = 0 ⇒ I 2 = (0.23∠51°) I1
Solving yields
Finally
I1 = 4.17∠ − 68° A and I 2 = 0.96∠ − 17°A
i1 ( t ) = 4.2 cos(100t −68°) A and i 2 ( t ) = 1.0 cos(100t −17°) A
11-34
P11.9-4
(10+ j5) I 1 − j50 I 2 = 10
− j 50 I1 +( 400+ j 500 ) I 2 = 0
Mesh equations:
Solving the mesh equations using Cramer’s rule:
I2 =
Then
(10+ j5)( 0 )−( − j50 )(10 )
2
(10+ j5) ( 400+ j500 ) − ( − j50 )
= 0.062 ∠29.7° A
V2 400 I 2
=
= 40 I 2 = 40 ( 0.062 ∠29.7° ) = 2.5 ∠29.7°
V1 10∠0°
P11.9-5
Mesh equations:
−10∠0° − j 5 I1 + j 9 I1 + j 3 I 2 = 0
28 I 2 + j 6 I 2 + j 3 I1 + j 9 I 2 − j 3 I 2 = 0
Solving the mesh equations yields
I1 = 0.25∠161° A and I 2 = 2.55∠ − 86° A
then
Finally
V = j 9(I1 − I 2 ) = j 9 ( 2.6 ∠−81° ) = 23 ∠9° V
v (t ) = 23cos (30t + 9°) V
11-35
P11.9-6
(a)
I 2 = 0 ⇒ I1 = 10 ∠0° A ⇒ i1 (0) = 10 A
w=
2
L1 i1 (0)
2
=
(0.3) (10) 2
= 15 J
2
Mesh equations:
(b)
j 6 I 2 - j 3 I1 = 0 ⇒ I1 = 2 I 2
I1 = 10∠0° A ⇒ I 2 = 5∠0° A
Then
w=
1
1
1
1
2
2
L1i1 (0) + L 2i1 (0) − M i1 (0) i 2 (0) = (0.3)(10) 2 + (1.2)(5) 2 − (0.6) (10)(5) = 0
2
2
2
2
(7 + j 6) I 2 − j 3 I1 = 0
(c)
I 2 = 3.25 ∠49.4° A
i 2 (t ) = 3.25cos(5t + 49.4°) A
i 2 (0) = 2.12 A
Finally
1
1
w = (0.3) (10) 2 + (1.2) (2.12) 2 − (0.6) (10) (2.12) = 5.0 J
2
2
P11.9-7
Mesh equations:
−VT + j8 I1 + j 5(I1 − I 2 ) − j 6 I1 + j 6 (I1 − I 2 ) + j5 I1 = 0
3 I 2 + j 6 (I 2 − I1 ) − j 5 I1 = 0
11-36
I 2 = (1.64 ∠27° ) I1
Solving yields
I1 ( j 18) + I 2 (− j 11) = VT
Then
Z =
VT
= 8.2 + j 2 = 8.4 ∠14° Ω
I1
P11.9-8
The coil voltages are given by
V1 = j 6 I1 − j 2 (I1 − I 2 ) − j 4 I 2 = j 4 I1 − j 2 I 2
V2 = j 4 ( I1 − I 2 ) − j 2 I1 + j 2 I 2 = j 2 I1 − j 2 I 2
V3 = j8 I 2 − j 4 I1 + j 2 ( I1 − I 2 ) = − j 2 I1 + j 6 I 2
The mesh equations are
5 I1 + V1 + 6 (I1 − I 2 ) + V2 = 10∠0°
−V2 + 6 (I 2 − I1 ) + 2 I 2 + V3 − j 5 I 2 = 0
Combining and solving yields
11 + j 6 10
−6 − j 4 0
60 + j 40
=
= 1.2 ∠0.28° A
I2 =
11 + j 6 −6 − j 4
50 + j 33
−6 − j 4 8 + j 3
Finally
V = − j 5 I 2 = 6.0 ∠ − 89.72° A ⇒ v(t ) = 6sin(2 t − 89.7°) V
11-37
Section 11-10: The Ideal Transformer
P11.10-1
Z = (2 + j 3) +
I1 =
(100− j 75)
=6 Ω
52
12∠0° 12∠0°
=
=2A
6
Z
 100− j 75 
 100− j 75 
V1 = I1 
 = (2) 
 = 10∠ − 36.9° V
2
n


 25 
V2 = nV1 = 5 (10∠−36.9°) = 50 ∠−36.9° V
I2 =
I1 2
=
A
n 5
P11.10-2
(a)
V0 = (5 ×10−3 )(10, 000) = 50 V
n=
V
N2
50
= 0 =
= 5
V1
N1
10
(b)
Rab =
(c)
Is =
1
1
R2 = (10 × 103 ) = 400 Ω
2
n
25
10
10
=
= 0.025 A = 25 mA
Rab 400
11-38
P11.10-3
Z1 =
1
Z 2 = 9 Z 2 = 9(5 − j8) = 45 − j 72 Ω
n2
Using voltage division, the voltage across Z 1 is


45− j 72
V1 = ( 80∠ − 50° ) 
 = 74.4 ∠ − 73.3° V
 45− j 72+ 30 + j 20 
then
V2 = nV1 =
74.4 ∠ − 73.3°
= 24.8 ∠ − 73.3° V
3
Using voltage division again yields
 − j8 
 8∠−90° 
= ( 24.8∠−73.3° ) 
Vc = V2 

 = 21.0∠ − 105.3° V
 89∠−58° 
 5− j 8 
P11.10-4
n = 5, Z1 =
( 5)
200
2
= 8 Ω ⇒ V1 =
8
( 50∠0° ) = 40∠0° V ⇒ V2 = n V1 = 200∠0° V
8+ 2
11-39
P11.10-5
Z=
320 jω L
+ 2
n2
n
Maximum power transfer requires
jω L
= j160 kΩ and
n2
320
= 80
n2
so n = 2. Then ω L = 640 kΩ so
L =
640×103
= 6.4 H
105
P11.10-6
Z=
1
( 2 + 6) = 2 Ω
22

 6   2 
Voc = 
 ( 2)  
16∠0°  = 12∠0° V
 6 + 2   2 + 2 

Z=
1
1
2 = Ω
2 ( )
2
2



1
1 16∠0° 
I sc = −I 2 = I1 = 
 = 3.25∠0° A
2
2 2+ 1 

2 
Then
Zt =
12∠0°
= 3.75∠0° Ω
3.2∠0°
11-40
P11.10-7
V2 =
1
V1
2
V − V2 V1
I3 = 1
=
2
4
V
V
I 2 = I3 − 2 = 1
6
6
V
1
I1 = − I 2 = − 1
2
12
V
I T = I 3 − I1 = 1
6
V
Z= 1 =6
IT
P11.10-8
*
Maximum power transfer requires Z L = Z t . First
 1 
1
2
1
=
X C1  2  = X L1 , 2 =
10 5
n2
 n2 
then
  1   1 
 RL  2  + 1  2  = 100 Ω ⇒
  n2    n1 
⇒
n2 = 5
1
100
3
=
⇒ n1 =
2
3
10
n1
11-41
P11.10-9
ZL =
1  20 (1+ j 7.54)  8.1∠23°
=
= 0.3 + j 0.13 Ω
52  20+10+ j 7.54 
25
V
V
( 230 ) = 88 kW/home
PL = L = 2 =
2 R 2 2 R L 2( 0.3)
2
2
2
Therefore, 529 kW are required for six homes.
P11.10-10
(a)
Coil voltages:
V1 = j16 I1
V2 = j12 I 2
Mesh equations:
8 I1 + V1 − 5∠45° = 0
−12 I 2 − V2 = 0
Substitute the coil voltages into the mesh equations and do some algebra:
11-42
8 I1 + j16 I1 = 5∠45° ⇒ I1 = 0.28∠ − 18.4°
12 I 2 + j12 I 2 = 0 ⇒ I 2 = 0
V2 = −12 I 2 = 0
(b)
V1 = j16 I1 + j8 I 2
Coil voltages:
V2 = j12 I 2 + j8 I1
Mesh equations:
8 I1 + V1 − 5∠45° = 0
−12 I 2 − V2 = 0
Substitute the coil voltages into the mesh equations and do some algebra:
8 I1 + ( j16 I1 + j8 I 2 ) = 5∠45°
12 I 2 + ( j12 I 2 + j8 I1 ) = 0
I1 = −
12 + j12
3
I 2 = ( j − 1) I 2
j8
2


3
( 8 + j16 )  2  ( j − 1) + j8 I 2 = 5∠45° ⇒ I 2 = 0.138∠ − 141°
 


V2 = −12 I 2 = 1.656∠39°
11-43
(c )
Coil voltages and currents:
V1 =
10
V2
8.66
8.66
I1 = −
I2
10
Mesh equations:
8 I1 + V1 − 5∠45° = 0
−12 I 2 − V2 = 0
Substitute into the second mesh equation and do some algebra:
 10  8.66
 10 
−12  −
I1  =
V1 ⇒ V1 = 12 
 I1
 8.66  10
 8.66 
2
 10 
8 I1 + 12 
 I1 = 5∠45° ⇒ I1 = 0.208∠45°
 8.66 
2
 10  12 (10 )
0.208∠45° = 2.88∠45°
V2 = −12 I 2 = −12  −
I1  =
8.66
 8.66 
11-44
PSpice Problems
SP 11-1
The coupling coefficient is k =
3
= 0.94868 .
2×5
11-45
SP 11-2
Here is the circuit with printers inserted to measure the coil voltages and currents:
Here is the output from the printers, giving the voltage of coil 2 as 2.498∠107.2°, the
current of coil 1 as 0.4484∠-94.57°, the current of coil 2 as 0.6245∠-72.77° and the
voltage of coil 1 as 4.292∠-58.74°:
FREQ
7.958E-01
VM(N00984)
2.498E+00
VP(N00984)
1.072E+02
FREQ
IM(V_PRINT1)IP(V_PRINT1)
7.958E-01
4.484E-01 -9.457E+01
FREQ
IM(V_PRINT2)IP(V_PRINT2)
7.958E-01
6.245E-01 -7.277E+01
FREQ
7.958E-01
VM(N00959) VP(N00959)
4.292E+00 -5.874E+01
The power received by the coupled inductors is
p=
( 4.292 )( 0.4484 ) cos
2
= 0.78016 − .78000 ≈ 0
( −58.74 − ( −94.57 ) ) +
( 2.498)( 0.6245) cos
2
(107.2 − ( −72.77 ) )
11-46
SP 11-3
The inductance are selected so that
L2
L1
=
N2
N1
=
3
and the impedance of these inductors
2
are much larger that other impedance in the circuit. The 1 GΩ resistor simulates an open
circuit while providing a connected circuit.
Here is the output from the printers, giving the voltage of coil 2 as 24.00∠114.1°, the
current of coil 1 as 4.000∠114.0°, the current of coil 2 as 2.667∠-65.90° and the voltage
of coil 1 as 16.00∠114.1°:
FREQ
6.366E-01
VM(N00984)
2.400E+01
VP(N00984)
1.141E+02
FREQ
IM(V_PRINT1)IP(V_PRINT1)
6.366E-01
4.000E+00
1.140E+02
FREQ
IM(V_PRINT2)IP(V_PRINT2)
6.366E-01
2.667E+00 -6.590E+01
FREQ
6.366E-01
VM(N00959)
1.600E+01
The power received by the transformer is
p=
(16 )( 4 ) cos
(114 − 114 ) +
2
= 32 − 32.004 ≈ 0
VP(N00959)
1.141E+02
( 24 )( 2.667 ) cos
2
(114 − ( −66 ) )
11-47
SP 11-4
The inductance are selected so that
L2
L1
=
N2
N1
=
2
and the impedance of these inductors
5
are much larger that other impedance in the circuit. The 1 GΩ resistor simulates an open
circuit while providing a connected circuit.
FREQ
6.366E-01
VM(N00921)
1.011E+02
VP(N00921)
7.844E+01
VR(N00921)
2.025E+01
VI(N00921)
9.903E+01
The printer output gives the voltage across the current source as
20.25 + j 99.03 = 101.1∠78.44° V
The input impedance is
Zt =
20.25 + j 99.03
= 20.25 + j 99.03 Ω = 101.1∠78.44° Ω
1
 52 
(We expected Z t = 8 +  2  ( 2 + j ( 4 )( 4 ) ) = 20.5 + j 100 Ω . That’s about 1% error.)
2 
11-48
Verification Problems
VP 11-1
The average power supplied by the source is
Ps =
(12 )( 2.327 ) cos
2
( 30° − ( −25.22° ) ) = 7.96
W
Capacitors and inductors receive zero average power, so the sum of the average powers
received by the other circuit elements is equal to the sum of the average powers received
by the resistors:
2.327 2
1.1292
PR =
( 4) +
( 2 ) = 10.83 + 1.27 = 12.10 W
2
2
The average power supplied by the voltage source is equal to the sum of the average
powers received by the other circuit elements. The mesh currents cannot be correct.
(What went wrong? It appears that the resistances of the two resistors were interchanged
when the data was entered for the computer analysis. Notice that
PR =
2.327 2
1.1292
( 2) +
( 4 ) = 5.41 + 2.55 = 7.96 W
2
2
The mesh currents would be correct if the resistances of the two resistors were
interchanged. The computer was used to analyze the wrong circuit.)
VP 11-2
The average complex supplied by the source is
Ss =
(12∠30° )(1.647∠ − 17.92° ) * = (12∠30° )(1.647∠17.92° ) = 9.88∠47.92° = 6.62 + j 7.33
2
2
The complex power received by the 4 Ω resistor is
S4Ω =
( 4 ×1.647∠ − 17.92° )(1.647∠ − 17.92° ) * = 5.43 + j 0
2
VA
The complex power received by the 2 Ω resistor is
S2Ω =
( 2 ×1.094∠ − 13.15° )(1.094∠ − 13.15° ) * = 1.20 + j 0
2
VA
11-49
W
The current in the 2 H inductor is
(1.647∠ − 17.92° ) − (1.094∠ − 13.15° ) = 0.5640∠ − 27.19°
The complex power received by the 2 H inductor is
S 2H =
( j 8 × 0.5640∠ − 27.19° )( 0.5640∠ − 27.19° ) * = 0 + j 1.27
VA
2
The complex power received by the 4 H inductor is
S 4H =
( j 16 ×1.094∠ − 13.15° )(1.094∠ − 13.15° ) * = 0 + j 9.57
2
VA
S 4 Ω + S 2 Ω + S 2H + S 4H = ( 5.43 + j 0 ) + (1.20 + j 0 ) + ( 0 + j 1.27 ) + ( 0 + j 9.57 ) = 6.63 + j 10.84 ≠ Ss
The complex power supplied by the voltage source is equal to the sum of the complex
powers received by the other circuit elements. The mesh currents cannot be correct.
(Suppose the inductances of the inductors were interchanged. Then the complex power
received by the 4 H inductor would be
S 4H =
( j 16 × 0.5640∠ − 27.19° )( 0.5640∠ − 27.19° ) * = 0 + j 2.54
2
VA
The complex power received by the 2 H inductor would be
S 2H =
( j 8 ×1.094∠ − 13.15° )(1.094∠ − 13.15° ) * = 0 + j 4.79
2
VA
S 4 Ω + S 2 Ω + S 2H + S 4H = ( 5.43 + j 0 ) + (1.20 + j 0 ) + ( 0 + j 2.54 ) + ( 0 + j 4.79 ) = 6.63 + j 7.33 ≈ S s
The mesh currents would be correct if the inductances of the two inductors were
interchanged. The computer was used to analyze the wrong circuit.)
VP 11-3
The voltage across the right coil must be equal to the voltage source voltage. Notice that
the mesh currents both enter the undotted ends of the coils. In the frequency domain, the
voltage across the right coil is
11-50
( j 16 )(1.001∠ − 47.01°) + ( j12 )( 0.4243∠ − 15° ) = 16.016∠42.99° + 5.092∠75°
= (11.715 + j 10.923) + (1.318 + j 4.918 )
= 13.033 + j 15.841
= 20.513∠50.55°
This isn’t equal to the voltage source voltage so the computer analysis isn’t correct.
What happened? A data entry error was made while doing the computer analysis. Both
coils were described as having the dotted end at the top. If both coils had the dot at the
top, the equation for the voltage across the right coil would be
( j 16 )(1.001∠ − 47.01°) − ( j12 )( 0.4243∠ − 15° ) = 16.016∠42.99° − 5.092∠75°
= (11.715 + j 10.923) − (1.318 + j 4.918 )
= 10.397 + j 6.005
= 12.007∠30.01°
This is equal to the voltage source voltage. The computer was used to analyze the wrong
circuit.
VP 11-4
First check the ratio of the voltages across the coils.
n1 2
12∠30°
= 2.5 ≠
=
n2 5
( 75 )( 0.064∠30° )
The transformer voltages don’t satisfy the equations describing the ideal transformer. The
given mesh currents are not correct.
That’ enough but let’s also check the ratio of coil currents. (Notice that the reference
direction of the i2(t) is different from the reference direction that we used when
discussing transformers.)
n1 2
0.064∠30°
= 2.5 ≠
=
n2 5
0.0256∠30°
The transformer currents don’t satisfy the equations describing the ideal transformer.
n1
to be 2.5 instead of 0.4 =
1
. This suggests that a data
2.5
n2
entry error was made while doing the computer analysis. The numbers of turns for the
two coils was interchanged.
In both case, we calculated
11-51
11-52
11-53
Design Problems
DP 11. 1
(a)
P 100

S= =
=125 kVA
P =100 W 

pf 0.8
 ⇒ 
pf = 0.8 
Q = S sin (cos −1 0.8) =125sin (36.9°) =75 kVAR

Now pf = 0.95 so
S=
P 100
=
=105.3 kVA
pf 0.95
Q = S sin (cos −1 0.95) =105.3sin (18.2°) =32.9 kVAR
so an additional 125 − 105.3 = 19.7 kVA is available.
(b)
Now pf = 1 so
S=
P 100
=
=100 kVA
pf 1
Q = S sin (cos −1 1) = 0
and an additional 125-100= 25 kVA is available.
(c)
In part (a), the capacitors are required to reduce Q by 75 – 32.9 = 42.1 kVAR. In
part (b), the capacitors are required to reduce Q by 75 – 0 = 0 kVAR.
(d)
Corrected power factor
Additional available
apparent power
Reduction in reactive power
0.95
19.7 kVA
1.0
25 kVA
42.1 kVAR
75 kVAR
11-54
DP 11-2
This example demonstrates that loads can be specified either by kW or kVA. The
procedure is as follows:
First load:
Second load:
Total load:
P1 = S1 pf =( 50 )( 0.9 ) = 45 W
S1 =50 VA 

 ⇒ 
−1
pf =0.9 
Q1 = S1 sin (cos 0.9) =50sin (25.8°) = 21.8 kVAR
P 45

= 49.45 kVA
S2 = 2 =
P2 = 45 W 

pf 0.91
 ⇒ 
pf = 0.91 
Q = S sin (cos −1 0.91) = 49.45sin (24.5°) = 20.5 kVAR
 2 2
S L = S1 + S 2 = (45 + 45) + j(21.8+20.5) = 90 + j 42.3 kVA
P 90

=92.8 kVA
Ss = s =
Ps =90 W 

pf 0.97
 ⇒ 
pf = 0.97 
Q = S sin (cos −1 0.97) =92.8sin (14.1°) = 22.6 kVAR
 s s
Specified load:
The compensating capacitive load is Qc = 42.3 − 22.6 = 19.7 kVAR .
The required capacitor is calculated as
Xc =
Vc
(7.2 × 103 ) 2
1
=
= 2626 Ω ⇒ C =
= 1.01 µ F
3
Qc
19.7 × 10
377 (2626)
2
11-55
DP 11-3
Find the open circuit voltage:
−10 + 5I + j10I − 0.5 Voc = 0
and
I=
10 − Voc
5
so
Voc = 8∠36.9° = 6.4 + j 4.8 V
Find the short circuit current:
I sc =
10∠0°
= 2∠0° A
5
The the Thevenin impedance is:
Zt =
The short circuit forces the controlling voltage to be
zero. Then the controlled voltage is also zero.
Consequently the dependent source has been replaced
by a short circuit.
Voc
= 3.2 + j 2.4 Ω
I sc
(a)
Maximum power transfer requires Z L = Z t * = 3.2 − j 2.4 Ω .
(c)
ZL can be implemented as the series combination of a resistor and a capacitor with
R = 3.2 Ω and C =
(b)
Pmax =
1
= 4.17 mF .
(100 ) (2.4)
| Voc |2
64
=
= 2.5 W
8R
8 ( 3.2 )
11-56
DP 11-4
Using an equation from section 11.8, the power is given as
 R  Vs
 2
n  2
P=
2
2
R  3


3 + 2  +  2 + 4
n  n


2
When R = 4 Ω,
n 2 R Vs
P=
25n 4 + 48n 2 + 25
2
4
2
2
3
dP
2  2 n(25n + 48n + 25) − n (100n + 96n ) 
= R Vs 

dn
(25n 4 + 48n 2 + 25) 2


5
4
⇒ − 50n + 50n = 0 ⇒ n = 1 ⇒ n = 1
0=
When R = 8 Ω, a similar calculation gives n = 1.31.
11-57
DP 11-5
Maximum power transfer requires
10 + j 6.28
= Z1 = (1 + j 0.628 ) *
n2
Equating real parts gives
10
= 1 ⇒ n = 3.16
n2
Equating imaginary parts requires
jX
= − j 0.628 ⇒
3.162
X = −6.28
This reactance can be realized by adding a capacitance C in series with the resistor and
inductor that comprise Z2. Then
−6.28 = X = −
1
1
+ 6.28 ⇒ C =
= 0.1267 µ F
5
5
( 2π ×10 ) C
( 2π ×10 ) (12.56 )
11-58
DP 11-6
Maximum power transfer requires
1
|| R = (100 + j107 ×10−6 ) *
j107 C
R
= 100 − j10
1 + j107 R C
R = (100 − j10) (1 + j107 R C ) = 100 + 108 R C + j (109 R C − 10 )
Equating real and imaginary parts yields
R = 100 + 108 R C and 109 R C − 10 = 0
then
RC = 10−8
 10−8 
10−8
99
C
⇒ R = 100 + 108 R 
=
Ω
⇒
=
= 0.101 nF

99
 R 
11-59
Chapter 12: Three-Phase Circuits
Exercises
Ex. 12.3-1
VC = 120∠ − 240° so VA = 120∠0° and VB = 120∠ − 120°
Vbc = 3 (120 ) ∠ − 90°
Ex. 12.4-1
Four-wire Y-to-Y Circuit
Mathcad analysis (12v4_1.mcd):
Vp := 120
Describe the three-phase source:
Va := Vp⋅ e
j⋅
π
180
⋅0
Vb := Va⋅ e
Describe the three-phase load:
Calculate the line currents:
IaA = 1.079 − 0.674i
IaA = 1.272
180
π
⋅ arg( IaA ) = −32.005
IaA :=
j⋅
π
180
⋅ − 120
ZA := 80 + j⋅ 50
IbB :=
Va
ZA
Vc := Va⋅ e
180
π
π
180
⋅ 120
ZB := 80 + j⋅ 80
Vb
ZB
IbB = −1.025 − 0.275i
IbB = 1.061
j⋅
⋅ arg( IbB) = −165
IcC :=
ZC := 100 − j⋅ 25
Vc
ZC
IcC = −0.809 + 0.837i
IcC = 1.164
180
π
⋅ arg( IcC) = 134.036
12-1
INn := IaA + IbB + IcC
Calculate the current in the neutral wire:
INn = −0.755 − 0.112i
Calculate the power delivered to the load:


SA := IaA ⋅ IaA ⋅ ZA
SB := IbB⋅ IbB⋅ ZB
SA = 129.438 + 80.899i

SC := IcC⋅ IcC⋅ ZC
SB = 90 + 90i
SC = 135.529 − 33.882i
Total power delivered to the load: SA + SB + SC = 354.968 + 137.017i
Calculate the power supplied by the source:

Sa := IaA ⋅ Va

Sb := IbB⋅ Vb
Sa = 129.438 + 80.899i

Sc := IcC⋅ Vc
Sb = 90 + 90i
Sc = 135.529 − 33.882i
Sa + Sb + Sc = 354.968 + 137.017i
Total power delivered by the source:
Ex. 12.4-2
Four-wire Y-to-Y Circuit
Mathcad analysis (12x4_2.mcd):
Vp := 120
Describe the three-phase source:
Va := Vp⋅ e
j⋅
π
180
⋅0
Vb := Va⋅ e
Describe the three-phase load:
Calculate the line currents:
IaA = 1.92 − 1.44i
IaA = 2.4
180
π
⋅ arg( IaA ) = −36.87
IaA :=
j⋅
π
180
⋅ − 120
ZA := 40 + j⋅ 30
IbB :=
Va
ZA
Vc := Va⋅ e
180
π
π
180
⋅ 120
ZB := ZA
Vb
ZB
IbB = −2.207 − 0.943i
IbB = 2.4
j⋅
⋅ arg( IbB) = −156.87
IcC :=
ZC := ZA
Vc
ZC
IcC = 0.287 + 2.383i
IcC = 2.4
180
π
⋅ arg( IcC) = 83.13
12-2
INn := IaA + IbB + IcC
Calculate the current in the neutral wire:
INn = 0
Calculate the power delivered to the load:


SA := IaA ⋅ IaA ⋅ ZA
SB := IbB⋅ IbB⋅ ZB
SA = 230.4 + 172.8i

SC := IcC⋅ IcC⋅ ZC
SB = 230.4 + 172.8i
SC = 230.4 + 172.8i
Total power delivered to the load: SA + SB + SC = 691.2 + 518.4i
Calculate the power supplied by the source:

Sa := IaA ⋅ Va

Sb := IbB⋅ Vb
Sa = 230.4 + 172.8i

Sc := IcC⋅ Vc
Sb = 230.4 + 172.8i
Sc = 230.4 + 172.8i
Sa + Sb + Sc = 691.2 + 518.4i
Total power delivered by the source:
Ex. 12.4-3
Three-wire unbalanced Y-to-Y Circuit with line impedances
Mathcad analysis (12x4_3.mcd):
Vp := 120
Describe the three-phase source:
Va := Vp⋅ e
j⋅
π
180
⋅0
Vb := Va⋅ e
Describe the three-phase load:
j⋅
π
180
⋅ − 120
ZA := 80 + j⋅ 50
Vc := Va⋅ e
j⋅
π
180
⋅ 120
ZB := 80 + j⋅ 80
ZC := 100 − j⋅ 25
Calculate the voltage at the neutral of the load with respect to the neutral of the source:
VnN :=
ZA ⋅ ZC⋅ e
4
j⋅ ⋅ π
3
+ ZA ⋅ ZB⋅ e
2
j⋅ ⋅ π
3
+ ZB⋅ ZC
ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC
VnN = −25.137 − 14.236i
VnN = 28.888
⋅ Vp
180
π
⋅ arg( VnN) = −150.475
12-3
IaA :=
Calculate the line currents:
Va − VnN
ZA
IaA = 1.385 − 0.687i
π
Check:
Vb − VnN
IcC :=
ZB
IbB = −0.778 − 0.343i
IaA = 1.546
180
IbB :=
180
π
IaA + IbB + IcC = 0
IcC = 1.195
⋅ arg( IbB) = −156.242
180
π
Calculate the power delivered to the load:


SA := IaA ⋅ IaA ⋅ ZA
SB := IbB⋅ IbB⋅ ZB
SA = 191.168 + 119.48i
⋅ arg( IcC) = 120.475

SC := IcC⋅ IcC⋅ ZC
SB = 57.87 + 57.87i
Total power delivered to the load:
ZC
IcC = −0.606 + 1.03i
IbB = 0.851
⋅ arg( IaA ) = −26.403
Vc − VnN
SC = 142.843 − 35.711i
SA + SB + SC = 391.88 + 141.639i
Ex. 12.4-4
Three-wire balanced Y-to-Y Circuit with line impedances
Mathcad analysis (12x4_4.mcd):
Vp := 120
Describe the three-phase source:
Va := Vp⋅ e
j⋅
π
180
⋅0
Describe the three-phase load:
Vb := Va⋅ e
j⋅
π
180
⋅ − 120
ZA := 40 + j⋅ 30
Vc := Va⋅ e
ZB := ZA
j⋅
π
180
⋅ 120
ZC := ZA
Calculate the voltage at the neutral of the load with respect to the neutral of the source:
VnN :=
ZA ⋅ ZC⋅ e
4
j⋅ ⋅ π
3
+ ZA ⋅ ZB⋅ e
2
j⋅ ⋅ π
3
+ ZB⋅ ZC
ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC
⋅ Vp
12-4
− 14
VnN = −1.31 × 10
− 14
+ 1.892i× 10
Calculate the line currents:
IaA :=
− 14
VnN = 2.301 × 10
Va − VnN
ZA
IaA = 1.92 − 1.44i
π
ZB
IbB = 2.4
⋅ arg( IaA ) = −36.87
Check:
Vb − VnN
IbB = −2.207 − 0.943i
IaA = 2.4
180
IbB :=
180
180
π
− 15
IaA + IbB + IcC = 1.055 × 10
IcC :=
ZC
IcC = 2.4
180
π
− 15
SA + SB + SC = 691.2 + 518.4i
Vc − VnN
IcC = 0.287 + 2.383i
− 2.22i × 10
SB = 230.4 + 172.8i
Total power delivered to the load:
⋅ arg( VnN) = 124.695
⋅ arg( IbB) = −156.87
Calculate the power delivered to the load:


SA := IaA ⋅ IaA ⋅ ZA
SB := IbB⋅ IbB⋅ ZB
SA = 230.4 + 172.8i
π
⋅ arg( IcC) = 83.13

SC := IcC⋅ IcC⋅ ZC
SC = 230.4 + 172.8i
Ex. 12.6-1
(See Table 12.5-1)
Z ∆ = 180∠ − 45°
Balanced delta
load:
phase currents:
I AB =
I BC =
I CA
line currents:
360∠0°
VAB
=
= 2∠45° A
Z ∆ 180∠− 45D
VBC 360∠−120°
=
= 2∠− 75° A
°
Z ∆ 18∠− 45
VCA 360∠120°
=
=
= 2∠165° A
°
Z ∆ 180∠− 45
I A = I AB − I CA = 2∠45° − 2∠165° = 2 3∠15° A
I B = 2 3∠−105° A
I C = 2 3∠135° A
12-5
Ex. 12.7-1
Three-wire Y-to-Delta Circuit with line impedances
Mathcad analysis (12x4_4.mcd):
Vp := 110
Describe the three-phase source:
Va := Vp⋅ e
j⋅
π
180
⋅0
Vb := Va⋅ e
j⋅
π
180
⋅ − 120
Z1 := 150 + j⋅ 270
Describe the delta connected load:
Vc := Va⋅ e
Z1⋅ Z3
Z1 + Z2 + Z3
ZA = 50 + 90i
Describe the three-phase line:
ZB :=
Z2⋅ Z3
Z1 + Z2 + Z3
ZB = 50 + 90i
ZaA := 10 + j⋅ 25
π
180
Z2 := Z1
Convert the delta connected load to the equivalent Y connected load:
ZA :=
j⋅
ZC :=
⋅ 120
Z3 := Z1
Z1⋅ Z2
Z1 + Z2 + Z3
ZC = 50 + 90i
ZbB := ZaA
ZcC := ZaA
12-6
Calculate the voltage at the neutral of the load with respect to the neutral of the source:
VnN :=
( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e
− 14
IaA :=
IaA = 0.392 − 0.752i
IaA = 0.848
Check:
− 14
− 14
+ 1.784i× 10
Calculate the line currents:
π
+ ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e
⋅ arg( IaA ) = −62.447
VnN = 2.135 × 10
Va − VnN
IbB :=
ZA + ZaA
IbB = −0.847 + 0.036i
Vb − VnN
ZB + ZbB
IbB = 0.848
180
π
⋅ arg( IbB) = 177.553
VAN := IaA ⋅ ZA
VAN = 87.311
π
⋅ arg( VAN) = −1.502
VBN := IbB⋅ ZB
VBN = 87.311
180
π
⋅ arg( VBN) = −121.502
Calculate the line-to-line voltages at the load:
VAB := VAN − VBN
VAB = 151.227
180
π
+ ( ZbB + ZB) ⋅ ( ZcC + ZC)
180
π
⋅ Vp
⋅ arg( VnN) = 123.304
IcC :=
Vc − VnN
ZC + ZcC
IcC = 0.455 + 0.716i
IcC = 0.848
180
π
⋅ arg( IcC) = 57.553
IaA + IbB + IcC = 0
Calculate the phase voltages of the Y-connected load:
180
2
j⋅ ⋅ π
3
( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC)
VnN = −1.172 × 10
180
4
j⋅ ⋅ π
3
⋅ arg( VAB) = 28.498
VBC:= VBN − VCN
VBC = 151.227
180
π
⋅ arg( VBC) = −91.502
VCN := IcC⋅ ZC
VCN = 87.311
180
π
⋅ arg( VCN) = 118.498
VCA := VCN − VAN
VCA = 151.227
180
π
⋅ arg( VCA) = 148.498
Calculate the phase currents of the ∆ -connected load:
IAB :=
VAB
Z3
IAB = 0.49
180
π
⋅ arg( IAB) = −32.447
IBC :=
VBC
Z1
IBC = 0.49
180
π
⋅ arg( IBC) = −152.447
ICA :=
VCA
Z2
ICA = 0.49
180
π
⋅ arg( ICA) = 87.553
12-7
Ex. 12.8-1
Continuing Ex. 12.8-1:
Calculate the power delivered to the load:


SA := IaA ⋅ IaA ⋅ ZA
SB := IbB⋅ IbB⋅ ZB
SA = 35.958 + 64.725i
SB = 35.958 + 64.725i
Total power delivered to the load:
Ex. 12.9-1
SA + SB + SC = 107.875 + 194.175i

SC := IcC⋅ IcC⋅ ZC
SC = 35.958 + 64.725i
P1 = VAB I A cos(θ +30° ) + VCB I C cos(θ − 30° )= P + P2
pf = .4 lagging ⇒ θ = 61.97
°
1
So P T = 450(24) cos 91.97° + cos 31.97°  = 8791 W
∴ P 1 = − 371 W P2 = 9162 W
Ex. 12.9-2
Consider Fig. 12.9-1 with P1 = 60 kW P2 = 40 kW .
(a.) P = P1 + P2 = 100 kW
(b.) use equation 12.9-7 to get
tan θ = 3
then
P2 − P1
40− 60
= 3
= − .346 ⇒ θ = − 19.11°
PL + P2
100
pf = cos ( − 19.110°) = 0.945 leading
12-8
Problems
Section 12-3: Three Phase Voltages
P12.3-1
Given VC = 277 ∠45° and an abc phase sequence:
VA = 277 ∠ ( 45−120 ) ° = 277 ∠ − 75°
VB = 277 ∠( 45° +120 )° = 277 ∠165°
VAB = VA − VB =( 277 ∠− 75° )−( 277 ∠165° )
=( 71.69 − j 267.56 ) −( −267.56+ j 71.69 )
=339.25− j 339.25 = 479.77 ∠− 45° 480 ∠− 45°
Similarly:
VBC = 480 ∠ − 165° and VCA = 480 ∠75°
P12.3-2
VAB = VA × 3∠30° ⇒ VA =
VAB = −VBA = − (12470 ∠−35° )
In our case:
VA =
So
VAB
3∠30°
= 12470 ∠145° V
12470 ∠145°
= 7200∠115°
3∠30°
Then, for an abc phase sequence:
VC = 7200 ∠ (115 + 120 ) ° = 7200 ∠235° = 7200 ∠ − 125°
VB = 7200 ∠ (115 − 120 ) ° = 7200 ∠ − 5° V
P12.3-3
Vab = Va × 3∠30° ⇒ Va =
Vab
3∠30°
Vab = 1500 ∠30° V
1500 ∠30°
Va =
= 866∠0° V
3∠30°
In our case, the line-to-line voltage is
So the phase voltage is
12-9
Section 12-4: The Y-to-Y Circuit
P12.4-1
Balanced, three-wire, Y-Y circuit:
where Z A = Z B = Z C = 12∠30 = 10.4 + j 6
MathCAD analysis (12p4_1.mcd):
Vp :=
Describe the three-phase source:
Va := Vp⋅ e
j⋅
π
180
⋅0
208
3
Vb := Va⋅ e
Describe the balanced three-phase load:
j⋅
π
180
⋅ − 120
ZA := 10.4 + j⋅ 6
Vc := Va⋅ e
j⋅
π
180
⋅ 120
ZB := ZA
ZC := ZB
Check: The voltage at the neutral of the load with respect to the neutral of the source should be zero:
VnN :=
ZA ⋅ ZC⋅ e
4
j⋅ ⋅ π
3
+ ZA ⋅ ZB⋅ e
IaA :=
Va − VnN
ZA
IaA = 8.663 − 4.998i
Check:
− 14
⋅ Vp
IbB :=
VnN = 2.762 × 10
Vb − VnN
IbB = −8.66 − 5.004i
IaA = 10.002
π
+ ZB⋅ ZC
ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC
Calculate the line currents:
180
2
j⋅ ⋅ π
3
IbB = 10.002
⋅ arg( IaA ) = −29.982
180
π
− 15
IaA + IbB + IcC = 4.696 × 10
⋅ arg( IbB) = −149.982
− 14
ZB
IcC :=
Vc − VnN
ZC
−3
IcC = −3.205 × 10
IcC = 10.002
180
π
+ 10.002i
⋅ arg( IcC) = 90.018
− 1.066i× 10
12-10
Calculate the power delivered to the load:


