February 5, 2006
CHAPTER 1
P.P.1.1
A proton has 1.602 x 10-19 C. Hence, 2 million protons have
+1.602 x 10-19 x 2 x 106 = 3.204 x 10–13 C
P.P.1.2
i = dq/dt = -10(–2)e-2t mA
At t = 0.5 sec, i = 20e-1 = 7.358 mA
P.P.1.3
q=
∫ idt =
1
∫ 2dt +
∫
0
2
1
2 1
2
2t dt = 2t
0
+ ( 2 / 3) t
3 2
1
= 2 + 14/3 = 6.667 C
P.P.1.4
(a)
Vab = w/q = -30/2 = –15 V
The negative sign indicates that point a is at higher potential than point b.
P.P.1.5
(b)
Vab = w/q = -30/-6 = 5 V
(a)
v = 2 i = 10 cos (60 π t)
p = v i = 50 cos2 (60 π t)
At t = 5 ms, p = 50 cos2 (60 π 5x10-3) = 50 cos2 (0.3 π)
= 17.27 watts
(b)
t
v = 10 + 5 ∫ idt = 10 +
0
∫
t
0
25 cos 60 π t dt = 10+
25
sin 60 π t
60π
p = vi = 5 cos (60 πt)[10 + (25/(60 π)) sin (60 π t)]
At t = 5 ms, p = 5 cos (0.3π){10 + (25/(60 π)) sin (0.3 π)}
= 29.7 watts
P.P.1.6
p = v i = 15 x 120 = 1800 watts; w = p x t
therefore, t = w/p = (30x103)/1800 = 16.667 seconds
P.P.1.7
p1 = 5(-8) = –40w
p2 = 2(8) = 16w
p3 = 0.61(3) = 0.6(5)(3) = 9w
p4 = 3(5) = 15w
P.P.1.8
i=
= e
dn
= -1.6 x 10-19 x 1013 = -1.6 x 10-6 A
dt
p = v0 i = 30 x 103 x (1.6 x 10-6) = 48mW
P.P.1.9
Minimum monthly charge
= $12.00
First 100 kWh @ $0.16/kWh
= $16.00
Next 200 kWh @ $0.10/kWh
= $20.00
Remaining 100 kWh @ $0.06/kWh = $6.00
Total Charge = $54.00
Average cost = $54/[100+200+100] = 13.5 cents/kWh
P.P.1.10
This assigned practice problem is to apply the detailed problem solving
technique to some of the more difficult problems of Chapter 1.
February 5, 2006
CHAPTER 2
P.P.2.1
i = V/R = 110/12 = 9.167 A
P.P.2.2
(a)
v = iR = 2 mA[10 kohms] = 20 V
(b)
G = 1/R = 1/10 kohms = 100 µS
(c)
p = vi = 20 volts[2 mA] = 40 mW
P.P.2.3
p = vi which leads to i = p/v = [20 cos2 (t) mW]/[10cos(t) mA]
or i = 2cos(t) mA
R = v/i = 10cos(t)V/2cos(t)mA = 5 kΩ
P.P.2.4
5 branches and 3 nodes. The 1 ohm and 2 ohm resistors are in parallel.
The 4 ohm resistor and the 10 volt source are also in parallel.
P.P.2.5
Applying KVL to the loop we get:
-10 + 4i – 8 + 2i = 0 which leads to i = 3A
v1 = 4i = 12V and
P.P.2.6
v2 = -2i = –6V
Applying KVL to the loop we get:
-35 + 10i + 2vx + 5i = 0
But, vx = 10i and v0 = -5i. Hence,
-35 + 10i + 20i + 5i = 0 which leads to i = 1A.
Thus, vx = 10V and v0 = –5V
P.P.2.7
Applying KCL, 6 = i0 + [i0 /4] + [v0 /8], but i0 = v0/2
Which leads to: 6 = (v0/2) + (v0/8) + (v0/8) thus, v0 = 8V and i0 = 4A
P.P.2.8
2Ω
+
-
5V
i1
+ V1 Loop 1
i3
4Ω
+ V3 -
i2
+
V2
-
8Ω
+
Loop 2
3V
At the top node,
i1 = i 2 + i 3
(1)
For loop 1
or
-5 + V1 + V2 = 0
V1 = 5 - V2
(2)
For loop 2
or
- V2 + V3 -3 = 0
V3 = V2 + 3
(3)
Using (1) and Ohm’s law, we get
(V1/2) = (V2/8) + (V3/4)
and now using (2) and (3) in the above yields
[(5- V2)/2] = (V2/8) + (V2+3)/4
V2 = 2 V
or
V1 = 5- V2 = 3V, V3 = 2+3 = 5V, i1 = (5-2)/2 = 1.5A,
i2 = 250 mA, i3 = 1.25A
2Ω
P.P.2.9
Req
6Ω
1Ω
3Ω
4Ω
5Ω
4Ω
3Ω
Combining the 4 ohm, 5 ohm, and 3ohm resistors in series gives 4+3+5 = 12.
But, 4 in parallel with 12 produces [4x12]/[4+12] = 48/16 = 3ohm.
So that the equivalent circuit is shown below.
2Ω
Req
3Ω
3Ω
6Ω
1Ω
Thus, Req = 1 + 2 + [6x6]/[6+6] = 6 Ω
20 Ω
P.P.2.10
8Ω
Req
5Ω
20 Ω
18 Ω
1Ω
9Ω
2Ω
Combining the 9 ohm resistor and the 18 ohm resistor yields [9x18]/[9+18] = 6 ohms.
Combining the 5 ohm and the 20 ohm resistors in parallel produces [5x20/(5+20)] = 4
ohms We now have the following circuit:
8Ω
4Ω
6Ω
1Ω
20 Ω
2Ω
The 4 ohm and 1 ohm resistors can be combined into a 5 ohm resistor in parallel with a
20 ohm resistor. This will result in [5x20/(5+20)] = 4 ohms and the circuit shown below:
8Ω
4Ω
6Ω
2Ω
The 4 ohm and 2 ohm resistors are in series and can be replaced by a 6 ohm resistor.
This gives a 6 ohm resistor in parallel with a 6 ohm resistor, [6x6/(6+6)] = 3 ohms. We
now have a 3 ohm resistor in series with an 8 ohm resistor or 3 + 8 = 11ohms. Therefore:
Req = 11 ohms
P.P. 2.11
8S
4S
8||4 = 8+4 = 12S
12 S
Geq
Geq
2S
4S
2||4 = 2+4 = 6S
12 S in series with 6 S = {12x6/(12+6)] = 4 or:
Geq = 4 S
6S
12 Ω
P.P.2.12
+ v1
i1
15V
+
-
-
i2
6Ω
+ v1
+
10 Ω
40 Ω
v2
15V
+
-
-
4Ω
+
8Ω
v2
-
-
6||12 = [6x12/(6+12)] = 4 ohm and 10||40 = [10x40/(10+40)] = 8 ohm.
Using voltage division we get:
v1 = [4/(4+8)] (15) = 5 volts, v2 = [8/12] (15) = 10 volts
i1 = v1/12 = 5/12 = 416.7 mA, i2 = v2/40 = 10/40 = 250 mA
P1 = v1 i1 = 5x5/12 = 2.083 watts, P2 = v2 i2 = 10x0.25 = 2.5 watts
P.P.2.13
1k Ω
i1
+
3k Ω
+
v1 10mA
-
i2
5k Ω
20k Ω
v2
-
4k Ω
4k Ω
10mA
Using current division, i1 = i2 = (10 mA)(4 kohm/(4 kohm + 4 kohm)) = 5mA
(a)
v1 = (3 kohm)(5 mA) = 15 volts
v2 = (4 kohm)(5 mA) = 20 volts
(b)
For the 3k ohm resistor, P1 = v1 x i1 = 15x5 = 75 mw
For the 20k ohm resistor, P2 = (v2)2 /20k = 20 mw
The total power supplied by the current source is equal to:
P = v2 x 10 mA = 20x10 = 200 mw
(c)
P.P.2.14
Ra = [R1 R2 + R2 R3 + R3 R1]/ R1 = [10x20 + 20x40 + 40x10]/10 = 140 ohms
Rb = [R1 R2 + R2 R3 + R3 R1]/ R2 = 1400/20 = 70 ohms
Rc = [R1 R2 + R2 R3 + R3 R1]/ R3 = 1400/40 = 35 ohms
P.P.2.15
We first find the equivalent resistance, R. We convert the delta
sub-network to a wye connected form as shown below:
i
a
13 Ω
24 Ω
100V
+
-
13 Ω
30 Ω
a
20Ω
10 Ω
24 Ω
10 Ω
a’
6Ω
50 Ω
b’
10 Ω
n
15 Ω
b
b
c’
Ra’n = 20x30/[20 + 30 + 50] = 6 ohms, Rb’n = 20x50/100 = 10 ohms
Rc’n = 30x50/100 = 15 ohms.
Thus, Rab = 13 + (24 + 6)||(10 + 10) + 15 = 28 + 30x20/(30 + 20) = 40 ohms.
i = 100/ Rab = 100/40 = 2.5 amps
P.P.2.16
For the parallel case, v = v0 = 110volts.
p = vi
i = p/v = 40/110 = 364 mA
For the series case, v = v0/N = 110/10 = 11 volts
i = p/v = 40/11 = 3.64 amps
P.P.2.17
(a)
We use equation (2.61)
R1 = 50x10-3/ (1-10-3) = 0.05/999 = 50 mΩ (shunt)
(b)
R2 = 50x10-3/(100x10-3 – 10-3) = 50/99 = 505 mΩ (shunt)
(c)
R3 = 50x10-3/(10x10-3-10-3) = 50/9 = 5.556 Ω (shunt)
February 5, 2006
CHAPTER 3
P.P.3.1
1A 1
6 Ω i1
i1
i2
1A
2 4A
i3
2Ω
4A
7Ω
At node 1,
1 = i1 + i2
1=
or 6 = 4v1 - v2
(1)
v1 − v 2 v1 − 0
+
6
2
At node 2,
v1 − v 2
v −0
= 4+ 2
6
7
i1 = 4 + i 3
or 168 = 7v1 - 13v2
(2)
Solving (1) and (2) gives
v1 = –2 V, v2 = –14 V
2Ω
i1
P.P.3.2
4ix
v1
i2
i2
3Ω
10 A
v2
v3
ix
4Ω
i3
6Ω
At node 1,
v1 − v 3 v1 − v 2
+
2
3
or 60 = 5v1 - 2v2 - 3v3
10 = i1 + i2 =
(1)
At node 2,
v1 − v 2
v
+3 2 = 0
3
4
i 2 + 4i x = i x
or 4v1 + 5v2 = 0
(2)
At node 3,
v1 − v 3 v 3 − 0
v
=
+4 2
2
6
4
i1 = i3 + 4ix
or -3v1 + 6v2 + 4v3 = 0
(3)
Solving (1) to (3) gives
v1 = 80 V, v2 = –64 V, v3 = 156 V
P.P.3.3
4Ω
7V
+
3Ω
v
+
-
-+
v1
2Ω
v
6Ω
-
At the supernode in Fig. (a),
7 − v v v1 v1
= + +
4
3 2
6
or 21 = 7v + 8v1
(1)
Applying KVL to the loop in Fig. (b),
- v - 3 + v1 = 0
+
+
v
v1
-
(b)
(a)
v1 = v + 3
(2)
3V
Solving (1) and (2),
v = – 200 mV
v1 = v + 3 = 2.8, i1 =
v1
= 1.4
2
i1 = 1.4 A
P.P.3.4
v1
v2
3V
v3
+-
- +
+
+
+
v1
v2
v3
-
-
-
(a)
(b)
From Fig. (a),
v1 v 2 v 3
+
+
=0
2
4
3
6v1 + 3v2 + 4v3 = 0
(1)
- v1 + 10 + v2 = 0
v1 = v2 + 10
(2)
- v2 - 5i + v3 = 0
v3 = v2 + 5i
(3)
From Fig. (b),
Solving (1) to (3), we obtain
v1 = 3.043V, v2 = –6.956 V, v3 = 652.2 mV
P.P.3.5 We apply KVL to the two loops and obtain
- 12 + 18ii - 12i2 = 0
8 + 24i2 - 12i1 = 0
From (1) and (2) we get
i1 = 666.7 mA, i2 = 0A
3ii - 2i2 = 2
(1)
- 3i1 + 6i2 = -2 (2)
P.P.3.6 For mesh 1,
- 20 + 6i1 – 2i2 - 4i3 = 0
3i1 - i2 - 2i3 = 10
(1)
For mesh 2,
10i2 - 2i1 - 8i3 - 10i0 = 0 = -i1 + 5i2 – 9i3
(2)
But i0 = i3,
18i3 - 4i1 - 8i2 = 0
- 2i1 - 4i2 + 9i3 = 0
From (1) to (3),
⎡ 3 − 1 − 2⎤ ⎡ i1 ⎤ ⎡10⎤
⎢ − 1 5 − 9⎥ ⎢i ⎥ = ⎢ 0 ⎥
⎥ ⎢ 2⎥ ⎢ ⎥
⎢
⎢⎣− 2 − 4 9 ⎥⎦ ⎢⎣i 3 ⎥⎦ ⎢⎣ 0 ⎥⎦
3 −1 − 2
−1 5 − 9
Δ = − 2 − 4 9 = 135 - 8 - 18 - 20 - 108 - 9 = - 28
3 −1 − 2
−1 5 − 9
10 − 1 − 2
0
5 −9
Δ 1 = 0 − 4 9 = 450 − 360 = 90
10 − 1 − 2
0
5 −9
3 10 − 2
−1 0 − 9
Δ 2 = − 2 0 9 = 180 + 90 = 270
3 10 − 2
−1 0 − 9
(3)
3 − 1 10
−1 5 0
Δ 3 = − 2 − 4 0 = 40 + 100 = 140
3 − 1 10
−1 5 0
i1 =
Δ
Δ 2 270
140
Δ1
90
=
= −9.643 , i3 = 3 =
= −5A
=
= −3.214, i2 =
Δ − 28
Δ − 28
Δ − 28
i0 = i3 = –5A
P.P.3.7
2Ω
2Ω
i3
i1
6V
i1
2Ω
i3
2Ω
4Ω
+
-
4Ω
+
3A
3A
8Ω
i2
i2
1Ω
i1
0 i2
(a)
(b)
For the supermesh,
- 6 + 2i1 - 2i3 + 12i2 - 4i3 = 0
i1 + 6i2 - 3i3 = 3
(1)
For mesh 3,
8i3 - 2i1 - 4i2 = 0
- i1 - 2i2 + 4i3 = 0
At node 0 in Fig. (a),
i1 = 3 + i2
i1 - i2 = 3
Solving (1) to (3) yields
i1 = 3.474A, i2 = 473.7 mA, i3 = 1.1052A
(2)
8Ω
P.P.3.8 G11 = 1/(1) + 1/(10) + 1/(5) = 1.3, G12 = -1/(5) = -0.2,
G33 = 1/(4) + 1 = 1.25, G44 = 1/(2) + 1/(4) = 0.75,
G12 = -1/(5) = - 0.2, G13 = - 1, G14 = 0,
G21 = -0.2, G23 = 0 = G26,
G31 = -1, G32 = 0, G34 = - 1/4 = - 0.25,
G41 = 0, G42 = 0, G43 = 0.25,
i1 = 0, i2 = 2 - 1 = 1, i3 = - 1, i4 = 3.