SA := IaA ⋅ IaA ⋅ ZA
SB := IbB⋅ IbB⋅ ZB
SA = 1.04 × 10 + 600.222i
3
Total power delivered to the load:
SB = 1.04 × 10 + 600.222i
3

SC := IcC⋅ IcC⋅ ZC
SC = 1.04 × 10 + 600.222i
3
SA + SB + SC = 3.121 × 10 + 1.801i× 10
3
3
Consequently:
(a) The phase voltages are
Va =
208
∠0° = 120∠0° V rms, Vb = 120∠ − 120° V rms and Vc = 120∠120° V rms
3
(b) The currents are equal the line currents
(c)
I a = I aA = 10∠ − 30° A rms, I b = I bB = 10∠ − 150° A rms
and
I c = I cC = 10∠90° A rms
(d) The power delivered to the load is S = 3.121 + j1.801 kVA .
P12.4-2
Balanced, three-wire, Y-Y circuit:
where
Va = 120∠0° Vrms, Vb = 120∠ − 120° Vrms and Vc = 120∠120° Vrms
Z A = Z B = Z C = 10 + j ( 2 × π × 60 ) (100 × 10−3 ) = 10 + j 37.7 Ω
and
Z aA = Z bB = Z cC = 2 Ω
Mathcad Analysis (12p4_2.mcd):
12-11
Vp := 120
Describe the three-phase source:
Va := Vp⋅ e
j⋅
π
180
⋅0
Vb := Va⋅ e
j⋅
π
180
⋅ − 120
Vc := Va⋅ e
ZA := 10 + j⋅ 37.7
Describe the three-phase load:
180
⋅ 120
ZB := ZA
ZaA := 2
Describe the three-phase line:
π
j⋅
ZC := ZB
ZbB := ZaA
ZcC := ZaA
Calculate the voltage at the neutral of the load with respect to the neutral of the source:
( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e
VnN :=
4
j⋅ ⋅ π
3
+ ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e
2
j⋅ ⋅ π
3
+ ( ZbB + ZB) ⋅ ( ZcC + ZC)
( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC)
− 15
VnN = −8.693 × 10
− 14
+ 2.232i× 10
Calculate the line currents:
IaA = 0.92 − 2.89i
π
Va − VnN
ZA + ZaA
⋅ arg( IaA ) = −72.344
180
π
Calculate the phase voltages at the load:
VA = 118.301
180
π
⋅ arg( VA) = 2.801
ZB + ZbB
⋅ arg( IbB) = 167.656
− 15
− 15
− 3.109i× 10
VA := ZA ⋅ IaA
VB = 118.301
180
π
180
Vb − VnN
IbB = 3.033
IaA + IbB + IcC = −1.332 × 10
Check:
IbB :=
IbB = −2.963 + 0.648i
IaA = 3.033
180
IaA :=
− 14
VnN = 2.396 × 10
⋅ arg( VB) = −117.199
π
⋅ Vp
⋅ arg( VnN) = 111.277
IcC :=
Vc − VnN
ZC + ZcC
IcC = 2.043 + 2.242i
IcC = 3.033
180
π
⋅ arg( IcC) = 47.656
VB := ZB⋅ IbB
VC := ZC⋅ IcC
VC = 118.301
180
π
⋅ arg( VC) = 122.801
Consequently, the line-to-line voltages at the source are:
Vab = Va × 3∠30° = 120∠0°× 3∠30° = 208∠30° Vrms,
Vbc = 208∠ − 120° Vrms and Vca = 208∠120° Vrms
The line-to-line voltages at the load are:
VAB = VA × 3∠30° = 118.3∠3°× 3∠30° = 205∠33° Vrms,
Vbc = 205∠ − 117° Vrms and Vca = 205∠123° Vrms
and the phase currents are
I a = I aA = 10∠ − 72° A rms, I b = I bB = 3∠168° A rms and I c = I cC = 3∠48° A rms
12-12
P12.4-3
Balanced, three-wire, Y-Y circuit:
where
Va = 10∠0° V = 7.07∠0° V rms, Vb = 7.07∠ − 120° V rms and Vc = 7.07∠120° V rms
Z A = Z B = Z C = 12 + j (16 )(1) = 12 + j16 Ω
and
MathCAD analysis (12p4_3.mcd):
Vp :=
Describe the three-phase source:
Va := Vp⋅ e
j⋅
π
180
⋅0
10
2
Vb := Va⋅ e
Describe the balanced three-phase load:
j⋅
π
180
⋅ − 120
Vc := Va⋅ e
ZA := 12 + j⋅ 16
j⋅
π
180
⋅ 120
ZB := ZA
ZC := ZB
Check: The voltage at the neutral of the load with respect to the neutral of the source should be zero:
VnN :=
ZA ⋅ ZC⋅ e
4
j⋅ ⋅ π
3
+ ZA ⋅ ZB⋅ e
IaA :=
IaA = 0.212 − 0.283i
IaA = 0.354
π
+ ZB⋅ ZC
ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC
Calculate the line currents:
180
2
j⋅ ⋅ π
3
⋅ arg( IaA ) = −53.13
Va − VnN
ZA
− 15
⋅ Vp
IbB :=
VnN = 1.675 × 10
Vb − VnN
IbB = −0.351 − 0.042i
IbB = 0.354
180
π
⋅ arg( IbB) = −173.13
Calculate the power delivered to the load:


SB := IbB⋅ IbB⋅ ZB
SA := IaA ⋅ IaA ⋅ ZA
SA = 1.5 + 2i
Total power delivered to the load:
SB = 1.5 + 2i
SA + SB + SC = 4.5 + 6i
ZB
IcC :=
Vc − VnN
ZC
IcC = 0.139 + 0.325i
IcC = 0.354
180
π
⋅ arg( IcC) = 66.87

SC := IcC⋅ IcC⋅ ZC
SC = 1.5 + 2i
12-13
Consequently
(a) The rms value of ia(t) is 0.354 A rms.
(b) The average power delivered to the load is P = Re {S} = Re {4.5 + j 6} = 4.5 W
P12.4-4
Unbalanced, three-wire, Y-Y circuit:
where
Va = 100∠0° V = 70.7∠0° V rms, Vb = 70.7∠ − 120° V rms and Vc = 7.07∠120° V rms
Z A = 20 + j ( 377 ) ( 60 ×10−3 ) = 20 + j 22.6 Ω, Z B = 40 + j ( 377 ) ( 40 × 10−3 ) = 40 + j 15.1 Ω
Z C = 60 + j ( 377 ) ( 20 ×10−3 ) = 60 + j 7.54 Ω
and
Z aA = Z bB = Z cC = 10 + j ( 377 ) ( 5 × 10−3 ) = 10 + j 1.89 Ω
Mathcad Analysis (12p4_4.mcd):
Describe the three-phase source:
Va := Vp⋅ e
j⋅
π
180
⋅0
Vp := 100
Vb := Va⋅ e
j⋅
π
180
⋅ 120
Enter the frequency of the 3-phase source: ω := 377
Describe the three-phase load:
Describe the three-phase line:
ZA := 20 + j⋅ ω⋅ 0.06
Vc := Va⋅ e
j⋅
π
180
⋅ − 120
ZB := 40 + j⋅ ω⋅ 0.04
ZaA := 10 + j⋅ ω⋅ 0.005 ZbB := ZaA
ZC := 60 + j⋅ ω⋅ 0.02
ZcC := ZaA
12-14
Calculate the voltage at the neutral of the load with respect to the neutral of the source:
VnN :=
( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e
4
j⋅ ⋅ π
3
+ ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e
2
j⋅ ⋅ π
3
+ ( ZbB + ZB) ⋅ ( ZcC + ZC)
( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC)
VnN = 12.209 − 24.552i
Calculate the line currents:
VnN = 27.42
IaA :=
IaA = 2.156 − 0.943i
Va − VnN
ZA + ZaA
IaA = 2.353
180
π
⋅ arg( IaA ) = −23.619
(
)
IbB :=
IbB = −0.439 + 2.372i
Total power delivered to the load:
Vb − VnN
ZB + ZbB
IbB = 2.412
180
π
⋅ arg( IbB) = 100.492
(
)
Calculate the power delivered to the load:


IbB⋅ IbB
IaA ⋅ IaA
⋅ ZA
SB :=
⋅ ZB
SA :=
2
2
SA = 55.382 + 62.637i
180
SB = 116.402 + 43.884i
π
⋅ Vp
⋅ arg( VnN) = −63.561
IcC :=
Vc − VnN
ZC + ZcC
IcC = −0.99 − 0.753i
IcC = 1.244
180
π
⋅ arg( IcC) = −142.741
SC :=

(IcC
⋅ IcC)
2
⋅ ZC
SC = 46.425 + 5.834i
SA + SB + SC = 218.209 + 112.355i
The average power delivered to the load is P = Re {S} = Re {218.2 + j112.4} = 218.2 W
P12.4-5
Balanced, three-wire, Y-Y circuit:
where
Va = 100∠0° V = 70.7∠0° V rms, Vb = 70.7∠ − 120° V rms and Vc = 7.07∠120° V rms
Z A = Z B = Z C = 20 + j ( 377 ) ( 60 ×10−3 ) = 20 + j 22.6 Ω
and
Z aA = Z bB = Z cC = 10 + j ( 377 ) ( 5 × 10−3 ) = 10 + j 1.89 Ω
12-15
Mathcad Analysis (12p4_5.mcd):
Describe the three-phase source:
Va := Vp⋅ e
j⋅
π
180
⋅0
Vp := 100
Vb := Va⋅ e
j⋅
π
180
⋅ 120
Vc := Va⋅ e
Enter the frequency of the 3-phase source: ω := 377
ZA := 20 + j⋅ ω⋅ 0.06
Describe the three-phase load:
180
⋅ − 120
ZB := ZA
ZaA := 10 + j⋅ ω⋅ 0.005
Describe the three-phase line:
π
j⋅
ZC := ZA
ZbB := ZaA
ZcC := ZaA
Calculate the voltage at the neutral of the load with respect to the neutral of the source:
VnN :=
( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e
4
j⋅ ⋅ π
3
+ ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e
2
j⋅ ⋅ π
3
+ ( ZbB + ZB) ⋅ ( ZcC + ZC)
( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC)
− 15
VnN = −8.982 × 10
− 14
Calculate the line currents:
IaA :=
IaA = 1.999 − 1.633i
IaA = 2.582
180
π
− 14
+ 1.879i× 10
⋅ arg( IaA ) = −39.243

(IaA
⋅ IaA )
VnN = 2.083 × 10
Va − VnN
ZA + ZaA
IbB = 0.415 + 2.548i
⋅ ZA
2
SA = 66.645 + 75.375i
Total power delivered to the load:
Vb − VnN
ZB + ZbB
IbB = 2.582
180
π
⋅ arg( IbB) = 80.757
Calculate the power delivered to the load:
SA :=
IbB :=
SB :=

(IbB
⋅ IbB)
⋅ ZB
2
SB = 66.645 + 75.375i
180
π
⋅ Vp
⋅ arg( VnN) = 115.55
IcC :=
Vc − VnN
ZC + ZcC
IcC = −2.414 − 0.915i
IcC = 2.582
180
π
⋅ arg( IcC) = −159.243
SC :=

(IcC
⋅ IcC)
⋅ ZC
2
SC = 66.645 + 75.375i
SA + SB + SC = 199.934 + 226.125i
The average power delivered to the load is P = Re {S} = Re {200 + j 226} = 200 W
12-16
P12.4-6
Unbalanced, three-wire, Y-Y circuit:
where
Va = 10∠ − 90° V = 7.07∠ − 90° V rms, Vb = 7.07∠150° V rms and Vc = 7.07∠30° V rms
and
Z A = 4 + j ( 4 )(1) = 4 + j 4 Ω, Z B = 2 + j ( 4 )( 2 ) = 2 + j 8 Ω and Z C = 4 + j ( 4 )( 2 ) = 4 + j 8 Ω
Mathcad Analysis (12p4_6.mcd):
Vp := 10
Describe the three-phase source:
Va := Vp⋅ e
j⋅
π
180
⋅ − 90
Vb := Vp⋅ e
j⋅
π
180
⋅ 150
Enter the frequency of the 3-phase source: ω := 4
ZA := 4 + j⋅ ω⋅ 1
Describe the three-phase load:
Vc := Vp⋅ e
j⋅
π
180
⋅ 30
ZB := 2 + j⋅ ω⋅ 2
ZC := 4 + j⋅ ω⋅ 2
Calculate the voltage at the neutral of the load with respect to the neutral of the source:
VnN :=
ZA ⋅ ZC⋅ Vb + ZA ⋅ ZB⋅ Vc + ZB⋅ ZC⋅ Va
ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC
VnN = 1.528 − 0.863i
Calculate the line currents:
VnN = 1.755
IaA :=
IaA = −1.333 − 0.951i
IaA = 1.638
180
π
⋅ arg( IaA ) = −144.495
Va − VnN
ZA
IbB :=
180
Vb − VnN
IbB = 0.39 + 1.371i
IbB = 1.426
180
π
⋅ arg( IbB) = 74.116
ZB
π
⋅ arg( VnN) = −29.466
IcC :=
Vc − VnN
ZC
IcC = 0.943 − 0.42i
IcC = 1.032
180
π
⋅ arg( IcC) = −24.011
12-17
(
)
(
)
Calculate the power delivered to the load:


IaA ⋅ IaA
IbB⋅ IbB
SA :=
⋅ ZA
SB :=
⋅ ZB
2
2
SA = 5.363 + 5.363i
SC :=
SB = 2.032 + 8.128i
SA + SB + SC = 9.527 + 17.754i
Total power delivered to the load:

(IcC
⋅ IcC)
2
⋅ ZC
SC = 2.131 + 4.262i
The average power delivered to the load is P = Re {S} = Re {9.527 + j17.754} = 9.527 W
P12.4-7
Unbalanced, three-wire, Y-Y circuit:
where
Va = 10∠ − 90° V = 7.07∠ − 90° V rms, Vb = 7.07∠150° V rms and Vc = 7.07∠30° V rms
and
Z A = Z B = Z C = 4 + j ( 4 )( 2 ) = 4 + j 8 Ω
Mathcad Analysis (12p4_7.mcd):
Describe the three-phase source:
Va := Vp⋅ e
j⋅
π
180
⋅ − 90
Vp := 10
Vb := Vp⋅ e
j⋅
π
180
⋅ 150
Enter the frequency of the 3-phase source: ω := 4
Describe the three-phase load:
ZA := 4 + j⋅ ω⋅ 2
Vc := Vp⋅ e
j⋅
π
180
⋅ 30
ZB := ZA
ZC := ZA
The voltage at the neutral of the load with respect to the neutral of the source should be zero:
VnN :=
ZA ⋅ ZC⋅ Vb + ZA ⋅ ZB⋅ Vc + ZB⋅ ZC⋅ Va
ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC
− 15
VnN = 1.517 × 10
12-18
Calculate the line currents:
IaA :=
IaA = −1 − 0.5i
π
⋅ arg( IaA ) = −153.435
(
IbB :=
ZA
Vb − VnN
IbB = 0.067 + 1.116i
IaA = 1.118
180
Va − VnN
)
ZB
IbB = 1.118
180
π
⋅ arg( IbB) = 86.565
(
)
Calculate the power delivered to the load:


IaA ⋅ IaA
IbB⋅ IbB
SA :=
⋅ ZA
SB :=
⋅ ZB
2
2
SA = 2.5 + 5i
Total power delivered to the load:
SB = 2.5 + 5i
SA + SB + SC = 7.5 + 15i
IcC :=
Vc − VnN
ZC
IcC = 0.933 − 0.616i
IcC = 1.118
180
π
⋅ arg( IcC) = −33.435
SC :=

(IcC
⋅ IcC)
2
⋅ ZC
SC = 2.5 + 5i
The average power delivered to the load is P = Re {S} = Re {7.5 + j15} = 7.5 W
Section 12-6: The ∆- Connected Source and Load
P12.5-1
Given I B = 50∠ − 40° A rms and assuming the abc phase sequence
we have
I A = 50∠80° A rms and I C = 50∠200° A rms
From Eqn 12.6-4
I A = I AB × 3∠ − 30° ⇒ I AB =
so
IA
3∠ − 30°
50∠80°
= 28.9∠110° A rms
3∠−30°
= 28.9∠ − 10° A rms and ICA = 28.9∠ − 130° A rms
I AB =
I BC
12-19
P12.5-2
The two delta loads connected in parallel are equivalent to a single delta
load with
Z ∆ = 5 || 20 = 4 Ω
The magnitude of phase current is
480
Ip =
= 120 A rms
4
The magnitude of line current is
I L = 3 I p = 208 A rms
Section 12-6: The Y- to ∆- Circuit
P12.6-1
We have a delta load with Z = 12∠30° . One phase current is
I AB
 208
  208

∠−30°  −
∠−150° 

V
V −V
3
  3
 = 208∠0° = 17.31∠ − 30° A rms
= AB = A A = 
12∠30°
12∠30°
Z
Z
The other phase currents are
I BC = 17.31∠ − 150° A rms and I CA = 17.31∠90° A rms
One line currents is
I A = I AB × 3∠ − 30° = (17.31∠ − 30° ) ×
(
)
3∠ − 30° = 30∠0° A rms
The other line currents are
I B = 30∠ − 120° A rms and I C = 30∠120° A rms
The power delivered to the load is
P = 3(
208
) (30) cos ( 0 − 30° ) = 9360 W
3
12-20
The balanced delta load with Z ∆ = 39∠− 40° Ω is
equivalent to a balanced Y load with
P12.6-2
ZY =
Z∆
= 13∠ − 40° = 9.96 − j 8.36 Ω
3
Z T = Z Y + 4 = 13.96 − j 8.36 = 16.3∠ − 30.9 Ω
480
∠−30°
3
then I A =
= 17∠0.9° A rms
°
16.3 ∠−30.9
P12.6-3
Vab = Va × 3∠30° ⇒ Va =
Vab
3∠30°
In our case, the given line-to-line voltage is
Vab = 380 ∠30° V rms
380 ∠30°
= 200∠0° V rms
So one phase voltage is Va =
3∠30°
So
VAB = 380∠30° V rms
VA = 220∠0° V rms
VBC = 380∠-90° V rms
VCA = 380∠150° V rms
One phase current is
IA =
VB = 220∠−120° V rms
VC = 220∠120° V rms
VA 220∠0°
= 44∠ − 53.1° A rms
Z
3+ j4
The other phase currents are
I B = 44∠−173.1° A rms amd I C = 44∠66.9° A rms
12-21
P12.6-4
Vab = Va × 3∠30° ⇒ Va =
Vab
3∠30°
In our case, the given line-to-line voltage is
Vab = 380 ∠0° V rms
Va =
So one phase voltage is
380 ∠0°
= 200∠ − 30° V rms
3∠30°
Vab = 380∠0° V rms
So
Vbc = 380∠-120° V rms
Vca = 380∠120° V rms
One phase current is
IA =
Va = 220∠ − 30° V rms
Vb = 220∠−150° V rms
Vc = 220∠90° V rms
Va 220∠−30°
=
= 14.67∠ − 83.1° A rms
Z
9 + j12
The other phase currents are
I B = 14.67∠ − 203.1° A rms and I C = 14.67∠36.9° A rms
12-22
12-23
Section 12-7: Balanced Three-Phase Circuits
P12.7-1
Va =
IA
25
×103 ∠0° Vrms
3
25
×103 ∠0°
Va
=
= 3
= 96∠ − 25° A rms
Z
150 ∠25°
 25

×103  96 cos(0 − 25°) = 3.77 mW
P = 3 Va I A cos (θ v -θ I ) = 3 
 3

P12.7-2
Convert the delta load to an equivalent Y connected load:
Z ∆ = 50 Ω
ˆ = 50 Ω
Z∆ ⇒ Z
Y
3
To get the per-phase equivalent circuit shown to the right:
The phase voltage of the source is
Va =
45×103
∠0° = 26∠0° kV rms
3
The equivalent impedance of the load together with the line is
(10 + j 20 )
50
3 + 2 = 12 + j 5 = 13∠22.6° Ω
Z eq =
50
10 + j 20 +
3
The line current is
Ι aA =
Va 26 × 103 ∠0°
=
= 2000∠ − 22.6° A rms
Z eq
13∠22.6°
The power delivered to the parallel loads (per phase) is
50 

 (10 + j 20 ) 3 
2
6
PLoads = I aA × Re 
 = 4 ×10 × 10 = 40 MW
50
10 + j 20 + 
3 

The power lost in the line (per phase) is
12-24
PLine = I aA × Re {Z Line } = 4 × 106 × 2 = 8 MW
2
The percentage of the total power lost in the line is
PLine
8
×100% =
× 100% = 16.7%
PLoad + PLine
40 +8
P12.7-3
Ia =
Va 5∠30°
=
= 0.5∠ − 23° A ∴ I a = 0.5 A
Z T 6 + j8
I
= 3 a Re {Z Load } = 3 × 0.125 × 4 = 1.5 W
2
2
PLoad
also (but not required) :
PSource = 3
(5) (0.5)
cos(−30 − 23) = 2.25 W
2
I
= 3 a Re{Z Line } = 3×0.125× 2 = 0.75 W
2
2
Pline
12-25
Section 12-8: Power in a Balanced Load
P12.8-1
Assuming the abc phase sequence:
VCB = 208∠15° V rms ⇒ VBC = 208∠195° V rms ⇒ VAB = 208∠315° V rms
VAB
208∠315° 208
=
=
∠285° V rms
3∠30°
3∠30°
3
Then
VA =
also
I B = 3∠110° A rms ⇒ I A = 3∠230° A rms
Finally
P = 3 VAB I A cos (θ V − θ I ) = 3(
P12.8-2
Assuming a lagging power factor:
208
) (3) cos(285° − 230°) = 620 W
3
cos θ = pf = 0.8 ⇒
θ = 36.9°
The power supplied by the three-phase source is given by
Pin =
η
Pout
=
Pin = 3 I A VA pf
20 ( 745.7 )
= 17.55 kW where 1 hp = 745.7 W
0.85
⇒
Pin
17.55 ×103
IA =
=
= 26.4 A rms
3 VA pf
 480 
3
 ( 0.8 )
 3
480 °
I A = 26.4∠ − 36.9° A rms when VA =
∠0 V rms
3
12-26
P12.8-3
(a) For a ∆-connected load, Eqn 12.8-5 gives
PT = 3 VP I L pf ⇒ I L =
1500
PT
=
= 4.92 A rms
3 VP I L pf 3( 220 )(.8)
3
The phase current in the ∆-connected load is given by
IP =
I
IL
4.92
⇒ IP = L =
= 2.84 A rms
3
3
3
The phase impedance is determined as:
Z=
V
220
VL VL
=
∠ (θ V − θ I ) = L ∠ cos −1 pf =
∠ cos −1 0.8 = 77.44∠36.9° Ω
IP
IP
IP
2.84
(b) For a ∆-connected load, Eqn 12.8-4 gives
PT = 3 VP I L pf ⇒ I L =
1500
PT
=
= 4.92 A rms
3 VP I L pf 3( 220 )(.8)
3
The phase impedance is determined as:
220
V
V
V
Z = P = P ∠ (θ V − θ I ) = P ∠ cos −1 pf = 3 ∠ cos −1 0.8 = 25.8∠36.9° Ω
IP
IP
IP
4.92
P12.8-4
Parallel ∆ loads
Z1Z 2
(40∠30° ) (50∠−60° )
=
= 31.2 ∠−8.7° Ω
Z∆ =
°
°
Ζ1 + Ζ 2
40∠30 + 50∠− 60
VL = VP , Ι P =
VP
600
=
= 19.2 A rms,
Z∆
31.2
IL =
3 Ι P = 33.3 A rms
So P = 3 VL I L pf = 3 (600) (33.3) cos ( − 8.7° ) = 34.2 kW
12-27
P12.8-5
We will use
In our case:
S = S ∠θ = S cosθ + S sin θ = S pf + S sin ( cos −1 pf )
S 1 = 39 (0.7) + j 39 sin ( cos −1 ( 0.7 ) ) = 27.3 + j 27.85 kVA
15
sin ( cos −1 ( 0.21) ) = 15 − j 69.84 kVA
S 2 = 15 +
0.21
S
S 3φ = S 1 +S 2 = 42.3 − j 42.0 kVA ⇒ S φ = 3φ = 14.1− j 14.0 kVA
3
The line current is
 S  (14100+ j 14000)
S = Vp I L ⇒ I L =   =
= 117.5 + j 116.7 A rms = 167 ∠45° A rms
V 
208
 p
3
208
∠0° = 120∠0° V rms. The source must
The phase voltage at the load is required to be
3
provide this voltage plus the voltage dropped across the line, therefore
*
*
= 120∠0° + (0.038 + j 0.072)(117.5 + j 116.7) = 115.9 + j 12.9 = 116.6 ∠6.4° V rms
V
Sφ
Finally
V
= 116.6 V rms
Sφ
P12.8-6
The required phase voltage at the load is VP =
4.16
∠0° = 2.402∠0° kVrms .
3
Let I1 be the line current required by the ∆-connected load. The apparent power per phase
500 kVA
required by the ∆-connected load is S1 =
= 167 kVA . Then
3
S1 = S1 ∠θ = S1 ∠ cos −1 ( pf ) = 167 ∠ cos−1 ( 0.85) = 167∠31.8° kVA
and
S1 = VP I1
*
*
3
 S1   (167 ×10 ) ∠31.8° 
 = 69.6∠ − 31.8° = 59 − j36.56 A rms
⇒ I1 =   = 
3
 VP   ( 2.402 ×10 ) ∠0° 
*
12-28
Let I2 be the line current required by the first Y-connected load. The apparent power per phase
75 kVA
required by this load is S 2 =
= 25 kVA . Then, noticing the leading power factor,
3
S 2 = S 2 ∠θ = S 2 ∠ cos −1 ( pf ) = 25 ∠ cos −1 ( 0 ) = 25∠ − 90° kVA
and
S 2 = VP I 2
*
3
 S 2   ( 25 ×10 ) ∠ − 90° 

 = 10.4∠90° = j10.4 A rms
⇒ I2 =   =
3


V
×
∠
°
2.402
10
0
) 
 P (
*
*
Let I3 be the line current required by the other Y-connected load. Use Ohm’s law to determine I3
to be
2402∠0° 2402∠0°
I3 =
+
= 16 − j 10.7 A rms
150
j 225
The line current is
I L = I1 + I 2 + I 3 = 75− j 36.8 A rms
The phase voltage at the load is required to be VP =
4.16
∠0° = 2.402∠0° kVrms .The source
3
must provide this voltage plus the voltage dropped across the line, therefore
VSφ = 2402∠0° + (8.45 + j 3.9) (75 − j 36.8) = 3179 ∠−0.3° Vrms
Finally
VSL = 3 (3179) = 5506 Vrms
P12.8-7
The required phase voltage at the load is VP =
4.16
∠0° = 2.402∠0° kVrms .
3
Let I1 be the line current required by the ∆-connected load. The apparent power per phase
1.5 MVA
required by the ∆-connected load is S1 =
= 0.5 MVA . Then
3
S1 = S1 ∠θ = S1 ∠ cos −1 ( pf ) = 0.5 ∠ cos −1 ( 0.75) = 0.5∠41.4° MVA
and
S1 = VP I1
*
*
6
 S1   ( 0.5 ×10 ) ∠41.4° 

 = 2081.6∠ − 41.4° = 1561.4 − j1376.6 A rms
⇒ I1 =   =
3

V
×
∠
°
2.402
10
0
) 
 P (
*
12-29
Let I2 be the line current required by the first Y-connected load. The complex power, per phase,
is
0.67
S 2 = 0.67 +
sin ( cos −1 ( 0.8 ) ) = 0.67 + j 0.5 MVA
0.8
*
6
 S 2   ( 0.67 + j 0.5 ) ×106   ( 0.833 ×10 ) ∠ − 36.9° 
 =

I2 =   = 
3
3
 VP   ( 2.402 ×10 ) ∠0°   ( 2.402 ×10 ) ∠0° 
= 346.9∠ − 36.9° = 277.4 − j 208.3 A rms
The line current is
I L = I1 + I 2 = 433.7 − j 345.9 = 554.7∠ − 38.6 A rms
*
*
The phase voltage at the load is required to be VP =
4.16
∠0° = 2.402∠0° kVrms .The source
3
must provide this voltage plus the voltage dropped across the line, therefore
VSφ = 2402∠0° + (0.4 + j 0.8) (433.7 − j 345.9) = 2859.6 ∠ − 38.6° Vrms
Finally
VSL = 3 (2859.6) = 4953 Vrms
The power supplied by the source is
PS =
3 (4953) (554.7) cos (4.2° + 38.6° ) = 3.49 MW
The power lost in the line is
PLine = 3 × ( 554.7 2 ) × Re {0.4+ j 0.8} = 0.369 MW
The percentage of the power consumed by the loads is
3.49 − 0.369
×100% = 89.4%
3.49
12-30
P12.8–8
The required phase voltage at the load is VP =
600
∠0° = 346.4∠0° Vrms .
3
θ = cos −1 (0.8)
= 37 °
Let I be the line current required by the load. The complex power, per phase, is
S = 160 +
The line current is
160
sin ( cos −1 ( 0.8 ) ) = 160 + j 120 kVA
0.8
 S   (160 + j 120 ) × 103 
I=  =
 = 461.9 − j 346.4 A rms
346.4∠0°
 VP  

*
*
The phase voltage at the load is required to be VP =
600
∠0° = 346.4∠0° Vrms .The source
3
must provide this voltage plus the voltage dropped across the line, therefore
VSφ = 346.4∠0° + (0.005 + j 0.025) (461.9 − j 346.4) = 357.5 ∠1.6° Vrms
Finally
VSL = 3 (357.5) = 619.2 Vrms
The power factor of the source is
pf = cos (θ V − θ I ) = cos (1.6 − ( − 37)) = 0.78
12-31
Section 12-9: Two-Wattmeter Power Measurement
P12.9-1
Pout = 20 hp × 746
Pin = 3 VL I L cos θ
W
= 14920 W
hp
P
14920
= 20 kW
Pin = out =
0.746
η
20 × 103
Pin
=
= 0.50
3 VL I L
3 (440) (52.5)
⇒ cos θ =
⇒ θ cos -1 ( 0.5 ) = 60°
The powers read by the two wattmeters are
P1 = VL I L cos (θ + 30° ) = (440) (52.5)cos ( 60° + 30° ) = 0
and
P2 = VL I L cos (θ − 30° ) = (440) (52.5)cos ( 60° − 30° ) = 20 kW
P12.9-2
VP = VL = 4000 V rms
IP =
VP
4000
=
= 80 A rms
Z∆
50
Z
∆
= 40 + j 30 = 50 ∠36.9°
Ι L = 3 I P = 138.6 A rms
pf = cos θ = cos (36.9° ) = 0.80
P1 = VL I L cos (θ + 30° ) = 4000 (138.6) cos 66.9° = 217.5 kW
P2 = VL I L cos (θ −30° ) = 4000 (138.6) cos 6.9° = 550.4 kW
PT = P1 + P2 = 767.9 kW
Check : PT = 3 Ι L VL cos θ =
3 (4000) (138.6) cos 36.9°
= 768 kW which checks
12-32
P12.9–3
Vp = Vp =
200
= 115.47 Vrms
3
VA =115.47∠0° V rms, VB = 115.47∠−120° V rms
and VC = 115.47∠120° V rms
IA =
VA
115.47∠0°
=
= 1.633∠ − 45° A rms
Z
70.7∠45°
I B = 1.633 ∠ − 165° A rms and I C = 1.633 ∠75° A rms
PT =
3 VL I L cos θ = 3 (200) (1.633) cos 45° = 400 W
PB = VAC I A cos θ1 = 200 (1.633) cos (45° − 30° ) = 315.47 W
PC = VBC I B cos θ 2 = 200 (1.633) cos (45° + 30° ) = 84.53 W
P12.9-4
ZY = 10∠ − 30° Ω and Z ∆ = 15∠30° Ω
Convert Z ∆ to Z Yˆ → Z Yˆ =
then Zeq =
Z∆
= 5∠30° Ω
3
(10∠−30° ) ( 5∠30° ) =
10∠−30°+5∠30°
208
Vp = Vp =
= 120 V rms
3
VA = 120∠0° V rms ⇒ I A =
50∠0°
= 3.78∠10.9° Ω
13.228 ∠−10.9°
120∠0°
= 31.75 ∠−10.9°
3.78 ∠10.9°
I B = 31.75∠−130.9°
I C = 31.75∠109.1°
PT = 3VL I L cos θ = 3 ( 208 ) ( 31.75 ) cos (10.9 ) =11.23 kW
W1 = VL I L cos (θ −30°) = 6.24 kW
W2 = VL I L cos (θ + 30°) = 4.99 kW
12-33
P12.9-5
PT = PA + PC = 920 + 460 = 1380 W
tan θ = 3
( −460 ) = −0.577 ⇒ θ = −30°
PA − PC
= 3
1380
PA + PC
PT = 3 VL I L cos θ so I L =
IP =
P12.9-6
IL
= 4.43 A rms
3
∴ Z∆ =
Z = 0.868 + j 4.924 = 5∠80°
VL = 380 V rms, VP =
I L = I P and I P =
1380
PT
=
=7.67 A rms
3 VL cos θ
2 ×120×cos( −30 )
⇒
120
= 27.1 Ω ο r Z ∆ = 27.1 ∠−30°
4.43
θ = 80°
380
= 219.4 V rms
3
VP
= 43.9 A rms
Z
P1 = ( 380 ) ( 43.9 ) cos (θ −30° ) = 10,723 W
P2 = ( 380 ) ( 43.9 ) cos (θ + 30° ) = −5706 W
PT = P1 + P2 = 5017 W
12-34
PSpice Problems
SP 12-1
FREQ
6.000E+01
IM(V_PRINT3)IP(V_PRINT3)IR(V_PRINT3)II(V_PRINT3)
3.142E+00 -1.644E+02 -3.027E+00 -8.436E-01
FREQ
6.000E+01
IM(V_PRINT1)IP(V_PRINT1)IR(V_PRINT1)II(V_PRINT1)
3.142E+00 -4.443E+01
2.244E+00 -2.200E+00
FREQ
6.000E+01
VM(N01496)
2.045E-14
FREQ
6.000E+01
IM(V_PRINT2)IP(V_PRINT2)IR(V_PRINT2)II(V_PRINT2)
3.142E+00
7.557E+01
7.829E-01
3.043E+00
VP(N01496)
2.211E+01
VR(N01496)
1.895E-14
VI(N01496)
7.698E-15
I A = 3.142∠ − 43.43° A and RA = 20 Ω ⇒ PA =
3.1422
20 = 98.7 W
2
3.1422
I B = 3.142∠75.57° A and RB = 20 Ω ⇒ PB =
20 = 98.7 W
2
3.1422
I C = 3.142∠ − 164.4° A and RC = 20 Ω ⇒ PC =
20 = 98.7 W
2
P = 3 ( 98.7 ) = 696.1 W
12-35
SP 12-2
FREQ
6.000E+01
IM(V_PRINT3)IP(V_PRINT3)IR(V_PRINT3)II(V_PRINT3)
1.612E+00 -1.336E+02 -1.111E+00 -1.168E+00
FREQ
6.000E+01
IM(V_PRINT1)IP(V_PRINT1)IR(V_PRINT1)II(V_PRINT1)
2.537E+00 -3.748E+01
2.013E+00 -1.544E+00
FREQ
6.000E+01
VM(N01496) VP(N01496)
1.215E+01 -1.439E+01
FREQ
6.000E+01
IM(V_PRINT2)IP(V_PRINT2)IR(V_PRINT2)II(V_PRINT2)
2.858E+00
1.084E+02 -9.023E-01
2.712E+00
I A = 2.537∠ − 37.48° A and
VR(N01496) VI(N01496)
1.177E+01 -3.018E+00
RA = 20 Ω ⇒ PA =
2.537 2
20 = 64.4 W
2
2.8582
I B = 2.858∠108.4° A and RB = 30 Ω ⇒ PB =
30 = 122.5 W
2
1.6122
I C = 1.612∠ − 133.6° A and RC = 600 Ω ⇒ PC =
60 = 78 W
2
P = 64.4 + 122.5 + 78 = 264.7 V
12-36
Verification Problems
VP 12-1
416
= 240 V = VA
3
Z = 10 + j4 = 10.77 ∠21.8° Ω
VA =
VA
240
=
= 22.28 A rms ≠ 38.63 A rms
Z
10.77
38.63
= 22.3 . It appears that the line-to-line voltage was
The report is not correct. (Notice that
3
mistakenly used in place of the phase voltage.)
IA =
VP 12-2
VL = VP = 240∠0° Vrms
Z = 40 + j 30 = 50 ∠36.9° Ω
IP =
240∠0°
VP
=
= 4.8 ∠−36.9° A rms
50∠36.9°
Z
The result is correct.
Design Problems
DP 12-1
P = 400 W per phase,
0.94 = pf = cos θ
400 =
⇒
θ = cos-1 ( 0.94 ) =20°
208
I L 0.94 ⇒ I L = 3.5 A rms
3
I
I ∆ = L = 2.04 A rms
3
V
208
= 101.8 Ω
Z = L =
2.04
I∆
Z = 101.8 ∠20° Ω
12-37
DP 12-2
VL = 240 V rms
PA = VL I L cos (30° + θ ) = 1440 W
PC = VL I L cos (30° − θ ) = 0 W ⇒
then 1440 = 240 I L cos (−30° )
IL
= IP =
VP
Z
⇒
⇒
30−θ = 90° or θ = −60°
I L = 6.93 A rms
240
V
Z = P = 3 = 20 Ω
IP
6.93
Finally, Z = 20 ∠ − 60° Ω
DP 12-3
Pin =
η
Pout
100 hp × (746
=
W
)
hp
0.8
= 93.2 kW, P =
φ
Pin
= 31.07 kW
3
VL = 480 V rms, pfc = 0.9 and pf = 0.75. We need the impedance of the load so that we can use
Eqn 11.6-7 to calculate the value of capacitance needed to correct the power factor.
0.75 = pf = cos θ
⇒
θ = cos-1 ( 0.75) = 41.4°
31070 =
480
I P 0.75 ⇒ I P = 149.5 A rms
3
480
VP
3 = 1.85 Ω
Z =
=
IP
149.5
Z = 1.85 ∠41.4° Ω = 1.388 + j1.223 Ω
The capacitance required to correct the power factor is given by
 tan (cos −1 0.75) − tan (cos −1 0.9) 
1.365
C=
= 434 µ F
×
1.3652 +1.2042
377
(Checked using LNAPAC 6/12/03)
12-38
DP 12-4
VL = 4∠0° kV rms
25
n2
VL = 4000∠0° = 100∠0° kVrms
n1
1
Try n2 = 25 then V2 =
VL 4×103∠0°
= 3∠0° kA rms
=
IL =
4
ZL
3
3000∠0°
The line current in 2.5 Ω is I =
= 120∠0° A rms
25
Thus V1 = ( R + j X ) I + V2
= (2.5 + j 40) (120∠0°) + 100×103 = 100.4 ∠2.7° kV
Step need : n1 =
Ploss = I
2
100.4 kV
= 5.02 ≅ 5
20 kV
R = 120
2
(2.5) = 36 kW, P = (4×103 ) (3× 103 ) = 12 MW
12 − .036
× 100% = 99.7 % of the power supplied by the source
12
is delivered to the load.
∴η =
12-39
Chapter 13: Frequency Response
Exercises
Ex. 13.3-1
Vo (ω )
1
=
Vs (ω )
1 + jω C R
1
H (ω ) =
gain =
1 + (ω C R) 2
phase shift = − tan −1 ω C R
1
When R = 104 , ω = 100, and C = 10−6 , then gain =
= 0.707 and phase shift = − 45o
2
Ex. 13.3-2
H (ω ) =
gain =
R
Vo (ω )
=
R + jω L
Vs (ω )
R
R 2 + (ω L) 2
 30 
2
  − 30
 .6 
⇒ ω=
= 20 rad s
2
2
0.6 =
30
30 + (2ω ) 2
2
Ex. 13.3-3
H (ω ) =
gain =
I (ω )
1
=
Vs (ω )
R + jω L
1
R 2 + (ω L) 2
phase shift = − tan −1
When R = 30 Ω, L = 2 H, and ω = 20 rad/s, then
gain =
1
30 + 40
2
2
= 0.02
ωL
R
A
 40 
and phase shift = − tan −1   = − 53.1°
V
 30 
13-1
Ex. 13.3-4
H (ω ) =
gain =
1
Vo (ω )
=
Vs (ω ) 1 + jω C R
1
1 + (ω C R) 2
phase shift = − tan −1 ω C R
−45° = − tan −1 (20 ⋅10−6 ⋅ R )
⇒ R=
tan (45° )
= 50 ⋅103 Ω
−6
20⋅10
Ex. 13.3-5
H (ω ) =
gain =
Vo (ω )
1
=
1 + jω C R
Vs (ω )
1
1 + (ω C R) 2
ω , C , and R are all positive, or at least nonnegative, so gain ≤ 1. These specifications cannot be met.
Ex. 13.4-1
(a) dB = 20 log (.5) = −6.02 dB
(b) dB = 20 log 2 = 6.02 dB
Ex. 13.4-2
 1 
20 log H = 20 log  2  = 20 log (ω )-2 = −40 log ω
ω 
ω 
slope = 20 log H (ω 2 ) − 20 log H (ω1 ) = 40 log ω 2 + 40 log ω1 = -40 log 2 
 ω1 
let ω 2 = 10 ω1 to consider 1 decade, then slope = − 40 log10 = − 40 dB
decade
13-2
Ex. 13.4-3
When ω C >> B, H (ω ) −
(d )
(b)
jω A A
=
jω C C
 A
H (ω ) in dB = 20 log10 H (ω ) = 20 log10  
C
H (ω ) does not depend on ω so slope = 0
When ω C << B, H (ω ) −
jω A
 A
= jω  
B
 B
 A
H (ω ) in dB = 20 log10 H (ω ) = 20 log10ω +20 log10  
 B
(c) The slope is the coefficient of 20 log10 ω , that is, slope = 20 dB
decade
B
(a) The break frequency is the frequency at which ω C = B, that is, ω =
C
Ex. 13.4-4
 R 
Vo (ω ) =  1+ 1  Vc (ω )
 R2 
 R 

1
=  1+ 1  
 Vs (ω )
 R2   1 + jω C R 
H (ω ) =

Vo (ω )  R1 
1
=1+ 

Vs (ω )  R2  1+ jω C R 
When R C = 0.1 and
then H (ω ) =
4
1+j
R1
= 3,
R2
ω
10
13-3
Ex. 13.4-5
Zo = R2 +
a)
jω C
1
ω
Vo
Zo
ω1
jω C
=
=
=
1
ω
Vs
R1 + Z o
R1 + R2 +
1+j
jω C
ω2
where ω1 =
and ω 2 =
vs ( t ) = 10 cos 20 t or Vs = 10∠0°
(
(
1
R2 C
R2 +
1
1+ j
= 16.7 rad/s
1
= 5.56 rad/s
( R1 + R2 )C
)
)
1+ j 20
Vo
16.7
∴
=
20
Vs
1+ j
5.56
1+ j 1.20
=
= 0.417 ∠− 24.3°
1+ j 3.60
b) So
Vo = 4.17 ∠ − 24.3°
vo (t ) = 4.17 cos(20t − 24.3°) V
13-4
Ex. 13.4-6
Ex. 13.5-1
a) Q ωo RC = R C
b) BW =
Ex. 13.5-2
Q=
ω0
ωo
BW
Now ωo =
Ex. 13.5-3
ωo = 1
Q =
ωo
LC
BW
ωo L
Q
=
7
= 10
1
1
=
= 500 rad s
−3
−7
Q LC
20 (40×10 ) (2.5×10 )
2 × 105
= 50
⇒ L =
1
LC
=
2.5 × 10−7
= 8000
= 20
L
40 × 10−3
1
ωo C
2
 (10−3 )(10−5 ) 


4
= 10
2π (15.9)
1
= 1 mH
(10 ) (10 × 10−12 )
7 2
= 104 rad s
1
1
=
2
= 100
(104 )(10−3 )
R=
=
= 0.1 Ω
Q
100
13-5
Ex. 13.5-4
a)
ωo = 1
ω
Q= o
b)
Q =
H=
H=
ωo
⇒ C =
LC
BW
1
(106 ) 2 (0.01)
ωo RC ⇒ R = ω 2
o
BW
103
ω ω 
1+ j Q − o 
 ωo ω 
1
103
= 10 Ω
=
(106 ) 2 (10−10 )
C
BW
= 1
6
= 10
= 100 pF
=
= 1000
1
1.05×106
106 
−
1+ j 1000 

6
1.05×104 
 10
1
1+ j 97.6
13-6
Problems
Section 13-3: Gain, Phase Shift, and the Network Function
P13.3-1
R 2 ||
H (ω ) =
jω C
1
=
Vo (ω )
=
Vi (ω )
=
1 + jω C R 2
R2
1 + jω C R 2
R2
R1 +
1 + jω C R 2
R2
R1 + R 2
1 + jω C R p
where Rp = R1 || R2.
When R1 = 40 W, R2 = 10 W and C = 0.5 F
H (ω ) =
R2
0.2
1 + j 4ω
(checked using ELab on 8/6/02)
P13.3-2
H (ω ) =
R2 +
Vo (ω )
jω C
=
Vi (ω ) R + R + 1
1
2
jω C
1 + jω C R 2
=
1 + jω C R1 + R 2
1
(
)
When R1 = 40 kW, R2 = 160 kW and C = 0.025 µF
H (ω ) =
1 + j ( 0.004 ) ω
1 + j ( 0.005 ) ω
(checked using ELab on 8/6/02)
13-7
P13.3-3
H (ω ) =
R2
Vo (ω )
=
Vi (ω ) R1 + R 2 + jω L
=
When R1 = 4 W, R2 = 6 W and L = 8 H
H (ω ) =
R2
R1 + R 2
L
1 + jω
R1 + R 2
0.6
1 + j ( 0.8 ) ω
(checked using ELab on 8/6/02)
P13.3-4
H (ω ) =
R 2 + jω L
Vo (ω )
=
Vi (ω ) R + R 2 + jω L
L

1 + jω

 R2 
R2
=

 R + R2 
L

  1 + jω
R + R2







Comparing the given and derived network functions, we require
L

1 + jω

 R2 
R2




L
 R + R 2   1 + jω
R + R2

Since R2 = 60 W, we have L =


1+
 = ( 0.6 )

1+


j
j
ω
ω
12
20
 R2
= 0.6

 R + R2
 R

2
⇒ 
= 12
 L
 R + R2
= 20

L

60
= 5 H , then R = ( 20 )( 5 ) − 60 = 40 Ω .
12
(checked using ELab on 8/6/02)
13-8
P13.3-5
R 2 ||
H (ω ) =
jω C
1
=
R2
1 + jω C R 2
Vo (ω )
=
Vi (ω )
=
R2
1 + jω C R 2
R2
R+
1 + jω C R 2
R2
R + R2
1 + jω C R p
where Rp = R || R2.
Comparing the given and derived network functions, we require
R2
R + R2
0.2
=
1 + jω C R p 1 + j 4 ω
Since R2 = 2 W, we have
Finally, C =
4
= 2.5 F .
1.6
 R2
= 0.2

⇒  R + R2
 CR =4
p

( 2 )(8 ) = 1.6 Ω .
2
= 0.2 ⇒ R = 8 Ω . Then R p =
2+8
R+2
(checked using ELab on 8/6/02)
P13.3-6
Vi (ω )


R + jω L

 ⇒
1
Vo (ω ) =
( A Ia (ω ) )
jω C

I a (ω ) =
A
Vo (ω )
CR
=
L
Vi (ω )
( j ω ) 1 + j ω 
R

13-9
When R = 20 W, L = 4 H, A = 3 A/A and C = 0.25 F
H (ω ) =
0.6
( j ω ) (1 + j ( 0.2 ) ω )
(checked using LNAP on 12/29/02)
P13.3-7
In the frequency domain, use voltage division on the left side of the circuit to get:
1
1
jω C
VC (ω ) =
Vi (ω ) =
Vi (ω )
1
1
j
C
R
ω
+
1
R1 +
jω C
Next, use voltage division on the right side of the circuit to get:
2
A
R3
2
3
Vo (ω ) =
A VC (ω ) = A VC (ω ) =
Vi (ω )
R 2 + R3
3
1 + jω C R1
Compare the specified network function to the calculated network function:
2
2
A
A
1
2
3
3
=
=
⇒ 4 = A and
= 2000 C
ω 1 + jω C R1 1 + jω C 2000
3
100
1+ j
100
4
Thus, C = 5 µF and A = 6 V/V.
(checked using ELab on 8/6/02)
13-10
P13.3-8
H (ω ) =
Vo (ω )
=−
Vi (ω )
R2
jω C
R1
1
 R2 
−
 R1 


=
1+ jω C R2
When R 1 = 10 kΩ, R 2 = 50 kΩ , and C = 2 µ F, then
R2
R1
= 5 and R2 C =
1
−5
so H (ω ) =
ω
10
1+ j
10
P13.3-9
H (ω ) =
Vo (ω )
=−
Vi (ω )
R2
1
j ω C2
R1
1
j ω C1
R2
1 + jω C2 R2
= −
R1
1 + jω C1R1
 R   1 + jω C1R1 
H (ω ) = −  2  

 R1   1 + jω C2 R2 
When R1 = 10 kΩ, R2 = 50 kΩ , C1 = 4 µ F and C2 = 2µ F,
then
so
R2
R1
= 5 , C1R1 =
1
1
and C2 R 2 =
25
10
ω 

 1 + j 25 
H (ω ) = − 5 
ω 
 1 + j

10 

13-11
gain = H(ω ) = ( 5 )
1+
1+
ω2
625
ω2
100
ω 
ω 
phase shift = ∠Η (ω ) = 180 + tan −1   − tan −1  
 25 
 10 
P13.3-10
1
R1
jω C
=
1
1 + jω C R3
R3 +
jω C
1
=
jω C
R3
H (ω ) = −
R3
R2 +
R3
1+ jω C R3
R + R + jω R2 R3C
=− 2 3
R1
R1 + jω R1 R3C
5 = lim H (ω ) =
ω →0
2 = lim H (ω ) =
ω →∞
then
P13.3-11
H (ω ) = −
R2 + R3
R1
R2
⇒ R2 = 2 R1 = 20 kΩ
R1
R3 = 5 R1 − R2 = 30 kΩ
R2 +
jω C
1
R1
=−
1 + jω C R 2
jω C R1
∠H (ω ) = 180° + tan −1 ω C R 2 − 90°
∠H (ω ) = 135° ⇒ tan −1 ω CR2 = 45° ⇒ ω C R 2 = 1
⇒ R2 =
10 = lim H (ω ) =
ω →∞
R2
R
⇒ R1 = 2 = 1 kΩ
10
R1
1
= 10 kΩ
10 10−7
3
13-12
−R2
jω C R 2
= −
1
1+ jω C R1
jω C
R
10 = lim H(ω ) = 2 ⇒ R2 = 10 R1
R1
ω →∞
H (ω ) =
P13.3-12
∠ H(ω ) = 180°+90°− tan −1 ω C R1
⇒ R1 =
tan (270°−∠ H(ω ))
= 104 ⋅tan(270°−∠H(ω )) = 104 = 10 kΩ
ωC
⇒ R2 = 100 kΩ
P13.3-13
1
jω C2
V (ω )
=−
H (ω ) = o
1
Vs (ω )
R1 +
jω C1
R2
( −C1R2 ) jω
=
(1 + jω R1C1 ) (1 + jω R2C2 )
When R1 = 5 kΩ, C1 = 1 µ F,
R2 = 10 kΩ and C2 = 0.1 µ F,
( −0.01) jω
then
H (ω ) =
so
ω  
ω 


1 + j
 1 + j
200  
1000 

ω
0
500
2500
H (ω )
0
1.66
0.74
∠H (ω )
−90°
175°
116°
Then
v ( t ) = (0) 50 + (1.66) ( 30 ) cos(500t + 115° + 175°) − (0.74) ( 20 ) cos(2500t + 30° + 116°)
o
=49.8cos(500t −70°) −14.8cos (2500t +146°) mV
When R1 =5 kΩ, C1 =1 µ F, R 2 =10 kΩ and C2 = 0.01 µ F, then
13-13
H (ω )= − 0.01
ω
So
0
500
2500
jω
ω 
ω 

 1+ j
  1+ j
200  
10,000 

H (ω )
0
1.855
1.934
∠H (ω )
−90°
−161°
170°
Then
v (t ) = (0) ( 50 ) + (1.855) ( 30 ) cos(500t + 115° − 161°) − (1.934) ( 20 ) cos(2500t + 30° + 170°)
o
= 55.65 cos(500t − 46°) − 38.68cos(2500t + 190°) mV
P13.3-14
a)
b)
2 V
(8 div) 

 div  = 8 V
Vs =
2
2 V
(6.2 div) 

 div  = 6.2 V
Vo =
2
V 6.2
= 0.775
gain = o =
8
Vs
H (ω ) =
Vo (ω )
1
jω C
=
=
1
1 + jω C R
Vs (ω )
R+
jω C
1
1
Let g = H (ω ) =
then C =
ωR
1 + ω 2C 2 R 2
1
In this case ω = 2π ⋅ 500 = 3142 rad s ,
1
  −1
g
2
H (ω ) = 0.775 and R = 1000 Ω so C = 0.26 µ F.
c)
tan( −∠ H (ω ))
RC
Recalling that R = 1000 Ω and C=0.26µF, we calculate
∠ H (ω )= − tan −1 ω R C so ω =
13-14
ω
H (ω )
0.95
0.26
2π (200)
2π (2000)
∠H (ω )
−18°
−73°
( )
tan  − −45° 

 = 3846 rad s
∠ H (ω )= − 45° requires ω =
1000 .26×10−6
(
∠ Η (ω ) = −135° requires ω =
)
(
)
tan ( −(−135°))
= − 3846 rad s
(1000)(0.26×10−6 )
A negative frequency is not acceptable. We conclude that this circuit cannot
produce a phase shift equal to −135 .
°
d)

tan (−(60° ))
=
= 0.55µ F
C

(2π ⋅ 500) (1000)
tan (−∠H (ω ))

⇒ 
C=
°
ωR
C = tan (−(−300 )) = −0.55µ F

(2π ⋅ 500 ) (1000)
A negative value of capacitance is not acceptable and indicates that this
circuit cannot be designed to produce a phase shift at −300 at a frequency of
500 Hz.
°
e)
tan( − (−120° ))
C =
= −0.55 µ F
(2π ⋅ 500)(100)
This circuit cannot be designed to produce a phase shift of −120 at 500 Hz.
°
13-15
Section 13-4: Bode Plots
P13.4-1
ω

20  1 + j 
5
H (ω )= 
ω

1 + j 
50 

P13.4-2
H1 (ω ) =
1+ j
1+ j



20



 20  j ω 



≈
 5


  ω
 20  j 5 
 = 200
 
 jω
  50 
ω
ω
5
50
H 2 (ω ) = 10
ω <5
5 < ω < 50
50 < ω
1+ j
1+ j
ω
ω
5
50
Both H1(ω) and H2(ω) have a pole at ω = 50 rad/s and a zero at ω = 5 rad/s. The slopes
of both magnitude Bode plots increase by 20 dB/decade at ω = 5 rad/s and decrease by 20
dB/decade at ω = 50rad/s. The difference is that for ω < 5rad/s
H1 (ω ) 1 = 0 dB and
H 2 (ω ) 10 = 20 dB
13-16
P13.4-3
R2
jω
1+ jω C2 R2
H (ω ) = −
= −C1 R2
1
(1+ jω R1C1 )(1+ jω R2C2 )
R1 +
jω C1
This network function has poles at
p1 =
so
1
1
= 2000 rad s and p2 =
= 1000 rad s
R1C1
R2C2

(C R ) jω
 1 2

jω
R
H (ω ) (C1 R2 )
= 2 =2
jω C1 R1 R1


1
jω
=
(C1 R2 )
( jω C1 R1 )( jω C2 R2 ) jω C2 R1

ω < p1
p1 < ω < p2
ω > p2
13-17
P13.4-4
R2
1
1
R (1+ jω C1 R1 )
R
1+ jω C2 R2
so K = − 2 , z =
and p =
H (ω ) = −
=− 2
R1
R1 (1+ jω C2 R2 )
R1
C1 R1
C2 R2
1+ jω C1 R1
When z < p
When z > p
P13.4-5
Using voltage division twice gives:
V2 (ω )
=
Vi (ω )
and
jω L R 2
R 2 + jω L
jω L R 2
L
jω
=
=
jω L R 2
R1 R 2 + jω L ( R1 + R 2 ) R1
L ( R1 + R 2 )
R1 +
ω
j
+
1
R 2 + jω L
R1 R 2
Vo (ω )
=
V2 (ω )
R4
A R4
R 3 + jω C R 4
A R4
R3 + R 4
A=
=
R4
C R3 R 4
R3 + R 4 + jω C R3 R 4
R3 +
1 + jω
R 3 + jω C R 4
R3 + R 4
Combining these equations gives
13-18
H (ω ) =
ALR 4
Vo (ω )
jω
=
Vi (ω ) R1 ( R 3 + R 4 ) 
L ( R1 + R 2 )  
CR 3 R 4 
1 + jω
  1 + jω


 
R1 R 2
R 3 + R 4 


The Bode plot corresponds to the network function:
H (ω ) =
k jω
k jω
=
ω 
ω 

ω 
ω  
1+ j

1 + j 1 + j  1 + j 200 
20000 

p1 
p2 




 k jω
= k jω

1 ⋅1

 k jω
H (ω ) ≈ 
= k p1
ω
j

⋅1
p1

 k jω
k p1 p2

=
jω
 jω ⋅ jω
 p p
 1
2
ω ≤ p1
p1 ≤ ω ≤ p2
ω ≥ p2
This equation indicates that |H(ω)|=k p1 when p1 ≤ ω ≤ p2. The Bode plot indicates that
|H(ω)|=20 dB = 10 when p1 ≤ ω ≤ p2. Consequently
k=
H (ω ) =
Finally,
10 10
=
= 0.05
p1 200
0.05 jω
ω 
ω 


1 + j
1 + j
200 
20000 

Comparing the equation for H(ω) obtained from the circuit to the equation for
H(ω)obtained from the Bode plot gives:
0.05 =
R1 ( R 3 + R 4 )
ALR 4
, 200 =
L ( R1 + R 2 )
R1 R 2
and 20000 =
R3 + R 4
C R3 R 4
Pick L = 1 H, and R1 = R2 , then R1 = R2 = 400 Ω. Let C = 0.1 µF and R3 = R4 , then R3 =
R4 = 1000 Ω. Finally, A=40. (Checked using ELab 3/5/01)
13-19
P13.4-6
From Table 13.4-2:
R2
R1
= k = 32 dB = 40 R 2 = 40 (10 ×103 ) = 400 kΩ
1
1
= p = 400 rad/s ⇒ C 2 =
= 6.25 nF
C 2 R2
( 400 ) ( 400 ×103 )
1
1
= z = 4000 rad/s ⇒ C 1 =
= 25 nF
C 1 R1
( 4000 ) (10 ×103 )
P13.4-7
H (ω ) =
H (ω ) =
R 2 + jω L
Vo (ω )
=
Vi (ω ) R + R 2 + jω L
L

1 + jω

 R2 
R2
=

 R + R2 
L

  1 + jω
R + R2

( 0.2 ) (1 + j ( 0.25 ) ω )
1 + j ( 0.05 ) ω







 k = 0.2

1

⇒  z=
=4
0.25

1

 p = 0.05 = 20
P13.4-8
• The slope is 40dB/decade for low frequencies, so the numerator will include the
factor (jω)2 .
• The slope decreases by 40 dB/decade at ω = 0.7rad/sec. So there is a second order
pole at ω 0 = 0.7 rad/s. The damping factor of this pole cannot be determined from the
asymptotic Bode plot; call it δ 1. The denominator of the network function will
contain the factor
1 + 2 δ1 j
•
ω
 ω 
−

0.7  0.7 
2
The slope increases by 20 dB/decade at ω = 10 rad/s, indicating a zero at 10 rad/s.
13-20
•
•
The slope decreases by 20 dB/decade at ω = 100 rad/s, indicating a pole at 100 rad/s.
The slope decreases by 40 dB/decade at ω = 600 rad/s, indicating a second order pole
at ω 0 = 600rad/s. The damping factor of this pole cannot be determined from an
asymptotic Bode plot; call it δ 2. The denominator of the network function will
contain the factor
ω
 ω 
−
1 + 2δ 2 j

600  600 
H (ω ) =
K (1+ j
ω
2
)( jω ) 2
2
2

ω  ω  
ω  ω  
ω 
−
−
 1+ 2δ1 j
1+ 2δ 2 j

 
 1+ j
0.7  0.7  
600  600  
100 

10
To determine K , notice that H (ω ) = 0 dB=1
1=
K (1)( jω ) 2
 ω 
−
 (1)(1)
 0.7 
2
when 0.7 < ω < 10. That is
= K (0.7) 2 ⇒ K = 2
ω

K 1+ j 
z

H (ω ) =
jω
P13.4-9
(a)
H (ω ) =
ω
K
ω 
1+  
z
H (ω ) dB = 20 log10
ω
K
2
ω 
1+ 
z
2
= 20 log10 K − 20 log10 ω + 20 log10
Let
H L (ω ) dB = 20 log10 K − 20 log10 ω
and H H (ω ) dB = 20 log10
K
z
 H L (ω ) dB
Then H (ω ) dB ~_ 
 H H (ω ) dB
ω 
1+ 
z
2
ω << z
ω >> z
13-21
So H L (ω ) dB and H H (ω ) dB are the required low and high-frequency asymptotes.
The Bode plot will be within 1% of |H(ω)| dB both for ω << z and for ω >> z. The range
when ω << z is characterized by
H L (ω ) = 0.99 H (ω )
(gains not in dB)
or equivalently
20 log10 ( 0.99 ) = H L (ω ) dB − H (ω ) dB
= 20 log10 K − 20 log10 ω − 20 log10
= − 20 log10
Therefore
0.99 =
2
ω 
1+ 
ω
z
K
2
1
ω 
1+  
z
2
z
 1 
⇒ ω= z 
 −1 = 0.14 z −
7
 .99 
2
1
ω 
1+ 
z
ω 
1+  = 20 log10
z
(gains in dB)
2
The range when ω >> z is characterized by
H H (ω ) = .99 H (ω )
(gains not in dB)
or equivalently
13-22
20 log10 0.99 = H H (ω ) dB − H (ω ) dB
(gains in dB)
ω 
= 20 log10 K − 20 log10 z − 20 log10
1+  
ω
z
K
2
1
ω 
= − 20 log10
1+  = 20 log10
2
ω
z
z
  +1
ω 
2
z
Therefore
 1 
= 
 −1 ⇒
ω
 .99 
The error is less than 1% when ω <
P13.4-10
ω=
2
z
H (ω ) =
=
V0 (ω )
=
Vs (ω )
z
 1 

 −1
 .99 
2
=
z
− 7 z
0.14
z
and when ω > 7z.
7
Rt
R t + R1
R t (1+ jω C R1 )
jω C
1
=
Rt +
 Rt
=
R1 + R t + jω C R1 R t  R1 + R t
When R1 = 1 kΩ, C = 1 µ F and R t = 5 kΩ
ω 

1+ j

5
1000  ⇒
H (ω ) = 

6  1+ j ω 
1200 

Rt
R1
1+ jω C R1
 1+ jω C R1

 1+ jω  C R1 R t

 R1 + R t



5
ω <1000
6

 5  ω
1000<ω <1200
H (ω ) ≅   j
 6  1000
1
ω <1200


13-23
P13.4-11
Mesh equations:
Vin (ω ) = I (ω ) [ R1 + ( jω L1 − jω M ) + (− jω M + jω L2 ) + R2 ]
Vo (ω ) = I (ω ) [(− jω M + jω L2 ) + R2 ]
Solving yields:
H (ω ) =
V0 (ω )
R2 + jω ( L2 − M )
=
Vin (ω ) R1 + R2 + jω ( L1 + L2 − 2M )
Comparing to the given Bode plot yields:
K1 = ωlim
→∞ |H (ω )| =
z =
L2 − M
R2
= 0.75 and K 2 = lim | H (ω ) | =
= 0.2
ω →0
L1 + L2 − 2 M
R1 + R2
R2
R1 + R2
= 333 rad s and p =
=1250 rad s
L2 − M
L1 + L2 − 2 M
13-24
P13.4-12
1
1+ jω R1 C1
1 (1+ jω R1 C1 )
jω C2
H (ω ) = −
=−
=−
jω R1 C2
R1 C2
jω
1
R1
jω C1
 1  1 
−


 R C jω 
H (ω ) −  1 2 
 − 1 ( R C ) = − C1
 R1 C2 1 1
C2
With the given values:
C1 1
= = −6 dB,
C2 2
ω<
ω>
1
R1 C1
1
R1 C1
1
= 4000 rad / s
R1 C1
13-25
P13.4-13
Pick the appropriate circuit from Table 13.4-2.
We require
200 = z =
1
1
p C
, 500 = p =
and 14 dB = 5 = k = 1
C 1 R1
C2 R 2
z C2
Pick C1 = 1 µ F, then C2 = 0.2 µ F, R1 = 5 kΩ and R 2 = 10 kΩ.
P13.4-14
Pick the appropriate circuit from Table 13.4-2.
We require
500 = p =
R2
1
and 34 dB = 50 =
C R2
R1
Pick C = 0.1 µ F, then R 2 = 20 kΩ and R1 = 400 Ω.
13-26
P13.4-15
Pick the appropriate circuit from Table 13.4-2.
We require
500 = z =
1
1
p C
, 200 = p =
and 14 dB = 5 = k = 1
C 1 R1
C2 R 2
z C2
Pick C1 = 0.1 µ F, then C2 = 0.05 µ F, R1 = 20 kΩ and R 2 = 100 kΩ.
13-27
P13.4-16
Pick the appropriate circuit from Table 13.4-2.
We require
200 = p 1 =
1
1
, 200 = p 2 =
and 34 dB = 50 = k = C 1 R 2
C 1 R1
C2 R 2
Pick C1 = 1 µ F, then C2 = 0.04 µ F, R1 = 5 kΩ and R 2 = 50 kΩ.
P13.4-17
H (ω ) =
10(1+ jω 50)
(1+ jω 2)(1+ jω 20)(1+ jω 80)
13-28
ϕ = ∠H (ω ) = tan −1 (ω 50 ) − ( tan −1 (ω 2 ) + tan −1 (ω 20 ) + tan −1 (ω 80 ) )
P13.4-18
(a)
H (ω ) =
Vo (ω )
R2 R1
=−
Vs (ω )
1+ jω R2C
=−
(b)
10 = 20 dB
(c)
10,000 rad/s
1+ j
ω
10
10,000
13-29
P13.4-19




 ⇒ Vo (1 + jω C 1 R1 )(1 + jω C 2 R 2 ) = jω C 1 R1Vo + Vs


Va (ω ) − Vs (ω )
0=
+ jω C 1 (Va (ω ) − Vo (ω )) 
R1

1
jω C 2
Vo (ω ) =
Va (ω )
1
R+
jω C 2
T(ω ) =
Vo (ω )
1
1
=
=
2
2
−ω + 0.8 jω +1
Vs (ω ) 1+ C 2 R 2 jω −ω C 1C 2 R1 R 2
This is a second order transfer function with ωo = 0 and δ = 0.4 .
13-30
Section 13-5: Resonant Circuits
P13.5-1
ω0 =
ω0
1
1
=
= 60 k rad sec
LC
 1  1

x 10−6 


120
30



1
× 10−6
C
= 10, 000 30
= 20
Q= R
1
L
120
ω 
ω 
ω
2
2
+  0  +ω 0 = 58.52 k rad s and ω 2 = 0 +  0  +ω 0 = 61.52 k rad s
ω1 = −
2Q
2Q
 2Q 
 2Q 
1
1
=
= 3 krad s
BW =
RC
 1
−6 
(10000 ) ×10 
 30

2
2
Notice that BW = ω 2 − ω1 =
ω0
Q
.
P13.5-2
H (ω ) =
k
ω ω 
1+ Q  − 0 
 ω0 ω 
2
2
so
R = k = H (ω 0 ) =
At ω = 897.6 rad s , H (ω ) =
4
= 200, so
20.10−3
200 =
Then
8
= 400 Ω and ω 0 = 1000 rad s
20⋅10−3
400
 897.6 1000 
−
1+ Q 2 

 1000 897.6 
2
⇒ Q =8
1

= ω 0 = 1000 
LC

C = 20 µ F
 ⇒ L = 50 mH
C
400
=Q=8 

L
13-31
P13.5-3
P13.5-4
P13.5-5
ω0 =
1
1 L
R
= 105 rad s , Q =
= 10, BW = = 104 rad s
R C
L
LC
ω0 =
1
1 L
R
= 104 rad s , Q =
= 10, BW = = 103 rad s
R C
L
LC
R = Z (ω 0 ) = 100 Ω
1
= BW = 500 ⇒ C = 20 µ F
100 C
1
= ω 0 = 2500 ⇒ L = 8 mH
−6
20
10
L
⋅
(
)
P13.5-6
R=
Y (ω 0 )
1
= 100 Ω
100
= BW = 500 ⇒ L = 0.2 H
L
1
= ω 0 = 2500 ⇒ C = 0.8 µ F
( 0.2 )C
13-32
P13.5-7
Y (ω ) = jω C +
1
1
+
R1 + jω L R 2
=
(R +R
=
R1 ( R1 + R 2 −ω 2C L R 2 ) −ω 2 L( L +C R1 R 2 )+ jω R1 ( L C R1 R 2 ) − jω L( R1 + R 2 −ω 2C L R 2 )
1
2
−ω 2C L R 2 ) + jω ( L +C R1 R 2 ) R1 − jω L
×
R1 − jω L
R 2 ( R1 + jω L )
R 2 ( R1 −ω 2 L2 )
ω = ω 0 is the frequency at which the imaginary part of Y (ω ) is zero :
R1 ( L C R1 R 2 ) − L ( R1 + R 2 −ω C L R 2 ) = 0
2
0
⇒ ω0 =
L R 2 −C R12 R 2
C L2 R 2
= 12.9 M rad sec
13-33
P13.5-8
(a) Using voltage division yields
(
Vo = 1000∠0°
(
(100 )( − j100 )
) (100)(100− j−100j100) + j100
)
100 − j100
105
100 2 ∠−135°
= 1000∠0°
= 2∠−135° = 1000∠90° V
100 2 ∠−135°+ j100 50 2∠−135°
∴|Vo | = 1000 V
(b) Do a source transformation to obtain
This is a resonant circuit with ω 0 = 1
LC = 400 rad/s. That’s also the frequency of the input, so
this circuit is being operated at resonance. At resonance the impedances of the capacitor and
inductor cancel each other, leaving the impedance of the resistor. Increasing the resistance by a
factor of 10 will increase the voltage Vo by a factor of 10. This increased voltage will cause
increased currents in both the inductance and the capacitance, causing the sparks and smoke.
13-34
P13.5-9
Let G 2 =
1
. Then
R2
Z = R1 + jω L +
(R G
=
1
2
1
G 2 + jω C
+ 1 − ω 2 L C ) + j (ω LG 2 + ω C R1 )
At resonance, ∠Z = 0° so
tan −1
so
G 2 + jω C
ω L G2 +ω C R1
ωC
= tan −1
2
G2
( R1G 2 +1−ω L C )
ω L G 2 +ω C R1
C − L G 22
ωC
2
=
⇒ ω =
LC2
( R1 G 2 +1−ω 2 L C ) G 2
and C > G 22 L
C−L
. Then choose C and calculate L:
LC2
C = 10 mF ⇒ L = 5 mH
With R1 = R 2 = 1 Ω and ω 0 = 100 rad s , ω0 = 104 =
2
Since C > G 22 L , we are done.
13-35
P13.5-10
R
R −ω 2 R L C ) + jω L
(
jω C
=
Z in = jω L +
1
1+ jω R C
R+
jω C
Consequently,
(a)
| Zin | =
(b)
( R −ω
R L C ) + (ω L )
2
1+(ω R C )
2
2
2
(c)
ω=
1
LC
⇒ | Zin | =
1
C  R2 C 
1 +

L
L 
P13.5-11
Let V (ω ) = A∠0 and V2 (ω ) = B∠θ . Then
I (ω )
=
Y (ω ) =
V (ω )
V (ω ) − V2 (ω )
A − B∠θ A − B cos θ − j B sin θ
R
=
=
AR
AR
V (ω )
| Y (ω ) | =
( A − B cosθ ) + ( B sin θ )
2
2
AR
13-36
PSpice Problems
SP13.1
Here are the magnitude and phase frequency response plots:
From the magnitude plot, the low frequency gain is k = 200m = 0.2.
From the phase plot, the angle is -45° at p = 2π ( 39.891) = 251 rad/s .
13-37
SP13-2
Here is the magnitude frequency response plot:
The low frequency gain is 0.6 = lim H (ω ) = k ⇒ k = 0.6 .
ω →0
The high frequency gain is 1 = lim H (ω ) = k
ω →∞
At ω = 2π ( 2.8157 ) = 17.69 rad/s ,
 17.69 
1+ 