Hence,
−1
0 ⎤
⎡ 1.3 − 0.2
⎢− 0.2 0.2
0
0 ⎥⎥
⎢
⎢ −1
0
1.25 − 0.25⎥
⎥
⎢
0
− 0.25 0.75 ⎦
⎣ 0
P.P.3.9
⎡ v1 ⎤ ⎡ 0 ⎤
⎢v ⎥ ⎢ 3 ⎥
⎢ 2⎥ = ⎢ ⎥
⎢ v3 ⎥ ⎢− 1⎥
⎢ ⎥ ⎢ ⎥
⎣v4 ⎦ ⎣ 3 ⎦
R11 = 50 + 40 + 80 = 170, R22 = 40 + 30 + 10 = 80,
R33 = 30 + 20 = 50, R44 = 10 + 80 = 90,
R55 = 20 + 60 = 80, R12 = -40, R13 = 0, R14 = -80,
R15 = 0, R21 = -40, R23 = -30, R24 = -10, R25 = 0,
R31 = 0, R32 = -30, R34 = 0, R35 = -20,
R41 = -80, R42 = -10, R43 = 0, R45 = 0,
R51 = 0, R52 = 0, R53 = -20, R54 = 0,
v1 = 24, v2 = 0, v3 = -12, v4 = 10, v5 = -10
Hence the mesh-current equations are
0
0 ⎤ ⎡ i1 ⎤
− 80
⎡ 24 ⎤
⎡ 170 − 40
⎥
⎢
⎥
⎢ 0 ⎥
⎢− 40 80 − 30 − 10
0 ⎥ ⎢i 2 ⎥
⎥
⎢
⎢
⎢ 0
0
− 30 50
− 20⎥ ⎢i 3 ⎥ = ⎢− 12⎥
⎥
⎥ ⎢ ⎥
⎢
⎢
0
90
0 ⎥ ⎢i 4 ⎥
⎢ 10 ⎥
⎢ − 80 − 10
⎢⎣− 10⎥⎦
⎢⎣ 0
0
0
80 ⎥⎦ ⎢⎣i 5 ⎥⎦
− 20
P.P.3.10
The schematic is shown below. It is saved and simulated by selecting
Analysis/Simulate. The results are shown on the viewpoints:
v1 = –40 V, v2 = 57.14 V, v3 = 200 V
-40.0000
P.P.3.11
57.1430
200.0000
The schematic is shown below. After saving it, it is simulated by choosing
Analysis/Simulate. The results are shown on the IPROBES.
i 1 = –428.6 mA, i2 = 2.286 A, i3 = 2 A
-4.286E-01
2.286E+00
2.000E+00
P.P.3.12
For the input loop,
-5 + 10 x 103 IB + VBE + V0 = 0
(1)
For the outer loop,
-V0 - VCE - 500 I0 + 12 = 0
(2)
But
V0 = 200 IE
(3)
Also
IC = βIB = 100 IB, α = β/(1 + β) = 100/(101)
IC = αIE
IE = IC/(α) = βIB/(α)
IE = 100 (101/(100)) IR = 101 IB
(4)
From (1), (3) and (4),
10,000 IB + 200(101) IR = 5 - VBE
IB =
5 − 0.7
= 142.38μA
10,000 + 20,000
V0 = 200 IE = 20,000 IB = 2.876 V
From (2),
VCE = 12 - V0 - 500 IC = 9.124 - 500 x 100 x 142.38 x 10-6
VCE = 1.984 V {often, this is rounded to 2.0 volts}
P.P.3.13
20 kΩ
i1
i0
iC
30 kΩ
iB
+
20 kΩ
+
1V
+
-
v0
-
VBE
-
+
-
22V
First of all, it should be noted that the circuit in the textbook should have a 22V
source on the right hand side rather than the 10 V source.
iB =
B
1 − 0 .7
= 10μA, iC = βiB = 0.8 mA
30k
i1 = iC + i0
Also,
-20ki0 – 20ki1 + 22 = 0
(1)
i1 = 1.1 mA – i0
Equating (1) and (2),
1.1mA – i0 = 0.8 mA + i0
v0 = 20 ki0 = 20 x 0.15 = 3 V
i0 = 150 μA
(2)
February 5, 2006
CHAPTER 4
P.P.4.1
6Ω
i2
i1
+
2Ω
iS
4Ω
vo
−
2
1
is = is
2+6+4
6
2
v 0 = 4i 2 = i s
3
2
When is = 15A, v 0 = (15) = 10V
3
2
When is = 30A, v 0 = (30) = 20V
3
By current division, i 2 =
P.P.4.2
12 Ω
v1
+
VS = 10 V
Let v0 = 1. Then i =
+
−
5Ω
1
1
and v 1 = (12 + 8) = 2.5
8
8
giving vs = 2.5V.
If vs = 10V, then v0 = 4V
vo
−
8Ω
P.P.4.3
Let v0 = v1 + v2, where v1 and v2 are contributions to the 20-V and 8-A
sources respectively.
3Ω
5Ω
i
+
v1
+
−
2Ω
−
(a)
3Ω
i2
i1
5Ω
+
v2
8A
2Ω
−
(b)
To get v1, consider the curcuit in Fig. (a).
(2 + 3 + 5)i = 20
v1 = 2i = 4V
i = 20/(10) = 2A
To get v2, consider the circuit in Fig. (b).
i1 = i2 = 4A, v2 = 2i2 = 8V
Thus,
v = v1 + v2 = 4 + 8 = 12V
20 V
P.P.4.4 Let vx = v1 + v2, where v1 and v2 are due to the 10-V and 2-A sources
respectively.
20 Ω
10 V
v1
+
−
20 Ω
4Ω
(a)
v2
2A
4Ω
(b)
To obtain v1, consider Fig. (a).
0.1v1 +
10 − v1 v1
=
20
4
v1 = 2.5
For v2, consider Fig. (b).
2 + 0.1v2 +
0 − v2 v2
=
20
4
vx = v1 + v2 = 12.5V
0.1v1
v2 = 10
0.1v2
P.P.4.5
Let i = i1 + i2 + i3
where i1, i2, and i3 are contributions due to the 16-V, 4-A, and 12-V sources respectively.
2Ω
6Ω
2Ω
8Ω
6Ω
8Ω
4A
i1
16V
+
−
i2
(a)
(b)
6Ω
2Ω
8Ω
i3
12V
+
−
(c)
For i1, consider Fig. (a), i1 =
16
= 1A
6+2+8
For i2, consider Fig. (b). By current division, i 2 =
2
( 4 ) = 0 .5
2 + 14
− 12
= −0.75A
16
Thus, i = i1 + i2 + i3 = 1 + 0.5 - 0.75 = 750mA
For i3, consider Fig. (c), i 3 =
6x3
= 2Ω .
9
Adding the 1-Ω and 4-Ω resistors in series gives 1 + 4 = 5Ω. Transforming the left
current source in parallel with the 2-Ω resistor gives the equivalent circuit as shown in
Fig. (a).
P.P.4.6
Combining the 6-Ω and 3-Ω resistors in parallel gives 6 3 =
2Ω
5V
− +
10V
io
+
−
7Ω
5Ω
3A
7Ω
5Ω
3A
(a)
io
2Ω
7.5A
(b)
io
10.5A
(10/7) Ω
7Ω
(c)
Adding the 10-V and 5-V voltage sources gives a 15-V voltage source. Transforming the
15-V voltage source in series with the 2-Ω resistor gives the equivalent circuit in Fig. (b).
Combining the two current sources and the 2-Ω and 5-Ω resistors leads to the circuit in
Fig. (c). Using circuit division,
10
i o = 7 (10.5) = 1.78 A
10
+7
7
P.P.4.7
We transform the dependent voltage source as shown in Fig. (a). We combine
the two current sources in Fig. (a) to obtain Fig. (b). By the current division principle,
ix =
5
(4 − 0.4i x )
15
ix = 1.176A
ix
4A
10 Ω
5Ω
0.4ix
(a)
ix
10 Ω
4 – 0.4ix A
5Ω
(b)
P.P.4.8
To find RTh, consider the circuit in Fig. (a).
6Ω
6Ω
4Ω
RTh
(a)
6Ω
+
6Ω
2A
2A
4Ω
VTh
−
(b)
R Th = (6 + 6) 4 =
12 x 4
= 3Ω
18
To find VTh, we use source transformations as shown in Fig. (b) and (c).
6Ω
24 V
6Ω
+
4Ω
+
−
VTh
−
(c)
Using current division in Fig. (c),
VTh =
i=
4
(24) = 6V
4 + 12
VTh
6
=
= 1.5A
R Th + 1 3 + 1
P.P.4.9 To find VTh, consider the circuit in Fig. (a).
5Ω
6V
+
−
3Ω
Ix
a
+
i2
VTh
4Ω
i1
−
1.5Ix
i2
i1
o
b
(a)
0.5Ix
5Ω
3Ω
Ix
i
a
+
−
4Ω
1.5Ix
(b)
b
1V
Ix = i2
i2 - i1 = 1.5Ix = 1.5i2
i2 = -2i1
(1)
For the supermesh, -6 + 5i1 + 7i2 = 0
(2)
From (1) and (2), i2 = 4/(3)A
VTh = 4i2 = 5.333V
To find RTh, consider the circuit in Fig. (b). Applying KVL around the outer loop,
5(0.5I x ) − 1 − 3I x = 0
1
i = − I x = 2.25
4
1
1
R Th = =
= 444.4 mΩ
2.25
i
P.P.4.10
Ix = -2
Since there are no independent sources, VTh = 0
4vx
10 Ω
+ −
+
vx
+
5Ω
15 Ω
vo
−
io
−
(a)
4vx
10 Ω
15 Ω
+ −
+
vx
+
5Ω
vo
i
−
−
+
–
15io
(b)
To find RTh, consider Fig.(a). Using source transformation, the circuit is transformed to
that in Fig. (b). Applying KVL, ).
But vx = -5i. Hence, 30i - 20i + 15io = 0
vo = (15i + 15io) = 15(-1.5io + io) = -7.5io
RTh = vo/(io) = –7.5Ω
10i = -15io
P.P.4.11
3Ω
3Ω
6Ω
RN
(a)
3Ω
5A
3Ω
4A
(b)
From Fig. (a), RN = (3 + 3) 6 = 3 Ω
From Fig. (b), IN =
1
(5 + 4) = 4.5A
2
IN
P.P.4.12
2vx
i
+ −
+
vx
+
6Ω
2Ω
ix
vx
−
−
1V
+
−
(a)
2vx
+ −
+
6Ω
2Ω
10 A
vx
Isc
−
(b)
To get RN consider the circuit in Fig. (a). Applying KVL, 6 i x − 2 v x − 1 = 0
But vx = 1,
6ix = 3
ix = 0.5
v
i = i x + x = 0.5 + 0.5 = 1
2
1
R N = R Th = = 1Ω
i
To find IN, consider the circuit in Fig. (b). Because the 2Ω resistor is shorted, vx = 0 and
the dependent source is inactive. Hence, IN = isc = 10A.
P.P.4.13
Fig. (a).
We first need to find RTh and VTh. To find RTh, we consider the circuit in
+
vx
− v0 4 Ω
2Ω
+
vx
4Ω
−
2Ω
i
1Ω
1Ω
+
−
+
−
1V
9V
VTh
io
+
−
3vx
(a)
(b)
Applying KCL at the top node gives
1 − v o 3v x − v o v o
+
=
4
1
2
But vx = -vo. Hence
v
1 − vo
− 4v o = o
vo = 1/(19)
2
4
1
1−
1 − vo
19 = 9
i=
=
4
4
38
RTh = 1/i = 38/(9) = 4.222Ω
To find VTh, consider the circuit in Fig. (b),
-9 + 2io + io + 3vx = 0
But vx = 2io. Hence,
9 = 3io + 6io = 9io
io = 1A
VTh = 9 - 2io = 7V
RL = RTh = 4.222Ω
Pmax
+
−
2
v Th
49
=
=
= 2.901W
4R L 4(4.222)
+
−
3vx
P.P.4.14
find VTh and Rth.
We will use PSpice to find Voc and Isc which then can be used to
Clearly Isc = 12 A
Clearly VTh = Ioc = 5.333 volts. RTh = Voc/Isc = 5.333/12 = 444.4 m-ohms.
P.P.4.15 The schematic is the same as that in Fig. 4.56 except that the 1-kΩ resistor is
replaced by 2-kΩ resistor. The plot of the power absorbed by RL is shown in the figure
below. From the plot, it is clear that the maximum power occurs when RL = 2kΩ and it is
125μW.
P.P.4.16
VTh = 9V, R Th = (v oc − VL )
RL
20
= (9 − 1)
= 2.5Ω
VL
8
2.5 Ω
+
9V
VL =
10
(9) = 7.2V
10 + 2.5
+
−
VL
−
10 Ω
P.P.4.17
R1 = R3 = 1kΩ, R2 = 3.2kΩ
R
R x = 3 R 2 = R 2 = 3.2kΩ
R1
P.P.4.18
We first find RTh and VTh. To get RTh, consider the circuit in Fig. (a).
R Th = 20 30 + 60 40 =
20 x 30 60 x 40
+
50
100
= 12 + 24 = 36Ω
20 Ω
30 Ω
20 Ω
a
+
−
v2
30 Ω
a +
VTh
RTh
60 Ω
b
40 Ω
+ v
1
b −
60 Ω
−
10 V
(a)
+ −
(b)
To find VTh, we use Fig. (b). Using voltage division,
v1 =
60
(16) = 9.6,
100
v2 =
But − v 1 + v 2 + v Th = 0
20
(16) = 6.4
50
vTh = v1 - v2 = 9.6 - 6.4 = 32V
IG =
VTh
3.2
=
= 64mA
R Th + R m 3.6 + 1.4
40 Ω
February 5, 2006
CHAPTER 5
P.P.5.1
The equivalent circuit is shown below:
vd
vs
+
+
5 kΩ
1
2 MΩ
40 kΩ
+
v1
2
20 kΩ
-
At node 1,
v s − v1
v − v0
v1
=
+ 1
6
3
2x10
5x10
40x10 3
At node 2,
Av d − v 0 v1 − v 0
v0
+
=
3
50
40x10
20x133
i0
50 Ω
+
v0
+
Avd
-
-
v1 =
v S + 50v 0
451
(1)
But vd = v1 - vS.
[2 x 105 (v1 - vS) - v0] 4000/(5) + v1 - v0 = 2v0
1600 x 105 (vS - v1) + 803v0 ≅ 0
Substituting v1 in (1) into (2) gives
1.5914523 x 108 vS - 17737556v0 = 0
v 0 1.5964523x10 8
=
= 9.00041
vS
17737556
If vS = 1 V, v0 = 9.00041 V, v1 = 1.0000455
vd = vS - v1 = - 4.545 x 10-5
Av d − v 0
= 657 μA
Avd = - 9.0909, i0 =
50
(2)
P.P.5.2
20 kΩ
i
V1
10 kΩ
VS
+
+
V2
-
+
V0
-
At node 1,
v S − v1 v1 − v 0
=
10
20
But v1 = v2 = 0,
vS
v
=− 0
10
20
i0 =
v0
= –2
vS
0 − v0
v0
=−
3
20x10
20x10 3
When vs = 2V, v0 = -4, i0 =
P.P.5.3
v0 = −
i=
P.P.5.4
4 x10 −3
= 200 μA
20
R2
− 15
(40mV) = –120 mV
vi =
R1
5
0 − v0
= 8 μA
15k
(a) iS =
0 − v0
R
v0
= −R
iS
iS
R1
0V
V1
R2
R3
20 kΩ
+
+
V2
V0
-
(b)
At node 2, iS =
At node 1,
0 − v1
R1
v1 = -iSR1
(1)
0 − v 1 v 1 − 0 v1 − v 0
=
+
R1
R2
R3
⎛ 1
1
1 ⎞ − v0
⎟⎟ =
+
+
-v1 ⎜⎜
R
R
R
R3
2
3 ⎠
⎝ 1
⎛ 1
1
1 ⎞
⎟⎟
+
+
v0 = -iSR1R3 ⎜⎜
R
R
R
2
3 ⎠
⎝ 1
⎛ R
v0
R ⎞
= −R 1 ⎜⎜1 + 3 + 3 ⎟⎟
iS
⎝ R1 R 2 ⎠
P.P.5.5
By voltage division
v1 =
8
(3) = 2V
4+8
where v1 is the voltage at the top end of the 8kΩ resistor. Using the formula for
noninverting amplifier,
⎛ 5⎞
v0 = ⎜1 + ⎟(2) = 7 V
⎝ 2⎠
P.P.5.6
This is a summer.
8
8
⎡8
⎤
v 0 = − ⎢ (1.5) + (2) + (1.2)⎥ = –3.8 V
10
6
⎣ 20
⎦
i0 =
P.P.5.7
v0 v0
3.8 3.8
+
=−
−
= –1.425 mA
8
4
8
4
If the gain is 4, then
R2
=4
R1
But
R2 = 4R1
R2 R4
=
R1 R 3
R4 = 4R3
If we select R1 = R3 = 10kΩ, then R2 = R4 = 40kΩ
P.P.5.8
v0 =
R2
R1
⎛ 2R 3 ⎞
⎜⎜1 +
⎟ (v2 - v1)
R 4 ⎟⎠
⎝
R3 = 0, R4 = ∞, R2 = 40kΩ, R1 = 20kΩ
40
(8.01 − 8) = 0.02
20
v
0.02
= 2μA
i0 = 0 =
10k 10x10 3
v0 =
P.P.5.9
Due to the voltage follower
va = 4V
For the noninverting amplifier,
⎛
v0 = ⎜1 +
⎝
i0 =
6⎞
⎟ va = (1 + 1.5) (4) = 10V
4⎠
vb
mA
4
-
a
+
+
+
+
vS
b
-
4 kΩ
v0
6 kΩ
-
But vb = va = 4
i0 =
P.P.5.10
4
= 1mA
4
As a voltage follower,
va = v1 = 2V
where va is the voltage at the right end of the 20 kΩ resistor.