 0.6 p 
0.8 = 0.6
2
 17.69 
1+ 

 p 
p
z
2
⇒ z = ( 0.6 ) p
⇒
16 p 2 + 869
= 2
9
p + 313
⇒
16 2
p + 313 = p 2 + 869
9
⇒
p = 20 rad/s
⇒
⇒
(
)
( 0.77778 ) p 2 = 312.56
z = 12 rad/s
13-38
SP13-3
From the magnitude plot, the low frequency gain is k = 4.0.
From the phase plot, the angle is -45° at p = 2π (15.998 ) = 100.5 rad/s .
13-39
SP13-4
From the magnitude plot, the low frequency gain is k = 5.0.
From the phase plot, the angle is 180°-45°=135° at p = 2π (1.5849 ) = 9.958 rad/s .
13-40
SP13-5
104
104
R
R
H (ω ) = −
=
∠ − tan −1 (ω C 104 )
4
2
1 + jω C 10
1 + (ω C 104 )
When ω = 200 rad/sec = 31.83 Hertz
1.8565∠158° =
104
R
1 + (ω C 10
)
4 2
∠ − tan −1 (ω C 104 )
Equating phase shifts gives
ω C 104 = 103
C R 104
= tan(22°) = 0.404 ⇒ C = 0.2 µ F
R + 104
Equating gains gives
1.8565 =
104
R
1 + (ω C 10
)
4 2
=
104
R
1 + ( 0.404 )
2
⇒ R = 5 kΩ
SP13-6
104
1 + jω C R 2
104
104
C R 104 
R + 104
R + 104
−1 
ω
=
=
∠
−
tan
H (ω ) =

4 
C R 104
104
4 2
 R + 10 


C
R
10
R
j
1
ω
+
+
1 + ω
R + 104
1 + jω C 104
4 
 R + 10 
When ω = 1000 rad/sec = 159.1 Hertz
104
 C R 104 
R + 104
∠ − tan −1  ω
0.171408∠ − 59° =
4 
2
 R + 10 
 C R 104 
1 + ω
4 
 R + 10 
Equating phase shifts gives
4
C R 104
3 C R 10
ω
=
= tan(59°) = 1.665
10
R + 104
R + 104
Equating gains gives
13-41
0.171408 =
104
R + 104
 C R 104 
1 + ω
4 
 R + 10 
2
=
104
R + 104
1 + (1.665 )
2
⇒ R = 20 kΩ
Substitute this value of R into the equation for phase shift to get:
C ( 20 ×103 ) 104
C R 104
3
1.665 = 10
= 10
R + 104
( 20 ×103 ) + 104
3
⇒ C = 0.25 µ F
Verification Problems
VP13-1
When ω < 6300 rad/s, H(ω) ≅ 0.1, which agrees with the tabulated values of | H(ω)|
corresponding to ω = 200 and 400 rad/s.
When ω > 6300 rad/s, H(ω) ≅ 0.1, which agrees with the tabulated values of | H(ω)|
corresponding to ω = 12600, 25000, 50000 and 100000 rad/s.
At ω = 6300 rad/s, we expect | H(ω)| = −3 dB = 0.707. This agrees with the tabulated value of |
H(ω)| corresponding to ω = 6310 rad/s.
At ω = 630 rad/s, we expect | H(ω)| = −20 dB = 0.14. This agrees with the tabulated values of |
H(ω)| corresponding to ω = 400 and 795 rad/s.
This data does seem reasonable.
VP13-2
BW =
ω0
Q
=
10,000
= 143 ≠ 71.4 rad s . Consequently, this report is not correct.
70
VP13-3
ω0 =
1
1 L
R
= 10 k rad s = 1.59 kHz, Q =
= 20 and BW = = 500 rad s = 79.6 Hz
R C
L
LC
The reported results are correct.
13-42
VP13-4
The network function indicates a zero at 200 rad/s and a pole at 800 rad/s. In contrast, the Bode
plot indicates a pole at 200 rad/s and a zero at 800 rad/s. Consequently, the Bode plot and
network function don’t correspond to each other.
Design Problems
DP13-1
Pick the appropriate circuit from Table 13.4-2.
We require
2π ×1000 < z =
R2
p C
1
1
, 2π ×10000 > p =
, 2=k =
and 5 = k = 1
C 1 R1
C2 R 2
z C2
R1
Try z = 2π × 2000. Pick C1 = 0.05 µ F. Then
R1 =
Check: p =
1
C
C
= 1.592 kΩ, R 2 = 2 R1 = 3.183 kΩ and C2 = 1 = 1 = 0.01 µ F
p 2
C1 z
k
z
1
= 31.42 k rad s < 2π ⋅10, 000 rad s.
C 2 R2
13-43
DP13-2
1
V (ω )
1+ jω C R
jω C
LC
H (ω ) = o
=
=
=
R
1
1
Vs (ω )
 1

+
−ω 2 + jω
|| R  jω L +
jω L + 
1+ jω C R
RC LC
 jω C 
1
Pick
|| R
R
1
= ω 0 = 2π (100 ⋅103 ) rad s . When ω = ω 0
LC
So H (ω 0 ) = R
1
LC
H 0 (ω ) =
1
1
1
1
−
+j
+
LC
LC RC LC
C
. We require
L
−3 dB = 0.707 = H (ω 0 ) = R
Finally
C
C
= 1000
L
L
1

= 2π (100⋅103 ) 
LC
C =1.13 nF

⇒
C
 L = 2.26 mH
0.707 =1000

L
13-44
DP13-3
R1 = 10 kΩ
R 2 = 866 kΩ
R 3 = 8.06 kΩ
R 4 = 1 MΩ
R 5 = 2.37 MΩ
R 6 = 499 kΩ
C 1 = 0.47 µ F
C 2 = 0.1 µ F
Va = −
Circuit A
Circuit B
Vo = −
Circuit C
Vc = −
R3
R2
Vc −
R3
R1
Vs = −H 1 Vc − H 2 Vs
R5
R4
1 + jω C 1 R 5
Va = − H 3 Va
1
Vo = −H 4 Vo
jω C 2 R 6
Vc = H 3 H 4 Va
Then
Va = −H 2 Vs − H1 H 3 H 4 Va
Vo = −H 3 Va =
After some algebra
Vo =
jω
⇒ Va =
−H 2
Vs
1 + H1 H 3 H 4
H 2 H3
Vs
1 + H1 H 3 H 4
R3
R1 R 4 C 1
R3
R 2 R 4 R 6 C1 C 2
−ω + j
2
ω
Vs
R5 C1
This MATLAB program plots the Bode plot:
R1=10;
R2=866;
R3=8.060;
R4=1000;
% units: kOhms and mF so RC has units of sec
13-45
R5=2370;
R6=449;
C1=0.00047;
C2=0.0001;
pi=3.14159;
fmin=5*10^5;
fmax=2*10^6;
f=logspace(log10(fmin),log10(fmax),200);
w=2*pi*f;
b1=R3/R1/R4/C1;
a0=R3/R2/R4/R6/C1/C2;
a1=R5/C1;
for k=1:length(w)
H(k)=(j*w(k)*b1)/(a0-w(k)*w(k)+j+w(k)*a1);
gain(k)=abs(H(k));
phase(k)=angle(H(k));
end
subplot(2,1,1), semilogx(f, 20*log10(gain))
xlabel('Frequency, Hz'), ylabel('Gain, dB')
title('Bode Plot')
subplot(2,1,2), semilogx(f, phase*180/pi)
xlabel('Frequency, Hz'), ylabel('Phase, deg')
13-46
DP13-4
Pick the appropriate circuits from Table 13.4-2.
We require
10 = − k1k2 = R 2C1
Pick C 1 = 1 µ F. Then R1 =
Next
R4
R3
, 200 = p1 =
1
1
and 500 = p2 =
R1 C 1
C 2 R4
1
1
= 5 kΩ. Pick C 2 = 0.1 µF. Then R 4 =
= 20 kΩ.
p1 C1
p2C2
10 =
Let R 2 = 500 kΩ and R 3 = 1 kΩ.
R2
R3
(10−6 )(20 ⋅103 ) ⇒
R2
R3
= 500
13-47
DP13-5
Pick the appropriate circuits from Table 13.4-2.
We require
20 dB = 10 = − k1k2 = R 2C1
Pick C 1 = 20 µ F. Then R1 =
Next
R4
R3
, 0.1 = p1 =
1
1
and 100 = p2 =
R1 C 1
C 2 R4
1
1
= 500 kΩ. Pick C 2 = 1 µF. Then R 4 =
= 10 kΩ.
p1 C 1
p2C2
10 =
Let R 2 = 200 kΩ and R 3 = 4 kΩ.
R2
R3
(20 ⋅10−6 )(10 ⋅103 ) ⇒
R2
R3
= 50
13-48
DP13-6
The network function of this circuit is H (ω ) =
1+
R2
R3
1+ jω R1C
The phase shift of this network function is θ = − tan −1 ω R1C
The gain of this network function is G = H (ω ) =
1+
R3
R2
1+ (ω R1C ) 2
=
1+
R3
1+ ( tan θ )
R2
2
Design of this circuit proceeds as follows. Since the frequency and capacitance are known, R1 is
tan(−θ )
calculated from R1 =
. Next pick R2 = 10kΩ (a convenient value) and calculated R3 using
ωC
R 3 = (G ⋅ 1+ (tan θ ) 2 − 1) ⋅ R 2 . Finally
θ = −45 deg, G = 2, ω = 1000 rad s ⇒ R1 = 10 kΩ, R 2 = 10 kΩ, R 3 = 18.284 kΩ, C = 0.1 µF
DP13-7
From Table 13.4-2 and the Bode plot:
800 = z =
1
⇒ R1 = 2.5 kΩ
R1 (0.5×10−6 )
32 dB = 40 =
200 = p =
(Check: 20 dB = 10 = k
DP13-8
R2
R1
⇒ R 2 = 100 kΩ
1
1
⇒ C =
= 0.05µ F
R2C
(200)(100×103 )
p 0.5×10−6
0.5×10−6
=
=
)
z
C
0.05×10−6
−R2
jω C R 2
= −
1
1+ jω C R1
1+
jω C
tan(270°−195°)
= 37.3 kΩ
195° = 180 + 90 − tan −1 ω C R1 ⇒ R1 =
(1000)(0.1×10−6 )
R
10 = lim H (ω ) = 2 ⇒ R 2 = 10 R1 = 373 kΩ
ω →∞
R1
H (ω ) =
13-49
Chapter 14: The Laplace Transform
Exercises
Ex. 14.3-1
f ( t ) = cos ω t =
e + jω t + e − jωt
1
and L  e at  =
s −a
2
= [cos ω t ] =
F ( s ) ฀L
Ex. 14.3-2
Ex. 14.4-1
Ex. 14.4-2
Ex. 14.4-3
Ex. 14.4-4
F ( s ) = L [e
−2 t
s
1 1
1 
+
= 2 2


2  s − jω s + jω  s +ω
s2 + s + 3
1
1
−2 t


+ sin t ] = L e  + L [sin t ] =
+
=
s + 2 s 2 +1
( s + 2)( s 2 + 1)
F ( s ) = L [2u (t ) + 3e −4t u (t )] = 2L [u (t )] + 3L [e−4t u (t )] =
2
3
+
s s+4
 1 
F ( s ) = L [sin(t − 2)u (t − 2)] = e −2 s L [sin t ] = e −2 s  2 
 s +1 
F ( s ) = L[t e − t ] = L[t ] s → s +1=
1
s2
s → s +1
=
1
( s + 1) 2
 5

 5

f ( t ) =  − t + 5  u ( t ) −  − ( t − 4.2 )  u ( t − 4.2 )
 3

 3

−4.2 s
− 1)
 5 5  −4.2 s  5  15 s + 5 ( e
−
=
F (s) =  − 2 +  − e

2 
s
3 s2
 3s
 3s 
14-1
Ex. 14.4-5
F (s) = ∫0
∞
3 e − st
f (t ) e dt = ∫ 0 3 e dt =
−s
− st
2
− st
2
0
=
3(1−e −2 s )
s
Ex. 14.4-6
5 2 t 0<t < 2
f (t ) = 
otherwise
0
f (t ) =
5
5
5
5
t u ( t ) −u ( t − 2 )  = t u ( t ) − t u ( t − 2 ) = t u ( t ) −( t − 2 )u ( t − 2 )− 2u (t − 2) 
2
2
2
2
−2 s
−2 s
5 1 e
2e  5 1
1− e −2 s − 2 se −2 s 
∴F ( s ) = L  f ( t )  =  2 − 2 −
=

2 
2s
s
s  2 s
Ex. 14.5-1
F (s) =
c + jd
c − jd
me jθ
me jθ
where m = c 2 + d 2 , θ = tan −1 d
+
=
+
c
s + a − jω s + a + jω s + a − jω s + a + jω
∴ f ( t ) = e − at [ c cos ω t − d sin ω t ] u ( t ) = e− at  c 2 + d 2 cos(ω t +θ )  u ( t ) = m e− at cos(ω t +θ ) u ( t )


Ex. 14.5-2
(a)
F (s) =
8 s −3
1 2( 8s −3)
= ×
s + 4s +13 2 ( s + 2 )2 + 9
2
∴ a = 2, c =8, ω =3 & ca −ω d =−3 ⇒ d =
−3( 8 )( 2 )
= 6.33
−3
2
2
 6.33 
°
∴ θ = tan −1 
 =38.4 , m = ( 8 ) + ( 6.33) =10.2
 8 
⇒ f ( t ) =10.2 e −2 t cos( 3 t +38.40 ) u ( t )
14-2
(b)
Given F ( s ) =
( 2( 3 ) ) .
3e − s
3
1
= ×
, first consider F1 ( s ) = 2
2
s + 2s +17
s + 2s +17 2 ( s +1)2 +16
Identify a =1, c = 0, ω = 4 and −ω d =3 ⇒ d =−3 4. Then m =|d |=3 4, θ = tan −1 ( −3 ( 4 0 ) )=−90°
So f1 (t ) = (3 4)e −t sin 4t u ( t ) . Next, F ( s ) = e − s F1 ( s ) ⇒ f ( t ) = f1 ( t −1) . Finally
∴ f ( t ) =(3 4)e − (t −1) sin  4( t −1)  u ( t −1)
Ex. 14.5-3
(a)
where
F (s) =
A = sF ( s ) |s = 0 =
Multiply both sides by s ( s + 1)
s 2 −5
s ( s +1)
=
2
A B
C
+
+
s s +1 ( s +1)2
1−5
−5
2
= −5 and C = ( s +1) F ( s ) |s = −1 =
=4
1
−1
2
s 2 − 5 = −5 ( s +1) + Bs ( s +1) + 4 s ⇒
2
F (s) =
Then
F (s) =
(b)
and
4
−5 6
+
+
s s +1 ( s +1)2
f ( t ) = ( −5 + 6 e − t + 4 t e − t ) u ( t )
Finally
Where
B=6
A=
( s +3)
4s 2
3
=
A
B
C
+
+
2
( s +3) ( s +3) ( s+3)3
1 d2 
d 
2
3
s + 3) F ( s ) 
= 4, B =
( s +3) F ( s ) s =−3 = −24
2 (


3
s
=−
2 ds
ds
C = ( s + 3) F ( s ) s =−3 = 36
3
Then
Finally
F (s) =
4
36
−24
+
+
2
( s + 3 ) ( s + 3 ) ( s + 3 )3
f ( t ) = ( 4 − 24 t + 18t 2 ) e −3t u ( t )
14-3
Ex. 14.6-1
(a) F ( s ) =
(b) F ( s ) =
6s +5
s + 2s +1
2
6
s − 2s +1
2
 s( 6s +5) 
f ( 0 ) = lim sF ( s ) = lim  2
=6
s →∞
s →∞  s + 2 s +1 
 s( 6s +5 ) 
f ( ∞ ) = lim sF ( s ) = lim  2
=0
s →0
s →0  s + 2 s +1 
 6s 
f ( 0 ) = lim  2
=0
s →∞
 s − 2s +1 
 6s 
= undefined ⇒ no final value
f ( ∞ ) = lim  2
s →0
 s − 2s +1 
Ex. 14.7-1
KCL:
v1
5
+ i = 7 e −6 t
di
di
+ 3 i − v1 = 0 ⇒ v1 = 4 + 3 i
dt
dt
di
4 + 3i
35
di
Then dt
+ i = 7 e −6 t ⇒
+ 2 i = e −6 t
dt
5
4
KVL: 4
Taking the Laplace transform of the differential equation:
s I ( s ) − i (0) + 2 I ( s ) =
35 1
35
1
⇒ I (s) =
4 s +6
4 ( s + 2)( s + 6)
Where we have used i (0) = 0 . Next, we perform partial fraction expansion.
1
1
A
B
where A =
=
+
( s + 2) ( s + 6) s + 2 s + 6
s+6
Then
I ( s) =
s =−2
=
1
1
1
and B =
=−
4
4
s + 2 s = −6
35
35
35 1
35 1
−
⇒ i (t ) = e −2t − e −6t
16 s + 2 16 s + 6
16
16
14-4
Ex. 14.7-2
Apply KCL at node a to get
1 d v1 v 2 − v1
=
48 dt
24
⇒ 2 v1 +
d v1
dt
= 2 v2
Apply KCL at node b to get
v 2 − 50 cos 2 t
20
+
v 2 − v1
24
+
v2
30
+
d v2
1 d v2
= 0 ⇒ − v1 + 3 v 2 +
= 60 cos 2t
dt
24 dt
Take the Laplace transforms of these equations, using v1 (0) = 10 V and v2 (0) = 25 V , to get
( 2+ s ) V1 ( s) − 2V2 ( s) = 10 and − V1 ( s) + ( 3+ s ) V2 ( s) =
25s 2 + 60s +100
s2 + 4
Solve these equations using Cramer’s rule to get
 25s 2 + 60s +100 
( 2+ s ) 
 +10 ( 2+ s ) ( 25s 2 + 60 s +100 ) +10 ( s 2 + 4 )
s2 +4


V2 ( s ) =
=
( 2+ s ) (3+ s) − 2
( s 2 + 4 ) ( s +1)( s + 4 )
=
25s 3 +120s 2 + 220 s + 240
( s 2 + 4 ) ( s +1)( s + 4 )
Next, partial fraction expansion gives
V2 ( s ) =
where
A =
A
A*
B
C
+
+
+
s + j 2 s − j 2 s +1 s + 4
25s 3 +120 s 2 + 220 s + 240
( s +1) ( s + 4 ) ( s − j 2 )
A* = 6 − j 6
B =
C =
25s 3 +120 s 2 + 220 s + 240
( s2 +4) ( s+4)
25s 3 +120 s 2 + 220 s + 240
( s 2 + 4 ) ( s +1)
s =− j 2
s =−1
s =−4
=
=
=
−240− j 240
= 6 + j6
−40
115 23
=
15
3
−320 16
=
3
−60
Then
14-5
V2 ( s ) =
Finally
6+ j 6 6− j 6 23 3 16 3
+
+
+
s + j 2 s − j 2 s +1 s + 4
v2 (t ) = 12 cos 2 t + 12 sin 2 t +
23 − t 16 −4t
e + e V t≥0
3
3
Ex. 14.7-3
Taking Laplace Transform of the differential equation:
s 2 F ( s ) = s f ( 0 ) − f ' ( 0 ) + 5  s F ( s )− f ( 0 )  + 6 F ( s ) =
10
s +3
Using the given initial conditions
( s 2 + 5s + 6) F ( s ) =
10
2 s 2 +16s + 40
+ 2s + 10 =
s +3
s +3
2 s 2 +16s + 40
A
B
C
=
+
F (s) =
+
2
( s + 3) ( s + 2 ) ( s + 3) ( s + 3) ( s + 3) ( s + 2 )
where A = − 10, B = − 14, and C = 16 . Then
F ( s) =
−10
( s + 3)
2
+
−14 16
+
⇒
s +3 s + 2
f (t ) = − 10te−3t − 14e −3t + 16e−2t
for t ≥ 0
Ex. 14.8-1
KCL at top node:
VC ( s )
3
s
2
+ VC ( s ) = + 2
s
2
VC ( s ) =
2
VC ( s )
−2= 3
I C (s) =
2
2
s+
3
s
(
6
2
−
s s+ 2
3
v C (t ) = 6 − 2 e
⇒ iC (t ) =
− ( 2 / 3) t
) u(t ) V
2 − ( 2 / 3) t
e
u (t ) A
3
14-6
Ex. 14.8-2
Mesh Equations:
4 1
4 
1 
− − I C ( s ) − 6 ( I ( s ) − I C ( s )) = 0 ⇒ − =  6 +  I C ( s ) + 6 I ( s )
2s 
s 2s
s 
10
6 ( I ( s ) − I C ( s )) + 3 I ( s ) + 4 I C ( s ) = 0 ⇒ I ( s ) = − I C ( s )
9
Solving for Ic(s):
4  2 1 
6
− =  − +  I C (s) ⇒ I C (s) =
3
s  3 2s 
s−
4
So Vo(s) is
24
Vo ( s ) = 4 I C ( s ) =
3
s−
4
Back in the time domain:
v o ( t ) = 24 e0.75t u (t ) V
Ex. 14.8-3
KVL:
so
8
 20

+ 4 =  + 8 + 4s  I L ( s )
s
 s

I L ( s) =
( s + 1) + 1
2+ s
=
s + 2 s + 5 ( s + 1) 2 + 4
2
Taking the inverse Laplace transform:
1


i ( t ) =  e − t cos 2 t + e − t sin 2 t  u ( t ) A
2


14-7
10 
 5
(a) impulse response = L−1 
−
= ( 5 e − 5 t − 10 e −10 t ) u (t )

 s + 5 s + 10 
1 
 1
−
= ( e−10 t − e − 5 t ) u (t )
(b) step response = L−1 

 s + 10 s + 5 
Ex. 14.9-1
Ex. 14.9-2
H ( s ) = L 5 e − 2 t sin ( 4 t ) u ( t )  =
5 ( 4)
( s + 2)
2
+4
=
20
s + 4 s + 20
2
 H (s) 

s+4
1
-1  1
= (1 − e − 2 t (cos 4t − sin 4t )) u (t )
step response = L-1 
=L  − 2

2
 s s + 4 s + 20 
 s 
2
Ex. 14.10-1
Voltage division yields
(8 )
2

s
2
8+
V ( s)
s
=
H (s) = c
2
V1 ( s )
( 8) 
s
2+  
2
8+
s
16
16
1
s
=
=
=
4 16 16 s + 20 s +1.25
16 + +
s s
so
h ( t ) = L-1  H ( s )  = 1.25 e −1.25 t u ( t )
14-8
Ex. 14.10-2
h ( t ) = e −2 t
f (t ) = u(t )
⇒
⇒
H (s) =
1
s+2
1
F(s) =
s
 1  −1 1 2 −1 2   1 −2t 
+
h( t ) ∗ f ( t ) = L−1  H ( s ) F ( s )  = L−1 
= L 
 =  2 −e  u ( t )
2
2
s
s
s
s
+
+
(
)



− (3 − k ) ±
Ex. 14.11-1
The poles of the transfer function are p1,2 =
a.) When k = 2 V/V, the poles are p1,2 =
(3 − k )
2
−8
2
.
−1 ± −7
so the circuit is stable. The transfer function is
2
H (s) =
Vo ( s )
Vi ( s )
=
2s
s +s+2
2
The circuit is stable when k =2 V/V so we can determine the network function from the transfer
function by letting s = jω.
V o (ω )
V i (ω )
= H (ω ) = H ( s ) s = j ω =
2s
2 jω
=
s + s + 2 s = j ω ( 2 − ω 2 ) + jω
2
The input is v i ( t ) = 5cos 2 t V . The phasor of the steady state response is determine by
multiplying the phasor of the input by the network function evaluated at ω = 2 rad/s.

2 jω
V o (ω ) = H (ω ) ω = 2 × V i (ω ) = 
 2 − ω 2 ) + jω
(

 ( 5∠0° ) =  j 4  ( 5∠0° ) = 7.07∠ − 45°

 −2 + j 2 
ω =2 
The steady state response is vo ( t ) = 7.07 cos ( 2 t − 45° ) V .
b. When k = 3 − 2 2 , the poles are p1,2 =
transfer function is
H (s) =
−2 2 ± 0
= − 2, − 2 so the circuit is stable. The
2
(s + 2) (s + 2) (s + 2)
0.17 s
2
=
0.17
−
0.17 2
2
The impulse response is
14-9
h ( t ) = L -1 H ( s )  = 0.17 e −
2t
(1 −
)
2 t u (t )
We see that when k = 3 − 2 2 the circuit is stable and lim h(t ) = 0 .
c. When, k = 3 + 2 2 the poles are p1,2 =
transfer function is
The impulse response is
H (s) =
t →∞
2 2± 0
= 2, 2 so the circuit is not stable. The
2
(s − 2) (s − 2) (s − 2)
5.83s
2
=
+
5.83
h ( t ) = L -1 H ( s )  = 5.83 e
2t
5.83 2
(1 +
2
)
2 t u (t )
We see that when k = 3 + 2 2 the circuit is unstable and lim h(t ) = ∞ .
t →∞
Ex. 14.12-1
For the poles to be in the left half of the s-plane, the s-term needs to be positive.
2 

s +5
10
2s + 20
2

+
V0 ( s ) = 0.1  − 2
  = 0.1  − 2
2
2

2
2 

 s s +10 s + 125 
 s  ( s +5 ) +10
( s +5) +10  
= 0.1
2( s 2 +10 s +125 ) −( 2s + 20 ) s
s 2 +10s +125
250
25
= 0.1 2
= 2
s +10s +125
s +10s +125
These specifications are consistent.
14-10
Problems
Section 14-3: Laplace Transform
P14.3-1
L  A f1 ( t )  = A F1 ( s )
f1 ( t ) = cos (ω t )
P14.3-2
P14.3-3

As

s  ⇒ F ( s ) = s2 + ω 2
⇒ F1 ( s ) = 2 2 
s +ω 
L−1 t n  =
n!
s n +1
F ( s ) = L−1 t1  =
Linearity: L  a1 f1 ( t ) + a2 f 2 ( t )  = a1 F1 ( s ) + a2 F2 ( s )
Here a1 = a2 = 1
L  f1 ( t )  = L e −3t  =
L  f 2 ( t )  = L[t ] =
so F ( s ) =
P14.3-4
1
1!
= 2
1+1
s
s
1
1
+ 2
s +3 s
1
= F1 ( s )
s+3
1
= F2 ( s )
s2
f ( t ) = A (1−e −bt ) u ( t ) = L [ Af1 (t ) ] = AF1 ( s )
f1 ( t ) = A(1− e − bt ) u ( t ) = 1u ( t ) −e − bt u ( t ) = f 2 ( t ) + f3 ( t )
−1
1
, F3 ( s ) =
s
s +b
1 
Ab
1
∴ F (s) = A  −
=

 s s +b  s( s +b )
F2 ( s ) =
14-11
Section 14-4: Impulse Function and Time Shift Property
f ( t ) = A u ( t ) − u ( t −T ) 
P14.4-1
P14.4-2
(1−e
A Ae − sT
=A
F ( s ) = AL u ( t )  − AL u ( t −T )  = −
s
s
s
f ( t ) = 1 u ( t ) −u ( t −T )  eat
L u ( t ) −u ( t −T )  =
(b)
(c)
P14.4-4
F (s) =
( s +3)
1− e− sT
s


1− e( s − a )T
F
s
⇒
=
( )

( s −a )


2
3
f ( t ) = δ ( t −T ) u ( t −T ) ⇒ F ( s ) = e − sT L δ ( t )  = e − sT
F (s) =
( s + 4 ) +( 5 )
5
2
2
=
5
5
= 2
( s + 8 s + 16 ) + 25 s + 8 s + 41
2
g ( t ) = e − t u ( t −0.5 ) = e − (t + (0.5−0.5))u ( t − 0.5 ) = e−0.5 e− ( t − 0.5)u ( t − 0.5 )
L e −0.5 e − (t −0.5)u ( t − 0.5 )  = e−0.5 L e− (t − 0.5)u ( t −0.5 )  = e−0.5 e−0.5 s L  e− t u ( t )  =
P14.4-5
)
⇒ F ( s ) = L e at u ( t )−u ( t −T )  
L eat g ( t )  =G ( s − a )


P14.4-3
(a)
− sT
e0.5 e −0.5 s e0.5 − 0.5 s
=
s +1
s +1
− sT
e − sT
 t −T
 − sT  t
 e
L −
L  − t u ( t )  =
u( t − T ) = e L − u( t ) =
T
T s2
 T

 T

14-12
Section 14-5: Inverse Laplace Transform
P14.5-1
F (s) =
s +3
s +3
A
Bs + C
=
=
+ 2
2
2
s + 3s + 6s + 4 ( s +1) ( s +1) + 3 s +1 s + 2s + 4


3
where
Then
A=
s +3
( s +1)
2
+3
s =−1
=
2
3
( s + 3)
2
8
2
Bs + C
4

= 3 + 2
⇒ ( s + 3) = ( + B ) s 2 +  + B + C  s + + C
2
3
3
3

( s +1) ( s + 2s + 4 ) s +1 s + 2s + 4
Equating coefficient yields
Then
s2 : 0 =
2
2
+ B ⇒ B= −
3
3
4 2
1
s : 1= − + C ⇒ C =
3 3
3
1
2
2 1
2
2
3
− s+
− ( s +1)
3
3
3
3
3
3
F (s) =
+
=
+
+
s +1 ( s +1)2 + 3 s +1 ( s +1)2 + 3 ( s +1)2 + 3
Taking the inverse Laplace transform yields
f (t ) =
2 −t 2 −t
1 −t
e − e cos 3t +
e sin 3
3
3
3
14-13
P14.5-2
F (s) =
s 2 − 2s + 1
s 2 − 2s + 1
a
a*
b
=
=
+
+
3
2
s + 3s + 4s + 2 ( s +1) ( s +1− j )( s +1+ j ) s +1− j s +1+ j s +1
where
s 2 − 2s +1
b=
a=
( s +1)
2
+1 s =−1
=4
3− j 4 3
s 2 − 2s +1
=− + j 2
=
( s +1) ( s +1+ j ) s =−1+ j −2 2
3
a* = − − j 2
2
Then
3
− + j2
F (s) = 2
+
s +1− j
Next
m=
From Equation 14.5-8
P14.5-3
B=
Then
A=
F (s) =
2




5
2
=
= 126.9°
and θ = tan −1 

3
2
− 
 2
f ( t ) = 5 e − t cos ( t + 127° ) + 4 e −t  u ( t )
F (s) =
where
Finally
( -3 2 )2 + ( 2 )
3
− − j2
4
2
+
s +1+ j s +1
5 s −1
( s +1) ( s − 2 )
5 s −1
s −2
2
s =−1
=
A
B
C
+
+
2
s +1 ( s +1)
s −2
= 2 and C =
d 
2
( s +1) F ( s )

ds
2
1
−1
+
+
⇒
2
s +1 ( s +1)
s−2
s =−1
=
5 s −1
( s +1)
2
s =2
−9
( s −2)
2
s =−1
=1
= −1
f ( t ) =  −e − t + 2 t e −t + e 2t  u ( t )
14-14
P14.5-4
Y (s) =
1
1
A
Bs +C
=
=
+
2
( s +1) ( s + 2s + 2 ) ( s +1) ( s +1) + 1 s +1 ( s +1)2 +1
2
where
Next
A=
1
s + 2s + 2
2
s =−1
=1
1
1
Bs +C
=
+ 2
⇒ 1 = s 2 + 2 s + 2 + ( Bs + C ) ( s + 1)
( s +1) ( s + 2s + 2 ) s +1 s + 2s + 2
2
⇒ 1 = ( B +1) s 2 + ( B + C + 2 ) s + C + 2
s 2 : 0 = B + 1 ⇒ B = −1
s : 0 = B + C + 2 ⇒ C =−1
Equating coefficients:
Finally
Y (s) =
P14.5-5
F (s) =
1
s +1
−
⇒
s +1 ( s +1)2 +1
2( s +3)
( s +1) ( s
2
+ 2 s +5)
=
y ( t ) = e− t − e − t cos t  u ( t )
−( s +1)
1
2
+
+
2
s +1 ( s +1) + 4 ( s +1)2 + 4
f ( t ) = e− t − e − t cos ( 2t ) + e −t sin ( 2t )  u ( t )
P14.5-6
where
and
Finally
F (s) =
A = sF ( s ) s =0 =
2( s+3)
A
B
C
= +
+
s( s+1) ( s + 2 ) s s +1 s + 2
2( s + 3 )
( s +1) ( s + 2 )
s =0
= 3, B = ( s +1) F ( s ) s =−1 =
( s + 2 ) F ( s ) s = −2 =
3 −4
1
+
⇒
F (s) = +
s s +1 s + 2
2( s +3)
s( s +1)
s =−2
2( s + 3)
s( s + 2 )
s =−1
= −4
= C =1
f ( t ) = ( 3 − 4e− t + e −2t ) u ( t )
14-15
Section 14-6: Initial and Final Value Theorems
P14.6-1
(a)
(b)
2s 2 −3s + 4 2s 2
= 2 =2
f ( 0 ) = lim sF ( s ) = lim
2
s
s →∞
s →∞ s +3s + 2
4
f ( ∞ ) = lim sF ( s ) = = 2
2
s →0
P14.6-2
Initial value:
Final value:
s( s +16 )
s 2 + 16 s
= lim 2
=1
v ( 0 ) = lim sV ( s ) = lim 2
s →∞
s →∞ s + 4 s + 12
s →∞ s + 4 s + 12


s +16
s 2 + 16 s
=
=0
v ( ∞ ) = lim s  2
lim

2
s→ 0
 s + 4 s + 12  s → 0 s + 4 s +12
(Check: V(s) is stable because Re { pi } < 0 since pi = − 2 ± 2.828 j . We
expect the final value to exist.)
Initial value:
s 2 +10 s
= 0
v(0) = lim sV ( s ) = lim
s →∞
s →∞ 3 s 3 + 2 s 2 +1
Final value:
v ( ∞ ) = lim sV ( s ) = lim
s ( 3s 2 + 2s +1)
f ( 0 ) = lim s F ( s ) = lim
−2 s 2 −14 s
= −2
s 2 − 2 s + 10
P14.6-3
s →0
s →0
s ( s +10 )
= 10
(Check: V(s) is stable because pi = −0.333 ± 0.471 i . We expect the
final value to exist.)
P14.6-4
Initial value:
s →∞
Final value:
s →∞
F(s) is not stable because Re { p1} > 0 since pi = 1 ± 3i . No final value
exists.
14-16
Section 14-7: Solution of Differential Equations Describing a Circuit
P14.7-1
KVL:
50 i + 0.001
4
di
+ v = 2 e−2×10 t
dt
The capacitor current and voltage are related
by
i = ( 2.5 × 10−6 )
v1 = 2 e −2×10 V , i (0) = 1 A, v(0) = 8 V
dv
dt
4t
Taking the Laplace transforms of these equations yields
50 I ( s) + 0.001 [ s I ( s) − i (0) ] + V ( s) =
I ( s) = ( 2.5 × 10−6 )  sV ( s ) − v( 0 ) 
2
s + 2×104
Solving for I(s) yields
I (s) =
where
s 2 + 1.4× 104 s − 1.6 × 108
( s +104 ) ( s + 2 × 104 ) ( s + 4 ×104
A = ( s + 10 ) I ( s )
4
s = −10
B = ( s + 2×104 ) I ( s )
C = ( s + 4×104 ) I ( s )
4
)
=
A
B
C
+
+
4
4
s +10
s + 2× 10
s + 4×104
s 2 + 1.4× 104 s −1.6× 108
=
( s + 2 × 104 )( s + 4 × 104 )
s = − 2 ×10
s = − 4 ×10
4 =
4 =
s = −10
s 2 + 1.4× 104s −1.6 × 108
( s+ 104 ) ( s + 4×104 )
s 2 + 1.4× 104s −1.6 × 108
( s+104 ) ( s+ 2× 104 )
4
−2×108 −2
=
=
3× 108
3
s = − 2 ×10
=
4
s = − 4 ×10
=
4
.4 × 108
1
=
8
2 × 10
5
8.8 × 108
22
=
8
6 × 10
15
Then
I (s) = −
23
15
22 15
+
+
4
s +10
s + 2×10 s + 4 × 104
⇒ i (t ) =
4
4
4
1
−10 e−10 t + 3e−2x10 t + 22 e−4x10 t  u ( t ) A


15
14-17
We are given v ( t ) = 160 cos 400t .
P14.7-2
The capacitor is initially uncharged, so
v C ( 0 ) = 0 V . Then
i ( 0) =
KCL yields
10−3
dvC
+
160 cos ( 400 × 0 ) − 0
= 160 A
1
=i
vC
Apply Ohm’s law to the 1 Ω resistor to get
v −v C
i=
⇒ vC = v− i
1
Solving yields
di
+ 1010 i = 1600 cos 400t − ( 6.4 ×104 ) sin 400t
dt
dt
100
Taking the Laplace transform yields
s I ( s ) − i (0) + (1010 ) I ( s ) =
so
Next
I (s) =
Then
Finally
2
−
( 6.4×10 ) ( 400 )
s 2 + ( 400 )
2
2
160
1600s − 2.5×107
+
s + 1010 ( s + 1010 )  s 2 + (400) 2 
A
B
B*
1600s − 2.5×107
=
+
+
( s + 1010 )  s 2 + (400)2  s + 1010 s + j 400 s − j 400
where
B =
s 2 + ( 400 )
1600s
A =
1600 s − 2.5 x 107
( s +1010 ) ( s − j 400 )
1600s − 2.5×107
s 2 + ( 400 )
s = − j 400
I (s) =
=
2
s = −1010
2.56 x 107 ∠1.4°
8.69 x 10 ∠68.4
5
°
= − 23.1 ,
= 11.5 − j 27.2 and B* = 11.5 + j 27.2
136.9
11.5− j 27.2 11.5 + j 27.2
+
+
s + 1010
s + j 400
s − j 400
i ( t ) = 136.9e−1010t + 2 (11.5 ) cos 400t − 2 ( 27.2 ) sin 400t for t > 0
= 136.9e−1010t + 23.0 cos 400t − 54.4sin 400t for t > 0
14-18
P14.7-3
vc +15 i = 10 cos 2t 

 ⇒
1 d vc
i=

30 dt
vC (0) = 0
d vc
+ 2 vc = 20 cos 2t
dt
Taking the Laplace Transform yields:
sVc ( s ) − vc ( 0 ) + 2Vc ( s) = 20
where
A=
Then
P14.7-4
20 s
s2 +4
s = −2
=
s
A
B
B*
20s
⇒
=
=
+
+
V
s
(
)
c
s2+ 4
( s + 2 )( s 2 + 4 ) s + 2 s + j 2 s − j 2
20s
−40
= −5, B =
8
( s + 2 )( s − j 2 )
5 5 5 5
+j
−j
−5 2 2 2 2
Vc ( s ) =
+
+
s+2 s+ j2 s− j2
vc + 12i L + 2
diL
dt
s = − j2
=
5
5
5
j5 5
= + j and B* = − j
1+ j 2
2
2
2
⇒ vc ( t ) = −5e −2t + 5 ( cos 2t + sin 2t ) V
= −8 and i L = C
d vc
dt
Taking the Laplace transform yields
Vc ( s ) + 12 I L ( s ) + 2  sI L ( s ) −iL ( 0 )  = −
I L ( s ) = C  sVc ( s ) −vc ( 0 ) 
8
s
vc (0) = 0, iL (0) = 0
Solving yields
14-19
−4 C
Vc ( s ) =
(a) C =
C

s  s 2 + 6s + 
2

Vc ( s ) =
1
F
18
−72
s ( s + 3)
2
=
c
a
b
+
+
s s + 3 ( s + 3) 2
a = − 8, b = 8, and c = 24 ⇒ Vc ( s ) =
24
8
−8
+
+
s
s +3 ( s + 3) 2
vc ( t ) = −8 + 8 e −3t + 24 t e −3t
(b) C =
Vc ( s ) =
1
F
10
c
−40
a
b
+
= +
s ( s +1) ( s +5 ) s s +1 s + 5
a = − 8, b = 10, and c = −2 ⇒ Vc ( s ) =
−2
−8 10
+
+
s
s +1 s + 5
vc ( t ) = − 8 + 10 e − t − 2 e −5 t
P14.7-5
vc (0− ) = 10 V, i L( 0− ) = 0 A
i = ( 5 × 10−6 )
d vc
di
and 400 i + 1 + vc = 0
dt
dt
 1 
 −  400

I ( s ) = ( 5 ×10 ) ( sVc ( s ) − 10 ) 
−10
40 
= 
 ⇒ I (s) = 2
2
5
s + 400 s + 2 ×10
400 I ( s ) + ( s I ( s ) − 0 ) + Vc ( s ) = 0 
( s + 200 ) + 4002
Taking Laplace transforms yields
−6
so
i (t ) = −
1 −200t
e sin ( 400t ) u ( t ) A
40
14-20
Section 14-8: Circuit Analysis Using Impedance and Initial Conditions
P14.8-1
6
6
− 0.010
− 0.002 s
.003
.005
s
s
=
=
=
I L ( s) =
5s + 2000 s ( s + 400)
s
s + 400
2 mA
iL (t ) = 
−400 t
mA
 3−5e
t <0
t >0
P14.8-2
10
8
(
)
.015
V
s
−
−
(
) ⇒ V ( s) = 3
s + VL ( s ) + L
0=
L
4000
2000
4000
5s
s+
15
8
0.005
0.002
V ( s ) + 0.15
15
I L ( s) = L
=
+ 0.015 =
−
4000
5s
5
 4000 
s+
s  s+

15
15 

VL ( s ) −
iL (t ) = 5 − 3e
−
4000
t
15
mA, t > 0
14-21
P14.8-3
Vc ( s ) −
8
s =0
−
0.006 Vc ( s )
+
+
s
2000
−
6000
8

+ 500 Vc ( s ) + 0.5s  Vc ( s ) −  = 0
s
s

Vc ( s ) =
106
.5s
8 s + 12000 12
4
=
−
s ( s +1000)
s
s +1000
Vc (t ) = 12 − 4e −1000t V, t > 0
P14.8-4
Vc ( s ) −
6
s + Vc ( s ) +  0.5s   V ( s) − 8  = 0

 c

s
2000
4000  106  
6
8


500  Vc ( s ) −  + 250Vc ( s ) + 0.5s  Vc ( s ) −  = 0
s
s


Vc ( s ) =
6000 + 8s 4
4
= +
s ( s + 1500 ) s s + 1500
vc (t ) = 4 + 4e−1500t V, t > 0
14-22
P14.8-5
Node equations:
Va ( s ) − VC ( s ) Va ( s ) 1
6
6
VC ( s ) +
+
=
⇒ Va ( s ) =
6
s
s
s+6
s+6
6 
 6
6
VC ( s ) − 
VC ( s ) +
VC ( s ) −

3
s+6 1 s
 s+6
s+3 +
+ + VC ( s ) − = 0
4
2
s
s 4
After quite a bite of algebra:
VC ( s ) =
6 s 2 + 56s + 132
( s + 2 )( s + 3)( s + 5)
Partial fraction expansion:
44
1
6 s + 56 s + 132
9
V (s) =
= 3 −
+ 3
c
( s +3)( s + 2 )( s +5) s + 2 s +3 s +5
Inverse Laplace transform:
v (t ) = 44 3 e −2t − 9e −3t + (1 3)e −5t V
c
2
14-23
P14.8–6
Write a node equation in the frequency domain:
10
s = Vo ( s ) − 5 C + Vo ( s ) ⇒ V ( s ) =
o
1
R1
R2
Cs
Inverse Laplace transform:
vo ( t ) = 10

R 2  −t
+  5 − 10
e
R1 
R1 
R2
R 2C
10
+5s
R1C

1 
ss +


R 2C 

=
10
R2
R1
s
+
5 − 10
R2
R1
1
s+
R 2C
= 10 − 5 e−1000t V for t > 0
14-24
P14.8-7
Here are the equations describing the coupled coils:
v1 (t ) = L1
di1
di
+M 2
dt
dt
di
di
v2 (t ) = L2 2 + M 1
dt
dt
⇒ V1 ( s) = 3 ( s I1 ( s ) − 2 ) + ( sI 2 ( s) − 3) = 3s I1 ( s ) + sI 2 ( s ) − 9
⇒ V2 ( s) = s ( I1 ( s ) − 2 ) + 2( sI 2 ( s ) −3) = sI1 ( s) + 2sI 2 ( s ) − 8
Writing mesh equations:
5
5
= 2 ( I1 ( s ) + I 2 ( s ) ) + V1 = 2 ( I1 ( s ) + I 2 ( s ) ) + 3s I1 ( s ) + sI 2 ( s ) − 9 ⇒ ( 3s + 2 ) I1 + ( s + 2 ) I 2 = 9 +
s
s
V1 ( s ) = V2 ( s ) + 1I 2 ( s ) ⇒ 3s I1 ( s ) + sI 2 ( s ) − 9 = sI1 ( s ) + 2 sI 2 ( s ) − 8+ I 2 ( s ) ⇒ 2s I1 − ( s +1) I 2 =1
Solving the mesh equations for I2(s):
I2 ( s ) =
15s + 8
3s + 1.6
0.64
2.36
=
+
=
s + 0.26
s + 1.54
5s 2 + 9s + 2 ( s + 0.26 )( s +1.54 )
Taking the inverse Laplace transform:
i2 (t ) = 0.64e −0.26t + 2.36e −1.54t A for t > 0
P14.8-8
t<0
time domain
frequency domain
Mesh equations in the frequency domain:
6 I 1 ( s ) + 6 ( I 1 ( s ) − I 2 ( s )) + 6 I 1 ( s ) +
12
2
2
= 0 ⇒ I1 (s) = I 2 (s) −
s
3
3s
14-25
2
6
2
6

I 2 ( s ) − − 6 ( I 1 ( s ) − I 2 ( s )) = 0 ⇒  6 +  I 2 ( s ) − 6 I 1 ( s ) =
s
s
s
s

Solving for I2(s):
1


2
2
2
6


⇒ I 2 (s) = 2
 6 +  I 2 (s) − 6 I 2 (s) −  =
1
s
s
s
3
3




s+
2
Calculate for Vo(s):
 1 
1
6 1 2  6
4
−2
Vo ( s ) = I 2 ( s ) − = 
−
− =

2
s 2 s+ 1  s s+ 1 s
2
2

Take the Inverse Laplace transform:
vo ( t ) = − ( 4 + 2 e− t / 2 ) V for t > 0
(Checked using LNAP, 12/29/02)
P14.8-9
t<0
frequency domain
time domain
2



−6  s + 


12
3
3
5
( 4 + 5 s ) I ( s ) + 30 + = 0 ⇒ I ( s ) =  4  = −  + 4 
s


 s s + 
ss + 
5

5

Take the Inverse Laplace transform:
Writing a mesh equation:
i ( t ) = −3 (1 + e−0.8 t ) A for t > 0
(Checked using LNAP, 12/29/02)
14-26
P14.8-10
Steady-state for t<0:
From the equation for vo(t):
vo ( ∞ ) = 6 + 12 e
vo ( ∞ ) =
From the circuit:
Therefore:
6=
− 2 (∞)
=6 V
Steady-state for t>0:
3
(18)
R+3
3
(18) ⇒ R = 6 Ω
R+3

−6
1  18 6
I (s) 2 +
 + − = 0 ⇒ I (s) =
1
Cs s s

s+
2C



6
1
18 1
12
18
12
−6  18 −12
+
+
+ =
Vo ( s ) =
I (s) + =

+ =
1
1
1
Cs
s Cs  s +
s
s s+
s
 s
s+

2C
2C 
2C

Taking the inverse Laplace transform:
vo ( t ) = 6 + 12 e − t / 2C V for t > 0
Comparing this to the given equation for vo(t), we see that 2 =
1
2C
⇒ C = 0.25 F .
(Checked using LNAP, 12/29/02)
14-27
Section 14-9: Transfer Function and Impedance
P14.9-1
H (s) =
R1
R2
C1s
and Z 2 =
=
R 1 R1C1s + 1
R2C2 s +1
R1 +
C1s
R 2(τ 1 s + 1)
Let τ 1 = R1C1 and τ 2 = R2C2 then H ( s ) =
R 1 (τ 2s + 1) + (τ 1 s + 1) R 2
Z2
Z1 + Z 2
When τ 2 = τ 2 = τ ⇒ H ( s ) =
∴ we require R1C 1 = R2C2
where
Z1 =
R1
R2(τ 1s + 1)
R2
=
= constant, as required.
( R1 + R2 ) (τ s + 1) R1 + R2
P14.9-2
Let Z1 = R +
1
and Z 2 = R + Ls then the input impedance is
Cs


L
1 


R + ( R + Ls )
LCs 2 +  RC +  s +1 


ZZ
Cs 
R


Z (s) = 1 2 = 
= R
2
1
Z1 + Z 2
LCs
RCs
2
1
+
+


R+
+ R + Ls


Cs


L
Now require : RC + = 2 RC ⇒ L = R 2C then Z = R
R
P14.9-3
The transfer function is
R2
H (s) =
1
R1 C
R1 + R 2
R 2 C s +1
=
R2
R1 +
s+
R 2 C s +1
R1 R 2 C
Using R1 = 2 Ω, R 2 = 8 Ω and C = 5 F
gives
H (s) =
0.1
s + 0.125
The impulse response is h (t ) = L -1  H ( s ) = 0.1 e−0.125 t u (t ) V .
The step response is
14-28


 H ( s)
0.1
 = L -1  0.8 − 0.8  = 0.8 1− e−0.125 t u (t ) V
 = L -1 
L -1 




s + 0.125 
 s
 s 
 s ( s + 0.125)
(
)
(Checked using LNAP, 12/29/02)
P14.9-4
The transfer function is:
H ( s ) = L 12 t e −4 t u ( t )  =
( s + 4)
12
2
=
12
s + 8s + 16
2
The Laplace transform of the step response is:
3
H (s)
12
k
−3
=
= 4+
+
2
2
s+4
s
s ( s + 4)
s ( s + 4)
2
The constant k is evaluated by multiplying both sides of the last equation by s ( s + 4) .
3

3
3
2
12 = ( s + 4) − 3s + ks ( s + 4) =  + k  s 2 + (3 + 4k ) s + 12 ⇒ k = −
 4

4
4
The step response is
 H ( s )   3 −4 t 

 =  − e 3 t + 3  u (t ) V
L −1 
 s   4

4 


P14.9-5
The transfer function can also be calculated form the circuit itself. The circuit can be represented
in the frequency domain as
We can save ourselves some work be noticing that the 10000 ohm resistor, the resistor labeled R
and the op amp comprise a non-inverting amplifier. Thus

R 
Va ( s ) = 1 +
V s
 10000  c ( )
Now, writing node equations,
14-29
Vc ( s ) −Vi ( s )
Vo ( s ) −Va ( s ) Vo ( s )
+ CsVc ( s ) = 0 and
+
=0
1000
5000
Ls
Solving these node equations gives
R  5000
1 

1 +

1000 C
10000  L
H (s) =

1 
5000 
  s +

 s +
L 
1000C 
Comparing these two equations for the transfer function gives


1 
1 
 = ( s + 2000) or  s +
 = ( s + 5000)
 s +
 1000C 
1000C 




 s + 5000  = ( s + 2000) or  s + 5000  = ( s + 5000)




L 
L 
R  5000
1 
= 15 × 10 6
1 +

1000C  10000  L
The solution isn’t unique, but there are only two possibilities. One of these possibilities is


 s + 1  = ( s + 2000) ⇒ C = 0.5 µ F
 1000C 
5000 

 = ( s + 5000) ⇒
s +

L 
L =1H


1 + R  5000 = 15×106 ⇒ R = 5 kΩ

1000 0.5×106  10000  1
1
(
)
(Checked using LNAP, 12/29/02)
14-30
P14.9-6
The transfer function of the circuit is
R2
1
1+ R2 C s
R1 C
H (s) = −
=−
1
R1
s+
R2 C
The give step response is vo ( t ) = −4 (1 − e−250 t ) u ( t ) V . The correspond transfer function is
calculated as
{
}
H (s)
−1000
−1000
4 
4
= L −4 (1 − e − 250 t ) u ( t ) = −  −
⇒ H (s) =
=
s + 250
s
 s s + 250  s ( s + 250 )
Comparing these results gives
1
1
1
= 250 ⇒ R 2 =
=
= 40 kΩ
250 C 250 ( 0.1×10 − 6 )
R2 C
1
1
1
= 1000 ⇒ R1 =
=
= 10 kΩ
1000 C 1000 ( 0.1×10 − 6 )
R1 C
(Checked using LNAP, 12/29/02)
P14.9-7
 12

Vo ( s ) =  s
12
 6 +
s

The transfer function is:
 4 
 2 
Va ( s ) = 
 Vi ( s ) = 
 Vi ( s )
 s+2
 4+ 2s 


 2 
 2 
 2   2 
 Vb ( s ) = 
 Vb ( s ) = 
 5 Va ( s ) = 
5 
 Vi ( s )
 s+2
 s+2
 s+2  s+2


H (s) =
Vo ( s )
20
=
Vi ( s ) ( s + 2 )2
14-31
The Laplace transform of the step response is:
20
5 −5
−10
Vo ( s ) =
= +
+
2
s s + 2 ( s + 2 )2
s ( s + 2)
Taking the inverse Laplace transform:
vo ( t ) = 5 − 5 e −2 t (1 + 2t )  u ( t ) V
(checked using LNAP 8/15/02)
P 14.9-8
 1 
 1   4 



 L 
6C 
 4 
Cs


=
H (s) = 
k
k
(
)
(
)





 4 + Ls 
s+ 1 
 6 + 1   s + 4 

Cs  
L
6 C 


From the given step response:
H (s)
2
4
6
12
= L ( 2 + 4 e −3t − 6 e −2 t ) u ( t )  = +
−
=
s
s s + 3 s + 2 s ( s + 3)( s + 2 )
so
12
H (s) =
s ( s + 3)( s + 2 )
From the circuit:
Comparing the two representations of the transfer functions let
4
= 2 ⇒ L = 2 H and 2 × 3 × k = 12 ⇒ k = 2 V/V .
L
1
1
=3 ⇒ C =
F,
6C
18
(Checked using LNAP, 12/29/02)
P 14.9.9
From the circuit:
14-32
R
V (s)
R+ Ls
L
=
=
H (s) = o
Vi ( s ) 12 + R + L s s + 12 + R
L
From the given step response:
s+
H (s)
s+2
0.5 0.5
= L  0.5 (1 + e −4 t ) u ( t )  =
+
=
⇒
s
s
s + 4 s ( s + 4)
H (s) =
s+2
s+4
Comparing these two forms of the transfer function gives:
R

=2 
12 + 2 L

L
= 4 ⇒ L = 6 H, R = 12 Ω
 ⇒
12 + R
L

=4

L
(Checked using LNAP, 12/29/02)
P14.9-10
Mesh equations:
1 1 
1

V ( s ) =  R1 + +  I1 ( s ) −
I2 ( s )
Cs Cs 
Cs

1 
1

I1 ( s )
0 =  R+ R+  I2 ( s ) −
Cs 
Cs

Solving for I2(s):
Then Vo ( s ) = R I 2 ( s ) gives
H (s) =
 1 
V (s)  
 Cs 
I2 ( s ) =
1
2 
1 

 R1 +   2 R +  −
(Cs) 2
Cs  
Cs 

V0 ( s )
RCs
=
=
V (s)
[ R1Cs + 2][ 2 RCs +1] − 1
s


4 RC + R1C
1
2


2 R1C s +
s+
2 2
2 RR1C 2

( 2RR1C ) 

14-33
P14.9-11
 1 
R 
R
Cs
Z2 =   =
1
RCs + 1
R+
Cs
Z1 = Rx + Lx s
Let
Then
R
RCs + 1
V2
Z2
=
=
V1 Z1 + Z 2
Rx + Lx s +
V2
V1
R
RCs + 1
=
R
Lx RCs + ( Lx + Rx RC ) s + Rx + R
2
1
LC
=
( L + R RC ) s + Rx + R
s2 + x x
Lx RC
Lx RC
P14.9-12
Node equations:
(V1 − Vin ) sC1 +
V1 − Vout
=0 ⇒
R1
( R C s + 1)V
1
1
1
= R1C1sVin + Vout
V0
+ Vout + sC2 = 0 ⇒ V1 = − R 2C2 sVout
R2
Solving gives:
H (s) =
− R1C1s
Vout
=
=
Vin R1 R 2C1C2 s 2 + R 2C2 s +1
s2 +
−
1
s
R 2 C2
1
1
s+
R1C1
R1 R 2C1C2
14-34
P14.9-13
Node equations in the frequency domain:
V1 − Vi V1 − V2 V1 − V0
+
+
=0
R1
R2
R3
 1
V
1
1  V
⇒ V1 +
+ − 0 = i
 R1 R2 R3  R3 R1
V2 − V1
− sC2V0 = 0 ⇒ V1 = − sC2 R2 V0
R2
After a little algebra:
H ( s )=
V0
− R3
=
Vi sC2 R2 R3 + sC2 R1 R3 + sC2 R1 R2 + R1
P14.9-14
1
1
Vo ( s )
Cs
LC
=
=
H (s) =
R
1
1
Vi ( s ) Ls + R +
s2 + s +
Cs
L
LC
L, H
C, F
R, Ω
2
0.025
18
2
0.025
8
1
0.391
4
2
0.125
8
H(s)
20
20
=
s + 9 s + 20 ( s + 4 )( s + 5 )
2
20
20
=
s + 4s + 20 ( s + 2 )2 + 44
2
2.56
2.56
=
s + 4 s + 2.56 ( s + 0.8 )( s + 3.2 )
2
20
20
=
s + 4s + 4 ( s + 2 )2
2
14-35
a) H ( s ) =
b) H ( s ) =
20
( s + 4 )( s + 5)
20
20
−
⇒ h ( t ) = ( 20e −4t − 20e −5t ) u (t )
s + 4 s +5
20
1 −5
4
H ( s)
L {step response} =
⇒
=
= +
+
s
s ( s + 4) ( s + 5) s s + 4 s +5
L {h(t )} = H ( s ) =
step response = (1+ 4e −5t −5e −4t ) u (t )
( s + 2)
20
2
+4
4
L {h( t )} = H (s) =
5(4)
⇒ h ( t ) = 5e −2t sin 4t u (t )
( s + 2) 2 + 42
H ( s)
K s + K2
20
1
=
= + 21
L {step response} =
2
s
s ( s + 4s + 20) s s + 4s + 20
20 = s 2 + 4s + 20 + s ( K1s + K 2 ) = s 2 (1+ K1 ) + s ( 4+ K 2 ) + 20
⇒ K1 = −1, K 2 = − 4
1
− ( 4)
−( s + 2 )
1
2
L {step response} = +
+
s ( s + 2 )2 + 42 ( s + 2 ) + 42
1



step response = 1− e −2t  cos 4t + sin 4t   u (t )
2



2.56
H (s) =
( s + 0.8)( s + 3.2 )
c)
d) H ( s ) =
L {h( t )} = H ( s ) =
1.07
1.07
−
⇒ h ( t ) = 1.07 ( e −.8t − e −3.2t ) u(t)
s + .8 s + 3.2
1
−4
2.56
1
H (s)
=
= + 3 + 3
L {step response} =
s
s ( s + .8) ( s + 3.2) s s + .8 s + 3.2
4
 1

step response = 1+ e −3.2t − e −.8t  u (t )
3
 3

( s + 2)
20
2
h( t ) = 4te −2t u (t )
step response = (1−(1+ 2t )e −2t ) u (t )
14-36
P14.9-15
For an impulse response, take V1 ( s ) = 1 . Then
3( s + 2 )
A
B
B*
= +
+
V0 ( s ) =
s ( s + 3− j 2 ) ( s + 3+ j 2 ) s s + 3− j 2 s + 3+ j 2
Where
A = sV0 ( s )
s =0
=.462, B = (s + 3 − j 2) V0 ( s )
Then
V0 ( s ) =
s =−3+ j 2
= 0.47∠ − 119.7 o and B* = 0.47 ∠119.7 o
0.462 0.47 ∠−119.7 o 0.47 ∠119.7 o
+
+
s
s +3− j 2
s +3+ j 2
The impulse response is
v0 (t ) = 0.462 + 2(0.47)e −3t cos ( 2 t − 119.7 o )  u ( t ) V
Section 14-10: Convolution Theorem
1 e− s 1 − e− s
f ( t ) = u ( t ) − u ( t − 1) ⇒ F ( s ) = L u ( t ) − u ( t − 1)  = −
=
s s
s
2
−s
 1 − e − s  
+ e−2 s 
−1 1 − 2 e
f ( t ) * f ( t ) = L −1 F 2 ( s )  = L −1
L
=




s2
 s  


= t u ( t ) − 2 ( t − 1) u ( t − 1) + ( t − 2 ) u ( t − 2 )
P14.10-1
P14.10-2
f ( t ) = 2 u ( t ) − u ( t − 2 )  ⇒ F ( s ) =
2 2e −2 s
−
s
s
8e −2 s 4e −4 s 
−1
−1  4
L
L
f ∗f =
 F ( s ) F ( s )  =
 s 2 − s 2 + s 2  = 4t u ( t ) − 8 ( t − 2 ) u ( t − 2 ) + 4 ( t − 4 ) u ( t − 4 )


14-37
P14.10-3
v1 ( t ) = t u ( t ) ⇒ V1 ( s ) =
V2 ( s )
1
s2
1
1
= Cs = RC
H (s) =
1
V1 ( s ) R + 1
s+
Cs
RC
v2 ( t ) = h ( t ) ∗ v1 ( t ) = L−1 V1 ( s ) H ( s ) 
 1 

 1
V2 ( s ) = V1 ( s ) H ( s ) =  2   RC 
 s   s+ 1 
RC 

v2 ( t ) = t − RC (1 − e − t / RC ), t ≥ 0
P14.10-4
1
1
h ( t ) ∗ f ( t ) = L−1  H ( s ) F ( s )  where H ( s ) = 2 and F ( s ) =
s
s+a
 1   1 
A
B
C
So H ( s ) F ( s ) =   
=
+ 2 +

s
s
s+a
 s 2   s +a
Solving the partial fractions yields: A = −1 a 2 , B = 1 a, C = 1 a 2
So h( t ) ∗ f ( t ) =
−1 t e − ( at )
+ + 2 ,
a2 a
a
t≥0
14-38
Section 14-11: Stability
P14.11-1
a. From the given step response:
H (s)
=L
s
3
−100 t
) u ( t ) = s ( s 7+5100 )
 4 (1 − e
From the circuit:
R
H (s) =
R + 5 + Ls
Comparing gives
R

= 75 
R = 15 Ω

L
 ⇒
R+5
L = 0.2 H
= 100 

L
b. The impulse response is
c.
⇒
R
H (s)
L
=
R+5
s

ss +

L 

 75 
h ( t ) = L -1 
= 75 e −100 t u ( t )

100
s
+


H (ω ) ω =100 =
75
3
=
∠45°
j 100 + 100 4 2
15
 3

∠45°  ( 5∠0° ) =
∠45° V
Vo (ω ) = 
4 2
4 2

vo ( t ) = 2.652 cos (100 t − 45° ) V
(Checked using LNAP, 12/29/02)
P14.11-2
The transfer function of this circuit is given by
H (s)
−10
5 −5
20
= L ( 5 − 5 e −2 t (1 + 2t ) ) u ( t )  = +
+
=
2
2
s
s s + 2 ( s + 2)
( s + 2)
⇒
H (s) =
s ( s + 2)
20
2
This transfer function is stable so we can determine the network function as
H (ω ) = H ( s ) s = j ω =
( s + 2)
20
2
s= jω
=
(2 + jω )
20
2
14-39
The phasor of the output is
Vo (ω ) =
( 2 + j 2)
20
2
( 5∠45° ) =
(2
20
2∠45°
)
2
( 5∠45° ) = 12.5∠ − 45°
V
vo ( t ) = 12.5cos ( 2 t − 45° ) V
The steady-state response is
(Checked using LNAP, 12/29/02)
The transfer function of the circuit is H ( s ) = L −130 t e −5t u (t )  =
P 14.11-3
( s + 5)
30
2
. The circuit is stable
so we can determine the network function as
H (ω ) = H ( s ) s = j ω =
( s + 5)
30
2
s= jω
=
(5 + j ω )
30
2
The phasor of the output is
Vo (ω ) =
( 5 + j 3)
30
The steady-state response is
2
(10∠0° ) =
( 5.83∠31° )
30
2
(10∠0° ) = 8.82∠ − 62°
V
vo ( t ) = 8.82 cos ( 3 t − 62° ) V
14-40
PSpice Problems
SP 14-1
14-41
SP 14-2
v(t ) = A + B e −t / τ
for t > 0
⇒ 7.2 = A + B 
 ⇒ B = −0.8 V
8.0 = v(∞) = A + B e −∞ ⇒ A = 8.0 V 
0.05
 8 − 7.7728 
7.7728 = v(0.05) = 8 − 0.8 e −0.05 / τ ⇒ −
= ln 
 = −1.25878
0.8
τ


0.05
⇒ τ=
= 39.72 ms
1.25878
7.2 = v(0) = A + B e 0
Therefore
v(t ) = 8 − 0.8 e −t / 0.03972 for t > 0
14-42
SP 14-3
i (t ) = A + B e −t / τ
0 = i (0) = A + B e 0
4 × 10−3 = i (∞) = A + B e −∞
for t > 0
⇒ 0 = A+ B
⇒
A = 4 × 10−3

−3
 ⇒ B = −4 ×10 A
A 
2.4514 ×10−3 = v(5 ×10−6 ) = ( 4 ×10−3 ) − ( 4 × 10−3 ) e
⇒ −
5 × 10−6
⇒ τ=
Therefore
τ
(
)
− 5×10−6 / τ
 ( 4 − 2.4514 ) × 10−3 
= ln 
 = −0.94894
−3
4
10
×


5 × 10−6
= 5.269 µ s
0.94894
i (t ) = 4 − 4 e −t / 5.269×10
−6
for t > 0
14-43
SP 14-4
Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure
to edit the labels of the parts so, for example, there is only one R1.)
14-44
V(C1:2), V(C2:2) and V(C3:2) are the capacitor voltages, listed from top to bottom.
14-45
SP 14-5
Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure
to edit the labels of the parts so, for example, there is only one R1.)
14-46
V(R2:2), V(R4:2) and V(R6:2) are the output voltages, listed from top to bottom.
14-47
Verification Problems
VP 14-1
v L (t ) = 3
iC (t ) =
d
i L ( t ) = −6 e − 2.1t − 2 e −15.9 t
dt
1 d
v C ( t ) = −0.092 e − 2.1t − 0.575 e −15.9 t
75 dt
v R1 ( t ) = 12 − v L ( t ) = 12 + 6 e − 2.1t + 2 e −15.9 t
i R2 (t ) =
12 − ( v L ( t ) + v C ( t ) )
i R3 (t ) =
Thus,
6
vC (t )
6
= 1 + 0.456 e − 2.1t − 0.123 e −15.9 t
= 1 + 0.548 e − 2.1t + 0.452 e −15.9 t
−12 + v L ( t ) + v R1 ( t ) = 0 and i R 2 ( t ) = i C ( t ) +i R 3 ( t )
as required. The analysis is correct.
14-48
VP 14-2
I1 (s) =
KVL for left mesh:
KVL for right mesh:
18
20
and I 2 ( s ) =
3
3
s−
s−
4
4

 

12 1  18   18
20 
+ 
−
 + 6
 = 0 (ok)
s 2s  s − 3   s − 3 s − 3 
4 
4
4


 
 

 18




20
20
18 
−6 
−
+ 3
− 4
= 0 (ok)


3
3
3
3
s−
s−  s−  s− 
4
4 
4 
4

The analysis is correct.
VP 14-3
Initial value of IL (s):
Final value of IL (s):
Initial value of VC (s):
Final value of VC (s):
lim
s+2
s 2
= 1 (ok)
s→∞
s +s+5
lim
s+2
s 2
= 0 (ok)
s→0
s +s+5
lim
−20 ( s + 2 )
s
= 0 (not ok)
s→∞
s ( s 2 + s + 5)
lim
−20 ( s + 2 )
= −8 (not ok)
s
s→0
s ( s 2 + s + 5)
14-49
Apparently the error occurred as VC (s) was calculated from IL (s). Indeed, it appears that VC (s)
was calculated as −
VC ( s ) = −
20
20
8
I L ( s ) instead of − I L ( s ) + . After correcting this error
s
s
s
20  s + 2  8

+ .
s  s2 + s+5 s
Initial value of VC (s):
Final value of VC (s):
 −20 ( s + 2 ) 8 
+  = 8 (ok)
s
s → ∞  s ( s 2 + s + 5) s 


lim
 −20 ( s + 2 ) 8 
+  = 0 (ok)
s
s → 0  s ( s 2 + s + 5) s 


lim
14-50
Design Problems
DP 14-1
Equating the Laplace transform of the step response of the give circuit to the Laplace transform
of the given step response:
kR
5
L
=
Vo ( s ) =
2
R
1
( s + 4)
s2 + s+
L
LC
Equating the poles:
R
4
R
− ±   −
L
 L  LC
s 1,2 =
= −4 ± 0
2
Summarizing the results of these comparisons:
2
R
= 4, R =
2L
2
kR
and
=5
L
LC
Pick L = 1 H, then k = 0.625 V/V, R = 8 Ω and C = 0.0625 F.
DP 14-2
Equating the Laplace transform of the step response of the give circuit to the Laplace transform
of the given step response:
14-51
kR
10
10
L
Vo ( s ) =
=
= 2
2
R
1
( s + 4 ) + 4 s + 8 s + 20
s2 + s+
L
LC
Equating coefficients:
R
1
kR
= 8,
= 20, and
= 10
L
LC
L
Pick L = 1 H, then k = 1.25 V/V, R = 8 Ω and C = 0.05 F.
DP 14-3
Equating the Laplace transform of the step response of the give circuit to the Laplace transform
of the given step response:
kR
5
5
10
L
Vo ( s ) =
=
−
= 2
R
1
( s + 2) ( s + 4) s + 6 s + 8
s2 + s+
L
LC
Equating coefficients:
R
1
kR
= 6,
= 8, and
= 10
L
LC
L
Pick L = 1 H, then k = 1.667 V/V, R = 6 Ω and C = 0.125 F.
DP 14-4
14-52
Comparing the Laplace transform of the step response of the give circuit to the Laplace
transform of the given step response:
kR
5
5
10 s + 30
L
Vo ( s ) =
≠
+
= 2
R
1
( s + 2) ( s + 4) s + 6 s + 8
s2 + s+
L
LC
These two functions can not be made equal by any choice of k, R, C and L because the
numerators have different forms.
DP 14-5
a) Use voltage division to get
Vo ( s )
=
V1 ( s )
R2
sC2 R 2 +1
R1
R2
+
sC1 R1 +1 sC2 R 2 +1
1


s
+


C1 R1
C1 

=
R1 + R2
C1 +C2  s +

 R1 R 2 ( C1 + C2 ) 


b) To make the natural response be zero, we eliminate the pole by causing it to cancel the zero.
−
R1 + R 2
1
=−
C1 R1
R1 R 2 (C1 +C2 )
⇒
C2 R1
=
C1 R 2
1
c) Let v1 (t ) = u (t ) ⇒ V1 ( s ) = . Then
s


1


s+
R1C1
C1 
K2
 K1
=
+
Vo ( s ) =


R1 + R 2
C1 + C2 
s

R1 + R 2
 s s +
s+

R1 R 2 (C1 +C2 )


  R1 R 2 (C1 +C2 )  
where K1 =
Then
R2
R1 + R 2
and K 2 =
R2
C1
−
C1 + C2 R1 + R 2
 R2
 C
R2  −t 
1
vo (t ) = 
+
−
 e τ  u (t )
 R1 + R 2  C1 + C2 R 1 + R 2 

14-53
where τ =
R1 R 2 (C1 + C2 )
R1 + R 2
. To make the step response be proportional to the step in put, we
require
R2
C1
=
C1 + C2 R1 + R 2
Then
vo (t ) =
R2
R1 + R 2
u (t )
DP 14-6
The initial conditions are vc (0) = −0.4 V and i (0) = 0 A . Consider the circuit after t = 0. A
source transformation yields
The mesh equations are
Solving for I 2 ( s ) yields
8
0.4
 8
 2 +  I1 ( s ) − I 2 ( s ) =
s
s
 s
8
8

− I1 ( s ) +  Ls +  I 2 ( s ) = 0
s
s

I2 ( s ) =
Therefore, the characteristic equation is
1.6
s ( Ls + 4 Ls +8)
s 2 + 4s +
2
8
=0
L
We require complex roots with significant damping. Try L = 1 H. Then
I ( s) =
Finally
1.6
0.2 −0.2( s + 4)
0.2
0.8
−0.2( s + 2)
=
+ 2
=
+
−
2
( s + 2) + 4 ( s + 2) 2 + 4
s ( s + 4 s +8)
s
s + 4 s +8
s
2
i (t ) = 0.2 − 0.2e −2t cos 2t − 0.4e−2t sin 2t  u ( t ) A
14-54
Chapter 15: – Fourier Series
Exercises
Ex. 15.3-1
Notice that
f (t − T ) = f1 (t − T ) + f 2 (t − T ) = f1 (t ) + f 2 (t ) = f ( t )
Therefore, f(t) is a periodic function having the same period, T. Next
f ( t ) = k1 f1 ( t ) + k2 f 2 ( t )
∞


= k1  a10 + ∑ ( a1n cos ( n ω 0 t ) + b1n sin ( n ω 0 t ) ) 
n =1


∞


+ k2  a20 + ∑ ( a2 n cos ( n ω 0 t ) + b2 n sin ( n ω 0 t ) ) 
n =1


= ( k1 a10 + k2 a20 ) + ∑ ( ( k1 a1n + k2 a2 n ) cos ( n ω 0 t ) + ( k1 b1n + k2 b2 n ) sin ( n ω 0 t ) )
∞
n =1
Ex. 15.3-1
f(t) = K is a Fourier Series. The coefficients are a0 = K; an = bn = 0 for n ≥ 1.
Ex. 15.3-2
f(t) = Acosw0t is a Fourier Series. a1 = A and all other coefficients are zero.
15-1
Ex. 15.4-1
2π
π  π
T = 4   = , ω0 =
= 4 rad
s
T
8 2
Set origin at t = 0, so have an odd function; then an = 0 for n = 0,1, . . . Also, f(t) has half wave
symmetry, so bn = 0 for n = even. For odd n, we have
bn =
2 T2
2 0
2 T2
(
)
sin
sin
sin ( n ω 0 t ) dt
f
t
n
t
dt
n
t
dt
ω
ω
=
−
+
(
)
(
)
0
0
T
T
∫
∫
T − 2
T − 2
T ∫0
4 T
= ∫0 2 sin ( n ω 0 t ) dt
T
4
4
(1−cos( n ω 0 T )) =
n = 1, 3, 5, . . .
=
2π nf 0T
nπ
Finally,
Ex. 15.4-2
f (t ) =
T = π , ω0 =
N 1
∑ sin nω 0t;
π n n
4
n odd and ω 0 = 4 rad s
2π
=2
T
 a0 = 0 , an = 0 for all n
odd function with quarter wave symmety ⇒ 
n = even
bn = 0
 −2t
0<t <π
 π
6
8 π4
bn = ∫0 f (t ) sin nω 0t dt where f (t ) =  6
π

π ≤t <π
−2
6
4
−24 1
 nπ 
Thus bn = 2 2 sin 

π n
 3 
−24 N 1
 nπ 
so f (t ) = 2 ∑ 2 sin 
 sin (2nt )
π n =1 n
3 

odd n
15-2
Ex. 15.4-3
a) is neither even nor odd. f(t) will contain both sine and cosine terms
1
b)
wave symmetry ⇒ no even harmonics
4
c) average value of f(t) = 0 ⇒ a0 = 0
Ex. 15.5-1
T = 2 s, ω 0 =
Cn =
− jnπ t
1 2
1 1 − jnπ t
1 2 − jnπ t
(
)
=
−
f
t
e
dt
e
dt
e
dt
2 ∫0
2 ∫0
2 ∫1
1
1
 −e − jnπ +1+ e − j 2 nπ −e− jnπ  =
(1− e− jnπ )
=
2 jnπ
jnπ
 2

Cn =  jnπ
 0

Finally,
f (t ) =
Ex. 15.5-3
Cn =
2π
= π rad/s
T
2
jπ
n odd
n even
 jπ t 1 j 3π t 1 j 5π t

− jπ t 1 − j 3π t 1 − j 5π t
...
e
e
e
e
+
+
+
−
− e
− e
−... 