As an inverter, vb = −
50
v 2 = −7 . 5 V
10
Where vb is the voltage at the right end of the 50kΩ resistor. As a summer
60 ⎤
⎡ 60
v0 = − ⎢ v a +
vb
30 ⎥⎦
⎣ 20
= [6 - 15] = 9V
P.P.5.11
The schematic is shown below. When it is saved and run, the results are
displayed on 1PROBE and VIEWPOINT as shown. By making vs = 1V, we obtain
v0 = 9.0027V and i0 = 650.2 µA
6.502E-04
9.0027
R
Rf
R
V1 + f V2 + f V3
R1
R2
R3
P.P.5.12
-V0 =
or
V0 = V1 + 0.5V2 + 0.25V3
(a)
If [V1V2V3] = [010], V0 = 0.5V
(b)
If [V1V2V3] = [110], V0 = 1 + 0.5 = 1.5V
(c)
If
(d)
V0 = 1.25, then V1 = 1, V2 = 0, V3 = 1, i.e.
[V1V2V3] = [101]
V0 = 1.75, then V1 = 1, V2 = 1, V3 = 1, i.e.
[V1V2V3] = [111]
P.P.5.13
Av = 1 +
2R
RG
RG =
RG =
2R
Av −1
2x 25x10 3
= 354.6 Ω
142 −1
February 5, 2006
CHAPTER 6
P.P.6.1
P.P.6.2
q 120 x10 −6
=
= 40V
C
3x10 −6
1
1
w = Cv 2 = x 3x10 −6 x1600 = 2.4mJ
2
2
v=
dv
d
= 10 x10 −6 (50 sin 2000 t )
dt
dt
= cos2000t A
i( t ) = C
t
1 t
10 −3
idt
=
50 sin 120πt dt V
−
3
∫
∫
C 0
0.1x10 0
500
50
=−
cos 120πt 0t =
(1 − cos 120πt )V
120π
12π
50
v(t = 1ms) =
(1 − cos 0.12π) = 93.14mV
12π
50
v(t = 5ms) =
(1 − cos 0.6π) = 1.7361 V
12π
P.P.6.3
v=
⎡50t , 0 ⟨ t ⟨ 2
i(t) = ⎢
⎣100, 2 ⟨ t ⟨ 6
1
1
v = ∫ idt = −3 ∫ idt ⋅ 10 −3 = ∫ idt
C
10
1
For 0< t <2, v = ∫ 0t 50 t dt = 25t2 x 103
C
1 t
For 2< t <6, v = ∫ 100dt + v(2) = (100 t − 0.2 + 0.1)
C 2
= (100t - 0.1)V
At t = 2ms, v = 100mV
At t = 5ms, v = (500 - 100)mV
= 400 mV
P.P.6.4
P.P.6.5
Under dc conditions, the capacitors act like open-circuits as shown below:
v2
+
−
i
1 kΩ
3 kΩ
+
10V
+
−
6 kΩ
v1
−
i=
10
= 1mA
1+ 3 + 6
v1 = (3k + 6k )i = 9V
v 2 = (3k )i = 3V
1
1
w 1 = C1 v12 = (10 x10 −6 )(9) 2 = 405μJ
2
2
1
1
w 2 = C 2 v 22 = (20 x10 −6 )(3) 2 = 90μJ
2
2
P.P.6.6
P.P.6.7
60 x120
= 40μF
180
40μF in parallel with 20μF = 40 + 20 = 60μF
50μF in parallel with 70μF = 50 + 70 = 120μF
60 x120
60μF in series with 120μF =
= 40μF
180
60 x 30
60μF in series with 30μF =
= 20μF
90
20μF in parallel with 20μF = 40μF
v2
+
−
Combining 60 and 120μF in series =
60V
+
−
40 μF +
v1
40 μF
−
From the Figure, v1 = v 2 =
60
= 30V
2
Hence v2 = 10V1, v4 = 20V
Note that q3 = q4 = 60 x 10μC. Thus v1 = v2 = 30V, v3 = 10V, v4 = 20V
di
d
= 10 −3 (20 cos 100 t ) ⋅ 10 −3
dt
dt
= –2 sin 100t mV
P.P.6.8 v = L
(
)
1 2 1
Li = x10 −3 400 cos 2 100 t ⋅ 10 −6
2
2
2
= 200 cos 100t ηJ
w=
1 t
1 t
v( t )dt + i( t 0 ) = ∫ 10(1 − t )dt + 2
∫
L t0
2 0
2
⎛ t ⎞
= 5⎜⎜ t − ⎟⎟ + 2
2⎠
⎝
At t = 4, i = 5(4 - 8) + 2 = –18A
P.P.6.9
i=
5
⎡
⎤
p = vi = 10(1 - t) ⎢5t − t 2 + 2⎥
2
⎣
⎦
2
3
= 20 + 30t - 75t + 25t
4
w = ∫ p dt = [20t + 15t2 - 25t3 + 25t 4 4]
0
4
0
= 80 + 15 x 16 - 1600 + 1600
w = 320J
P.P.6.10
Under dc conditions, the circuit is equivalent to that shown below
iL
+
3 kΩ
4A
3
( 4) = 3A
1+ 3
vC = 1iC = 3V
1
1⎛1⎞
wL = Li 2L = ⎜ ⎟(3) 2 = 1.125J
2
2⎝4⎠
1
1
wC = Cv C2 = (2)(3) 2 = 9J
2
2
iL =
1 kΩ
vC
−
P.P.6.11
40mH in series with 20mH = 40 + 20 = 60mH
60mH in parallel with 30mH = 30 x 60/(90) = 20mH
20mH in series with 100mH = 120mH
120mH in parallel with 40mH = 40 x 120/(160) = 30mH
30mH in series with 20mH = 50mH
50mH in parallel with 50mH = 25mH
Leq = 25mH
P.P.6.12
(a) i2 = i - i1
i2(0) = i(0) – i1(0) = 1.4 - 0.6 = 800 mA
di
(b) v1 = 6 1 = 6(0.6)(−2)e − 2 t = −7.2e − 2 t
dt
1 t
1 (−7.2) − 2 t t
i 2 = ∫ v1dt + i 2 (0) =
e 0 + 0.8
0
3
3 (−2)
= -0.4 + 1.2e–2tA
i = i1 + i2 = 0.4 + 1.8e–2tA
(c) From (b),
v1 = –7.2e–2tV
di
v2 = 8 = 8(−2)(1.8)e − 2 t = –28.8e-2tV
dt
v = v1 + v2 = –36e-2tV
P.P.6.13
RC = 25 x 103 x 10 x 10-6 = 0.25
1 t
1 t
vo = −
v i ( t ) + v o ( 0) = −
10dt mV + 0
∫
o
RC
0.25 ∫o
= 40t mV
P.P.6.14
RC = 10 x 103 x 2 x 10-6 = 2 x 10-2
dv
d
v o = −RC i = −2 x10 − 2 (3t )
dt
dt
vo = -60mV
P.P.6.15
dv o2
dv
= 4 cos 10t − 3 o − 2 v o
2
dt
dt
Using this we obtain the analog computer as shown below. We may let RC = 1s.
2V
− +
C
C
R
d2vo/dt2
t=0
R
R
−
+
R/2
−
+
-dvo/dt
−
+
vo
R
R
−
+
R/3
dvo/dt
R
R
cos10t
+
−
−
+
R/4
-cos10t
d2vo/dt2
February 5, 2006
CHAPTER 7
P.P.7.1
The circuit in Fig. (a) is equivalent to the one shown in Fig. (b).
8Ω
io
+
+
12 Ω
6Ω
vo
−
1/3 F
vx
−
+
+
Req
vc
v
−
−
(a)
(b)
R eq = 8 + 12 || 6 = 12 Ω
τ = R eq C = (12)(1 / 3) = 4 s
v c = v c (0) e - t τ = 30 e - t 4 = 30 e -0.25 t V
vx =
4
v = 10 e -0.25t V
4+8 c
vx = vo + vc
io =
P.P.7.2
⎯
⎯→
v o = v x − v c = -20 e -0.25t V
vo
= - 2.5 e -0.25t A
8
When t < 0, the switch is closed as shown in Fig. (a).
6Ω
+
24 V
+
−
vc(0)
12 Ω
4Ω
−
(a)
R eq = 4 || 12 = 3 Ω
v c ( 0) =
3
( 24) = 8 V
3+ 6
1/3 F
When t > 0, the switch is open as shown in Fig. (b).
6Ω
+
−
24 V
t=0
3Ω
1/6 F
(b)
τ = R eq C = (3)(1 / 6) = 1 / 2 s
v( t ) = v c (0) e - t τ = 8 e -2 t V
w c ( 0) =
1 2
1 1
Cv c (0) = × × 64 = 5.333J
2
2 6
P.P.7.3
This can be solved in two ways.
Method 1:
Find R th at the inductor terminals by inserting a voltage source.
3Ω
io
+
vo = 1 V
+
−
vx −
1Ω
i1
i2
+
−
Applying mesh analysis gives
4i1 − i 2 + 2v x − 1 = 0 ,
Loop 1:
10i1 − i 2 = 1
Loop 2:
6i 2 − i1 − 2v x = 0
7
i 2 = i1
6
From (1) and (2),
i o = i1 =
6
53
5Ω
2vx
where v x = 3i1
(1)
(2)
R th =
v o 53
=
,
io
6
τ=
16
L
1
=
=
R 53 6 53
i( t ) = 5e -53t A
Method 2:
We can obtain i using mesh analysis.
3Ω
i
1/6 H
+
vx −
1Ω
i1
i2
+
−
2vx
Applying KVL to the loops, we obtain
1 di1
+ 4i1 − i 2 + 2v x = 0
Loop 1:
6 dt
1 di1
+ 10i1 − i 2 = 0
6 dt
Loop 2:
6i 2 − i1 − 2v x = 0
7
i 2 = i1
6
Substituting (4) into (3) yields
1 di1
7
+ 10i1 − i1 = 0
6 dt
6
di1
+ 53 i1 = 0
or
dt
i1 = Ae -53 t
i = - i1 = Be -53 t
i ( 0) = 5 = B
i( t ) = 5e -53t A
Therefore,
i( t ) = 5e -53t A
and
v x ( t ) = -3i(t) = - 15e -53t V
5Ω
where v x = 3i1
(3)
(4)
P.P.7.4
For t < 0, the equivalent circuit is shown in Fig. (a).
i(t)
12 Ω
12 Ω
5A
5Ω
8Ω
8Ω
2H
(a)
(b)
i ( 0) =
8
(5) = 2 A
8 + 12
For t > 0, the current source is cut off and the RL circuit is shown in Fig. (b).
L
2
τ=
= = 0.5
R eq = (12 + 8) || 5 = 20 || 5 = 4 Ω ,
R eq 4
i( t ) = i(0) e - t τ = 2 e -2t A, t > 0
P.P.7.5
For t < 0, the switch is closed. The inductor acts like a short so the
equivalent circuit is shown in Fig. (a).
3Ω
i
i
io
1H
io
4Ω
6A
2Ω
(a)
i=
4
(6) = 4 A ,
4+2
4Ω
2Ω
(b)
io = 2 A ,
v o = 2i = 8 V
For t > 0, the current source is cut off so that the circuit becomes that shown in Fig. (b).
The Thevenin equivalent resistance at the inductor terminals is
L
1
R th = (4 + 2) || 3 = 2 Ω ,
τ=
=
R th 2
3 (-i) - 1
- 4 -2t
8
and
io =
= i=
e
v o = -2i o = e -2t
6+3 3
3
3
Thus,
⎧ 4A
i = ⎨ - 2t
⎩4 e A
P.P.7.6
t<0
2A
t<0
⎧
io = ⎨
- 2t
t>0
⎩- (4 3 ) e A t > 0
8V
t<0
⎧
vo = ⎨
-2t
⎩ (8 3) e V t > 0
⎧ 0
t<0
⎪
i( t ) = ⎨ 10 0 < t < 2
⎪- 10 2 < t < 4
⎩
i( t ) = 10 [u ( t ) − u ( t − 2)] − 10 [u ( t − 2) − u ( t − 4)]
i( t ) = 10 [ u(t ) − 2 u(t − 2) + u(t − 4)] A
Let I = ∫-∞ i dt .
t
For t < 0,
I = 0.
For 0 < t < 2, I = ∫0 10 dt = 10 t
t
For 2 < t < 4, I = ∫010 dt − 10 ∫2 dt = 20 − 10 t
2
For t > 4,
t
I = 20 − 10 t
4
2
t
2
= 40 − 10 t
=0
Thus,
⎧ 0
t<0
⎪⎪ 10 t
0<t<2
I=⎨
⎪ 40 − 10 t 2 < t < 4
⎪⎩ 0
t>4
or
I = 10 [ r(t ) − 2r(t − 2) + r(t − 4)] A
which is sketched below
∫ i dt
20
0
P.P.7.7
⎧ 2 − 2t 0 < t < 2
⎪
i( t ) = ⎨- 6 + 2t 2 < t < 3
⎪ 0
otherwise
⎩
2
4
t
i( t ) = (2 − 2t ) [ u ( t ) − u ( t − 2)] + (-6 + 2t) [ u(t - 2) - u(t - 3)]
i( t ) = 2 u ( t ) − 2 t u ( t ) + 4( t − 2) u ( t − 2) − 2( t − 3) u ( t − 3)
i ( t ) = 2 u( t ) − 2 r ( t ) + 4 r ( t − 2 ) − 2 r ( t − 3 ) A
h ( t ) = 4 [ u ( t ) − u ( t − 2)] + (6 − t ) [ u ( t − 2) − u ( t − 3)]
h ( t ) = 4 u ( t ) − ( t − 2) u ( t − 2) + r ( t − 6)
h ( t ) = 4 u( t ) − r ( t − 2 ) + r ( t − 6 )
P.P.7.8
P.P.7.9
(a)
(b)
∫ (t
∞
-∞
∫
10
0
3
+ 5t 2 + 10 ) δ( t + 3) dt = t 3 + 5t 2 + 10
t = -3
= -27 + 45 + 10 = 28
δ( t − π ) cos(3t ) dt = cos(3π ) = - 1
P.P.7.10
For t < 0, the capacitor acts like an open circuit.
−
v(0 ) = v(0 + ) = v(0) = 10
For t > 0,
R th
6
2
(10) −
(50) = -5
2+6
6+ 2
3
3 1 1
= 2 || 6 = Ω ,
τ = R th C = × =
2
2 3 2
v(∞) =
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
v( t ) = -5 + (10 + 5) e -2t
v( t ) = - 5 + 15 e -2t V
At t = 0.5,
v(0.5) = -5 + 15 e -1 = 518.2 mV
P.P.7.11
For t < 0, only the left portion of the circuit is operational at steady state.
−
v(0 ) = v(0 + ) = v(0) = 20 ,
i ( 0) = 0
For t > 0, 20 u (-t) = 0 so that the voltage source is replaced by a short circuit.
Transforming the current source leads to the circuit below.
10 Ω
i
10 Ω
+
v
−
0.2 F
+
−
30 V
5
(30) = 10
15
10
= 5 || 10 =
Ω,
3
v(∞) =
R th
τ = R th C =
10
2
× 0.2 =
3
3
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
v( t ) = 10 + (20 − 10) e -3t 2
v( t ) = 10 ( 1 + e -1.5 t )
- v( t )
= -2 ( 1 + e -1.5 t )
5
⎧
0
t<0
i( t ) = ⎨
-1.5t
⎩- 2 ( 1 + e ) A t > 0
i( t ) =
P.P.7.12
⎧
20 V
t<0
v( t ) = ⎨
- 1.5t
⎩10 ( 1 + e ) V t > 0
Applying source transformation, the circuit is equivalent to the one below.
i
1.5 H
10 Ω
t=0
5Ω
+
−
30 V
At t < 0, the switch is closed so that the 5 ohm resistor is short circuited.
30
i ( 0 − ) = i ( 0) =
=3A
10
For t > 0, the switch is open.
R th = 10 + 5 = 15 ,
i(∞) =
τ=
L
1 .5
=
= 0 .1
R th 15
30
=2A
10 + 5
i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ
i( t ) = 2 + (3 − 2) e -10t
i( t ) = ( 2 + e-10t ) A, t > 0
P.P.7.13
For 0 < t < 2, the given circuit is equivalent to that shown below.
10 Ω
20 Ω
i(t)
6A
15 Ω
5H
Since switch S1 is open at t = 0 − , i(0 − ) = 0 . Also, since i cannot jump, i(0) = i(0 − ) = 0 .
90
i(∞) =
=2A
15 + 10 + 20
L
5 1
R th = 45 Ω , τ =
=
=
R th 45 9
i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ
i( t ) = 2 + (0 − 2) e -9 t
i( t ) = 2 (1 − e -9 t ) A
When switch S 2 is closed, the 20 ohm resistor is short-circuited.
i(2 + ) = i(2 − ) = 2 (1 − e -18 ) ≅ 2
This will be the initial current
90
i(∞) =
= 3 .6 A
15 + 10
5 1
R th = 25 Ω , τ =
=
25 5
i( t ) = i(∞) + [ i(2 + ) − i(∞)] e -( t − 2 ) τ
i( t ) = 3.6 + (2 − 3.6) e -5( t − 2 )
i( t ) = 3.6 − 1.6 e -5( t − 2 )
⎧
0
t<0
⎪
0<t<2
Thus, i( t ) = ⎨ 2 (1 − e -9t ) A
⎪ 3.6 − 1.6 e -5( t − 2 ) A
t>2
⎩
At t = 1 ,
i(1) = 2 (1 − e -9 ) = 1.9997 A
At t = 3 ,
i(3) = 3.6 − 1.6 e -5 = 3.589 A
P.P.7.14
The op amp circuit is shown below.
C
+
v
−
Rf
1
2
R1
−
+
+
vo
−
Since nodes 1 and 2 must be at the same potential, there is no potential difference across
R 1 . Hence, no current flows through R 1 . Applying KCL at node 1,
v
dv
dv
v
+C
=0 ⎯
⎯→
+
=0
Rf
dt
dt CR f
which is similar to Eq. (7.4).
Hence,
v( t ) = v o e - t τ , τ = R f C
v(0) = v o = 4 ,
τ = (50 × 10 3 )(10 × 10 -6 ) = 0.5
v( t ) = 4 e -2 t V, t > 0
Alternatively, since no current flows through R 1 , the feedback loop forms a first order
RC circuit with v(0) = 4 and τ = R f C = 0.5 . Hence,
v( t ) = 4 e -2 t V, t > 0
To get to v o from v, we notice that v is the potential difference between node 1 and the
output terminal, i.e.
0 − vo = v ⎯
⎯→ v o = -v
v o = - 4 e -2t V , t > 0
P.P.7.15
Let v1 be the potential at the inverting terminal.
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
where τ = RC = 100 × 10 3 × 10 -6 = 0.1 ,
v ( 0) = 0
v1 = 0 for all t
v1 − v o = v
(1)
For t > 0, the switch is closed and the op amp circuit is an inverting amplifier with
- 100
v o (∞) =
( 4 mV) = -40 mV
10
From (1),
v(∞) = 0 − v o (∞) = 40 mV
Thus, v( t ) = 40 ( 1 − e -10 t ) mV
v o = v1 − v = -v
v o = 40 ( e -10t − 1 ) mV
P.P.7.16
This is a noninverting amplifier so that the output of the op amp is
⎛ Rf ⎞
⎟ vi
v a = ⎜1 +
⎝ R1 ⎠
⎛ Rf ⎞
⎛ 40 ⎞
⎟ v i = ⎜1 + ⎟ 2 u ( t ) = 6 u ( t )
v th = v a = ⎜1 +
⎝ 20 ⎠
⎝ R1 ⎠
To get R th , consider the circuit shown in Fig. (a), where R o is the output resistance of
the op amp. For an ideal op amp, R o = 0 so that
R th = R 3 = 10 kΩ
R3
Ro
Rth
Rth
R2
(a)
τ = R th C = 10 × 10 3 × 2 × 10 -6 =
Vth
+
−
C
(b)
1
50
The Thevenin equivalent circuit is shown in Fig. (b), which is a first order circuit.
Hence,
v o (t) = 6 ( 1 − e -t τ ) u(t)
v o ( t ) = 6 ( 1 − e -50 t ) u(t ) V
P.P.7.17
The schematic is shown in Fig. (a). Construct and save the schematic.
Select Analysis/Setup/Transient to change the Final Time to 5 s. Set the Print Step
slightly greater than 0 (20 ns is default). The circuit is simulated by selecting Analysis/
Simulate. In the Probe menu, select Trace/Add and display V(R2:2) as shown in Fig. (b).
(a)
(b)
P.P.7.18
The schematic is shown in Fig. (a). While constructing the circuit, rotate
L1 counterclockwise through 270° so that current i(t) enters pin 1 of L1 and set IC = 10
for L1. After saving the schematic, select Analysis/Setup/Transient to change the Final
Time to 1 s. Set the Print Step slightly greater than 0 (20 ns is default). The circuit is
simulated by selecting Analysis/ Simulate. After simulating the circuit, select Trace/Add
in the Probe menu and display I(L1) as shown in Fig. (b).
(a)
(b)
P.P.7.19
v(0) = 0 . When the switch is closed, we have the circuit shown below.
10 kΩ
R
a
+
80 μF
9V
−
b
We find the Thevenin equivalent at terminals a-b.
10 ( R + 4)
R th = (R + 4) || 10 =
R + 14
v th = v(∞) =
R+4
(9)
R + 14
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ ,
v( t ) = v(∞) ( 1 − e
-t τ
)
τ = R th C
Since v(0) = 0 ,
v( t )
9
( 1 − e -t τ ) mA
i( t ) =
=
R+4 R+4
Assuming R is in kΩ,
9
( 1 − e -t 0 τ ) × 10 -3
R + 14
R + 14
(0.12)
= 1 − e -t 0 τ
9
0.12R + 1.68 7.32 − 0.12R
=
e -t 0 τ = 1 −
9
9
120 × 10 -6 =
or
⎛
⎞
9
⎟
t 0 = τ ln ⎜
⎝ 7.32 − 0.12R ⎠
10 (R + 4)
⎛
⎞
9
⎟
t0 =
× 80 × 10 -6 × ln ⎜
⎝ 7.32 − 0.12R ⎠
R + 14
When R = 0,
⎛ 9 ⎞
40 × 80 × 10 -6
⎟ = 0.04723 s
× ln ⎜
t0 =
⎝ 7.32 ⎠
14
4 kΩ
When R = 6 kΩ,
100
⎛ 9 ⎞
× 80 × 10 - 6 × ln ⎜
t0 =
⎟ = 0.124 s
20
⎝ 6 .6 ⎠
The time delay is between 47.23 ms and 124 ms.
P.P.7.20
(a)
(b)
(c)
(d)
(e)
q = CV = (2 × 10 -3 )(80) = 0.16 C
1
1
CV 2 = ( 2 × 10 -3 )(6400) = 6.4 J
2
2
Δq
0.16
ΔI=
=
= 200 A
Δ t 0.8 × 10 -3
Δw
6 .4
=
= 8 kW
p=
Δ t 0.8 × 10 -3
Δq
0.16
Δt =
=
= 32 s
Δ I 5 × 10 -3
W=
L 500 × 10 -3
τ= =
= 2.5 ms
P.P.7.21
R
200
110
i ( 0) = 0 ,
i(∞) =
= 550 mA
200
i( t ) = 550 ( 1 − e - t τ ) mA
350 mA = i( t 0 ) = 550 ( 1 − e - t 0 τ ) mA
35
20
= 1 − e -t 0 τ ⎯
⎯→ e - t 0 τ =
55
55
e t0 τ =
55
20
⎛ 55 ⎞
⎛ 55 ⎞
t 0 = τ ln ⎜ ⎟ = 2.5 ln ⎜ ⎟ ms
⎝ 20 ⎠
⎝ 20 ⎠
t 0 = 2.529 ms
P.P.7.22
(a)
(b)
(c)
5L 5 × 20 × 10 -3
=
= 20 ms
R
5
2
⎛12 ⎞
1 2 1
-3 )
(
W = LI = 20 × 10 ⎜ ⎟ = 57.6 mJ
⎝5⎠
2
2
⎛ 12 5 ⎞
di
⎟ = 24 kV
V = L = 20 × 10 -3 ⎜
⎝ 2 × 10 -6 ⎠
dt
t = 5τ =
February 5, 2006
CHAPTER 8
P.P.8.1
(a)
At t = 0-, we have the equivalent circuit shown in Figure (a).
10 Ω
i
+
2Ω
v
vL
a
+
24V
+
−
2Ω
vC
−
−
(a)
(b)
+
i
50mF
+
−
i(0-) = 24/(2 + 10) = 2 A, v(0-) = 2i(0-) = 4 V
hence, v(0+) = v(0-) = 4V.
(b)
At t = 0+, the switch is closed.
L(di/dt) = vL, leads to (di/dt) = vL/L
But,
vC(0+) + vL(0+) = 24 = 4 + vL(0+), or vL(0+) = 20
(di(0+)/dt) = 20/0.4 = 50 A/s
C(dv/dt) = iC leading to (dv/dt) = iC/C
But at node a, KCL gives i(0+) = iC(0+) + v(0+)/2 = 2 = iC(0+) + 4/2
or iC(0+) = 0, hence (dv(0+)/dt) = 0 V/s
(c)
As t approaches infinity, the capacitor is replaced by an open circuit and the
inductor is replaced by a short circuit.
v(∞) = 24 V, and i(∞) = 12 A.
24V
P.P.8.2
(a)
At t = 0-, we have the equivalent circuit shown in (a).
5Ω
5Ω
a
iR
2A
iL
3A
+
vC
+
vR
10 μF
b
−
−
(a)
+
vL
2H
−
3A
(b)
iL(0-) = -3A, vL(0-) = 0, vR(0-) = 0
At t = 0+, we have the equivalent circuit in Figure (b). At node b,
iR(0+) = iL(0+) + 3, since iL(0+) = iL(0-) = -3A, iR(0+) = 0,
and vR(0+) = 5iR(0+) = 0. Thus, iL(0) = –3 A, vC(0) = 0, and vR(0+) = 0.
(b)
dvC(0+)/dt = iC(0+)/C = 2/0.2 = 10 V/s.
To get (dvR/dt), we apply KCL to node b, iR = iL + 3, thus diR/dt = diL/dt.
Since vR = 5iR, dvR/dt = 5diR/dt = 5diL/dt. But LdiL/dt = vL, diL/dt = vL/L.
Hence, dvR(0+)/dt = 5vL(0+)/L.
Applying KVL to the middle mesh in Figure (b),
-vC(0+) + vR(0+) + vL(0+) = 0 = 0 + 0 + vR(0+), or vR(0+) = 0
Hence, dvR(0+)/dt = 0 = diL(0+)/dt;
diL(0+)/dt = 0, dvC(0+)/dt = 10 V/s, dvR(0+)/dt = 0.
(c)
As t approaches infinity, we have the equivalent circuit shown below.
5Ω
2A
iL
3A
2
= 3 + iL(∞) leads to iL(∞) = -1A
vC(∞) = vR(∞) = 2x5 = 10V
Thus, iL(∞) = –1 A, vC(∞) = vR(∞) = 10 V
P.P.8.3
(a)
α = R/(2L) = 10/(2x5) = 1, ωo = 1
LC = 1
5x 2 x10 −2 = 10
s1,2 = − α ± α 2 − ωo2 = −1 ± 1 − 100 = -1 ± j9.95.
(b)
Since α < ωo, we clearly have an underdamped response.
P.P.8.4
For t < 0, the inductor is connected to the voltage source although it acts
like a short circuit.
i(0-) = 50/10 = 5 = i(0+) = i(0)
The voltage across the capacitor is 0 = v(0-) = v(0+) = v(0).
For t > 0, we have a source-free RLC circuit.
ωo = 1
LC = 1
1x
1
= 3
9
α = R/(2L) = 5/(2x1) = 2.5
Since α < ωo, we have an underdamped case.
s1,2 = − α ± α 2 − ωo2 = −2.5 ± 6.25 − 9 = -2.5 ± j1.6583
i(t) = e-2.5t[A1cos1.6583t + A2sin1.6583t]
We now determine A1 and A2.
i(0) = 5 = A1
di/dt = -2.5{e-2.5t[A1cos1.6583t + A2sin1.6583t]}
+ 1.6583e-2.5t[-A1sin1.6583t + A2cos1.6583t]
di(0)/dt = -(1/L)[Ri(0) + v(0)] = -2.5A1 + 1.6583A2
= -1[25] = -2.5(5) + 1.6583A2
A2 = -7.5378
Thus, i(t) = e-2.5t[5cos1.6583t – 7.538sin1.6583t] A
P.P.8.5
α = 1/(2RC) = 1/(2x2x25x10-3) = 10
ω0 = 1
LC = 1
0.4 x 25x10 −3 = 10
since α = ωo, we have a critically damped response. Therefore,
v(t) = [(A1 + A2t)e-10t]
v(0) = 0 = A1 + A2x0 = A1, which leads to v(t) = [A2te-10t].
dv(0)/dt = -(v(0) + Ri(0))/(RC) = -2x3/(2x25x10-3) = -120
dv/dt = [(A2 - 10A2t)e-10t]
At t = 0,
-120 = A2 therefore, v(t) = (–120t)e–10t volts
P.P.8.6
For t < 0, the switch is closed. The inductor acts like a short circuit while
the capacitor acts like an open circuit. Hence,
i(0) = 2 and v(0) = 0.
α = 1/(2RC) = 1/(2x20x4x10-3) = 6.25
ωo = 1
LC = 1 10 x 4 x10 −3 = 5
Since α > ωo, this is an overdamped response.
s1,2 = − α ± α 2 − ωo = −6.25 ± (6.25) 2 − 25 = -2.5 and –10
Thus, v(t) = A1e-2.5t + A2e-10t
v(0) = 0 = A1 + A2, which leads to A2 = -A1
dv(0)/dt = -(v(0) + Ri(0))/(RC) = -12.5(2x20) = -500
But, dv/dt = -2.5A1e-2.5t -10A2e-10t
At t = 0, -500 = -2.5A1 -10A2 = 7.5A1 since A1 = -A2
A1 = -66.67, A2 = 66.67
Thus, v(t) = 66.67(e–10t – e–2.5t) V
P.P.8.7
The initial capacitor voltage is obtained when the switch is in position a.
v(0) = [2/(2 + 1)]12 = 8V
The initial inductor current is i(0) = 0.
When the switch is in position b, we have the RLC circuit with the voltage source.
α = R/(2L) = 10/(2x2.5) = 2
ωo = 1
LC = 1 (5 / 2) x (1 / 40) = 4
Since α < ωo, we have an underdamped case.
s1,2 = − α ± α 2 − ωo = −2 ± (2) 2 − 16 = -2 ± j 3.464
Thus, v(t) = vf + [(A1cos3.464t + A2sin3.464t)e-2t]
where vf = v(∞) = 10, the final capacitor voltage. We now impose the initial
conditions to get A1 and A2.
v(0) = 8 = 10 + A1 leads to A1 = -2
The initial capacitor current is the same as the initial inductor current.
i(0) = C(dv(0)/dt) = 0 therefore, dv(0)/dt = 0
But, dv/dt = 3.464[(-A1sin3.464t + A2cos3.464t)e-2t]
-2[(A1cos3.464t + A2sin3.464t)e-2t]
dv(0)/dt = 0 – 2A1 + 3.464A2, which leads to A2 = -4/3.464 = -1.1547
Thus, v(t) = {10 + [(–2cos3.464t – 1.1547sin3.464t)e-2t]} V
i = C(dv/dt), vR = Ri = RC(dv/dt) = (1/4)dv/dt
= (1/4)[(4 – 4)cos3.464t + (2x1.1547 + 2x3.464)sin3.464t]e-2t
vR(t) = {[2.31sin3.464t]e-2t} V
P.P.8.8
When t < 0, v(0) = 0, i(0) = 0; for t > 0,
α = 0, ωo = 1 LC = 1 0.2x5 = 1
i(t) = is + A1cost + A2sint = 20 + A1cost + A2sint
i(0) = 0 = 20 + A1, therefore A1 = -20
Ldi(0)/dt = v(0) = 0
But di/dt = -A1sint + A2cost
At t = 0, di(0)/dt = 0 = 0 + A2 leading to i(t) = 20(1 – cost) A
v(t) = Ldi/dt = 5x20sint = 100sint V
P.P.8.9
At t = 0, the switch is open so that v(0) = 0, i(0-) = 0
(1)
For t > 0, the switch is closed. We have the equivalent circuit as in Figure (a).
iC
i
10 Ω
iC
4Ω
2A
(1/20)F
i
10 Ω
+
4Ω
2A
v
2H
−
(a)
(b)
v(0+) = 0, i(0+) = 0
(2)
-2 + iC + i = 0
(3)
From (3), i(0+) = 0 means that iC(0+) = 2, but iC(0+) = Cdv(0+)/dt
which leads to dv(0+)/dt = iC(0+)/C = 2/(1/20) = 40 V/s
As t approaches infinity, we have the equivalent circuit in (b).
i(∞) = 2 A, v(∞) = 4i(∞) = 8V
(5)
Next we find the network response by turning off the current source as shown in
Figure (c).
iC
i
10 Ω
4Ω
i
+
(1/20)F
v
2H
−
(c)
-v – 10iC + 4i + 2di/dt = 0
Applying KVL gives
Applying KCL to the top node,
(6)
i – iC = 0
Namely,
i = iC = -Cdv/dt = -(1/20)dv/dt
Combining (6) and (7),
-v – (10/20)dv/dt – (4/20)dv/dt – (2/20)d2v/dt2 = 0.
or
(7)
(d2v/dt2) + 7(dv/dt) + 10 = 0
The characteristic equation is s2 + 7s + 10 = 0 = (s + 2)(s + 5)
This means that vn = (Ae-2t + Be-5t) and vf = v(∞) = 8.