3
5
3
5

1 T 4 − jω0nt
1  T  − j 2π nt T
e
dt
e
=
T
∫
T − 4
T  − j 2π n 
(n −1)

 (−1) 2

 π n
Cn = 
0
 12



T
=
4
−T
4
1
e − jπ n / 2 − e jπ n / 2 
− j 2π n
n odd
n even , n ≠ 0
n =0
15-3
Ex. 15.6-1
Use the “stem plot” in Matlab to plot the required Fourier spectra:
% Fourier Spectrum of a Pulse Train
A = 8;
T = 4;
d = T/8;
pi = 3.14159;
w0 = 2*pi/T;
% pulse amplitude
% period
% pulse width
%fundamental frequency
N = 49;
n = linspace(-N,N,2*N+1);
x = n*w0*d/2;
% Eqn.15.6-3. Division by zero when n=0 causes Cn(N+1) to be NaN.
Cn = (A*d/T)*sin(x)./x;
Cn(N+1)=A*d/T;
% Fix Cn(N+1); sin(0)/0 = 1
% Plot the spectrum using a stem plot
stem(n,Cn,'filled');
xlabel('n');
ylabel('|Cn|');
title('Fourier Spectrum of Pules Train with d = T/8');
15-4
Ex. 15.8-1
ω 0 = 4 rad/s
From Example 15.4-1:
vs (t ) = 3.24 ∑
N
n =1
odd n
1  nπ
 sin
2
n2 
1
1
1



 sin nω 0t = 3.24  sin 4 t − sin12 t + sin 20 t − sin 28 t" 
9
25
49



The network function of the circuit is
Vo (ω )
1
1
jω C
=
=
Vs (ω ) R + 1
1 + jω C R 1 + jω 4
jω C
Evaluating the network function at the frequencies of the input series
1
H ( n4 ) =
n = 1,3,5...
1 + j 16 n
n
H(n4)
1
0.062∠-86°
3
0.021∠-89°
5
0.012∠-89°
7
0.0009∠-89°
Using superposition
0.021
0.012
0.0009


vo (t ) = 3.24  ( 0.062 ) sin ( 4 t −86° )−
sin (12 t −89° ) +
sin ( 20 t −89° )−
sin ( 28 t −89° )" 
9
25
49


H (ω ) =
=
1
vo (t ) = ( 0.2009 ) sin ( 4 t − 86° ) − (.00756 ) sin (12 t − 89° )
+ ( 0.00156 ) sin ( 20 t − 89° ) − ( 5.95 × 10−5 ) sin ( 28 t − 89° )"
Discarding the terms that are smaller than 25 of the fundamental term leaves
vo (t ) = ( 0.2009 ) sin ( 4 t − 86° ) − (.00756 ) sin (12 t − 89° )
15-5
Ex. 15.9-1
f (t ) = e − at u (t )
F (ω ) = ∫
+∞
−∞
Ex. 15.10-1
f (t ) e
F { f ( at )} = ∫
∞
−∞
jω t
∞
−∞
Ex. 15.10-2
f (t ) =
F -1 {δ (ω − ω 0 )} =
Ex. 15.11-1
1
2π
1
2π
∫
∞
−∞
− at
jω t
0
e( )
1
=
dt =
− ( a + jω ) 0 a + jω
− a + jω t
∞
f ( at ) e− jω t dt
Let τ = at ⇒ t =
F { f ( at )} = ∫
dt = ∫ e e
∞
τ
a
f (τ ) e − jωτ a d
τ
a
=
1 ∞
1 ω 
−j ω a τ
f (τ ) e ( ) dτ = F  
∫
a −∞
a a
1
ω
∫ ( 2πδ (ω ) A) e dt = 2π ∫ ( 2πδ (ω ) A) dt = A
∞
−∞
δ (ω − ω 0 ) e jω t dt =
Take the Fourier Transform of both sides to get:
jω t
− jω t
  e 0 + e 0
F { A cos ω 0t} = F  A 
2
 
0+
j t
(
0−
1 jω0t
e
2π
)
F e jω0t = 2πδ (ω − ω 0 )
( (
)
(
))
  A
A
F e jω0t + F e − jω0t = ( 2πδ (ω − ω 0 ) + 2πδ (ω + ω 0 ) )
 =
2
  2
= Aπδ (ω − ω 0 ) + Aπδ (ω + ω 0 )
15-6
Ex. 15.12-1
a)
b)
Vin (ω ) =
Win =
Wout =
∴η =
∫
π
1
∞
0
π
1
∫
120
24 + jω
⇒ Vin (ω ) =
1202
14400
=
2
2
24 + ω
576 + ω 2
14400
14000  1
 ω 
tan −1    = 300 J
dω =

2
576 + ω
π  24
 24  0
∞
14400
14000  1
 ω 
dω =
tan −1    = 61.3 J

2
576 + ω
π  24
 24  24
48
48
24
Wout
61.3
×100% =
×100% = 20.5%
Win
300
Ex. 15.13-1
f + ( t ) = te − at
f − ( t ) = te at
∴ F + (s) =
Then F (ω ) = F + ( s )
s = jω
+ F − (s)
f − ( −t ) = −te− at
⇒
(s + a)
1
s =− jω
=
=
2
and F − ( s ) =
(s + a)
1
2
s = jω
( a + jω )
1
2
−
+
−1
(s + a)
2
−1
(s + a)
( a − jω )
2
1
2
s =− jω
=
− j 4aω
(a
2
+ω2 )
2
15-7
Problems
Section 15.3: The Fourier Series
P15.3-1
T = 2 s ⇒ ω0 =
2π
= π rad/s and f (t ) = t 2 for 0 ≤ t ≤ 2 . The coefficients of the Fourier
2
series are given by:
1 2 2
t dt = 4
3
2 ∫0
2 2
4
an = ∫0 t 2 cos nπ t dt =
2
2
( nπ )
a0 =
bn =
∴ f (t ) =
2 2 2
−4
t sin nπ t dt =
∫
0
nπ
2
4 4 N 1
4 ∞ 1
+ 2 ∑ 2 cos nπ t − ∑ sin nπ t
π n =1 n
3 π n =1 n
15-8
P15.3-2
an =
T
 2π 
 2π  
2  T4
2
 ∫0 cos  n t  dt + ∫T 2 cos  n t  dt 
 T 
 T  
T
4
1
=
nπ
=

 2π 
sin  n T t 



T
4
0
T

 2π  2 
+ 2sin  n t 
 T T 
4
1 

 nπ 
 nπ
(sin 
 −0) + 2 (sin nπ ) −sin 

nπ 
 2 
 2

 
 
 
 (−1) n +1
1
2 odd n

 nπ 
sin 
= −
 =  nπ
nπ
 2 

even n
 0
( )
2  T4
2π
 ∫ sin n T t dt +
T 0
∫
 2π  
2 sin  n t  dt 
 T  
T
T

π
1 
2

4
 2π  2 
n
t
n
t
cos
2
cos
= −
+




nπ 
 T 0
 T T 
4
nπ 
1 
(2 cos (nπ ) −1) − cos 
=−

nπ 
2 
 3
n is odd
 nπ

 2
n = 2,6,10,…
= −
 nπ
n = 4,8,12,…
0

bn =
T
2
T
4
15-9
P15.3-3
a 0 = average value of f ( t ) =
t

f ( t ) = A 1 − 
 T
an =
bn =
2
T
2
T
∫
∫
T
0
T
0
t

 2π
A 1 −  cos  n
 T
 T
t

 2π
A  1 −  sin  n
 T
 T
A
2
for 0 ≤ t ≤ T
2A T

 2π
cos  n
t  dt =

∫
T  0

 T
1 T

 2π
t  dt − ∫ t cos  n
T 0

 T
 
t  dt 
 
T


π
π
π
2
2
2







cos  n
t+n
t  sin  n
t 
2A
1
 T   T 
 T  
=
0−
2


T
T
 2π 


n


 T 
0

−A
= 2 2 cos ( 2nπ ) − cos ( 0 ) + 2nπ sin ( 2nπ ) − 0 
2n π
=0
2 A  T  2π

t  dt =
sin  n
T  ∫ 0

 T
1 T

 2π
t  dt − ∫ t sin  n
0
T

 T
 
t  dt 
 
T


2
2
2
π
π
π







t−n
t  cos  n
t 
sin  n
2A
1
 T   T 
 T  
=
−
0
2


T
T
 2π 


n



 T 
0

−A
= 2 2 ( sin ( 2nπ ) − sin ( 0 ) ) − ( 2nπ cos ( 2nπ ) − 0 ) 
2n π
A
=
nπ
f (t ) =
A ∞ A
 2π
sin  n
+∑
2 n =1 n π
 T

t

15-10
P15.3-4
T = 2 s, ω 0 =
2π
= π rad/s , a 0 = average value of f ( t ) = 1 ,
2
f (t ) = t
2
an =
2
∫
2
0
t cos ( n π t ) dt =
=
2
bn =
2
∫
2
0
t sin ( n π t ) dt =
for 0 ≤ t ≤ 2
cos ( n π t ) + ( n π t ) sin ( n π t )
( nπ )
2
2
 cos ( 2nπ ) − cos ( 0 ) + 2nπ sin ( 2nπ ) − 0 
n π2 
0
1
=0
2
sin ( n π t ) − ( n π t ) cos ( n π t )
(nπ )
2
2
( sin ( 2nπ ) − sin ( 0 ) ) − ( 2nπ cos ( 2nπ ) − 0 ) 
n π2 
−2
=
nπ
=
0
1
2
f (t ) = 1 − ∑
∞
n =1
2
 2π
sin  n
nπ
 T

t

Use Matlab to check this answer:
% P15.3-4
pi=3.14159;
A=2;
T=2;
% input waveform parameters
% period
w0=2*pi/T;
tf=2*T;
dt=tf/200;
t=0:dt:tf;
%
%
%
%
a0=A/2;
v1=0*t+a0;
% avarage value of input
% initialize input as vector
fundamental frequency, rad/s
final time
time increment
time, s
for n=1:1:51
% for each term in the
an=0;
% specify
series
bn=-A/pi/n;
cn=sqrt(an*an + bn*bn);
% convert
thetan=-atan2(bn,an);
v1=v1+cn*cos(n*w0*t+thetan); % add the
Fourier series
end
Fourier series ...
coefficients of the input
to magnitude and angle form
next term of the input
15-11
plot(t, v1,'black')
% plot the Fourier series
grid
xlabel('t, s')
ylabel('f(t)')
title('P15.3-4')
15-12
Section 15-4: Symmetry of the Function f(t)
2π π
= rad/s .
4
2
The coefficients of the Fourier series are:
15.4-1
T = 4 s ⇒ ωo =
a0 = average value of vd ( t ) = 0
an = 0 because vd(t) is an odd function of t.
bn =
1 4
π
( 6 − 3 t ) sin  n t  dt
∫
0
2
 2 
4
3 4
 π 
 π 
= 3∫ sin  n t  dt − ∫ t sin  n t  dt
0
2 0
 2 
 2 

 π 


 − cos  n 2 t   3  1

π
π
π







 −
= 3
 2 2 sin  n t  −  n t  cos  n t  
π
 2   2 
 2 

 2n π
n


2
 4
 0
0
6
6
=
−1 + cos ( 2nπ ) ) − 2 2 ( sin ( 2nπ ) − 0 ) − ( 2 n π cos ( 2nπ ) )
(
nπ
nπ
12
=
nπ
4
4
(
)
The Fourier series is:
vd ( t ) = ∑
∞
n =1
P15.4-2
12
 π 
sin  n t 
nπ
 2 
vc ( t ) = vd ( t − 1) − 6 = −6 + ∑
∞
n =1
∞
π
12
12
 π

 π
sin  n ( t − 1)  = −6 + ∑
sin  n t − n 
nπ
2
 2

 2
n =1 n π
15-13
π
2π 1000 π
rad/s = krad/s
=
.006
3
3
The coefficients of the Fourier series are:
3× 2
1
V
a0 = average value of va ( t ) = 2 =
6
2
bn = 0 because va(t) is an even function of t.
P15.4-3
T = 6 ms = 0.006 s ⇒ ω o =
 2  0.001
 1000π 
an = 2 
t  dt
 ∫0 ( 3 − 3000 t ) cos  n
3
 0.006 


0.001
1
 1000π 
 1000π
= 2000 ∫
t  dt − ( 2 ×106 ) ∫ t cos  n
cos  n
0
0
3
3



  1000π
 sin  n 3
= 2000  
 n 1000π

3

t
 − 1000
n 2106 π 2
9

t  dt



 1000π   1000π   1000π   
t+n
t  sin  n
t  
cos
n

3
3
3

 
 
 


0
0.001


3
9
  π 
 π     π   π   
sin  n  − 0  − 2 3 2   cos  n  − 1 +   n  sin  n  − 0   
= 2000 

 3     3   3   
 n1000 π   3   n 10 π  
6
 π   18  
 π  6
 π
sin  n  −  2 2   cos  n  − 1 −
sin  n 
=
nπ
 3   n π 
 3   nπ
 3
 18  
 π 
= −  2 2   cos  n  − 1
 n π 
 3 
The Fourier series is
va ( t ) =
P15.4-4
1 ∞ 18 
 nπ  
 1000 π
+ ∑ 2 2 1 − cos 
  cos  n
3
2 n =1 n π 
 3 


t

1 ∞ 18 
 nπ  
 π

vb ( t ) = va ( t − 2 ) − 1 = −1 + + ∑ 2 2  1 − cos 
  cos  n ( t − 2 ) 
2 n =1 n π 
 3 
 3

1 ∞ 18 
 nπ
= − + ∑ 2 2  1 − cos 
2 n =1 n π 
 3
2π 

 1000 π
t−n
  cos  n

3
3 


15-14
Choose t 0 = − π
P15.4-5
2π
=1
2π
average value: a0 = 0
T = 2π , ω 0 =
−π < t < π
f (t ) = t
2 T
f ( t ) sin nω 0 t dt
T ∫0
an = 0 since have odd function
bn =
bn =
2
2π
∫ π t sin nt dt
π
−
1  sin nt t cos nt 
=  2 −

n
π n
−π
1 π π 
b1 =  +  = 2
π 1 1
b2 = −1
π
b3 = 2 3
P15.4-6
T = 8 s, ω 0 = π 4 rad/s
bn = 0 because f ( t ) is an even functon
a0 = average =
( 2×2 ) − 2×1 = 1 4
8
4 T2
f ( t ) cos n ω 0 t dt
T ∫0
π
π
2

4  1
=  ∫0 2 cos n t dt − ∫1 cos n t dt 
4
4
8 

2 
nπ
nπ 
3 sin −sin
=

nπ 
4
2 
a1 = .714, a2 = .955, a3 = .662
an =
15-15
P15.4-7
ω 0 = 2ω , T =
π
ω
ω 2ω
2A
a0 = ∫−π A cos ω t dt =
π
π 2ω
π
an =
2ω
π
∫
π
2ω
−π
2ω
A cos ω t cos 2nω t dt
2ω A  sin ( 2n −1)ω t sin ( 2n +1)ω t  2ω
=
+


2( 2n +1)ω  − π
π  2( 2n −1)ω
π
2ω
π
π

sin 2n −1)
sin ( 2n +1) 
2A  (
2+
2
=


2n +1 
π  2n −1


2A 
π
π
=
( 2n +1) sin ( 2n −1) −( 2n −1) sin ( 2n −1) 
2

2
2
π ( 4n −1) 
=−
4A
cos( nπ )
π ( 4n 2 −1)
=−
4 A( −1)
π ( 4n 2 −1)
n
bn = 0 due to symmetry
P15.4-8
2π
= 5 π rad s
T
0 ≤ t ≤ .1
 A cos ω 0t

.1 ≤ t < .3
f (t ) = 0
 A cos ω t
.3 ≤ t ≤ .4
0

T = 0.4 s, ⇒ ω 0 =
Choose period − .1 ≤ t ≤ .3 for integral
1 .1
A cos ω 0t = A π
T ∫−.1
2 .1
an = ∫−.1 A cos ω 0t cos nω 0t dt
T
a0 =
15-16
a1 = 5 A∫−.1 cos 2 ω 0t dt =
.1
A
2
an = 5 A∫−.1 cos ω 0t cos nω 0t dt
.1
= 5 A∫−.1 1 [ cos 5π (1+ n)t + cos 5π (1− n)t ] dt
2
2 A cos (nπ / 2)
n ≠1
=
1− n 2
π
bn =0 because the function is even.
.1
P15.4-9
a0 = 0 because the average value is zero
an = 0 because the function is odd
bn = 0 for even due to
1
wave symmetry
4
Next:
 nπ
8 sin 
T 4
 2
bn = ∫−T 4 t sin ( nω 0 t ) dt =

 nπ   8
 − 4nπ cos 

2
2

 2  =  n π

n 2π 2
− 8
 n 2 π 2
for n = 1,5,9, ...
for n = 3, 7,11, ...
Section 15.5: Exponential Form of the Fourier Series
P15.5-1
2π
= 2π , the coefficients of the complex Fourier series are given by:
1
1  e jπ t − e − jπ t  − j 2π nt
1 1
dt
Cn = ∫ A sin (π t ) e − j 2π nt dt = ∫ A 
 e
0 
1 0
2j


A 1 − jπ ( 2 n−1) t − jπ ( 2 n+1) t
e
dt
=
−e
2 j ∫0
T = 1 ⇒ ωo =
(
)
1
− jπ 2 n −1) t
− jπ 2 n +1) t 
Ae (
e (
−2 A
=
−

 =
2
2 j  − jπ ( 2n − 1) − jπ ( 2n + 1) 

 0 π (4n − 1)
where we have used e± j 2π n = 1 and e j π = e− j π .
15-17
P15.5-2
2π
1 1 A  − j
Cn = ∫  t  e T
T 0 T 
Recall the formula for integrating by parts:
dv
2π
− j nt
= e T dt .
∫t
t2
1
When n ≠ 0 , we get

2π
− j nt

A te T
Cn = 2 
T  − j 2π n

T

nt
dt =
A
∫0 t e
1
T2
−j
2π
nt
T
dt
u dv = u v t2 − ∫ v du . Take u = t and
t2
t
t1
1
T
+
0
1
e
2 π ∫0
j
n
T
T −j

2π

− j nt
π
j
n
2
−
AT e
e T
= 
+
T  − j 2 π n  2 π 2
n
j

T 



2π
nt 
T
dt 






0



T


 − j 2π n

AT e
e − j 2π n − 1 
=
+
T  − j 2 π n  2 π 2 

n  
j
 T  

A
= j
2π n
Now for n = 0 we have
Finally,
C0 =
f (t ) =
1 TA
A
t dt =
∫
2
T 0 T
A
A
+j
2
2π
1 jn
∑ ne
n =−∞
n =∞
2π
t
T
n≠0
15-18
P15.5-3
2π
− jn
t
d /2
T dt
e
−d / 2
A
Cn = ∫
T
π
 − j n 2π t 
 j nπ d
− jn d
T 
A e
A e T
e T
= 
= 
−

2π
T  − j n 2π 
T  j n 2π
jn
T
T  −d / 2
T


d /2





π
 jnπ d
− jn d
T
A e
−e T
=

nπ 
2j

A
 nπ d 
sin 
=

nπ
 T 
 nπ d 
sin 

 Ad   T 
=

 T  nπ d
T





m = 1 t0 +T ( a f ( t − t ) + b ) e − j nωot dt
C
n
d
T ∫t0
Let τ = t − td , then t = τ + td .
P15.5-4
m=1
C
n
T
1
=
T
∫t −t
t0 +T −td
∫t −t
t0 +T −td
0
d
e − j n ω o td
=
T
0
(
d
( a f (τ ) + b ) e− j nω (τ +t )dτ
o
( a f (τ ) + b ) e− j nω τ e− j nω t dτ
o
∫t −t
t0 +T −td
= a e − j n ω o td
But
∫t −t
t0 +T −td
0
d
be
− j n ω oτ
d
0
) T1 ∫
d
o d
( a f (τ ) + b ) e− j nω τ dτ
o
t0 +T −td
t 0 −t d
(
f (τ ) e− j n ωoτ dτ + e− j n ωotd
t +T −td
 e − j nωoτ  0
dτ = b 

 − j n ωo  t −t
0
d
) T1 ∫
t0 +T −td
t0 − t d
b e− j n ωoτ dτ
0 n ≠ 0
so
=
b = 0
m = aC +b
C
0
0
and
m = a e − j n ω o td C
C
n
n
n≠0
15-19
P15.5-5
T = 8 s, ω 0 =
2 × 2 − 2 (1× 1) 1
2π π
= rad/s, C 0 = average value =
=
T
4
8
4
The coefficients of the exponential Fourier series are calculated as
nπ
nπ
nπ
−j
−j
−j
t
t
t
1
2

1  −1
C n =  ∫ − 1× e 4 dt + ∫ 2 × e 4 dt + ∫ − 1× e 4 dt 
−1
1
8  −2

−1
1
2


nπ
nπ
nπ
−j
−j
−j
t
t
t


4
4
4
1
e
e
e

=  −1 ×
+ 2×
+ ( −1) ×
nπ
nπ
nπ 
8
−j
−j
−j

4 −2
4 −1
4 1 
−j
=
2 nπ
nπ
nπ
nπ
 j n4π
−j
j
j

 − j n4π
  − j n2π
2
4
−e +e
−e 4
 e − e  − 2  e



 


 
and
C −n
− nπ
− nπ
− nπ
t
t
t
−j
−j
−j
1
2

1  −1
4
4
=  ∫ − 1× e
dt + ∫ 2 × e
dt + ∫ − 1× e 4 dt 
−
−
2
1
1
8

−1
1
2


nπ
nπ
nπ
j
t
j
t
j
t

4
4
4
1
e
e
e

=  −1×
+ 2×
+ ( −1) ×
nπ
nπ
nπ 
8
j
j
j

4 −2
4 −1
4 1 
=
j
2 nπ
= −C n
nπ
 − j n4π
−j
2
e
e
−


nπ
−j

 j n4π
4
2
e
e
−
−




nπ
j
  j n2π
4
e
e
+
−
 
 


 
The function is represented as
f ( t ) = C0 + ∑ Cn e j n ω0 t + ∑ C− n e − j n ω0 t
∞
−∞
n =1
n =−1
15-20
This result can be checked using MATLAB:
pi = 3.14159;
N=100;
T = 8;
t = linspace(0,2*T,200);
c0 = 1/4;
w0 = 2*pi/T;
%
%
%
%
period
time
average value
fundamental frequency
for n = 1: N
C(n) = -j*((exp(+j*n*pi/4)-exp(+j*n*pi/2))-2*(exp(-j*n*pi/4)exp(+j*n*pi/4))+(exp(-j*n*pi/2)-exp(-j*n*pi/4)))/(2*pi*n);
end
for i=1:length(t)
f(i)=c0;
for n=1:length(C)
f(i)=f(i)+C(n)*exp(j*n*w0*t(i))+C(n)*exp(-j*n*w0*t(i));
end
end
plot(t,f,'black');
xlabel('t, sec');
ylabel('f(t)');
15-21
Alternately, this result can be checked using Mathcad:
N := 15
n := 1 , 2 .. N
T := 8
d :=
ω := 2
200
C :=
π
T
i := 1 , 2 .. 400
T
⌠

⌡
m := 1 , 2 .. N
−1
t := d ⋅ i
i
⌠
−1⋅ exp( −j⋅ n ⋅ ω⋅ t ) dt + 
⌡
1
2
−1
−2
⌠
2⋅ exp( −j ⋅ n ⋅ ω⋅ t) dt +  −1 exp( −j ⋅ n ⋅ ω⋅ t ) dt
⌡
1
n
C :=
T
⌠

⌡
−1
−2
−1⋅ exp( j ⋅ m⋅ ω⋅ t ) dt + 
⌠
⌡
1
−1
2⋅ exp( j ⋅ m⋅ ω⋅ t) dt + 
⌠
⌡
−1 exp( j ⋅ m⋅ ω⋅ t ) dt
1
m
f ( i) :=
2
T
∑
N
n =1
(
) ∑
C ⋅ exp j ⋅ n ⋅ ω⋅ t +
n
i
N
m =1
(
C ⋅ exp −1⋅ j ⋅ m⋅ ω⋅ t
m
)
i
C
0.357
0.357
1.685
0.477
0.477
1.745
0.331
0.331
1.807
0
0
1.856
0.199
0.199
1.88
0.159
0.159
1.872
0.051
0.051
1.831
0
0
1.767
0.04
0.04
1.693
0.095
0.095
1.628
0.09
0.09
1.589
0
0
1.589
0.076
0.076
1.633
0.068
0.068
1.717
0.024
0.024
1.825
n
2
f ( i)
0
2
100
200
i
300
400
m
=
f ( i) =
C =
1.643
15-22
P15.5-6
The function shown at right is related to the
given function by
v ( t ) = −v1 ( t + 1) − 6
(Multiply by –1 to flip v1 upside-down; subtract 6
to fix the average value; replace t by t+1 to shift
to the left by 1 s.)
From Table 15.5-1
v1 ( t ) =
∑
∞
n =−∞
Therefore
∞
j A ( −1) j n ω 0 t
j 6 ( −1) j n π2 t
e
e
= ∑
nπ
nπ
n =−∞
n
v ( t ) = −6 −
n
∑
∞
n =−∞
n
n
∞ 
j 6 ( −1) j n π2 (t +1)
j 6 ( −1) j n π2  j n π2 t
e
e
= −6 − ∑ 
e
 nπ

nπ
n =−∞


j 6 ( −1) j n π2
C0 = −6 and Cn = −
e
nπ
This result can be checked using Matlab:
The coefficients of this series are:
n
pi = 3.14159;
N=100;
A = 6;
T = 4;
t = linspace(0,2*T,200);
c0 = -6;
w0 = 2*pi/T;
% amplitude
% period
% time
% average value
% fundamental frequency
for n = 1: N
C(n) = (-j*A*(-1)^n/n/pi)*exp(+j*n*pi/2);
D(n) = (+j*A*(-1)^n/n/pi)*exp(-j*n*pi/2);
end
for i=1:length(t)
f(i)=c0;
for n=1:length(C)
f(i)=f(i)+C(n)*exp(j*n*w0*t(i))+D(n)*exp(-j*n*w0*t(i));
end
end
plot(t,f,'black');
xlabel('t, sec');
ylabel('f(t)');
title('p15.5-6')
15-23
15-24
P15.5-7
Represent the function as
(Check: f ( 0 ) = 0,
 1 − e −5 t
0 ≤ t ≤1
f ( t ) =  −5 ( t −1) −5
1≤ t ≤ 2
−e
e
f (1) = 1 − e −5 1,
T = 2 s, ω 0 =
f ( 2 ) = e−5 − e −5 = 0 )
2π
1
= π , also C0 = average value =
2
2
The coefficients of the exponential Fourier series are calculated as
(
)
2
1 1
−5 t −1
1 − e −5 t ) e − j nπ t dt + ∫ e ( ) − e −5 e− j nπ t dt 
(
∫

0
1


2
1
1
2 − 5+ j n π t
2
1
)
dt − e −5 ∫ e− j nπ t dt 
=  ∫ e − j nπ t dt − ∫ e −5 t e − j nπ t dt + e5 ∫ e (

0
1
1
2  0
1
2
1
− 5+ j nπ ) t
−( 5+ j n π ) t

− j nπ t 2  
 
1  e − j nπ t
e(
e
−5 e
5
 +e

=
−
−e
2  − j n π 0 − ( 5 + j n π ) 0   − ( 5 + j n π ) 1
− j n π 1 
 


Cn =
=
(
)(
)
− 5+ j n π ) 2
− 5+ j n π )
e− j nπ 2 − e− j nπ
−e (
1  e − j nπ − 1 e −5 e − j nπ − 1   5 e (
−
− e −5

 +  e
2  − j n π
− (5 + j nπ )  
− (5 + j nπ )
− j nπ
− j nπ 2
1  e − j nπ − 1 e −5 e − j nπ − 1   e−5 e− j 2 nπ − e− j nπ
− e − j nπ
−5 e
= 
−
−e
+
2  − j n π
− ( 5 + j n π )   − ( 5 + j n π )
− j nπ
n
n
n
n
−5
−5

1  ( −1) − 1 e ( −1) − 1   e − ( −1)
−5 1 − ( −1)
= 
−
−e
+

2  − j n π
− ( 5 + j n π )   − ( 5 + j n π )
− j n π  



 
 

 
 
The terms that include the factor e −5 = 0.00674 are small and can be ignored.
n
  − ( −1)n  
1  ( −1) − 1
−1
−
C n = 

+
2  − j n π
− ( 5 + j n π )   − ( 5 + j n π )  


1
 1
odd n
−

=  j nπ 5 + j nπ

0
even n

5

 j nπ 5 + j nπ
)(
)
= (

0

odd n
even n
15-25
This result can be checked using Matlab:
pi = 3.14159;
N=101;
T = 2;
t = linspace(0,2*T,200);
c0 = 0.5;
w0 = 2*pi/T;
% period
% time
% average value
% fundamental frequency
for n = 1:2:N
if n == 2*(n/2)
C(n) = 5/((+j*pi*n)*(5+j*pi*n));
D(n) = 5/((-j*pi*n)*(5-j*pi*n));
else
C(n)=0;
D(n)=0
end
end
for i=1:length(t)
f(i)=c0;
for n=1:length(C)
f(i)=f(i)+C(n)*exp(j*n*w0*t(i))+D(n)*exp(-j*n*w0*t(i));
end
end
plot(t,f,'black');
xlabel('t, sec');
ylabel('f(t)');
title('p15.5-7')
15-26
Section 15-6: The Fourier Spectrum
P15.6-1
Average value = 0 ⇒ a0 = 0
half − wave symmetry

4 T 2  4A 
4A
 2π 
 an = T ∫0  − T t  cos  n T t  dt = − 2 2 (cos (nπ ) −1)




n π

⇒ 
f (t )

 bn = 4 ∫0T 2  − 4 A t  sin  n 2π t  dt = − 2 A (1− cos ( n π ) )
T
nπ

 T   T 
n Cn = a n + bn
2
1 1.509 ⋅ A
2
−57.5°
2 0
3 0.434 ⋅ A
0
4 0
5 0.257 ⋅ A
0
6 0
0
7 0.183 ⋅ A
 bn 

 an 
θ n = tan −1 
−78.0°
−82.7°
−84.8°
15-27
P15.6-2
Mathcad spreadsheet (p15_6_2.mcd):
N := 100
n := 1 , 2 .. N
ω0 := 2
T := 32
π
T
Calculate the coefficients of the exponential Fourier series:
⌠4
4 
C1 := ⋅ 
n
T

⌡3⋅ T
T
⌠2
4 
 2⋅ π ⋅ t  exp( −j⋅ n⋅ ω0⋅ t) dt
C2 := ⋅  sin 
n
T
 T 

T
⌡
T
 16⋅ t − 3 exp( −j⋅ n ⋅ ω0⋅ t) dt
 T


4
16
3⋅ T
⌠ 4
4 
C3 := ⋅ 
n
T

⌡11⋅ T
4 ⌠

 2⋅ π ⋅ t  exp( −j⋅ n ⋅ ω0⋅ t) dt
C4 := ⋅
sin 
n

T
 T 
 T
⌡3⋅
 11 − 16⋅ t  exp( −j⋅ n ⋅ ω0⋅ t) dt

T 

T
4
16
C := C1 + C2 + C3 + C4
n
n
n
n
n
Check: Plot the function using it's exponential Fourier series:
d :=
T
200
i := 1 , 2 .. 400
t := d ⋅ i
i
f ( i) :=
∑
N
n =1
(
∑
i)
C ⋅ exp j⋅ n ⋅ ω0⋅ t +
n
N
n =1
(

C ⋅ exp −j⋅ n ⋅ ω0⋅ t
n
)
i
5
f ( i)
0
5
10
20
30
40
50
60
ti
15-28
Plot the magnitude spectrum:
1.5
1
Cn
0.5
0
1
2
3
4
5
6
7
8
9
10
n
That’s not a very nice plot. Here are the values of the coefficients:
C
n
( n)
=
arg C ⋅
1.385
180
π
=
-115.853
0
-90
0.589
22.197
0
-24.775
0.195
-113.34
0
106.837
0.139
66.392
0
-78.232
0.082
-69.062
0
-48.814
0.039
109.584
0
90.415
0.027
-25.598
0
1.226·10 -3
63.432
0
163.724
78.14
15-29
P15.6-3
Use Euler’s formula to convert the trigonometric series of the input to an exponential
series:
vi ( t ) = 10 cos t + 10 cos 10 t + 10 cos 100 t V
e−t + e−t
e−10 t + e−10 t
e−100 t + e−100 t
= 10
+ 10
+ 10
2
2
2
10 t
−100 t
−10 t
−t
t
= 5e
+ 5 e + 5 e + 5 e + 5 e + 5 e100 t
The corresponding Fourier spectrum is:
Evaluating the network function at the frequencies of the input:
ω, rad/s
1
10
100
|H(ω)|
1.923
0.400
0.005
∠ H(ω), °
-23
-127
-174
Using superposition:
vo ( t ) = 19.23 cos ( t − 23° ) + 4.0 cos (10 t − 127° ) + 0.05 cos (100 t − 174° ) V
Use Euler’s formula to convert the trigonometric series of the output to an exponential
series:
)
)
)
)
+e ( )
+e (
+e (
e (
e (
+ 4.0
+ 0.05
vo ( t ) = 19.23
V
2
2
2
=19.23e j 174°e − j t + 4.0 e j 127° e − j 10 t + 19.23e j 23° e − j t + 19.23e− j 23° e j t + 4.0 e − j 127° e j 10 t 19.23e− j 174° e j t
e
j ( t − 23° )
− j t − 23°
j 10 t −127°
− j 10 t −127°
j 100 t −174°
− j 100 t −174°
15-30
2π
1
= 2 π rad/s, C 0 =
T
2
f (t ) = 1− t
when 0 ≤ t < 1 s
P15.6-4
T = 1 s, ω 0 =
The coefficients of the exponential Fourier series are given by
Cn =
1
1
1 1
1 − t ) e − j 2π nt dt = ∫ e − j 2π nt dt − ∫ t e − j 2π nt dt
(
∫
0
0
1 0
Evaluate the first integral as
∫
1 − j 2π nt
e
0
e− j 2π nt
dt =
− j 2π n
=
1
0
e − j 2π n − 1
=0
− j 2π n
To evaluate the second integral, recall the formula for integrating by parts:
∫t
t2
1
u dv = u v t2 − ∫ v du . Take u = t and dv = e − j 2π nt dt . Then
t
1
∫0 t e
1
t2
t1
− j 2π nt
t e− j 2π nt
dt =
− j 2π n
1
+
0
1 − j 2π nt
1
e
dt
∫
j 2π n 0
e − j 2π n
e − j 2π nt
=
+
− j 2 π n ( j 2 π n )2
Therefore
1
=
0
 1
 2
Cn = 
 −j
 2 π n
e − j 2π n e − j 2π n − 1
1
+
= j
2
2π n
− j 2π n ( j 2π n )
n=0
n≠0
To check these coefficients, represent the function by it’s Fourier series:
j − j 2π nt 
1 n=∞  − j j 2π nt
f (t ) = + ∑ 
e
e
+

2 n=1  2 π n
2π n

Next, use Matlab to plot the function from its Fourier seris (p15_6_4check.m):
pi = 3.14159;
N=20;
T = 1;
t = linspace(0,2*T,200);
% period
% time
15-31
c0 = 1/2;
w0 = 2*pi/T;
% average value
% fundamental frequency
for n = 1: N
C(n) = -j/(2*pi*n);
end
for i=1:length(t)
f(i)=c0;
for n=1:length(C)
f(i)=f(i)+C(n)*exp(j*n*w0*t(i))-C(n)*exp(-j*n*w0*t(i));
end
end
plot(t,f,'black');
xlabel('t, sec');
ylabel('f(t)');
This plot agrees with the given function, so we are confident that the coefficients are
correct. The magnitudes of the coefficients of the exponential Fourier series are:
 1
 2
Cn = 
 1
 2 π n
n=0
n≠0
Finally, use the “stemplot” in Matlab to plot the Fourier spectrum (p15_6_4spectrum.m):
pi = 3.14159;
N=20;
n = linspace(-N,N,2*N+1);
Cn = abs(1/(2*pi)./n); % Division by 0 when n=0 causes Cn(N+1)=
NaN.
15-32
Cn(N+1)=1/2;
% Fix Cn(N+1); C0=1/2
% Plot the spectrum using a stem plot
stem(n,Cn,'-*k');
xlabel('n');
ylabel('|Cn|');
15-33
Section 15.8: Circuits and Fourier Series
P15.8-1
The network function of the circuit is:
100
Vo (ω )
1
jω
H (ω ) =
=
=
ω
100
Vi (ω ) 10 +
1+ j
jω
10
Evaluating the network function at the harmonic frequencies:
1
20
20
 π
 nπ 
=
=
∠ − tan −1 
Hn  =

 2  1 + j n π 20 + j n π
 20 
400 + n 2 π 2
20
From problem 15.4-2, the Fourier series of the input voltage is
vc ( t ) = −6 + ∑
∞
n =1
12
π
 π
sin  n t − n 
nπ
2
 2
Using superposition, the Fourier series of the output voltage is
vo ( t ) = −6 + ∑
∞
n =1
 π
 π
 nπ   
sin  n t −  n + tan −1 

 20   
n π 400 + n 2 π 2
 2  2
240
P15.8-2
The network function of the circuit is:
R2
H (ω ) =
j ω C1 R 2
1 + j ω C2 R 2
Vo (ω )
=−
=−
1
Vi (ω )
(1 + j ω C1R1 )(1 + j ω C2 R 2 )
R1 +
j ω C1
=−
=−
j ω (10−6 ) ( 2000 )
(1 + j ω (10 ) (1000) ) (1 + j ω (10 ) ( 2000 ))
−6
j
ω
−6
500
ω 
ω 

1 + j
 1 + j

1000  
500 

Evaluating the network function at the harmonic frequencies:
15-34
2π
 1000 π 
3
Hn
=−
π 
2π 
3 



1 + j n  1 + j n
3 
3 

From problem 15.4-4, the Fourier series of the input voltage is
jn
1 ∞ 18 
2π 
 nπ  
 1000 π
vb ( t ) = − + ∑ 2 2 1 − cos 
t−n

  cos  n
3 
2 n =1 n π 
3
 3 

Using superposition, the Fourier series of the output voltage is
 1000 π 
18 × H  n

3 
1
2π
 1000 π
 nπ  
 1000 π  

1 − cos 
+ ∠H  n
vb ( t ) = − + ∑
t−n
  cos  n


2
2
2 n =1
3
3
3  
n π
 3 



∞
P15.8-3
H (ω ) =
When ω = 0 (dc)
When ω = 100 rad/s
−5 = −
Vo (ω )
Ho
=
Vi (ω ) 1 + j ω
p
R
( 2 ) ⇒ R = 25 kΩ
104
135° = ∠H (ω ) = 180° − tan −1 (ω C R ) ⇒ tan ( 45° ) = (100 ) C ( 25000 )
⇒ C = 0.4 µF
25000
104
= 3.032
c4 = ( 5 ) H ( 400 ) = ( 5 )
1 + j ( 400 ) ( 0.4 ×10−6 ) ( 25000 )
θ 4 = 45° + ∠H ( 400 ) = 45° + 180 − tan −1 ( 400 × 0.4 × 10−6 × 25000 ) = 149°
15-35
P15.8-4
When ω = 0 (dc)
5 = H o ( 2 ) ⇒ H o = 2.5 V/V
When ω = 25 rad/s
ω 
25
−45° = ∠H (ω ) = − tan −1   ⇒ tan ( 45° ) =
⇒
p
 p
p = 25 rad/s
c4 = ( 5 ) H (100 ) = ( 5 )
2.5
= 3.03
100
1+ j
25
 100 
θ 4 = 45° + ∠H (100 ) = 45° − tan −1 
 = −31°
 25 
P15.8-5
When ω = 0 (dc)
R2
R1 + R 2
When ω = 1000 rad/s
H (ω ) =
=
R2
Vo (ω )
=
Vi (ω ) R1 + R 2 + jω C R1 R 2
3.75
 2.25 
 2.25 
⇒ R1 = 
 R2 = 
 (500) = 300 Ω
6
 3.75 
 3.75 

R1 R 2 
 ( 300 )( 500 ) 
−20.5° = ∠H (ω ) = − tan −1  ω C
⇒ tan ( 20.5° ) = (1000 ) C 



800
R1 + R 2 



⇒ C = 2 µF


500
 ( 5∠45° ) = 2.076∠ − 3.4
c3∠θ 3 = 
 800 + j ( 3000 ) ( 2 ×10−6 ) ( 500 )( 300 ) 


15-36
P15.8-6
Rather than find the Fourier Series of v(t ) directly, consider the signal vˆ(t ) shown above.
These two signals are related by
v(t ) = vˆ (t − 1) − 6
since v(t ) is delayed by 1 ms and shifted down by 6 V.
The Fourier series of vˆ (t ) is obtained as follows:
2π radians
π
=
rad/ms
4 ms
2
aˆ n = 0 because the average value of vˆ (t ) = 0
1 4
 π 
bˆn = ∫0 ( 6−3t ) sin  n t  dt because vˆ (t ) is an odd function.
2
 2 
3 4
4
 π 
 π 
= 3∫0 sin  n t  dt − ∫0 t sin  n t  dt
2
 2 
 2 
T = 4 ms ⇒ ω 0 =
 π 
− cos  n t 
 2 
=3



4




3
1
 π   nπ   π 
−
 2 2  sin  n t  −  t  cos n t  
π
2  n π   2   2   2  
0
n
 4 
2
 0
12
6
6
=
−1+ cos( 2nπ ) ) − 2 2 ( sin ( 2nπ ) − 0 ) − ( 2nπ − cos( 2π ) − 0 ) =
(
nπ
nπ
nπ
4
(
)
π
12
sin n t
n =1 nπ
2
vˆ (t ) = ∑
∞
Finally,
The Fourier series of v (t ) is obtained from the Fourier series of vˆ (t ) as follows:
∞ 12
π
π
12
 π
sin n ( t −1) = − 6 + ∑
sin  n t − n 
n =1 nπ
2
2
 2
n =1 nπ
v(t ) = − 6 + ∑
∞
where t is in ms. Equivalently,
15-37
v(t ) = − 6 +
∞ 1
π
 π 3
∑ sin  n 10 t − n 
π n =1 n  2
2
12
where t is in s.
R
s
L
Next, the transfer function of the circuit is H ( s ) =
=
.
1
1
2 R
+ Ls + R
s + s+
Cs
L
LC
R
jω
104 jω
L
The network function of the circuit is H (ω ) =
=
.
8
2
4
R
 1
−
+
ω
j
ω
10
10
2
(
)
− ω  + jω

L
 LC

We see that H(0) = 0 and
R
j 20nπ
 π

H ( n ω 0 ) = H  n 103  =
=
2 2
 2
 ( 400 − n π ) + j 20nπ
Finally,
v0 ( t ) =
π
12
∞
∑
n =1
( 400−n π )
1
2
2 2
+ 400n 2π 2
e
20 nπ 

j  90 − tan −1

400 − n 2π 2 

 π
π
 20nπ  
sin  n 103 t − n + 90° − tan −1 
2 2 
2
 400 − n π  
 2
n
( 400−n π )
2
2 2
+ 400n 2π 2
15-38
P15.8-7
Rather than find the Fourier Series of v(t ) directly, consider the signal vˆ(t ) shown below.
These two signals are related by v(t ) = vˆ ( t − 2 ) − 1
Let's calculate the Fourier Series of vˆ(t ), taking advantage of its symmetry.
2π rad π
= rad ms
6 ms
3
3.2
1
ao = average value of vˆ(t ) = 2 = V
6
2
bn = 0 because vˆ(t ) is an even function
T = 6 ms ⇒ ω 0 =
π
2 1

an = 2  ∫0 ( 3−3t ) cos n t dt 
3
6

π
π
an = 2∫0 cos n t dt − 2∫0 t cos n t dt
3
3
1
1

π
 sin n
3− 1
= 2
π
n 2π 2
 n
3
9

=

π
π 

π
 cos n t + n t sin n t  
3
3
3 


0
1
6
18 
π  18 
π  6
π
π 
sin n −  2 2  cos n −1 +
sin n  = − 2 2  cos n −1
3 n π 
3  nπ
3
3 
nπ
nπ 
so
vˆ(t ) =
1 ∞
18 
π
π
1− cos n  cos n t
+ ∑
2 2 
2 n =1 n π 
3
3
v ( t ) = vˆ ( t − 2 ) − 1 = −
nπ
1 ∞
18 
1− cos
+ ∑
2 2 
2 n =1 n π 
3
2π 

 π
 cos  n t − n 
3 

 3
15-39
where t is in ms. Equivalently,
nπ
1 ∞ 18 
v(t ) = − + ∑ 2 2 1− cos
2 n =1 n π 
3
2π 

 π 3
 cos  n 10 t − n 
3 

 3
where t is in s.
The network function of the circuit is:
R2
−
jω C1 R2
1+ jω C2 R2
=
H (ω ) =
1
(1+ jω R1C1 )(1+ jω R2C2 )
R1 +
jω C1
Evaluate the network function at the harmonic frequencies of the input to get.
− jn
π
 π

3
H ( nω 0 ) = H  n 103  =
 3
 1+ jn π 1+ jn 2π 



3 
3 

The gain and phase shift are
H( n ω0 )
∠H( n ω 0 )
nπ

181−cos
3

The output voltage is
v0 ( t ) = ∑
∞
n =1
π
nπ
3
=
2 2
2 2
 n π  4n π 
9 + n π )( 9+ 4n 2π 2 )
(
1
1
+
+



9 
9 

2π 
π

= −90° −  tan −1 n + tan −1 n 
3
3 

=
n
2
2
2π
2π 
π

 π 3
− 90° − tan −1 n − tan −1 n 
 cos  n 10 t − n
3
3
3 

 3
2 2
2 2
2 2
n π ( 9+ n π )( 9 + 4n π )
At t = 4 ms =0.004 s
v0 (.004 )
2π
nπ 

 4π
°
−1 π
−1 2π 
181− cos
 cos  n − n − 90 − tan n − tan n 
3 
3
3
3 
 3
=∑ 
2
2
2
2
2
2
n =1
n π ( 9 + n π )( 9 + 4n π )
∞
15-40
Section 15.9 The Fourier Transform
P15.9-1
Let g ( t ) = e − at u ( t ) − e at u ( −t ) . Notice that f ( t ) = lim g ( t ) . Next
G (ω ) = ∫ e e
∞
− at − jω t
0
dt − ∫ e e
Finally F (ω ) = lim G (ω ) = lim
a →0
a →0
P15.9-2
F (ω ) = ∫ Ae u ( t ) e
∞
−∞
− at
− jω t
0
−∞
a →0
at − jω t
−2 jω
2
=
2
2
a +ω
jω
dt = ∫ Ae e
∞
0
− at − jω t
∞
e− ( a + jω )t
e( a − jω ) t
dt =
−
− ( a + jω ) 0 ( a − jω ) −∞
0

 
 −2 jω
1
1
=  0 −
− 0  = 2
 − 
2
− ( a + jω )   ( a − jω )

 a +ω
∞
Ae − ( a + jω ) t
A
A
= 0−
=
dt =
− ( a + jω ) 0
− ( a + jω ) a + jω
P15.9-3
First notice that
2
 AT   T  2  ωT   − AT  2  ωT 
Then, from line 6 of Table 15.10-2: F { f1( t )} =  −
=
Sa
Sa 
 



 2  2 
 4   4 
 4 
AT 2 2  ωT 
d

From line 7 of Table 15.10-2: F { f ( t )} = F  f1( t )  = jω F { f1( t )} = − jω
Sa 

4
 dt

 4 
 ωT 
sin 2 

2
AT
4  4 A 2  ωT 

sin 
=
This can be written as: F { f ( t )} = − jω

2
jω
4
 4 
 ωT 


 4 
15-41
{(
P15.9-4
First notice that: F −1 δ ω − ω
{
}
1
1
− jω t
=
δ (ω − ω ) e
dω =
e
0 )} 2π ∫
0
2π
∞
−∞
− jω t
0
Therefore F e − jω0t = 2πδ (ω − ω 0 ) . Next, 10 cos 50t = 5 e j 50t + 5 e − j 50t .
Therefore F {10 cos 50t} = F {5 e j 50t } + F {5 e − j 50t } = 10πδ (ω − 50) + 10πδ (ω + 50) .
P15.9-5
F (ω ) = −2 ∫ e
2
1
P15.9-6
− jω t
−2e − jω t
dt =
− jω
F (ω ) = ∫
1
2 − j 2ω − jω
2
e
−e ) =
( ( cos 2ω − j sin 2ω ) − ( cos ω − j sin ω ) )
(
jω
jω
=
ω
2j
( cos ω − cos 2ω ) + ( sin ω − sin 2ω )
ω
2

1 
A − jω t
A  e − jω t
A  e − jω B
t e dt = 
jω B − 1) − 2 
− jω t − 1)  = 
2 (
2 (
B
B  ( − jω )
ω 
 0 B  −ω
F (ω ) = ∫ e
2
−2
P15.12-1
=
B
B
0
P15.9-7
2
− jω t
dt − ∫ e
1
−1
− jω t
e − jω t
dt =
− jω
e − jω t
−
− jω
−2
2
1
−1
=
A  − Be − jω B e− jω B 1 
+ 2 − 2
ω
ω 
B  jω
=
1
1
e j 2ω − e− j 2ω ) −
(
( e jω − e− jω )
jω
jω
=
ω
2
( sin 2ω − sin ω )
is ( t ) = 40 signum ( t )
 2  80
I s (ω ) = 40 
=
 jω  jω
I (ω )
1
=
H (ω ) =
I s (ω ) 4 + jω
I (ω ) = H (ω ) I s (ω ) =
1
80 20
20
×
=
−
4 + jω jω jω 4 + jω
∴ i ( t ) = 10 signum ( t ) − 20 e −4t u ( t )
15-42
P15.12-2
is ( t ) = 100 cos 3t A
I s (ω ) = 100π δ (ω − 3) + δ (ω + 3) 
I (ω )
1
=
H (ω ) =
I s (ω ) 4 + jω
 δ (ω − 3 ) + δ ( ω + 3 ) 
I (ω ) = 100π 

4 + jω


 e − j 3t
e j 3t 
100π ∞  δ (ω − 3) + δ (ω + 3)  jω t
+
i (t ) =
e
d
50
=
ω


 4 − j3 4 + j3 
2π ∫−∞ 
4 + jω



= 10 e
− j ( 3t −36.9 )
+e
j ( 3t −36.9 )
= 10 cos ( 3t − 36.9 )
P15.12-3


v ( t ) = 10 cos 2t
V (ω ) = 10π δ (ω + 2 ) + δ (ω − 2 ) 
1
Y (ω ) =
2 + jω
I (ω ) = Y (ω ) V (ω )
=
i (t ) =
10π δ (ω + 2 ) + δ (ω − 2 ) 
2 + jω
10π
2π
 δ (ω + 2 ) + δ (ω − 2 )  jω t
 e − j 2t
e j 2t 
5
=
+
ω
e
d



∫−∞ 
2 + jω
 2 − j2 2 + j2 

∞
= 5 e
− j ( 2 t − 45 )
+e
j ( 2 t − 45)
 = 5 cos ( 2t − 45 ) A

15-43
P15.12-4
v ( t ) = et u ( −t ) + u ( t )
F {e u ( −t )} = ∫ e u ( −t ) e
∞
t
−∞
t
F {u ( t )} = πδ (ω ) +
∴V (ω ) =
− jω t
dt = ∫ e e
0
−∞
t − jω t
e( )
dt =
1 − jω
1− jω t
0
−∞
=
1
1 − jω
1
jω
1
1
+ πδ (ω ) +
jω
1 − jω
1
 1  1 
 

1
1
2 + jω
 2   jω 
, H (ω ) =
=
=
1 1
1
2 + jω
3 + jω
1+
+
2 jω
2 + jω
1
1
1
−
πδ (ω )
1  1
1 
Vo (ω ) =
+ πδ (ω ) +
= 12 + 4 + 3 +


3 + jω 1 − jω
jω  3 + jω 1 − jω jω 3 + jω
πδ (ω )  1 ∞ πδ (ω ) − jω t
1
e dω =
F −1 
=
∫
−∞ 3 + jω
6
 3 + jω  2π
∴ vo ( t ) = −
P15.12-5
1 −3t
1
1
1
e u ( t ) + et u ( −t ) + signum ( t ) +
12
4
3
6
vs ( t ) = 15e−5t u ( t ) V ⇒ V (ω ) =
Ws = ∫
∞
−∞
(15e u ( t ) ) dt = ∫ (15e ) dt = 22.5
−5t
2
∞
−5t 2
0
1
1
jω C
H (ω ) =
= RC
1
1
R+
+ jω
jω C RC
C =10 µ F. Try R =10 kΩ. Then
10
15
Vo (ω ) =
×
10 + jω 5 + jω
Wo =
π
1
15
5 + jω
∫
∞
0
 10
15 
1
×

 dω =
π
 10 + jω 5 + jω 
2
∫
∞
0
J
300 
 300
dω = 15 J
−

2
2 
 25 + ω 100 + ω 
2
15-44
P15.12-6
H (ω ) =
4
4 + jω

8  
8  − jω
Vs (ω ) = F {8u ( t ) − 8u ( t − 1)} =  8πδ (ω ) +
 −  8πδ (ω ) +
e
jω  
jω 

8
Vs (ω ) =
(1 − e− jω ) since δ (ω ) e− jω = δ (ω )
jω
Vo (ω ) =
Next use
 8
4
8
8   8
8  − jω
1 − e− jω ) = 
×
−
−
(
−
e
4 + jω jω
 jω 4 + jω   jω 4 + jω 
1
1
=
+ πδ (ω ) − πδ (ω ) to write
jω jω
  1


8    1
8  − jω
Vo (ω ) =  8 
+ πδ (ω )  − πδ (ω ) −
+ πδ (ω )  − πδ (ω ) −
 − 8
e
4 + jω    jω
4 + jω 


  jω
  1


8    1
8  − jω
= 8
+ πδ (ω )  −
+ πδ (ω )  −
 − 8
e
 4 + jω    jω
 4 + jω 
  jω
(
vo ( t ) = 8u ( t ) − 8e −4t u ( t ) − 8u ( t − 1) − 8e
= 8(1 − e −4t )u ( t ) − 8(1 − e
−4( t −1)
−4( t −1)
u ( t − 1)
)u ( t − 1) V
)
15-45
PSpice Problems
SP 15-1
Vin
R1
1
1
.tran
.four
.probe
.end
0
0
pulse
1
(25
5
0
0
0
4
5)
0.01 5
0.2 v(1)
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)
DC COMPONENT =
HARMONIC
NO
1
2
3
4
5
6
7
8
9
8.960000E+00
FREQUENCY
(HZ)
2.000E-01
4.000E-01
6.000E-01
8.000E-01
1.000E+00
1.200E+00
1.400E+00
1.600E+00
1.800E+00
FOURIER
COMPONENT
NORMALIZED
COMPONENT
7.419E+00
6.030E+00
4.061E+00
1.935E+00
8.000E-02
1.182E+00
1.704E+00
1.537E+00
8.954E−01
1.000E+00
8.127E-01
5.473E-01
2.609E-01
1.078E-02
1.593E-01
2.297E-01
2.072E-01
1.207E-01
PHASE
(DEG)
1.253E+02
1.606E+02
-1.642E+02
-1.289E+02
-9.360E+01
1.217E+02
1.570E+02
-1.678E+02
-1.325E+02
NORMALIZED
PHASE(DEG)
0.000E+00
3.528E+01
-2.894E+02
-2.542E+02
-2.189E+02
-3.600E+00
3.168E+01
-2.930E+02
-2.578E+02
SP 15-2
Vin
R1
1
1
.tran
.four
.probe
.end
0
0
pulse
1
(1 -1 -0.5 1 0 0 1)
0.1
1
1 v(1)
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)
DC COMPONENT =
HARMONIC
NO
1
2
3
4
5
6
7
8
9
1.299437E-02
FREQUENCY
(HZ)
1.000E+00
2.000E+00
3.000E+00
4.000E+00
5.000E+00
6.000E+00
7.000E+00
8.000E+00
9.000E+00
FOURIER
COMPONENT
NORMALIZED
COMPONENT
6.364E-01
3.180E-01
2.117E-01
1.585E-01
1.264E-01
1.051E-01
8.972E-02
7.817E-02
6.916E-02
1.000E+00
4.996E-01
3.326E-01
2.490E-01
1.987E-01
1.651E-01
1.410E-01
1.228E-01
1.087E-01
PHASE
(DEG)
-1.777E+02
4.679E+00
-1.730E+02
9.366E+00
-1.683E+02
1.407E+01
-1.636E+02
1.880E+01
-1.588E+02
NORMALIZED
PHASE (DEG)
0.000E+00
1.823E+02
4.682E+00
1.870E+02
9.376E+00
1.917E+02
1.409E+01
1.965E+02
1.883E+01
Verification Problems
15-46
VP 15-1
t
f (t ) = 2 + cos ⇒ a0 = 2 , a1 = 1 and all other coefficients are zero.
2
The computer printout is correct.
VP 15-2
Table 15.4-2 shows that the average value of a full wave rectified sinewave is
2A
2(400)
= 255.
where A is the amplitude of the sinewave. In this case a0 =
π
π
Unfortunately the report says, "half-wave rectified." The report is not correct.
Design Problems
DP 15-1
For sinusoidal analysis, shift horizontal axis to average, which is 6 V.
Now we have an odd function so an = 0
T = π s , ω 0 = 2π / π = 2 rad/s
2×2 T / 2
f (t ) sin nω 0t dt
T ∫0
Need third harmonic :
4 π /2
4
π /2
b3 = ∫0 sin 6t dt = − cos 6t 0 = 0.424
6π
T
v1 ( t ) = 0.424sin ( 6t ) =0.424 cos(6t −90°) V ⇒ V1 (ω ) = 0.424∠−90°
bn =
−j −j
=
for third harmonic
ω C 6C
16
∴ transfer function is H (3ω 0 ) =
j
16−
6C
Zc =
V2 (ω ) = H ( 3ω 0 ) V1 (ω ) =
( H(3ω
) ∠H (3ω 0 ) ) (0.424∠−90° )
Choose V2 (ω ) = 1.36 ⇒ so H ( 3ω 0 ) = 3.2
This requires C =
0
1
16
F. Then H ( 3ω 0 ) =
= 3.2∠64.9°
205
16− j 34
∴ third harmonic of v2 ( t ) = 1.36sin(6t + 64.9° ) V
15-47
DP 15-2
Refer to Table 15.4-2.
In our case:
2A N 4A  1 
− ∑
v (t ) =

 cos(2nω0t )
s
π n = 1 π  4n 2 − 1 
360 N 640  1 
− ∑
v (t ) =

 cos(2n377t )
s
π n = 1 π  4n 2 − 1 
Let vs (t ) = vs0 + ∑ vsn (t ) and v0 (t ) = vo0 + ∑ von (t )
N
(
N
) ≤ 0.04 ⋅ v
n =1
We require ripple ≤ 0.04 ⋅ dc output
max ∑ von (t )
N
n =1
o0
⇒
n =1
vo1 (t ) ≤ 0.04 vo0
but vo0 = vs0 because the inductor acts like a short at dc.


R
Next, using the network function of the circuit gives Von = 
 Vsn .
 R + jω 0 n L 
For n=1:
Vo1 =
R
1
640
1
 640 
so V01 =
Vs1 =
Vs1 , but Vs1 =


R + jω 0 L
1+ j 377 L
1+ j 377 L  3π 
π (3)
We require Vo1 ≤ 0.04 vo0 and vo0 = vs0 =
π
360
. Then
1
1+ (377) 2 L 2
⋅
640
 360 
≤ 0.04 

3π
 π 
Solving for L yields L > 1.54 mH
15-48
DP 15-3
From Table 15.5-1, the Fourier series can represent the input to the circuit as:
vs ( t ) =
π
1
+
∞
j jω 0 t j jω 0 t
1
+ e
+ ∑
e
e j nω 0 t
2
4
4
even n = 2 π (1 − n )
The transfer function of the circuit is calculated as Vo1 =
Zp
R
Vs1 where Z p =
Z L +Z p
1+ jω RC
So
1
LC
Vo
= =
1
1
Vs
+
( jω ) 2 + ( jω )
RC LC
The gain at dc, ω = 0 , is 1 so
1
v =v =
o0
For n = 1
Vo1 =
1
1
1
vo0 =
vs0 =
20
20
20 π
s0
⇒
π
=
1/ LC
 ω   1 
 +

 RC   LC 
ω 4 +
2
2
1
20 π
We are given ω =800π and R =75 kΩ. Choosing L =0.1 mH yields C =0.1 F
15-49
Chapter 16: Filter Circuits
Exercises
Ex. 16.3-1
Tn ( s ) =
1
s +1
 s 
T ( s ) = Tn 
 =
 1250 
1
1250
=
s
s +1250
+1
1250
Problems
Section 16.3: Filters
P16.3-1
Equation 16-3.2 and Table 16-3.2 provide a third-order Butterworth low-pass filter having a
cutoff frequency equal to 1 rad/s.
1
H n (s) =
( s +1)( s 2 + s +1)
Frequency scaling so that ω c = 2π 100=628 rad/s :
H L ( s) =
1
2

s
 s
  s 
+
+1
1

 
 +
 628   628  628 
=
6283
247673152
=
2
2
( s + 628)( s + 628s + 628 ) ( s + 628)( s 2 + 628s + 394384)
P16.3-2
Equation 16-3.2 and Table 16-3.2 provide a third-order Butterworth low-pass filter having a
cutoff frequency equal to 1 rad/s and a dc gain equal to 1.
H n (s) =
1
( s + 1)( s 2 + s + 1)
Multiplying by 5 to change the dc gain to 5 and frequency scaling to change the cutoff
frequency to ωc = 100 rad/s:
H L ( s) =
 s

+1

 100 
5
  s 2 s

+1
 
 +

  100  100 
=
5⋅1003
5000000
=
2
2
( s +100)( s +100s +100 ) ( s +100)( s 2 +100 s +10000)
16-1
P16.3-3
Use Table 16-3.2 to obtain the transfer function of a third-order Butterworth high-pass filter
having a cutoff frequency equal to 1 rad/s and a dc gain equal to 5.
5 s3
H n ( s) =
( s +1) ( s 2 + s +1)
Frequency scaling to change the cutoff frequency to ω c = 100 rad/s
 s 
5

5⋅s 3
5⋅s 3
 100 
=
=
H H ( s) =
2
 ( s +100)( s 2 +100 s +1002 ) ( s +100)( s 2 +100s +10000)
s
 s
  s 
+1  
+1

 +
 100    100  100 
3
P16.3-4
Use Table 16-3.2 to obtain the transfer function of a fourth-order Butterworth high-pass filter
having a cutoff frequency equal to 1 rad/s and a dc gain equal to 5.
H
n
(s) =
5⋅s 4
( s 2 +0.765s +1) ( s 2 +1.848s +1)
Frequency scaling can be used to adjust the cutoff frequency 500 hertz = 3142 rad/s:
 s 
5⋅

 3142 
HH (s) =
2
  s 2
 s    s 
 s  
 
 + 0.765
 +1  
 +1.848
 +1
 3142     3142 
 3142  
  3142 
4
=
5⋅s 4
( s 2 + 2403.6s +31422 ) ( s 2 +5806.4s +31422 )
16-2
P16.3-5
First, obtain the transfer function of a second-order Butterworth low-pass filter having a dc gain
equal to 2 and a cutoff frequency equal to 2000 rad/s:
HL (s) =
2
 s 
 s 

 +1.414
 +1
 2000 
 2000 
2
=
8000000
s + 2828s + 4000000
2
Next, obtain the transfer function of a second-order Butterworth high-pass filter having a
passband gain equal to 2 and a cutoff frequency equal to 100 rad/s:
HH (s)
 s 
2⋅

2⋅s 2
 100 
=
=
2
s 2 +141.4 s +10000
 s 
 s 

 +1.414
 +1
 100 
 100 
2
Finally, the transfer function of the bandpass filter is
HB (s) = HL (s)⋅ H
P16.3-6
s
H( )
=
16000000⋅s 2
( s 2 +141.4s +10000 ) ( s 2 + 2828s + 4000000 )
250


s


250000 s 2
1
HB (s) = 4
=

2
2
250
 s2 +
s + 2502  ( s + 250 s + 62500 )
1


2
P16.3-7
First, obtain the transfer function of a second-order Butterworth high-pass filter having a dc gain
equal to 2 and a cutoff frequency equal to 2000 rad/s:
 s 
2

2 s2
 2000 
HL (s) =
=
2
s 2 + 2828 s + 4000000
 s 
 s 
1.414
1
+
+




 2000 
 2000 
2
Next, obtain the transfer function of a second-order Butterworth low-pass filter having a passband gain equal to 2 and a cutoff frequency equal to 100 rad/s:
16-3
HH (s) =
2
 s 
 s 

 +1.414 
 +1
 100 
 100 
2
=
20000
s +141.4 s +10000
2
Finally, the transfer function of the band-stop filter is
HN (s) = HL (s) + HH (s) =
2s 2 ( s 2 +141.4 s +10000 )+ 20000( s 2 + 2828s + 4000000 )
(s
2
+141.4s +10000 )( s 2 + 2828s + 4000000 )
2s 4 + 282.8s 3 + 40000s 2 + 56560000s +8⋅1010
= 2
( s +141.4s +10000 )( s 2 + 2828s + 4000000 )
P16.3-8
250


2
2
s
s
4
62500
+


(
)
1
HN (s) = 4 − 4
 = 2
2
2 250
2
s +
s + 250  ( s + 250 s + 62500 )
1


2
P16.3-9




2502
4⋅2504
HL (s) = 4
=

250
 s2 +
s + 2502 
s 2 + 250 s + 62500


1
2
P16.3-10
(
)
2




s2
4⋅s 4
HH (s) = 4
=

2
2
250
 s2 +
s + 2502  ( s + 250 s + 62500 )
1


2
16-4
Section 16.4: Second-Order Filters
P16.4-1
The transfer function is
so
s
V0 ( s )
RC
=
T (s) =
Vs ( s ) s 2 + s + 1
RC LC
K = 1 , ω 02 =
Pick C =1 µ F. Then L =
Cω
1
2
0
ω
1
1
C
and
= 0 ⇒ Q = RCω 0 = R
LC
RC Q
L
= 1 H and R = Q
L
= 1000 Ω
C
P16.4-2
The transfer function is
1
I 0 ( s)
LC
T (s) =
=
1
s
I s (s)
s2 +
+
RC LC
so
K = 1 , ω 02 =
Pick C = 1µ F then L =
Cω
1
2
0
ω
1
1
C
= 0 ⇒ Q = RCω 0 = R
and
LC
RC Q
L
= 25 H and R = Q
L
= 3535 Ω
C
16-5
P16.4-3
The transfer function is
−
1
R1 R C 2
T (s) =
1  R
1
s2 +
 2+  s + 2 2
R C  R1  R C
Pick C = 0.01 µ F , then
1
= ω 0 = 2000 ⇒ R = 50000 = 50 k Ω
RC
ω0
R
R
1 
=
= 8333 = 8.33 k Ω
 2 +  ⇒ R1 =
Q RC 
R1 
Q−2
P16.4-4
Pick C = 0.02 µ F. Then R1 = 40 kΩ, R2 = 400 kΩ and R 3 =3.252 kΩ.
P16.4-5
Pick C1 = C2 = C = 1 µ F . Then
and
In this case R 2 = R1
106
=ω 0
R1 R2
ω
1
= 0 ⇒ Q=
R1C
Q
R1
R2
⇒ R2 =
R1
Q2
106
and R1 =
= 1000 = 1 kΩ
1000
16-6
P16.4-6
The node equations are
V0 ( s ) =
R2
R2 +
1
C 2s
Va ( s )
V0 ( s ) −Va ( s )
− C1s (Va ( s ) −Vi ( s ) ) = 0
R1
The transfer function is:
V (s)
=
T (s) = 0
Vi ( s ) s 2 +
Pick C1 = C2 = C = 1 µ F. Then
In this case R1 = R 2 = R and
s2
1
s
+
R 2 C 2 R1 R 2 C 1 C 2
ω
1
1
= ω and
= 0 ⇒Q=
0
R 2C Q
C R1 R 2
1
= ω 0 ⇒ R = 1000 Ω .
CR
R2
R1
⇒ R1Q 2 = R 2 .
16-7
P16.4-7
1
Cs
1
V ( s)
LC
=
=
T (s) = 0
R
1
Vi ( s ) L s + R + 1
s2 + s+
Cs
L LC
When R = 25 Ω, L = 10−2 H and C = 4 × 10−6 F, then the transfer function is
T (s) =
25×106
s 2 + 2500 s + 25×106
so
ω old = 25×106 = 5000
and
kf =
ω new 250
=
= 0.05
ω old 5000
The scaled circuit is
P16.4-8
The transfer function of this circuit is
16-8
T (s) =
1+ R 2 C 2 s
V0 ( s )
=−
=−
1
Vi ( s )
 1
1 
1
R1 +
+
s 2 +
s+

C 1s
 R1 C 1 R 2 C 2  R1 R 2 C 1 C 2


R2
1
s
R1 C 2
100 µ F
500 µ F
= 0.1 µ F and
= 0.5 µ F.
1000
1000
Before scaling ( R1 = 20 Ω, C 1 =100 µ F, R 2 = 10 Ω and C2 = 500 µ F )
Pick km = 1000 so that the scaled capacitances will be
T (s) =
−100s
s + 700s+105
T (s) =
−100s
s + 700 s +105
2
After scaling ( R1 = 20000 Ω = 20 kΩ, C 1 = 0.1 µ F, R 2 =10000 Ω = 10 kΩ, C 2 = 0.5 µ F )
2
P16.4-9
This is the frequency response of a bandpass filter, so
T (s) =
K
s2 +
ω0
ω0
Q
Q
s
s +ω 0
2
From peak of the frequency response
ω 0 = 2 p × 10 × 106 = 62.8 ×106 rad/s and k=10 dB = 3.16
Next
ω0
Q
= BW = (10.1×106 − 9.9 × 106 ) 2π = (0.2 × 106 )2π = 1.26 × 106 rad/s
So the transfer function is
T (s) =
3.16(1.26)106 s
(3.98)106 s
=
s 2 + (1.26)106 s + 62.82.1012
s 2 + (1.26)106 s + 3.944.1015
16-9
P16.4-10
H (ω ) =
(a)
Z
V0 (ω )
= − 2
Vs (ω )
Z1
∴ H (ω ) = −
(b)
(c)
ω1 =
where Z 1 = R1 −
jω R 2C 1
 jω 
1+

 ω1 
 jω 
1+

 ω2 


j
ω C1
R2×
1
jω C 2
and Z 2 =
1
R2 +
jω C 2
where ω1 =
1
1
, ω2 =
R1 C 1
R2 C 2
1
1
, ω2 =
R1 C 1
R2 C 2
H (ω ) = −
jω R 2C 1
 jω 
 0+
 (1+0 )
 ω1 
=
ω R 2C 1
R2
= ω 1 R 2C 1 =
ω
R1
ω1
P16.4-11
Voltage division:
1
Cs
V0 ( s ) =
V ( s ) , ⇒ V1 ( s ) = (1 + s R C )V0 ( s )
1 1
R+
Cs
V1 − Vs
V −V
+ 1 0 + (V1 −V0 ) n C s = 0
mR
R
Combining these equations gives:
KCL:
 1

sC 2
V
V0 
+s C+
+ s n R C2  = s
m
mR
m R

16-10
Therefore
where ω 0 =
H (ω ) =
V0 (ω )
1
=
=
Vs (ω )
1+ s ( m +1) R C + n m R 2 C 2 s 2
mn
1
and Q =
m +1
m n RC
1
ω 
ω
1−
+j
 ω 0 
Qω 0


2
P16.4-12
R2
1
1
s
−
C 2s
R 2 C 2 s +1
R1 C 2
V0 ( s )
H (s) =
=−
=
=
R1 C 1 s +1
1
VS ( s )
 1
1 
1
R1 +
s 2 +
s+
+

C s
 R1 C 1 R 2 C 2  R1 R 2 C 1 C 2
C1 s
1


R2
where
ω0 =
BW=
1
= 70.7 k rad sec = 2π (11.25 kHz )
R1 R 2 C 1 C 2
ω0
Q
=
1
1
+
= 150 k rad s =2π ( 23.9 kHz )
R1 C 1 R 2 C 2
16-11
P16.4-13
Va −V0

1
= 0
s
−
R1
R2 C 2
V0 ( s)