Thus, v = vf + vn
v = 8 + (Ae-2t + Be-5t)
(8)
v(0) = 0 = 8 + A +B, or A + B = -8
(9)
dv/dt = (-2Ae-2t -5Be-5t)
dv(0)/dt = 40 = -2A – 5B
2A + 5B = -40
(10)
From (9) and (10), A = 0 and B = -8.
Thus, v(t) = 8(1 – e-5t) V for all t > 0.
But, from (3), i = 2 – iC = 2 –(1/20)dv/dt = 2 –(1/20)(40)e-5t
i(t) = 2(1 – e-5t) A for all t > 0.
P.P.8.10
For t <0, 5u(t) = 0 so that v1(0-) = v2(0-) = 0
(1)
For t > 0, the circuit is as shown in Figure (a).
1Ω
1Ω
v1
1Ω
v2
1Ω
+
5V
+
−
0.5F
(1/3)F 5V
+
−
+
v1
v2
−
(a)
−
(b)
i1 = C1dv1/dt, or dv1/dt = i1/C1; likewise dv2/dt = i2/C2
i2(0+) = (v1(0+) – v2(0+))/1 = (0 – 0)/1 = 0
(5 – v1(0+))/1 = i1(0+) + i2(0+), or 5 = i1(0+)
Hence,
dv1(0+)/dt = 5/(1/2) = 10 V/s
(2a)
dv2(0+)/dt = 0
(2b)
As t approaches infinity, the capacitors can be replaced by open circuits as shown in
Figure (b). Thus,
v1(∞) = v2(∞) = 5V
(3)
Next we obtain the network response by considering the circuit in Figure (c).
1Ω
1Ω
v1
0.5F
(c)
v2
(1/3)F
Applying KCL at node 1 gives (v1/1) + (1/2)(dv1/dt) + (v1 – v2)/1 = 0
v2 = 2v1 + (1/2)dv1/dt
or
(4)
Applying KCL at node 2 gives (v1 – v2)/1 = (1/3)dv2/dt
or v1 = v2 + (1/3)dv2/dt
(5)
Substituting (5) into (4) yields,
v2 = 2v2 +(2/3)(dv2/dt) + (1/2)(dv2/dt) + (1/6)d2v2/dt2
or,
(d2v2/dt2) + (7dv2/dt) + 6v2 = 0
Now we have, s2 + 7s + 6 = 0 = (s + 1)(s + 6)
Thus, v2n = (Ae-t + Be-6t) and v2f = v2(∞) = 5V.
v2 = v2n + v2f = 5 + (Ae-t + Be-6t)
v2(0) = 0 which implies that A + B = -5
(6)
dv2/dt = (-Ae-t - 6Be-6t)
dv2(0) = 0 = -A – 6B
(7)
From (6) and (7), A = -6 and B = 1.
Thus, v2(t) = (5 – 6e-t + e-6t) V
From (5),
v1 = v2 + (1/3)dv2/dt
Thus, v1(t) = (5 – 4e-t – e-6t) V
Now we can find,
vo = v1 – v2 = (2e–t – 2e–6t)V, t > 0
P.P.8.11
Let v1 equal the voltage at non-inverting terminal of the op amp.
Then vo is equal to the output of the op amp.
At the non-inverting terminal, (vs – vo)/R1 = C1dv1/dt
(1)
At the output terminal of the op amp, (v1 – vo)/R2 = C2dvo/dt
(2)
We now eliminate v1 from (2), v1 = vo + R2C2dvo/dt
vs = v1 + R1C1dv1/dt
From (1)
(3)
(4)
Substituting (3) into (4) gives
vs = vo + R2C2dvo/dt + R1C1dvo/dt + R1C1R2C2d2vo/dt2
or
d2vo/dt2 + [(1/(R1C1)) + (1/(R2C2))]dvo/dt + vo/(R1R2C1C2) = vs/(R1R2C1C2)
With the given parameters,
(R1R2C1C2) = 104x104x20x10-6x100x10-6 = 2x10-2
1/(R1R2C1C2) = 5
[(1/(R1C1)) + (1/(R2C2))] = 10-4[(1/20x10-6) + (1/200x10-6)] = 6
Hence, we now have s2 + 6s + 5 = 0 = (s +1)(s + 5)
Therefore von = Ae-t + Be-5t, and vof = 4V
Thus,
vo = 4 + Ae-t + Be-5t
(5)
For t < 0, vs = 0, v1(0-) = 0 = vo(0-)
For t > 0, vs = 4, but
v1(0+) - vo(0+) =0
From (2),
dvo(0+)/dt = [v1(0+) – vo(0+)]/R2C2 = 0
(6)
(7)
Imposing these conditions on vo(t),
0 = 4+A+B
(8)
0 = -A – 5B or A = -5B
(9)
From (8) and (9), A = -5 and B = 1
vo(t) = (4 – 5e-t + e-5t) V, t > 0
P.P.8.12
We follow the same procedure as in Example 8.12. The schematic
is shown in Figure (a). The current marker is inserted to display the inductor current.
After simulating the circuit, the required inductor current is plotted in Figure (b).
P.P.8.13
When the switch is at position a, the schematic is as shown in
Figure (a). We carry out dc analysis on this to obtain initial conditions. It is evident that
vC(0) = 8 volts.
(a)
With the switch in position b, the schematic is as shown in Figure (b). A voltage marker
is inserted to display the capacitor voltage. When the schematic is saved and run, the
output is as shown in Figure (c).
(b)
P.P.8.14
The dual circuit is obtained from the original circuit as shown in
Figure (a). It is redrawn as shown in Figure (b).
3H
50mA
3F
4H
10 Ω
50mV
+
−
0.1 Ω
(a)
4 μF
3H
0.1 Ω
50mV
+
−
4F
(b)
P.P.8.15
The dual circuit is obtained in Figure (a) and redrawn in Figure (b).
5Ω
0.2F
4H
0.2 Ω
4F
0.2 H
2A
1/3 Ω
2V
+
−
+
−
3Ω
20A
(a)
20 V
1/3 Ω
4F
0.2 H
2V
0.2 Ω
+
−
20A
(b)
P.P.8.16
Since 12 = 4i + vL + vC
or vC = 12 – 4i - vL
-(vC – 12) = 4i + vL = e-250t(12cosωdt + 0.2684sinωdt – 268sinωdt)
vC(t) = [12 – 12e-250tcos11.18t + 267.7e-250tsin11.18t] V
P.P.8.17
We follow the same procedure as in Example 8.17. The schematic
is as shown in Figure (a) with two voltage markers to display both input and output
voltages. When the schematic is saved and run, the result is as displayed in Figure (b).
(a)
(a)
(b)
February 5, 2006
CHAPTER 9
P.P.9.1
amplitude = 5
phase = -60°
angular frequency (ω) = 4π = 12.57 rad/s
2π
period (T) =
= 0.5 s
ω
1
frequency (f) =
= 2 Hz
T
i1 = -4 sin(ωt + 25°) = 4 cos(ωt + 25° + 90°)
i1 = 4 cos(ωt + 115°) ,
ω = 377 rad/s
Compare this with
i 2 = 5 cos(ωt − 40°)
indicates that the phase angle between i1 and i 2 is
115° + 40° = 155°
Thus,
i1 leads i2 by 155°
P.P.9.2
P.P.9.3
P.P.9.4
(a)
(5 + j2)(-1 + j4) = -5 + j20 – j2 – 8 = -13 + j18
5∠60° = 2.5 + j4.33
(5 + j2)(-1 + j4) – 5∠60° = -15.5 + j13.67
[ (5 + j2)(-1 + j4) – 5∠60 ]* = -15.5 – j13.67 = 20.67∠221.41°
(b)
3∠40° = 2.298 + j1.928
10 + j5 + 3∠40° = 12.298 + j6.928 = 14.115∠29.39°
-3 + j4 = 5∠126.87°
10 + j5 + 3∠40° 14.115∠29.39°
=
= 2.823∠ - 97.48°
- 3 + j4
5∠126.87°
2.823∠-97.48° = -0.3675 – j2.8
10∠30° = 8.66 + j5
10 + j5 + 3∠40°
+ 10∠30° = 8.293 + j2.2
- 3 + j4
(a)
-cos(A) = cos(A – 180°) = cos(A + 180°)
Hence,
v = -7 cos(2t + 40°) = 7 cos(2t + 40° + 180°)
v = 7 cos(2t + 220°)
The phasor form is
V = 7∠220° V
Since sin(A) = cos(A – 90°),
(b)
i = 4 sin(10t + 10°) = 4 cos(10t+10° – 90°)
i = 4 cos(10t – 80°)
The phasor form is
I = 4∠-80° A
P.P.9.5
(a)
Since -1 = 1∠-180° = 1∠180°
V = -10∠30° = 10∠(30°+180°) = 10∠210°
The sinusoid is
v(t) = 10 cos(ωt + 210°) V
(b)
I = j (5 – j12) = 12 + j5 = 13∠22.62°
The sinusoid is
i(t) = 13 cos(ωt + 22.62°) A
P.P.9.6
Let V = -10 sin( ωt + 30°) + 20 cos( ωt − 45°)
Then, V = 10 cos(ωt + 30° + 90°) + 20 cos(ωt − 45°)
Taking the phasor of each term
V = 10∠120° + 20∠-45°
V = -5 + j8.66 + 14.14 – j14.14
V = 9.14 – j5.48 = 10.66∠-30.95°
Converting V to the time domain
v(t) = 10.66 cos(ωt – 30.95°)V
P.P.9.7
Given that
dv
2 + 5v + 10 ∫ v dt = 20 cos(5t − 30°)
dt
we take the phasor of each term to get
10
2jω V +5 V +
V = 20∠-30°,
ω = 5
jω
V [j10 + 5 – j(10/5)] = V (5 + j8) = 20∠-30°
20 ∠ - 30° 20 ∠ - 30°
=
V =
5 + j8
9.434 ∠58°
V = 2.12∠-88°
Converting V to the time domain
v(t) = 2.12 cos(5t – 88°)V
P.P.9.8
For the capacitor,
V = I / (jωC),where V = 6∠-30°, ω = 100
I = jωC V = (j100)(50x10-6)(6∠-30°)
I = 30∠60° mA
i(t) = 30 cos(100t + 60°) mA
P.P.9.9
Vs = 5∠0°,
ω = 10
Z = 4 + jωL = 4 + j2
5∠0° 5 (4 − j2)
=
= 1 – j0.5 = 1.118∠-26.57°
4 + j2
16 + 4
V = jωL I = j2 I = (2∠90°)(1.118∠-26.57°) = 2.236∠63.43°
I = Vs / Z =
Therefore,
v(t) = 2.236 sin(10t + 63.43°) V
i(t) = 1.118 sin(10t – 26.57°) A
P.P.9.10
Let
Z1 = impedance of the 2-mF capacitor in series with the 20-Ω resistor
Z2 = impedance of the 4-mF capacitor
Z3 = impedance of the 2-H inductor in series with the 50-Ω resistor
1
1
= 20 +
= 20 − j50
jωC
j (10)(2 × 10 -3 )
1
1
Z2 =
=
= -j25
jωC j (10)(4 × 10 -3 )
Z3 = 50 + jωL = 50 + j (10)( 2) = 50 + j 20
Z1 = 20 +
Zin = Z1 + Z2 || Z3 = Z1 + Z2 Z3 / (Z2 + Z3)
- j25x (50 + j20)
Zin = 20 − j50 +
- j25 + 50 + j20
Zin = 20 – j50 + 12.38 – j23.76
Zin = 32.38 – j73.76 Ω
P.P.9.11
In the frequency domain,
the voltage source is Vs = 10∠75°
the 0.5-H inductor is jωL = j (10)(0.5) = j5
1
1
1
-F capacitor is
the
=
= - j2
jωC j (10)(1 20)
20
Let
and
Z1 = impedance of the 0.5-H inductor in parallel with the 10-Ω resistor
Z2 = impedance of the (1/20)-F capacitor
(10)( j5)
= 2 + j4
and
Z2 = -j2
10 + j5
Vo = Z2 / (Z1 + Z2) Vs
− j (10 ∠75°) 10 ∠(75° − 90°)
− j2
Vo =
(10 ∠75°) =
=
2 + j4 − j2
1+ j
2 ∠45°
Vo = 7.071∠-60°
vo(t) = 7.071 cos(10t – 60°) V
Z1 = 10 || j5 =
P.P.9.12
We need to find the equivalent impedance via a delta-to-wye
transformation as shown below.
c
Zcn
n
Zan
30∠0° V
Zbn
+
−
b
a
5Ω
10 Ω
-j2 Ω
4 (-5 + j8)
j4 (8 + j5)
=
= 0.32 + j3.76
8 + j6
j4 + 8 + j5 − j3
- j3 (8 + j5)
3 (5 − j8)(8 − j6)
=
=
= -0.24 – j2.82
8 + j6
100
j4 (- j3) 12 (8 − j6)
=
=
= 0.96 – j0.72
8 + j6
100
Zan =
Zbn
Zcn
The total impedance from the source terminals is
Z = Zcn + (Zan + 5 – j2) || (Zbn + 10)
Z = Zcn + (5.32 + j1.76) || (9.76 – j2.82)
(5.32 + j1.76) (9.76 − j2.82)
Z = Zcn +
(5.32 + j1.76) + (9.76 − j2.82)
Z = 0.96 – j0.72 + 3.744 + j0.4074
Z = 4.704 – j0.3126 = 4.714∠-3.802°
Therefore,
30 ∠0°
4.714 ∠ − 3.802°
I = 6.364∠3.802° A
I = V/Z =
Let us now check this using PSpice. The solution produces the magnitude of I =
6.364E+00, and the phase angle = 3.803E+00, which agrees with the above answer.
P.P.9.13
To show that the circuit in Fig. (a) meets the requirement, consider the
equivalent circuit in Fig. (b).
Z = -j10 || (10 – j10) =
10 Ω
V1
10 Ω
+
Vi
- j10 (10 − j10)
- j (10 − j10)
=
= 2 – j6 Ω
10 − j20
1 − j2
10 Ω
+
+
-j10 Ω
-j10 Ω
−
Vo
Vi = 10 V
+
−
V1
−
−
(a)
(b)
Z = 2−j6 Ω
2 − j6
10
(10) =
(1 − j)
10 + 2 − j6
3
⎛ - j ⎞ ⎛ 10 ⎞
- j10
10
⎟ ⎜ ⎟ (1 − j) = - j
Vo =
V1 = ⎜
3
10 − j10
⎝1 − j ⎠⎝ 3 ⎠
10
Vo =
∠ - 90°
3
This implies that the RC circuit provides a 90° lagging phase shift.
10
= 3.333 V
The output voltage is
3
V1 =
P.P.9.14
the 1-mH inductor is jωL = j (2π)(5 × 10 3 )(1 × 10 -3 ) = j31.42
the 2-mH inductor is jωL = j (2π)(5 × 10 3 )(2 × 10 -3 ) = j62.83
Consider the circuit shown below.
j31.42 Ω
V1
j62.83 Ω
+
Vi
+
10 Ω
50 Ω
−
Vo
−
(10)(50 + j62.83)
60 + j62.83
Z = 9.205 + j0.833 = 9.243∠5.17°
Z = 10 || (50 + j62.83) =
9.243∠5.17°
(1) = 0.276∠-68.9°
9.205 + j32.253
50 (0.276 ∠ - 68.9°)
50
V1 =
Vo =
= 0.172∠-120.4°
80.297 ∠51.49°
50 + j62.83
V1 = Z / (Z + j31.42) Vi =
Therefore,
magnitude = 0.172
phase = 120.4°
phase shift is lagging
P.P.9.15
Zx = (Z3 / Z1) Z2
Z3 = 12 kΩ
Z1 = 4.8 kΩ
Z2 = 10 + jωL = 10 + j (2π )(6 × 10 6 ) (0.25 × 10 -6 ) = 10 + j9.425
12k
Zx =
(10 + j9.425) = 25+ j23.5625 Ω
4 .8 k
Rx = 25,
Xx = 23.5625 = ωLx
Xx
23.5625
Lx =
=
= 0.625 μH
2πf 2π (6 × 10 6 )
i.e. a 25-Ω resistor in series with a 0.625-μH inductor.
February 5, 2006
CHAPTER 10
P.P.10.1
10 sin( 2 t ) ⎯
⎯→ 10 ∠0°, ω = 2
jωL = j4
1
0.2 F ⎯
⎯→
= - j2.5
jωC
Hence, the circuit in the frequency domain is as shown below.
2H
⎯
⎯→
-j2.5 Ω
V1
V2
4Ω
+
10∠0° A
2Ω
Vx
j4 Ω
+
−
3Vx
−
At node 1,
At node 2,
V1 V1 − V2
+
2
- j2.5
100 = (5 + j4) V1 − j4V2
(1)
V2 V1 − V2 3Vx − V2
where Vx = V1
=
+
j4
- j2.5
4
- j2.5V2 = j4 (V1 − V2 ) + 2.5 (3V1 − V2 )
0 = - (7.5 + j4) V1 + (2.5 + j1.5) V2
(2)
10 =
Put (1) and (2) in matrix form.
⎡ 5 + j4
- j4 ⎤⎡ V1 ⎤ ⎡100 ⎤
⎢ - (7.5 + j4) 2.5 + j1.5⎥⎢ V ⎥ = ⎢ 0 ⎥
⎣
⎦⎣ 2 ⎦ ⎣ ⎦
where Δ = (5 + j4)( 2.5 + j.15) − (-j4)(-(7.5 + j4)) = 22.5 − j12.5 = 25.74 ∠ - 29.05 °
j4 ⎤
⎡2.5 + j1.5
⎥
⎢
⎡ V1 ⎤ ⎣ 7.5 + j4 5 + j4⎦ ⎡100⎤
⎢V ⎥ =
⎢ 0 ⎥
22.5 − j12.5
⎣ ⎦
⎣ 2⎦
2.5 + j1.5
2.915∠30.96°
V1 =
(100) =
(100) = 11.32 ∠60.01°
22.5 − j12.5
25.74 ∠ - 29.05°
7.5 + j4
8.5∠28.07°
V2 =
(100) =
(100) = 33.02 ∠57.12°
22.5 − j12.5
25.74 ∠ - 29.05°
In the time domain,
v1 ( t ) = 11.32 sin(2t + 60.01°) V
v 2 ( t ) = 33.02 sin(2t + 57.12°) V
P.P.10.2
The only non-reference node is a supernode.
15 − V1 V1 V2 V2
=
+
+
4
j4 - j
2
15 − V1 = - j V1 + j4V2 + 2V2
15 = (1 − j) V1 + (2 + j4) V2
(1)
The supernode gives the constraint of
V1 = V2 + 20∠60°
(2)
Substituting (2) into (1) gives
15 = (1 − j)(20∠60°) + (3 + j3) V2
15 − (1 − j)(20 ∠60°) 14.327 ∠210.72°
V2 =
=
= 3.376 ∠165.7°
3 + j3
4.243∠45°
V1 = V2 + 20∠60° = (-3.272 + j0.8327) + (10 + j17.32)
V1 = 6.728 + j18.154
Therefore,
V1 = 19.36∠69.67° V,
P.P.10.3
Consider the circuit below.
V2 = 3.376∠165.7° V
2∠0° A
I3
-j2 Ω
8Ω
For mesh 1,
I1
(8 − j2 + j4) I1 − j4 I 2 = 0
(8 + j2) I1 = j4 I 2
6Ω
j4 Ω
I2
+
−
10∠30° V
(1)
For mesh 2,
(6 + j4) I 2 − j4 I 1 − 6 I 3 + 10∠30° = 0
For mesh 3,
I 3 = -2
Thus, the equation for mesh 2 becomes
(6 + j4) I 2 − j4 I1 = -12 − 10∠30°
From (1),
I2 =
(2)
8 + j2
I = (0.5 − j2) I 1
j4 1
(3)
Substituting (3) into (2),
(6 + j4) (0.5 − j2) I 1 − j4 I1 = -12 − 10∠30°
(11 − j14) I 1 = -(20.66 + j5)
- (20.66 + j5)
I1 =
11 − j14
20.66 + j5 21.256 ∠13.6°
=
11 − j14
17.8∠ - 51.84°
I o = 1.194∠65.44° A
Hence,
I o = - I1 =
P.P.10.4
Meshes 2 and 3 form a supermesh as shown in the circuit below.
10 Ω
-j4 Ω
j8 Ω
I2
50∠0° V
+
−
I1
I3
5Ω
-j6 Ω
− 50 + (15 − j4) I 1 − (− j4) I 2 − 5 I 3 = 0
(15 − j4) I 1 + j4 I 2 − 5 I 3 = 50
(1)
For the supermesh,
( j8 − j4) I 2 + (5 − j6) I 3 − (5 − j4) I 1 = 0
(2)
Also,
I3 = I2 + 2
(3)
For mesh 1,
Eliminating I 3 from (1) and (2)
(15 − j4) I 1 + (-5 + j4) I 2 = 60
(-5 + j4) I1 + (5 − j2) I 2 = -10 + j12
(4)
(5)
From (4) and (5),
⎡15 − j4 - 5 + j4 ⎤⎡ I 1 ⎤ ⎡ 60 ⎤
⎢ - 5 + j4 5 - j2 ⎥⎢ I ⎥ = ⎢ - 10 + j12 ⎥
⎦
⎣
⎦⎣ 2 ⎦ ⎣
15 − j4 - 5 + j4
= 58 − j10 = 58.86 ∠ - 9.78°
- 5 + j4 5 - j2
Δ=
Δ1 =
60
- 5 + j4
= 298 − j20 = 298.67 ∠ - 3.84°
- 10 + j12 5 - j2
Δ1
= 5.074∠5.94° A
Δ
Thus,
I o = I1 =
P.P.10.5
Let I o = I 'o + I "o , where I 'o and I "o are due to the voltage source and
current source respectively. For I 'o consider the circuit in Fig. (a).
-j2 Ω
6Ω
Io'
8Ω
j4 Ω
I1
I2
+
−
10∠30° V
(a)
For mesh 1,
For mesh 2,
(8 + j2) I1 − j4 I 2 = 0
I 2 = (0.5 − j2) I1
(1)
(6 + j4) I 2 − j4 I 1 − 10∠30° = 0
(2)
Substituting (1) into (2),
(6 + j4)(0.5 − j2) I 1 − j4 I 1 = 10∠30°
10∠30°
I 'o = I 1 =
= 0.08 + j0.556
11 − j14
For I "o consider the circuit in Fig. (b).
2∠0° A
-j2 Ω
6Ω
Io"
8Ω
j4 Ω
(b)
Let
j24
= 1.846 + j2.769 Ω
6 + j4
Z2
(2)(1.846 + j2.769)
I "o =
(2) =
= 0.4164 + j0.53
Z1 + Z 2
9.846 + j0.77
Z1 = 8 − j2 Ω ,
Therefore,
Z 2 = 6 || j4 =
I o = I 'o + I "o = 0.4961 + j1.086
I o = 1.1939∠65.45° A
P.P.10.6
Let v o = v 'o + v "o , where v 'o is due to the voltage source and v "o is due to
the current source. For v 'o , we remove the current source.
30 sin(5t ) ⎯
⎯→ 30 ∠0°, ω = 5
1
1
0 .2 F ⎯
⎯→
=
= -j
jωC j (5)(0.2)
1H
⎯
⎯→
jωL = j (5)(1) = j5
The circuit in the frequency domain is shown in Fig. (a).
8Ω
+
30∠0° V
+
−
-j Ω
Vo'
−
(a)
j5 Ω
Note that
- j || j5 = -j1.25
By voltage division,
- j1.25
(30) = 4.631∠ - 81.12°
8 − j1.25
v 'o = 4.631 sin(5t − 81.12°)
Vo' =
Thus,
For v "o , we remove the voltage source.
2 cos(10 t ) ⎯
⎯→ 2 ∠0°, ω = 10
1
1
0 .2 F ⎯
⎯→
=
= - j0.5
jωC j (10)(0.2)
1H ⎯
⎯→
jωL = j (10)(1) = j10
The corresponding circuit in the frequency domain is shown in Fig (b).
+
8Ω
j10 Ω
Vo"
I
-j0.5 Ω
2∠0°
−
(b)
Let
Z1 = - j0.5 ,
Z 2 = 8 || j10 =
j80
= 4.878 + j3.9
8 + j10
By current division,
I=
Z2
(2)
Z1 + Z 2
Thus,
Z2
- j (4.877 + j3.9)
(2)(-j0.5) =
Z1 + Z 2
4.878 + j3.4
6.245∠ - 51.36°
Vo" =
= 1.051∠ - 86.24°
5.94 ∠34.88°
v "o = 1.051 cos(10 t − 86.24°)
Therefore,
v o = v 'o + v "o
Vo" = I (-j0.5) =
v o = 4.631 sin(5t – 81.12°) + 1.051 cos(10t – 86.24°) V
P.P.10.7
If we transform the current source to a voltage source, we obtain the
circuit shown in Fig. (a).
4Ω
-j3 Ω
2Ω
jΩ
Io
VS
+
−
1Ω
j5 Ω
-j2 Ω
(a)
Vs = I s Z s = ( j4)(4 − j3) = 12 + j16
We transform the voltage source to a current source as shown in Fig. (b).
V 12 + j16
Let
Z = 4 − j3 + 2 + j = 6 − j2 . Then,
Is = s =
= 1.5 + j3 .
Z
6 − j2
Io
6Ω
j5 Ω
IS
-j2 Ω
(b)
Note that
Z || j5 =
(6 − j2)( j5) 10
= (1 + j) .
6 + j3
3
By current division,
Io =
10
(1 + j)
3
(1.5 + j3)
10
(1 + j) + (1 − j2)
3
− 20 + j40 44.72∠116.56°
Io =
=
13 + j4
13.602∠17.1°
I o = 3.288∠99.46° A
1Ω
-j2 Ω
P.P.10.8
When the voltage source is set equal to zero,
Z th = 10 + (- j4) || (6 + j2)
(-j4)(6 + j2)
Z th = 10 +
6 - j2
Z th = 10 + 2.4 − j3.2
Z th = 12.4 – j3.2 Ω
By voltage division,
- j4
(- j4)(30∠20°)
(30∠20°) =
6 + j2 − j4
6 − j2
(4 ∠ - 90°)(30 ∠20°)
Vth =
6.324 ∠ - 18.43°
Vth = 18.97∠-51.57° V
Vth =
P.P.10.9
To find Vth , consider the circuit in Fig. (a).
8 + j4
8 + j4
+ Vo −
+ Vo −
5∠0°
V2
V1
4 – j2
0.2Vo
(a)
At node 1,
At node 2,
VS a
a
4 – j2
b
V − V2
0 − V1
= 5+ 1
4 − j2
8 + j4
- (2 + j)V1 = 50 + (1 − j0.5)(V1 − V2 )
50 = (1 − j0.5)V2 − (3 + j0.5)V1
5 + 0.2Vo +
V1 − V2
= 0,
8 + j4
Hence, the equation for node 2 becomes
Is
+
−
0.2Vo
(b)
b
(1)
where Vo = V1 − V2 .
1∠0°
5 + 0.2 (V1 − V2 ) +
V1 = V2 −
V1 − V2
=0
8 + j4
50
3 + j0.5
(2)
Substituting (2) into (1),
50 = (1 − j0.5)V2 − (3 + j0.5)V2 + (50)
3 + j0.5
3 − j0.5
50
(35 + j12)
37
- 2.702 + j16.22
V2 =
= 7.35∠72.9°
2+ j
Vth = V2 = 7.35∠72.9° V
0 = -50 − (2 + j) V2 +
To find Z th , we remove the independent source and insert a 1-V voltage source between
terminals a-b, as shown in Fig. (b).
At node a,
But,
So,
and
I s = -0.2Vo +
Vs
8 + j4 + 4 − j2
8 + j4
V
8 + j4 + 4 − j2 s
1
8 + j4
2.6 + j0.8
I s = (0.2)
+
=
12 + j2 12 + j2
12 + j2
V
1
12 + j2
12.166∠9.46°
=
=
Z th = s =
I s I s 2.6 + j0.8 2.72∠17.10°
Z th = 4.473∠–7.64° Ω
Vs = 1
and
– Vo =
P.P.10.10
To find Z N , consider the circuit in Fig. (a).
4Ω
j2 Ω
4Ω
j2 Ω
I3
8Ω
1Ω
-j3 Ω
8Ω
1Ω
-j3 Ω
a
ZN
20∠0°
+
−
b
(a)
-j4 Ω
I1
IN
I2
(b)
Z N = (4 + j2) || (9 − j3) =
a
b
(4 + j2)(9 − j3)
13 − j
Z N = 3.176 + j0.706 Ω
To find I N , short-circuit terminals a-b as shown in Fig. (b). Notice that meshes 1 and 2
form a supermesh.
For the supermesh,
- 20 + 8 I 1 + (1 − j3) I 2 − (9 − j3) I 3 = 0
(1)
Also,
I1 = I 2 + j4
(2)
For mesh 3,
(13 − j) I 3 − 8 I 1 − (1 − j3) I 2 = 0
(3)
Solving for I 2 , we obtain
50 − j62 79.65∠ - 51.11°
=
9 − j3
9.487 ∠ - 18.43°
I N = 8.396∠-32.68° A
IN = I2 =
Using the Norton equivalent, we can find I o as in Fig. (c).
Io
IN
ZN
(c)
10 – j5 Ω
By current division,
ZN
3.176 + j0.706
IN =
(8.396∠ - 32.68°)
Z N + 10 − j5
13.176 − j4.294
(3.254 ∠12.53°)(8.396 ∠ - 32.68°)
Io =
13.858∠ - 18.05°
I o = 1.971∠-2.10° A
Io =
P.P.10.11
1
1
=
= -j20 kΩ
3
jωC1 j (5 × 10 )(10 × 10 -9 )
1
1
=
= -j10 kΩ
3
jωC 2 j (5 × 10 )(20 × 10 -9 )
10 nF ⎯
⎯→
20 nF ⎯
⎯→
Consider the circuit in the frequency domain as shown below.
-j20 kΩ
10 kΩ
20 kΩ
+
−
V1
2∠0° V
V2
+
−
Io
Vo
-j10 kΩ
As a voltage follower, V2 = Vo
At node 1,
At node 2,
2 − V1 V1 − Vo V1 − Vo
=
+
10
- j20
20
4 = (3 + j)V1 − (1 + j)Vo
(1)
V1 − Vo Vo − 0
=
20
- j10
V1 = (1 + j2)Vo
(2)
Substituting (2) into (1) gives
4 = j6Vo
or
Vo =
2
∠ - 90°
3
Hence,
v o ( t ) = 0.667 cos(5000t − 90°) V
v o ( t ) = 0.667 sin(5000t) V
Now,
Io =
But from (2)
Vo − V1 = - j2Vo =
Io =
Vo − V1
- j20k
-4
3
-4 3
= - j66.66 μA
- j20k
Hence,
i o ( t ) = 66.67 cos(5000t − 90°) μA
i o ( t ) = 66.67 sin(5000t) μA
P.P.10.12
Let Z = R ||
1
R
=
jωC 1 + jωRC
Vs
R
=
Vo R + Z
The loop gain is
1/ G =
Vs
R
=
=
Vo R + Z
R
R+
R
1 + jωRC
where ωRC = (1000)(10 × 10 3 )(1 × 10 -6 ) = 10
1 + j10 10.05∠84.29°
=
2 + j10 10.2∠78.69°
G = 1.0147∠–5.6°
1/ G =
=
1 + jωRC
2 + jωRC
P.P.10.13
The schematic is shown below.