=
 ⇒ H ( s)=
1
Va
Vs ( s) s 2 + 1 s +

− −C 2 s V0 = 0
R1 C 1 R1 R 2 C 1 C 2

R2
C 1 s (Va −Vs ) +
Comparing this transfer function to the standard form of the transfer function of a second order
bandpass filter gives:
ω0 =
1
= 104 rad sec
R1 R 2 C 1 C 2
BW =
Q=
1
= 103 rad/sec
R1 C 1
ω0
BW
= 10
16-12
P16.4-14
Node equations:
Vc ( s ) −Vs ( s )
V ( s ) −V0 ( s )
+ c
=0
R
R
V ( s ) −Vc ( s )
C
s (Vs ( s ) −V0 ( s ) ) + s
=0
a
R
a C sVc ( s ) +
Solving these equations yields the transfer function:
1
2  1
s 2 + + a 
s+
2
 a  RC
V0 ( s )
RC)
(
=
H (s) =
1
Vs ( s )
2 1
s 2 + 
s+
2
 a  RC (RC)
1
. Pick C = 0.01µ F then R = 1000 Ω . Next at s = j ω 0
RC
2
+a
a2
a
H (ω 0 ) =
= 1+
2
2
a
The specifications require
a2
201 = H (ω 0 ) = 1 +
⇒ a = 20
2
We require 105 =
16-13
P16.4-15
Node equations:
Va Va −V0
+
=0
R2
R1
Va −Vb
=0
R
V −V
C s (Vb −V0 ) + b s = 0
R
C sVa +
 R1  1
 1+  2 2
V0 ( s )
 R2  R C
=
Vs ( s )
 R  1
1
s 2 + 2− 1 
s+ 2 2
 R2  R C
RC


Solving the node equation yields:
ω0 =
1
1
=
= 41.67 k rad sec
3
RC (1.2×10 ) ( 20×10−9 )
16-14
Section 16.5: High-Order Filters
P16.5-1
This filter is designed as a cascade connection of a Sallen-key low-pass filter designed as
described in Table 16.4-2 and a first-order low-pass filter designed as described in Table 16.5-2.
Sallen-Key Low-Pass Filter:
MathCad Spreadsheet (p16_5_1_sklp.mcd)
c
The transfer function is of the form T(s) =----------------- .
s^2 + bs + a
Determine the Filter Specifications:
ω0 := a
Pick a convenient value for the capacitance:
Calculate resistance values:
Calculate the dc gain.
R :=
A = 1.586
a := 628
2
Enter the transfer function coefficitents:
1
C⋅ ω0
Q :=
ω0
b
b := 628
ω0 = 628
Q = 0.707
−6
C := 0.1⋅ 10
A := 3 −
1
Q
R = 1.592 × 10
4
R⋅ ( A − 1) = 9.331 × 10
3
16-15
First-Order Low-Pass Filter:
MathCad Spreadsheet (p16_5_1_1stlp.mcd)
-k
The transfer function is of the form T(s) =-------- .
s+p
Enter the transfer function coefficitents:
p := 628
Pick a convenient value for the capacitance:
Calculate resistance values:
R2 :=
1
C⋅ p
R1 :=
k := 0.5p
−6
C := 0.1⋅ 10
1
C⋅ k
R1 = 3.185 × 10
4
R2 = 1.592 × 10
4
P16.5-2
This filter is designed as a cascade connection of a Sallen-key high-pass filter, designed as
described in Table 16.4-2, and a first-order high-pass filter, designed as described in Table 16.52.
The passband gain of the Sallen key stage is 2 and the passband gain of the first-order stage is
2.5 So the overall passband gain is 2 × 2.5 = 5
Sallen-Key High-Pass Filter:
16-16
MathCad Spreadsheet (p16_5_2_skhp.mcd)
A s^2
The transfer function is of the form T(s) =----------------- .
s^2 + bs + a
Enter the transfer function coefficitents:
a := 10000
ω0 := a
Determine the Filter Specifications:
Q :=
R :=
1
C⋅ ω0
A := 3 −
A=2
Calculate the passband gain.
b
b := 100
ω0 = 100
Q=1
−6
C := 0.1⋅ 10
Pick a convenient value for the capacitance:
Calculate resistance values:
ω0
1
Q
R⋅ ( A − 1) = 1 × 10
R = 1 × 10
5
5
First-Order High-Pass Filter:
MathCad Spreadsheet (p16_5_2_1sthp.mcd)
-ks
The transfer function is of the form T(s) =-------- .
s+p
Enter the transfer function coefficitents:
Pick a convenient value for the capacitance:
Calculate resistance values:
R1 :=
1
C⋅ p
p := 100
k := 2.5
−6
C := 0.1⋅ 10
R2 := k⋅ R1
R1 = 1 × 10
5
R2 = 2.5 × 10
5
16-17
P16.5-3
This filter is designed as a cascade connection of a Sallen-key low-pass filter, a Sallen-key highpass filter and an inverting amplifier.
Sallen-Key Low-Pass Filter:
MathCad Spreadsheet (p16_5_3_sklp.mcd)
c
The transfer function is of the form T(s) =----------------- .
s^2 + bs + a
Enter the transfer function coefficitents:
Determine the Filter Specifications:
ω0 := a
Pick a convenient value for the capacitance:
Calculate resistance values:
Calculate the dc gain.
R :=
A = 1.586
a := 4000000
1
C⋅ ω0
Q :=
ω0
b
b := 2828
ω0 = 2 × 10 Q = 0.707
3
−6
C := 0.1⋅ 10
A := 3 −
1
Q
R = 5 × 10
3
R⋅ ( A − 1) = 2.93 × 10
3
16-18
Sallen-Key High-Pass Filter:
MathCad Spreadsheet (p16_5_3_skhp.mcd)
c s^2
The transfer function is of the form T(s) =----------------- .
s^2 + bs + a
Enter the transfer function coefficitents:
a := 10000
ω0 := a
Determine the Filter Specifications:
Calculate the passband gain.
R :=
1
C⋅ ω0
ω0
b
ω0 = 100
Q = 0.707
−6
C := 0.1⋅ 10
Pick a convenient value for the capacitance:
Calculate resistance values:
Q :=
b := 141.4
A := 3 −
A = 1.586
Amplifier: The required passband gain is
1
Q
R = 1 × 10
5
R⋅ ( A − 1) = 5.86 × 10
4
1.6×106
= 4.00 . An amplifier with a gain equal to
141.4×2828
4.0
= 1.59 is needed to achieve the specified gain.
2.515
16-19
P16.5-4
This filter is designed as the cascade connection of two identical Sallen-key bandpass filters:
Sallen-Key BandPass Filter:
MathCad Spreadsheet (p16_5_4_skbp.mcd)
cs
The transfer function is of the form T(s) =----------------- .
s^2 + bs + a
Enter the transfer function coefficitents:
Determine the Filter Specifications:
ω0 := a
Pick a convenient value for the capacitance:
R :=
Calculate resistance values:
R = 4 × 10
4
1
C⋅ ω0
2⋅ R = 8 × 10
Q :=
ω0
b
b := 250
ω0 = 250
Q=1
−6
C := 0.1⋅ 10
A := 3 −
1
Q
R⋅ ( A − 1) = 4 × 10
4
Calculate the pass-band gain.
a := 62500
4
A⋅Q = 2
16-20
P16.5-5
This filter is designed using this structure:
Sallen-Key Low-Pass Filter:
MathCad Spreadsheet (p16_5_5_sklp.mcd)
c
The transfer function is of the form T(s) =----------------- .
s^2 + bs + a
Enter the transfer function coefficitents:
Determine the Filter Specifications:
ω0 := a
Pick a convenient value for the capacitance:
Calculate resistance values:
Calculate the dc gain.
R :=
A = 1.586
a := 10000
1
C⋅ ω0
Q :=
ω0
b
b := 141.4
ω0 = 100
Q = 0.707
−6
C := 0.1⋅ 10
A := 3 −
1
Q
R = 1 × 10
5
R⋅ ( A − 1) = 5.86 × 10
4
16-21
Sallen-Key High-Pass Filter:
MathCad Spreadsheet (p16_5_5_skhp.mcd)
c s^2
The transfer function is of the form T(s) =----------------- .
s^2 + bs + a
Enter the transfer function coefficitents:
ω0 := a
Determine the Filter Specifications:
Pick a convenient value for the capacitance:
Calculate resistance values:
Calculate the passband gain.
a := 4000000
R :=
1
C⋅ ω0
A = 1.586
Q :=
ω0
b
b := 2828
ω0 = 2 × 10 Q = 0.707
3
−6
C := 0.1⋅ 10
A := 3 −
1
Q
R = 5 × 10
3
R⋅ ( A − 1) = 2.93 × 10
3
Amplifier: The required gain is 2, but both Sallen-Key filters have passband gains equal to 1.586.
2
= 1.26 to make the passband gain of the entire filter equal to 2.
The amplifier has a gain of
1.586
16-22
P16.5-6
This filter is designed as the cascade connection of two identical Sallen-key notch filters.
Sallen-Key Notch Filter:
MathCad Spreadsheet (p16_5_6_skn.mcd)
c(s^2 + a)
The transfer function is of the form T(s) =----------------- .
s^2 + bs + a
Enter the transfer function coefficitents:
Determine the Filter Specifications:
ω0 := a
Pick a convenient value for the capacitance:
Calculate resistance values:
R = 4 × 10
= 2 × 10
2
Calculate the pass-band gain.
4
R
R :=
a := 62500
1
C⋅ ω0
4
Q :=
ω0
b
b := 250
ω0 = 250
−6
C := 0.1⋅ 10
A := 2 −
Q=1
−7
2⋅ C = 2 × 10
1
2⋅ Q
R⋅ ( A − 1) = 2 × 10
4
A = 1.5
Amplifier: The required passband gain is 4. An amplifier having gain equal to
4
= 1.78
(1.5)(1.5)
is needed to achieve the required gain.
16-23
P16.5-7
(a) Voltage division gives:
H a (s) =
(b) Voltage division gives:
H b (s) =
(c) Voltage division gives:
H c (s) =
Doing some algebra:
H c (s) =
V1 ( s )
Vs ( s )
=
V1 ( s )
=
V2 ( s)
V2 ( s )
Vs ( s )
V2 ( s )
Vs ( s )
=
=
=
=
=
(d)
=
R1
R1 +
1
Cs
R1 C s
1 + R1 C s
Ls
R2 + L s
R1 || ( R 2 + L s )
1
+ R1 || ( R 2 + L s )
Cs
R1 × ( R 2 + L s )
R1 + ( R 2 + L s )
×
R1 × ( R 2 + L s )
1
+
C s R1 + ( R 2 + L s )
Ls
R2 + L s
×
Ls
R2 + L s
R1 R 2 C s + R1 L C s 2
R1 R 2 C s + R1 L C s + R1 + R 2 + L s
R1 C s ( R 2 + L s )
2
R1 L C s + ( R1 R 2 C + L ) s + R1 + R 2
2
R1 L C s 2 + ( R1 R 2 C + L ) s + R1 + R 2
×
×
Ls
R2 + L s
Ls
R2 + L s
R1 L C s 2
H c ( s ) ≠ H a ( s ) × H b ( s ) because the R 2 , L s voltage divider loads the
1
, R1 voltage
Cs
divider.
16-24
P16.5-8
P16.5-9
(a)
H (s) =
100
20
×

 
s 
s
s 
s 

1 +
 1 +
 1 +
 1 +
 200 π   20, 000 π   20 π   2000 π 
2000
=


s 
s 
s 
s

1 +
 1 +
 1 +
 1 +
 20 π   200 π  2000 π   20, 000 π 
The transfer function of each stage is
R2 ×
1
Cs
R2
R2
1
1
R 2 ||
R2 +
1+ R2 C s
R1
Cs
Cs
=−
=−
=−
H i (s) = −
R1
R1
R1
1+ R2 C s
The specification that the dc gain is 0 db = 1 requires R 2 = R1 .
The specification of a break frequency of 1000 rad/s requires
Pick C = 0.1 µ F . Then R 2 = 10 kΩ so R1 = 10 kΩ .
(b)
H (ω ) =
−1
1+ j
ω
1000
×
−1
1+ j
ω
1000
⇒
1
= 1000 .
R2 C


1
1
H (10, 000 ) = 
= −40.1 dB
 =
2
 1 + 10  101
2
16-25
PSpice Problems
SP 16-1
16-26
SP 16-2
16-27
SP 16-3
16-28
SP 16-4
16-29
SP 16-5
16-30
SP 16-6
16-31
SP 16-7
Vs
R1
R2
R3
L1
L2
L3
C1
C2
C3
7
7
7
7
6
1
2
5
3
4
0
6
1
2
5
3
4
0
0
0
ac 1
200
100
50
10m
10m
10m
1u
1u
1u
.ac dec 100 100 10k
.probe
.end
SP 16-8
Vs
R1
1
1
0
2
ac
100
1
16-32
C1
R2
C2
2
3
3
Xoa5 3
3
4
4
0.2u
200k
50p
0
4
FGOA
.subckt FGOA 1 2 4
*nodes listed in order - + o
Ri
1
2 500k
E
3 0 1
2 100k
Ro
4
3 1k
.ends FGOA
.ac dec 100 1k 100k
.probe
.end
SP 16-9
Vs
L1
Rw
C2
L2
Rmr
C3
Rt
1
1
2
1
3
4
1
5
0
2
0
3
4
0
5
0
ac 1
2.5m
8
34.82u
0.364m
8
5u
8
.ac dec 100 10 100k
.probe
.end
Bw=4.07k - 493 HZ ∼ 3600HZ
Verification Problems
16-33
VP 16.1
ω 0 = 10000 = 100 rad s and
ω0
Q
= 25 ⇒ Q =
100
= 4 ≠ 5
25
This filter does not satisfy the specifications.
VP 16.2
ω0
ω 0 = 10000 = 100 rad s ,
Q
= 25 ⇒ Q =
100
75
= 4 and k =
= 3
25
25
This filter does satisfy the specifications.
VP 16.3
ω 0 = 400 = 20 rad s ,
ω0
Q
= 25 ⇒ Q =
20
600
= 0.8 and k =
= 1.5
25
400
This filter does satisfy the specifications.
VP 16.4
ω 0 = 625 = 25 rad s ,
ω0
Q
= 62.5 ⇒ Q =
25
750
= 0.4 and k =
= 1.2
62.5
625
This filter does satisfy the specifications.
VP 16.5
ω 0 = 144 = 12 rad/s and
ω0
Q
= 30 ⇒ Q =
12
= 0.4
30
This filter does not satisfy the specifications.
16-34
Design Problems
DP 16.1
−
s
RC
V0 ( s )
=
2
2
V1 ( s )
s+
s+
R3 C R R3 C 2
2π (100.103 ) = ω 0 =
C = 100 pF is specified so R3 =
ω
2
2
and 2π (10.103 ) = BW = 0 =
2
R R3 C
R3 C
Q
(100×10
−12
2
2
= 318 kΩ and R =
= 1.6 kΩ
3
) (2π ×10×10 )
R3C 2ω 02
DP 16.2
16-35
DP 16.3
Choose ω1 = 0.1 , ω 2 = 2 , ω3 = 5 , ω 4 = 100 rad s . The corresponding Bode magnitude plot
is:
H (s) =
( ω)
(1+ sω ) (1+ s ω )
1+ s
2
1
1+ s 

ω3 

2
2
2
2
4
Minimum gain is − 46.2 dB at f min = 0.505 Hz
16-36
Chapter 17- Two-Port and Three Port Networks
Exercises
Ex. 17.4-1
R1 =
R2 =
R3 =
R a Rc
=
25(100)
= 10 Ω
250
R b Rc
=
(125)(125)
= 12.5 Ω
250
R a Rb
=
100(125)
= 50 Ω
250
R a + Rb + Rc
R a + Rb + Rc
R a + Rb + Rc
Ex. 17.5-1
−Y12 = −Y21 =
Y11 + Y12 =
1
42
Y22 +Y21 = 10.5
Z 11 =
Z 22 =
V1
I1
V2
I2
I1 = 0
Z 12 = Z 21 =
Since I =
=
I2 = 0
=
V1
I2
42 (21+10.5)
= 18 Ω
42 +31.5
( )
( )
1
21
1
3
⇒ Y11 =
− −1 =
21
42
42
1
⇒ Y22 =
− − 1 = 1/ 7
21
10.5
1 −1
41
21 
Y = 
 1 1
 − 21 7 
10.5(63)
=9Ω
73.5
I1 = 0
=6Ω
10.5
42(10.5)
I 2 , then V1 =
I2 = 6 I2
73.5
73.5
18
⇒ Z= 
6
6
9 
17-1
Ex. 17.6-1
Y11 =
I1 1
=
V1 6
1
I
Y21 = 2 = − = −.167
V1
6
Y12 =
Y22 =
Ex. 17.7-1
I1
= 0.0567
V2
I2
= 0.944
V2
I 2 = 6 i, V2 = (9 + 1) i = 10 i, V1 = 1i
h 22 =
h12 =
I2
6i
=
= 0.6 S
V2 10 i
V1
i
=
= 0.1
10 i
V2
V1 = 1 i
I1 = i +
10
V1
=
i
9
9
44
V
I2 = 5 i − 1 =
i
9
9
Therefore
h11 =
V1
=
I1
h 21 =
I2
=
I1
( )
( )
( )
i
= 0.9 Ω
10 i
9
44 i
9
= 4.4
10 i
9
17-2
Ex. 17.8-1
−1 
2
 2 1  12 6

4 1
1
15
5
5 5  

Y=
=
−
=
and ∆Y =
S ⇒ Z = 30 

 −1
1
2 
75 50 30
2  3 4 
 10 5 
 10 15
Ex. 17.8-2
Ex. 17.9-1
2/5

 − ( −1/10 )
T=
 1/ 30
 − −1/10
)
 (
1
Ta = 
0
1 
( −1/10 )   4
=
2 /15  1/ 3
−
( −1/10 ) 
−
10 
4 / 3
12 
0
1
1
, Tb = 
and Tc = 


1
1/6 1 
0
3
1
21 
1 12  1 0 
3 12  1 3  3
=
=
Ta Tb Tc = 
T
c





 
0 1 1/ 6 1 
1/ 6 1   0 1 1/ 6 3 / 2 
Problems
Section 17-4: T-to-T1 Transformations
P17.4-1
17-3
P17.4-2
P17.4-3
I2 =
R in =
− z21 I1
z22 + R L
⇒
Ai =
−I2
z21
=
I1
z22 + R L
(forward current gain)
z A I
V1 z11 I1 + z12 I 2
z z
=
= z11 − 12 i 1 = z11 − 12 21
I1
I1
I1
( z22 + R L )
V2 = − I 2 R L = Ai R L I1 and V1 = R in I1 ⇒
Av =
V2 Ai R L
=
V1
R in
∴ Ap = Ai Av = Ai2
(input resistance)
(forward voltage gain)
RL
R in
17-4
P17.4-4
First, simplify the circuit using a ∆-Y transformation:
R eq = R1 ||
R
= 5 || 20 = 4 Ω
3
30 = 18 I −10 I 2
Mesh equations:
50 = 10 I − 20 I 2
30 −10
50 − 20
−100
I=
=
= 0.385 A
18(−20) − (−10)10 −260
Solving for the required current:
P17.4-5
17-5
Section 17-5: Equations of Two-Port Networks
P17-5-1
Z 12 = 6 Ω
Z 11 − Z 12 = 12 Ω ⇒ Z 11 = 18 Ω
Z 22 − Z 21 = 3 Ω ⇒ Z 22 = 9 Ω
Y11 =
Y12 =
Y 22 =
I1
1
=
S
V1 V =0
14
2
I1
V2
I2
V2
V1 = 0
V1 = 0
=
−6 I 2
1
= − S = Υ 21
(6+12) V2
21
=
V2 / 7
1
=
S
V2
7
1
−1 
14
21

Υ=
− 1
1 
 21 7 
P17.5-2
 2− j 4
Z = 
− j 4
− j4 
+ j 2 
17-6
P17.5-3
Y11 =
V1 =
so
Finally
I1
V1
and Y 21 =
V2 = 0
I2
V1
V2 = 0
I1 + I 2
I+I
I
and 1 2 + 2 = bV1
G1
G1
G2
I1 = (G1 − (b −1) G2 ) V1 = −1 V1
and I 2 = (b −1) G2 V1 = 3 V1
Y11 = −1 S and Y21 = 3 S
Next
I1+I 2
V2 =
Y12 =
Y 22 =
and V 2 =
G3
I1
V2
V1 = 0
I2
V2
V1 = 0
−I 2
G2
= −G 2 = − 1 S
= G2 + G3 = 4 S
P17.5-4
Using Fig. 17.5-2 as shown:
−Y12 = −Y21 = 0.1 S or Y12 = Y21 = −0.1 S
Y11 = 0.2 − Y12 = 0.3 S
Y22 = 0.05 − Y21 = 0.15 S
P17.5-5
Y12 = −10 µ S = Y 21
Y11 +Y12 =13.33 µ S
Y11 = 23.33 µ S
Y 22 + Y 21 = 20 µ S
⇒ Y 22 = 30 µ S
17-7
P17.5-6
Z 11 =
Z 21 =
Z 12 =
Z 22 =
V1
I2
V2
I2
I1 = 0
I1 = 0
V1
I1
V2
I1
I2 = 0
I2 = 0
= 3 + j 3 − j 2 = (3+ j ) Ω
=
− j 2 I1
= − j2 Ω
I1
= − j2 Ω
= − j2 Ω
P17.5-7
Z 11− Z 21 = 4 

 ⇒
Z 21 − Z 12 = 1 
s
Z 22 − Z 21 = 2 s ⇒
P17.5-8
Given:
4 s +1
Z 11 = 4 + 1 =
s
s
2 s 2 +1
Z 22 = 2 s + 1 =
s
s
 s +1

−1 
Y=  s


 −1 s +1
Try a π circuit as shown at the right.
Y12 = −1s
s +1
s +1
1
⇒ Y11 +Y12 =
−1 =
s
s
s
Y 22 + Y 21 = ( s + 1) − 1 = s
Y11 =
17-8
P17.5-9
Given:
 s + 2s + 2
 s 2 + s +1
Z= 
 1
 s 2 + s +1
2
Try :
1 
s + s +1 
s 2 +1 
s 2 + s +1 
2
From the circuit, we calculate:
1
( R2 + L s )
L C R1 s 2 + ( R1 R 2 C + L ) s + R1 + R 2
R2 + L s
Cs
z 11 = R1 +
= R1 +
=
1
L C s2 + R2 C s + 1
1 + R2 C s + L C s2
+ R2 + L s
Cs
Comparing to the given z 11 yields:
LC =1

 R1 = 1 Ω

R2 C = 1


 R2 = 1 Ω
L C R1 = 1  ⇒ 
 L =1 H
R1 R 2 C + L = 2 

 C = 1 F
R1 + R 2 = 2 
Then check z 12 , z 21 and z 22 . The are all okay. If they were not, we would have to try a different
circuit structure..
P17.5-10
It is sufficient to require that the input resistance of each section of the circuit is equal to Ro, that
is
Then
Ro =
R (2 R + Ro )
⇒ Ro = R ± 4 R 2 + 4 (2 R 2 ) = R ± 3 R = ( 3 −1) R
3 R + Ro
17-9
Section 17-6: Z and Y Parameters
P17.6-1
i=−
V1
R1
and I 2 = −
(b + R1 )
R1 R 2
V1
 b + R1 + R 2 
I1 = − I 2 − i = 
 V1

R1 R 2


Y11 =
I 2 = − I1
V2 = R 2 I 2
I1
V1
V2 = 0
=
⇒ I2 =
Y22 =
I2
V2
b + R1 + R 2
R1 R 2
and Y21 =
I2
V1
V2 = 0
=−
(b + R1 )
R1 R 2
V2
R2
V1 = 0
=
1
R2
and Y12 =
I1
V2
V1 = 0
=−
1
R2
P17.6-2
 v1 = (1 + 3) i1 = 4 i1
i2 = 0 ⇒ 
v 2 = 3 i1

therefore
z 11 = 4 Ω and z 21 = 3 Ω
 v1 = 3 (α i 2 + i 2 )

i1 = 0 ⇒ 
 v1 = 3 (α i 2 + i 2 ) + 2 i 2
therefore
z 12 = 3 (1 + α ) and z 21 = 5 + 3α
Finally,
 4 3(1+α ) 
Z= 

3 5+3α 
17-10
P17.6-3
Treat the circuit as the parallel connection of two 2-port networks:
The admittance matrix of the entire network
can be obtained as the sum of the admittance
matrices of these two 2-port networks
1 0   2 s − s  1+ 2 s − s 
Y=
+
=
1+ 2 s 
 2 1  − s 2 s   2− s
When i1 ( t ) = u ( t ) :
 2 s +1 s

1
1




V1 ( s )   
V1 ( s ) 
−1  
 s − 2 2 s +1
Y
Y
=
⇒
=
=
s
s

V ( s ) 
 
V2 ( s )  0 
 2 
3 s 2 + 6 s +1
0
 
 
so
V2 ( s ) =
( S − 2)
1  −6 −1.25
7.25 
=  S +
+

2
3
S (3 S + 6 S +1)
S + 1.82 S + 0.184 
1
s
 
0 
Taking the inverse Laplace transform
v2 (t) =
1
 −6−1.25 e −1.82 t + 7.25 e − 0.184 t 
3
t ≥0
17-11
P17.6-4
KVL:
KCL:
1
( i1 − v 1 ) + 2 v 2 + v 2 − v 1 = 0
2
i1 − v 1 = 4 v 1 + 2 v 2
i1 − 5 v 1
v1 2

i
v
z
=
−
⇒
=
= Ω
3
6
1
1
11

i1 9
2
i1 = 3v1 − 6 v 2 

 ⇒ 
i1 = 5 v1 + 2 v2 
i = 5 i1 + 6 v 2 + 2 v ⇒ z = v 2 = − 1 Ω
2
21
1
i1
3
18

KVL:
v 2 = 1 v1 +
KCL:
i2 =
v2
12
1
13
v1 + 5 v1 = v1
2
2
+ 5 v1 = 2 v 2 + 5 v1
1
 13 
Ω
i 2 = 2  v1  + 5 v1 = 18 v1 ⇒ z 12 =
18
2 
and
2 
i 2 = 2 v 2 + 5  v 2  = 2.769 v1 ⇒ z 22 = 0.361 Ω
 13 
17-12
P17.6-5
i1 + i 2 =
KCL:
KVL:
Then
v1
R1
− R 2 i 2 − b v1 + 0 − v1 = 0
i2 = −
so
b +1
v1
R2
y 21 =
i2
v1
and
=−
 1 b +1
R 2 + R1 ( b + 1)
i1 =  +
v1 =
v1

 R1 R 2 
R
R
2
1


b +1
and
R2
y 11 =
i1
v1
KVL:
=
R 2 + R1 ( b + 1)
R1 R 2
R 2 i1 + v 2 = 0 ⇒ i1 = −
1
v2
R2
 1

v 2  + R3 i 2
KCL: v 2 = R 3 ( i1 + i 2 ) = R 3  −
 R2 


Then
y 12 =
i1
v2
=−
1
R2
 R3 
= R3 i 2
and v 2 1 +
 R 2 


⇒
y 22 =
i2
v2
=
1
1
+
R3 R 2
17-13
Section 17-7: Hybrid Transmission Parameters
17.7-1
B=
D=
A=
C=
V1
−I2
I1
−I2
V1
V2
I1
V2
V2 = 0
V2 = 0
I2 =0
I2 =0
=
V1
34
5
= 6.8 Ω since − I 2 =
= V1
5
2+ 4||10 34
=
10+ 4
= 1.4 since
10
=
I2 = −
10
I1
10 + 4
12
10
= 1.2 since V2 =
V1
10
10+ 2
=
1
= 0.1 S
10
V2 = 0
17.7-2
V1 = ( R i + R1 || R 2 ) I 1
so
therefore
h11 =
KVL:
V1
I1
V2 = 0
I2 +
therefore
I
h 21 = 2
I1
V2 = 0
= R i + R1 || R 2 = 600 kΩ
R1
R1 + R 2
= −( A
I1 = −A
Ri
Ro
+
Ri
Ro
R1
I1
R1 + R 2
) = −106
17-14
I 1 = 0 ⇒ vi = 0 ⇒
so
I2 =
therefore
h 22 =
I2
V2
Next,
therefore
I 1 =0
V2
R o ( R1 + R 2 )
=
V1 =
h12 =
R o + R1 + R 2
R o ( R1 + R 2 )
R1
R1 + R 2
V1
V2
A vi = 0
I 1 =0
=
= 10−3
V2
R1
R1 + R 2
=
1
2
P17.7-3
Compare :
V 2 = n V1
to
I 1 =−n I 2
V1 = h11 I 1 + h12V 2
I 2 = h 21 I 1 + h 22V 2
Then h11 = 0, h 22 = 0, h12 = 1 and h 21 = 1
−n
n
P17.7-4
V1 = ( R1 + R 2 || R 3 ) I 1 ⇒ h11 = R1 +
I2 = −
I2 =
V1 =
R2
R 2 +R 3
I 1 ⇒ h 21 = −
V2
⇒ h 22 =
R2
V 2 ⇒ h12 =
R 2 +R 3
R 2 +R 3
R2 R3
R 2 +R 3
R2
R 2 +R 3
1
R 2 +R 3
R2
R 2 +R 3
17-15
P17.7-5
I 2 = 0.1 v and v = 950 I 1
so I 2 = 95 I 1
h11 =
V1
h 21 =
I2
I1
I1
V2 = 0
V2 = 0
= 50 + 950 = 1000 Ω
= 95
I1 = 0 ⇒ v = 0
h12 =
h 22 =
V1
V2
I1 = 0
I2
V2
I1 = 0
=0
= 10−4 S
Section 17-8: Relationships between Two-Port Parameters
P17.8-1
Start with
V1 = h11 I1 + h12 V2
 I = Y11 V1 + Y12 V2
Y parameters:  1
and H parameters: 
 I 2 = h 21 I1 + h 22 V2
 I 2 = Y21 V1 + Y22 V2
Solve the Y parameter equations for V1 and I 2 to put them in the same form as the H parameter
equations.
−Y11 V1 = I1 + Y12 V2
−Y21 V1 + I 2 = Y22 V2
 −Y 0  V   −1 Y12   − I1 
⇒  11   1  = 
Y22   V2 
 −Y21 1  I 2  0
 −Y 0 
∴ H =  11 
 −Y21 1 
−1
 1
−
 −1 Y12   Y11
0
 =  Y
Y
 21

22 
− Y
 11
V   −Y 0
⇒  1  =  11 
 I 2   −Y21 1 

1
0
 −1 Y12   Y11
 
=
Y22   Y21
 0
1
Y

 11
−1
 −1 Y12   − I1 
0
Y22   V2 




Y12Y21 
Y22 −
Y11 
−
Y12
Y11
17-16
Z 12   6
 Z 22
2 
 ∆Z − ∆Z  14 −14 
 =
First ∆Z = (3)(6) − (2)(2) = 14 . Then Y = 
.
3
 − Z 21 Z 11   − 2
 ∆Z ∆Z   14
14 
P17.8-2
Y12 
1
−
Y
Y11  10 −1 
 =
.
First ∆Y = (0.1)(0.5) − (0.4)(0.1) = .01 S . Then H =  11
0.1
∆Y   4
 Y21
Y
Y11 
 11
P17.8-3
Y12 
 1
 Y - Y   2 0.8 
11
 =
First ∆Y = (0.5)(0.6) − (−0.4)(−0.4) S . Then H =  11
∆Y   −0.8 0.28
 Y21
Y
Y11 
 11
P17.8-4
Section17-9: Interconnection of Two-Port Networks
P17.9-1
Y12 = Y21 = − 1 S
3
Y22 = 0 − Y21 = 1 S
3
Y11 + Y12 = 1 S ⇒
Y11 = 4
Y12 = Y21 = −1 S
3
S
Y11 + Y12 = 1 S ⇒ Y11 = 3 S
2
2
Y21 +Y22 = 1 S ⇒ Y22 = 4 S
3
3
4 − 1 
3
3
Ya = 
− 1 1 
 3 3
3
2
Yb = 
 −1

−1 

4 
3
−4 
 ( 4 + 3 ) − 4  17
3 2
3  6
3

=
Y=
 −4
5   −4
5 
3
3   3
 3
17-17
P17.9-2
Admittance parameters:
Transmission parameters:
 10 −6 


Y =  44 44 
 −6 8 
 44 44 
 8 44 
6 6 
T=

 1 10 
 6 6 
 20 −12 


Yp = Y + Y =  44 44 
 −12 16 
 44 44 
108 792 
 36 36 
TC = T ⋅ T' = 

 18 144 
 36 36 
P17.9-3
1

 s + s −s 
G1 + G 2
Y = 
 + 
1 
 −G 2
−s
+s


s 
−G 2 

G 2 + G 3 
17-18
Verification Problems
VP 17-1
 75 
V1 = 50 
 I 2 = 15 I 2
 175+ 75 
Z 12 =
V1
I2
I1 = 0
= 15 Ω
 1 1 
I1 =  +
 V1 = 0.028 V1
 50 125 
Y11 =
V1
I1
V2 = 0
= 28 mS
Y11 ≠ 24 mS, so the report is not correct.
VP 17-2
V1 = (2 + 0.2 s ) I1  Z 11 = 2 + 0.2 s =0.2 ( s +10)
⇒
Z 21 =0.1 s
V2 = (0.1 s) I1 
Z 22 = 2 + 0.2 s and Z 12 = 0.1 s
∆Z = (2 + 0.2 s )(2 + 0.2 s ) − (0.1 s)(0.1 s) = 0.01(3 s 2 + 80 s + 40)
17-19
 Z11
Z
21
T= 
 1
Z
 21
∆Z   2( s +10)
Z 21  
s
=
Z 22  
0.1 s
Z 21  
0.1(3 s 2 +80 s + 40) 

s

2 ( s +10)


s
This is not the transmission matrix given in the report.
Design Problems
DP 17-1
We will need to find R and R1 by trial and error. A Mathcad spreadsheet will help with
the calculations. Given the restrictions R ≤ 10 Ω and R1 ≤ 10 Ω we will start with
R = 10 Ω and R1 = 10 Ω :
R1 := 10
Ra :=
14⋅ 20
14 + 20 + R
Rin := R1 +
R := 10
Rb :=
( Rb + 2) ⋅ ( Rc + 20)
Rb + 2 + Rc + 20
14⋅ R
14 + 20 + R
Rc :=
R⋅ 20
14 + 20 + R
Rin = 14.279
The specifications cannot be satisfied. R and R1 are at their maximum values but
R in needs to be larger. Reducing either R or R1 will reduce R in .
17-20
DP 17-2
Need VA + VB for balance
R1 V
=
R3 V
=
R1 + R 3
R1 + R 3
R2 V
R2 + R4
R4 V
R2 + R4
(1)
(2)
Dividing (1) by (2) yields:
DP 17-3
V1 = h11 I1 + h12 V2
I 2 = h 21 I1 + h 22 V2
Next
V2 = − I 2 R L


I2
1
= h 21 
⇒

 1 + h 22 R L 
I1




1
⇒
79 = 80 

 1 + h 22 R L 


We require
Next
and
Ai =
R1 R 2
= .
R3 R 4
⇒ I 2 = h 21 I1 − h 22 R L I 2


IL
I
1
= − 2 = − h 21 

 1 + h 22 R L 
I1
I1


RL 
79 
1 +
 =1 ⇒
80  80 × 103 
R L = 1.013 kΩ ≅ 1 kΩ
I2 = −
V2
= h 21 I1 + h 22 V2 ⇒ V2 (h 22 + 1/ RL ) = − h 21 I1
RL
Substituting this expression into the second hybrid equation gives:
V1 = h11 I1 +
h12 (−h 21 )
I1
(h 22 + 1 )
RL
The input resistance is given by
R in ≈ h11 − h12 R L h 21 (since h 22 << 1
Finally
RL
)
R in = 45 − (5 ×10−4 )(103 )(80) = 5 Ω < 10 Ω
17-21
DP 17-4
Z 11 = 2 +
4(12)
8(8)
= 5 Ω and Z 22 =
=4Ω
4 + 12
8+8
V
 4

V2 = 8 
I1  = 2 I1 ⇒ Z 21 = 2
I1
 4 + 12 
I2 =0
=2Ω
Similarly Z 12 = 2 Ω
Thèvenin: Z T = Z 22 = 4 Ω so for maximum
power transfer, use R L =4 Ω
 Vs 
 
2
PRL =   = 89.3 W ⇒ Vs = 37.8 V
4
2
17-22
DP 17-5
The circuit consists of 4 cascaded stages. Represent each stage by a transmission matrix
using:
1
T=
Y ( s )
1 Z ( s ) 
T=

0 1 
1 

1

Ta =  C 1 s 
0 1 
1


T =  C3 s
 L 2 C 3 s + 1
0
1
L1 s 

1

Tb =  L1 C 2 s + 1 
0

1
0

1

L1 C 1 s
C3 s

×
1 +
2
L1 C 2 C 1 s + C 1 s R L L 2 C 3 s + R L
T = Ta Tb Tc Td = 

C3 s

RL L2 C 3 s + RL

 1
Td =  1
 R L
0

1



L1 C 2 C 1 s + C 1 s 


1

L1 C 1 s
2
17-23