Since ω = 2πf = 3000 rad / s ⎯
⎯→ f = 477.465 Hz . Setup/Analysis/AC Sweep as
Linear for 1 point starting and ending at a frequency of 447.465 Hz. When the schematic
is saved and run, the output file includes
Frequency
4.775E+02
IM(V_PRINT1)
5.440E-04
IP(V_PRINT1)
-5.512E+01
Frequency
4.775E+02
VM($N_0005)
2.683E-01
VP($N_0005)
-1.546E+02
From the output file, we obtain
Vo = 0.2682∠-154.6° V
and
I o = 0.544∠-55.12° mA
Therefore,
v o ( t ) = 0.2682 cos(3000t – 154.6°) V
i o ( t ) = 0.544 cos(3000t – 55.12°) mA
P.P.10.14
The schematic is shown below.
We select ω = 1 rad/s and f = 0.15915 Hz. We use this to obtain the values of
capacitances, where C = 1 ωX c , and inductances, where L = X L ω . Note that IAC does
not allow for an AC PHASE component; thus, we have used VAC in conjunction with G
to create an AC current source with a magnitude and a phase. To obtain the desired
output use Setup/Analysis/AC Sweep as Linear for 1 point starting and ending at a
frequency of 0.15915 Hz. When the schematic is saved and run, the output file includes
Frequency
1.592E-01
IM(V_PRINT1)
2.584E+00
IP(V_PRINT1)
1.580E+02
Frequency
1.592E-01
VM($N_0004)
9.842E+00
VP($N_0004)
4.478E+01
From the output file, we obtain
Vx = 9.842∠44.78° V
P.P.10.15
P.P.10.16
and
I x = 2.584∠158° A
⎛ R2 ⎞
⎛ 10 × 10 6 ⎞
-9 )
⎟ C = ⎜1 +
⎟(
= 10 μF
C eq = ⎜1 +
3 10 × 10
R
10
×
10
⎝
⎠
⎝
1 ⎠
C = C1 = C 2 = 1 nF
If R = R 1 = R 2 = 2.5 kΩ
and
1
1
fo =
=
= 63.66 kHz
2πRC (2π)(2.5 × 10 3 )(1 × 10 -9 )
February 5, 2006
CHAPTER 11
P.P.11.1
i( t ) = 15 sin(10 t + 60°) = 15 cos(10 t − 30°)
v( t ) = 80 cos(10 t + 20°)
p( t ) = v( t ) i( t ) = (80)(15) cos(10 t + 20°) cos(10 t − 30°)
1
p( t ) = ⋅ 80 ⋅ 15 [cos( 20 t + 20° − 30°) + cos( 20 − -30°)]
2
p( t ) = 600 cos( 20t − 10°) + 385.7 W
P=
P.P.11.2
1
V I cos(θ v − θi ) = 385.7 W
2 m m
V = I Z = 200 ∠8°
1
V I cos(θ v − θi )
2 m m
1
P = (200)(10) cos(8° − 30°) = 927.2 W
2
P=
P.P.11.3
3Ω
8∠45° V
I=
+
−
I
j1 Ω
8∠45°
= 2.53∠26.57°
3+ j
For the resistor,
I R = I = 2.53∠26.57°
VR = 3 I = 7.59∠26.57°
1
1
PR = Vm I m = (2.53)(7.59) = 9.6 W
2
2
For the inductor,
I L = 2.53∠26.57°
VL = j I L = 2.53∠(26.57° + 90°) = 2.53∠116.57°
1
PL = (2.53) 2 cos(90°) = 0 W
2
The average power supplied is
1
P = (8)(2.53) cos( 45° − 26.57°) = 9.6 W
2
P.P.11.4
Consider the circuit below.
8Ω
40 V
+
−
I1
j4 Ω
-j2 Ω
I2
+
−
j20 V
For mesh 1,
- 40 + (8 − j2) I1 + (- j2) I 2 = 0
(4 − j) I1 − j I 2 = 20
(1)
- j20 + ( j4 − j2) I 2 + (- j2) I1 = 0
- j I 1 + j I 2 = j10
(2)
For mesh 2,
In matrix form,
⎡ 4 − j - j⎤⎡ I 1 ⎤ ⎡ 20 ⎤
=
⎢ -j
j ⎥⎦⎢⎣ I 2 ⎥⎦ ⎢⎣ j10 ⎥⎦
⎣
Δ = 2 + j4 ,
I1 =
Δ 1 = -10 + j20 ,
Δ1
= 5∠53.14° and
Δ
I2 =
Δ 2 = 10 + j60
Δ2
= 13.6∠17.11°
Δ
For the 40-V voltage source,
Vs = 40∠0°
I 1 = 5∠53.14°
-1
Ps = (40)(5) cos(-53.14°) = - 60 W
2
For the j20-V voltage source,
Vs = 20∠90°
I 2 = 13.6∠17.11°
-1
Ps = (20)(13.6) cos(90° − 17.11°) = - 40 W
2
For the resistor,
I = I1 = 5
V = 8 I 1 = 40
1
P = (40)(5) = 100 W
2
The average power absorbed by the inductor and capacitor is zero watts.
P.P.11.5
We first obtain the Thevenin equivalent circuit across Z L . Z Th is
obtained from the circuit in Fig. (a).
-j4 Ω
j10 Ω
8Ω
Zth
5Ω
(a)
Z Th = 5 || (8 − j4 + j10) =
(5)(8 + j6)
= 3.415 + j0.7317
13 + j6
VTh is obtained from the circuit in Fig. (b).
-j4 Ω
j10 Ω
I
8Ω
5Ω
2A
+
Vth
−
(b)
By current division,
I=
8 − j4
(2)
8 − j4 + j10 + 5
VTh = 5 I =
(10)(8 − j4)
= 6.25∠ - 51.34°
13 + j6
Z L = Z *Th = 3.415 − j0.7317 Ω
Pmax =
P.P.11.6
Let
VTh
2
8RL
=
(6.25) 2
= 1.429 W
(8)(3.415)
We first find Z Th and VTh across R L .
Z1 = 80 + j60
(90)(- j30)
= 9 (1 − j3)
90 − j30
(80 + j60)(9 − j27)
= Z1 || Z 2 =
= 17.181 − j24.57 Ω
80 + j60 + 9 − j27
Z 2 = 90 || (- j30) =
Z Th
Z2
(9)(1 − j3)
(120∠60°) =
(120∠60°)
Z1 + Z 2
89 + j33
= 35.98∠ - 31.91°
VTh =
VTh
R L = Z Th = 30 Ω
The current through the load is
VTh
35.98∠ - 31.91°
I=
=
= 0.6764∠ - 4.4°
Z Th + R L 47.181 − j24.57
The maximum average power absorbed by R L is
1 2
1
Pmax = I R L = (0.6764) 2 (30) = 6.863 W
2
2
P.P.11.7
⎧ 4t
0 < t <1
i( t ) = ⎨
⎩8 − 4 t 1 < t < 2
[
2
1 T 2
1 1
2
i
dt
=
(
4
t
)
dt
+
(8 − 4 t ) 2 dt
∫
∫
∫
0
0
1
T
2
2
16 1 2
=
t
dt
+
( 4 − 4 t + t 2 ) dt
∫
∫
1
2 0
⎡1 ⎛
t 3 ⎞ 2 ⎤ 16
2
= 8 ⎢ + ⎜ 4t − 2t + ⎟ 1 ⎥ =
3⎠ ⎦ 3
⎣3 ⎝
I 2rms =
I 2rms
I 2rms
T=2
[
]
]
16
= 2.309 A
3
I rms =
⎛ 16 ⎞
P = I 2rms R = ⎜ ⎟(9) = 48 W
⎝3⎠
P.P.11.8
T = π , v( t ) = 8 sin( t ), 0 < t < π
1 T 2
1 π
v dt = ∫ (8 sin( t )) 2 dt
∫
T 0
π 0
π
64 1
2
Vrms
=
[1 − cos(2t )] dt = 32
π ∫0 2
2
Vrms
=
Vrms = 5.657 V
P=
P.P.11.9
2
Vrms
32
=
= 5.333 W
R
6
The load impedance is
Z = 60 + j40 = 72.11∠33.7° Ω
The power factor is
pf = cos(33.7°) = 0.832 (lagging)
Since the load is inductive
150 ∠10°
V
I= =
= 2.08∠ - 23.7° A
Z 72.11∠33.7°
The apparent power is
S = Vrms I rms =
P.P.11.10
1
(150)(2.08) = 156 VA
2
The total impedance as seen by the source is
( j4)(8 − j6)
Z = 10 + j4 || (8 − j6) = 10 +
8 − j2
Z = 12.69∠20.62°
The power factor is
pf = cos( 20.62°) = 0.936 (lagging)
I rms =
Vrms
40∠0°
=
= 3.152∠ - 20.62°
Z
12.69∠20.62°
The average power supplied by the source is equal to the power absorbed by the load.
P = I 2rms R = (3.152) 2 (11.88) = 118 W
or
P = Vrms I rms pf = (40)(3.152)(0.936) = 118 W
P.P.11.11
(a)
S = Vrms I *rms = (110 ∠85°)(0.4 ∠ - 15°)
S = 44 ∠70° VA
S = S = 44 VA
(b)
S = 44 ∠70° = 15.05 + j41.35
P = 15.05 W ,
(c)
Q = 41.35 VAR
pf = cos( 70°) = 0.342 (lagging)
Vrms 110∠85°
=
= 275∠70°
I rms 0.4∠ - 15°
Z = 94.06 + j258.4 Ω
Z=
P.P.11.12
(a)
(b)
(c)
If Z = 250∠ - 75° ,
pf = cos( -75°) = 0.2588
Q = S sin θ ⎯
⎯→ S =
S=
2
Vrms
Z
(leading)
10 kVAR
Q
=
= 10.35 kVA
sin θ sin(-75°)
⎯
⎯→ Vrms = S ⋅ Z = (10353)(250) = 1608.8
Vm = 2Vrms = 2.275 kV
P.P.11.13
Consider the circuit below.
I
20 Ω
I1
V
+
−
(30–j10)Ω
+
Vo
−
I2
(60+j20) Ω
Let I 2 be the current through the 60-Ω resistor.
P 240
P = I 22 R ⎯
⎯→ I 22 = =
=4
R
60
I 2 = 2 (rms)
Vo = I 2 (60 + j20) = 120 + j40
I1 =
Vo
= 3.2 + j2.4
30 − j10
I = I1 + I 2 = 5.2 + j2.4
V = 20 I + Vo = (104 + j48) + (120 + j40)
V = 224 + j88 = 240.7∠21.45˚ Vrms
For the 20-Ω resistor,
V = 20 I = 204 + j48 = 114.54 ∠ 24.8°
I = 5.2 + j2.4 = 5.727 ∠24.8°
S = V I * = (114.54 ∠24.8°)(5.727 ∠ - 24.8°)
S = 656 VA
For the (30 – j10)-Ω impedance,
Vo = 120 + j40 = 126.5∠18.43°
I 1 = 3.2 + j2.4 = 4∠36.87°
S 1 = Vo I 1* = (126.5∠18.43 °)(4 ∠ - 36.87 °)
S1 = 506∠ - 18.44° = 480 − j160 VA
For the (60 + j20)-Ω impedance,
I 2 = 2∠0°
S 2 = Vo I *2 = (126.5∠18.43°)(2 ∠ - 0°)
S 2 = 253∠18.43° = 240 + j80 VA
The overall complex power supplied by the source is
S T = V I * = (240.67 ∠21.45°)(5.727 ∠ - 24.8°)
S T = 1378.3∠ - 3.35° = 1376 − j80 VA
P.P.11.14
For load 1,
P1 = 2000 ,
P1 = S1 cos θ1
pf = 0.75 = cos θ1 ⎯
⎯→ θ1 = -41.41°
P1
⎯
⎯→ S1 =
= 2666.67
cos θ1
Q1 = S1 sin θ1 = -176.85
S1 = P1 + jQ1 = 2000 − j1763.85 (leading)
For load 2,
P2 = 4000 ,
pf = 0.95 = cos θ 2 ⎯
⎯→ θ 2 = 18.19°
P2
= 4210.53
S2 =
cos θ 2
Q 2 = S 2 sin θ 2 = 1314.4
S 2 = P2 + jQ 2 = 4000 + j1314.4 (lagging)
The total complex power is
S = S1 + S 2 = 6 − j0.4495 kVA
pf =
P.P.11.15
P
S
=
6000
= 0.9972 (leading)
6016.18
pf = 0.85 = cos θ ⎯
⎯→ θ = 31.79°
Q
140
Q = S sin θ ⎯
⎯→ S =
=
= 265.8 kVA
sin θ sin(31.79°)
P = S cos θ = 225.93 kW
For pf = 1 = cos θ1
⎯
⎯→ θ1 = 0°
Since P remains the same,
P = P1 = S1 cos θ1
⎯
⎯→ S1 =
P1
= 225.93
cos θ1
Q1 = S1 sin θ1 = 0
The difference between the new Q1 and the old Q is Q c .
2
Q c = 140 kVAR = ωCVrms
140 × 10 3
= 30.69 mF
C=
(2π )(60)(110) 2
P.P.11.16
Let
The wattmeter measures the average power from the source.
Z1 = 4 − j2
Z 2 = 12 || j9 =
(12)( j9)
= 4.32 + j5.76
12 + j9
Z = Z1 + Z 2 = 8.32 + j3.76 = 9.13∠24.32°
S = VI =
*
V
Z*
2
(120) 2
=
= 1577.2 ∠24.32° kVA
9.13∠ - 24.32°
P = S cos θ = 1437 kW
P.P.11.17
Demand charge = $5 × 32,000 = $160,000
Energy charge for the first 50,000 kWh = $0.08 × 50,000 = $4,000
The remaining energy = 500,000 − 50,000 = 450,000 kWh
Charge for this bill = $0.05 × 450,000 = $22,500
Total bill = $160,000 + $4,000 + $22,500 = $186,500
P.P.11.18
Energy consumed = 800 kW × 20 × 26 = 416,000 kWh
The power factor of 0.88 exceeds 0.85 by 3× 0.01 . Hence, there is a power factor credit
which amounts to an energy credit of
0 .1
416,000 ×
× 3 = 1248 kWh
100
Total energy billed = 416,000 − 1,248 = 414,752 kWh
Energy cost = $0.06 × 414,752 = $24,885.12
February 5, 2006
CHAPTER 12
P.P.12.1
For the abc sequence, Van leads Vbn by 120° and Vbn leads Vcn by 120°.
Hence,
Van = 110 ∠(30° + 120°) = 110∠150° V
Vcn = 110 ∠(30° − 120°) = 110∠–90˚V
P.P.12.2
(a)
Vab = Van − Vbn = 120∠30° − 120∠ - 90°
Vab = (103.92 + j60) + j120
Vab = 207.8∠60˚V
Alternatively, using the fact that Vab leads Van by 30° and has a
magnitude of
3 times that of Van ,
Vab = 3 (120) ∠(30° + 30°) = 207.85∠60°
Following the abc sequence,
Vbc = 207.8∠–60˚V
Vca = 207.8∠180˚V
(b)
Ia =
Van
Z
Z = (0.4 + j0.3) + ( 24 + j19) + (0.6 + j0.7)
Z = 25 + j20 = 32 ∠38.66 °
Ia =
120 ∠30°
= 3.75 ∠ - 8.66 ° A
32 ∠38.66°
Following the abc sequence,
I b = I a ∠ - 120° = 3.75 ∠ - 128.66 ° A
I c = I a ∠ - 240° = 3.75∠111.34˚A
P.P.12.3
The phase currents are
V
180∠ - 20°
I AB = AB =
= 9 ∠ - 60° A
ZΔ
20∠40°
I BC = I AB ∠ - 120° = 9 ∠ - 180 ° A
I CA = I AB ∠120° = 9∠60°
The line currents are
I a = I AB 3 ∠ - 30° = 9 3 ∠ - 90° = 15.59 ∠ - 90° A
I b = I a ∠ - 120° = 15.59∠150˚A
I c = I a ∠120° = 15.59 ∠ 30° A
P.P.12.4
magnitude
In a delta load, the phase current leads the line current by 30° and has a
1
3
times that of the line current. Hence,
I AB =
Ia
3
∠30° =
22.5
3
∠65° = 13 ∠65° A
Z Δ = 18 + j12 = 21.63∠33.69° Ω
VAB = I AB Z Δ = (13∠65°)(21.63∠33.69°)
VAB = 281.2 ∠98.69 ° V
P.P.12.5
Z Y = 12 + j15 = 19.21∠51.34°
After converting the Δ-connected source to a Y-connected source,
240
Van =
∠(150° − 30°) = 138.56 ∠ - 15°
3
Ia =
Van 138.56∠ - 15°
=
= 7.21∠ - 66.34 ° A
Z Y 19.21∠51.34°
I b = I a ∠ - 120° = 7.21∠ - 186.34 ° A
I c = I a ∠120° = 7.21∠53.66 ° A
P.P.12.6
For the source,
S = 3 Vp I*p = (3)(120∠30°)(3.75∠8.66°)
S = −1350∠38.66° = –1054.2 – j843.3 VA
For the load,
2
S = 3 Ip Z
where
Z = 24 + j19 = 30.61∠38.37 °
I p = 3.75∠ - 8.66°
S = (3)(3.75) 2 (30.61∠38.37°)
S = 1291.36∠38.37° = 1012 + j801.6 VA
P.P.12.7
P = S cos θ ⎯
⎯→ S =
S = 3 VL I L
P
30 × 10 3
=
= 35.29 kVA
cos θ
0.85
⎯
⎯→ I L =
S
3 VL
=
35.29 × 10 3
3 (440)
= 46.31 A
Alternatively,
Pp =
30 × 10 3
= 10 kW ,
3
Vp =
440
3
V
Pp = Vp I p cos θ
Ip =
P.P.12.8
(a)
Pp
Vp cos θ
=
(10 × 10 3 ) 3
= 46.31 A
(440)(0.85)
For load 1,
Vp =
I a1 =
S1 =
VL
3
=
840
3
Va 840 ∠0°
1
=
⋅
= 9.7 ∠ - 53.13°
Zp
30 + j40
3
2
Vrms
(840) 2
=
= 14.112 ∠53.13° kVA
Z*
50 ∠ - 53.15°
For load 2,
S2 =
P2
48
=
= 60 kVA
cos θ 2 0.8
Q 2 = S 2 sin θ2 = (60)(0.6) = 36 kVAR
S 2 = 48 + j36 kVA
S = S1 + S 2 = 56.47 + j47.29 kVA
S = 73.65 ∠39.94° kVA
with pf = cos(39.94°) = 0.7667
(b)
Q c = P (tan θold − tan θ new )
Q c = (56.47)(tan 39.94° − tan 0°) = 47.29 kVAR
For each capacitor, the rating is 15.76 kVAR
(c)
At unity pf, S = P = 56.47 kVA
56470
S
IL =
=
= 38.81 A
3 VL
3 (840)
P.P.12.9
The phase currents are
VAB 200∠0°
=
= 17.89∠26.56°
I AB =
Z AB 10 − j5
I BC =
VBC 200∠ - 120°
=
= 12.5∠ - 120° = -6.25 − j10.825
16
Z BC
I CA =
VCA 200∠120°
=
= 20∠83.13°
Z CA
8 + j6
The line currents are
I a = I AB − I CA = (16 + j8) − (2.392 + j19.856)
= 13.608 − j11.856 = 18.05 ∠ - 41.06 ° A
I b = I BC − I AB = (-6.25 − j10.825) − (16 + j8)
= -22.25 − j18.825 = 29.15 ∠ 220.2 ° A
I c = I CA − I BC = (2.392 + j19.856) − (-6.25 − j10.825)
I c = 8.642 + j30.681 = 31.87 ∠74.27 ° A
P.P.12.10
The phase currents are
220∠0°
I AB =
= j44
- j5
I BC =
220∠0°
= 22 ∠30°
j10
I CA =
220 ∠120°
= 22 ∠ - 120°
10
The line currents are
I a = I AB − I CA = ( j44) − (-11 − j19.05)
I a = 11 + j63.05 = 64 ∠ 80.1 ° A
I b = I BC − I AB = (19.05 + j11) − ( j44)
I b = 19.05 − j33 = 38.1∠ - 60° A
I c = I CA − I BC = (-11 − j19.05) − (19.05 + j11)
I c = -30.05 − j30.05 = 42.5 ∠ 225° A
The real power is absorbed by the resistive load
P = I CA
2
(10) = (22) 2 (10) = 4.84 kW
P.P.12.11
The schematic is shown below. First, use the AC Sweep option of the
Analysis Setup. Choose a Linear sweep type with the following Sweep Parameters :
Total Pts = 1, Start Freq = 100, and End Freq = 100. Once the circuit is saved and
simulated, we obtain an output file whose contents include the following results.
FREQ
1.000E+02
IM(V_PRINT1)
8.547E+00
IP(V_PRINT1)
-9.127E+01
FREQ
1.000E+02
VM(A,N)
1.009E+02
VP(A,N)
6.087E+01
From this we obtain,
I bB = 8.547 ∠ - 91.27 ° A ,
Van = 100.9 ∠60.87 ° V
P.P.12.12
The schematic is shown below.
In this case, we may assume that ω = 1 rad / s , so that f = 1 2π = 0.1592 Hz . Hence,
L = X L ω = 10 and C = 1 ωX c = 0.1 .
Use the AC Sweep option of the Analysis Setup. Choose a Linear sweep type with the
following Sweep Parameters : Total Pts = 1, Start Freq = 0.1592, and End Freq =
0.1592. Once the circuit is saved and simulated, we obtain an output file whose contents
include the following results.
FREQ
1.592E-01
IM(V_PRINT1)
3.724E+01
IP(V_PRINT1)
8.379E+01
FREQ
1.592E-01
IM(V_PRINT2)
1.555E+01
IP(V_PRINT2)
-7.501E+01
FREQ
1.592E-01
IM(V_PRINT3)
2.468E+01
IP(V_PRINT3)
-9.000E+01
From this we obtain,
I ca = 24.68 ∠ - 90° A
P.P.12.13
(a)
I cC = 37.25 ∠83.79° A
IAB 15.55∠–75.01˚A
If point o is connected to point B, P2 = 0 W
P1 = Re (VAB I *a )
P1 = (200)(18.05) cos(0° + 41.06°) = 2722 W
P3 = Re ( VCB I *c )
where VCB = -VBC = 200 ∠(-120° + 180°) = 200 ∠60°
P3 = (200)(31.87) cos(60° − 74.27°) = 6177 W
(b)
P.P.12.14
Total power is
PT = P1 + P2 + P3 = 2722 + 0 + 6177 = 8899 W
VL = 208 V ,
P1 = -560 W ,
P2 = 800 W
(a)
PT = P1 + P2 = -560 + 800 = 240 W
(b)
Q T = 3 ( P2 − P1 ) = 3 (800 + 560) = 2.356 kVAR
(c)
tan θ =
Q T 2355.6
=
= 9.815 ⎯
⎯→ θ = 84.18°
PT
240
pf = cos θ = 0.1014
(d)
(lagging / inductive)
It is inductive because P2 > P1
For a Y-connected load,
V
208
Vp = L =
Ip = IL ,
= 120 V
3
3
Pp = Vp I p cos θ ⎯
⎯→ I p =
Zp =
Vp
Ip
=
80
= 6.575 A
(120)(0.1014)
120
= 18.25
6.575
Z p = Z p ∠θ = 18.25 ∠ 84.18° Ω
The impedance is inductive.
P.P.12.15
Z Δ = 30 − j40 = 50∠ - 53.13°
The equivalent Y-connected load is
Z
Z Y = Δ = 16.67 ∠ - 53.13°
3
440
Vp =
= 254 V
3
Vp
254
IL =
=
= 15.24
16.67
ZY
P1 = VL I L cos(θ + 30°)
P1 = (440)(15.24) cos(-53.13° + 30°) = 6.167 kW
P2 = VL I L cos(θ − 30°)
P2 = (440)(15.24) cos(-53.13° − 30°) = 0.8021 kW
PT = P1 + P2 = 6.969 kW
Q T = 3 (P2 − P1 ) = 3 (802.1 − 6167 )
Q T = - 9.292 kVAR
February 5, 2006
CHAPTER 13
P.P. 13.1
For mesh 1,
j6 = 4(1 + j2)I1 + jI2
(1)
For mesh 2,
0 = jI1 + (10 + j5)I2
(2)
For the matrix form
j ⎤⎡ I1 ⎤
⎡ j6⎤ ⎡4 + j8
=
⎢0⎥ ⎢ j
10 + j5⎥⎦ ⎢⎣I 2 ⎥⎦
⎣ ⎦ ⎣
Δ = j100, Δ2 = 6
I2 = Δ2/Δ = 6/j100
Vo = 10I2 = 60/j100 = 0.6∠-90° V
P.P. 13.2
Since I1 enters the coil with reactance 2Ω and I2 enters the coil with
reactance 6Ω, the mutual voltage is positive. Hence, for mesh 1,
12∠60o = (5 + j2 + j6 – j 3x2)I1 – j6I2 + j3I2
12∠60o = (5 + j2)I1 – j3I2
or
For mesh 2,
0 = (j6 – j4)I2 – j6I1 + j3I1
or
I2 = 1.5I1
Substituting this into (1),
(1)
(2)
12∠60o = (5 – j2.5)I1
I1 = (12∠60o)/(5.59∠–26.57o) = 2.147∠86.57o A
I2 = 1.5I1 = 3.22∠86.57o A
P.P. 13.3
The coupling coefficient is, k = m/ L1 L 2 = 1 / 2x1 = 0.7071
To obtain the energy stored, we first obtain the frequency-domain circuit shown below.
20cos(ωt) becomes 20∠0o, ω = 2
1H becomes jω1 = j2
2H becomes jω2 = j4
(1/8) F becomes 1/jωC = -j4
4Ω
VS
+
–
-j4
I1
j4
j2
For mesh 1,
20 = (4 – j4 + j4)I1 – j2I2
or
10 = 2I1 – jI2
For mesh 2,
–j2I1 + (2 + j2)I2 = 0
or
I1 = (1 – j)I2
Substituting (2) into (1),
I2
2Ω
(1)
(2)
(2 – j3)I2 = 10
I2 = 10/((2 – j3) = 2.78∠56.31o
I1 = 3.93∠11.31o
In the time domain,
At t = 1.5,
i1 = 3.93cos(2t + 11.31o)
i2 = 2.78cos(2t + 56.31o)
2t = 3 rad = 171.9o
i1 = 3.93cos(171.9o + 11.31o) = –3.924 A
i2 = 2.78cos(171.9o + 56.31o) = -1.85 A
The total energy stored in the coupled inductors is given by,
W = 0.5L1(i1)2 + 0.5L2(i2)2 – 0.5M(i1i2)
= 0.5(2) (-3.924)2 + 0.5(1)(-1.85)2 – (1)(-3.924)(-1.85)
= 9.85 J
P.P. 13.4
Zin = 4 + j8 + [32/(j10 – j6 + 6 + j4)]
= 4 + j8 + 9/(6 + j8)
= 8.58∠58.05o ohms
The current from the voltage is,
I = V/Z = 10∠0o/8.58∠58.05o = 1.165∠–58.05o A
P.P. 13.5
L1 = 10, L2 = 4, M = 2
L1L2 – M2 = 40 – 4 = 36
LA = (L1L2 – M2)/(L2 – M) = 36/(4 – 2) = 18
LB = (L1L2 – M2)/(L1 – M) = 36/(10 – 2) = 4.5
LC = (L1L2 – M2)/M = 36/2 = 18
Hence, we get the π equivalent circuit as shown below.
18 H
18 H
4.5 H
P.P. 13.6
If we reverse the direction of i2 so that we replace I2 by –i2, we
have the circuit shown in Figure (a).
j3
-j4
+
–
j3
i1
j6
i2
12 Ω
o
12∠0
(a)
We now replace the coupled coil by the T-equivalent circuit and assume ω = 1.
La = 5 – 3 = 2 H
Lb = 6 – 3 = 3 H
Lc = 3 H
Hence the equivalent circuit is shown in Figure (b). We apply mesh analysis.
-j4
j2
j3
j3
12∠0o
+
–
I1
I2
(b)
12 Ω
12 = i1(-j4 + j2 + j3) + j3i2
or 12 = ji1 + j3i2
Loop 2 produces,
(1)
0 = j3i1 + (j3 + j3 + 12)i2
or i1 = (-2 + j4)i2
Substituting (2) into (1),
(2)
12 = (-4 + j)i2, which leads to i2 = 12/(-4 + j)
I2 = -i2 = 12/(4 – j) = 2.91∠14.04o A
I1 = i1 = (-2 + j4)i2 = 12(2 – j4)/(4 – j) = 13∠-49.4o A
P.P. 13.7
(a)
n = V2/V1 = 110/3300 = 1/30 (a step-down transformer)
(b)
S = V1I1 = 3300x3 = 9.9 kVA
(c)
I2 = I1/n = 3/(1/30) = 90 A
P.P. 13.8
resulting in
The 16 – j24-ohm impedance can be reflected to the primary
Zin = 2 + (16 – j24)/16 = 3 – j1.5
I1 = 100/(3 – j1.5) = 29.82∠26.57o
I2 = –I1/n = –7.454∠26.57o
Vo = -j24i2 = (24∠–90o)(–7.454∠26.57o) = 178.92∠116.57oV
S1 = V1I1 = (100)( 29.82∠26.57o) = 2.982∠-26.57okVA
P.P. 13.9
8Ω
+ v0 –
4Ω
i1
1
2Ω
1:2
2
i2
+
–
60∠0o
+
+
v1
v2
–
–
+
v3
10 Ω
–
Consider the circuit shown above.
At node 1,
(60 – v1)/4 = i1 + (v1 – v3)/8
(1)
At node 2,
[(v1 – v3)/8] + [(v2 – v3)/2] = (v3)/8
(2)
At the transformer terminals, v2 = -2v1 and i2 = -i1/2
(3)
But i2 = (v2 – v3)/2 = -i1/2 which leads to i1 = (v3 – v2)/1 = v3 + 2v1.
Substituting all of this into (1) and (2) leads to,
(60 – v1)/4 = v3 + 2v1 + (v1 – v3)/8 which leads 120 = 19v1 + 7v3
(4)
[(v1 – v3)/8] + [(-2v1 – v3)/2] = v3/8 which leads to v3 = -7v1/6
(5)
From (4) and (5),
120 = 10.833v1 or v1 = 11.077 volts
v3 = -7v1/6 = -12.923
vo = v1 – v3 = 24 volts
P.P. 13.10
We should note that the current and voltage of each winding of the
autotransformer in Figure (b) are the same for the two-winding transformer in Figure (a).
6A
0.5A
+
6A
6.5A
+
+
120V
10V
–
–
+
+
–
+
130V
120V
120V
–
10V
0.5A
–
–
(b)
(a)
For the two-winding transformer,
s1 = 120/2 = 60 VA
s2 = 6(10) = 60 VA
For the autotransformer,
s1 = 120(6.5) = 780 VA
s2 = 130(6) = 780 VA
P.P. 13.11
i2 = s2/v2 = 16,000/800 = 20 A
Since s1 = v1i1 = v2i2 = s2, v2/v1 = i1/i2, 800/1250 = i1/20,
or i1 = 800x20/1250 = 12.8 A.
At the top, KCL produces i1 + io = i2, or io = i2 – i1 = 20 – 12.8 = 7.2 A.
P.P. 13.12
(a)
sT = (√3)vLiL, but sT = pT/cosθ = 40x106/0.85 = 47.0588 MVA
iLS = sT/(√3)vLS = 47.0588x106/[(√3)12.5x103] = 2.174 kA
(b)
vLS = 12.5 kV, vLP = 625 kV, n = vLS/vLP = 12.5/625 = 0.02
(c)
iLP = niLS = 0.02x2173.6 = 43.47 A
or iLP = sT/[(√3)vLP] = 47.0588x106/[(√3)625x103] = 43.47 A
(d)
The load carried by each transformer is (1/3)sT = 15.69 MVA
P.P. 13.13
The process is essentially the same as in Example 13.13. We are
given the coupling coefficient, k = 0.4, and can determine the operating frequency from
the value of ω = 4 which implies that f = 4/(2π) = 0.6366 Hz.
Saving and then simulating produces,
io = 100.6cos(4t + 68.52o) mA
P.P. 13.14
Following the same basic steps in Example 13.14, we first assume
ω = 1. This then leads to following determination of values for the inductor and the
capacitor.
j15 = jωL leads to L = 15 H
-j16 = 1/(ωC) leads to C = 62.5 mF
The schematic is shown below.
FREQ
VM($N_0005,0)
VP($N_0005,0)
1.592E-01
7.652E+01
2.185E+00
FREQ
VM($N_0001,0)
VP($N_0001,0)
1.592E-01
1.151E+02
2.091E+00
Thus,
V1 = 76.52∠2.18° V
V2 = 115.1∠2.09° V
Note, if we divide V2 by V1 we get 1.5042∠–.09˚ which is in good agreement that the
transformer is ideal with a voltage ratio of 1.5:1!
P.P. 13.15
V2/V1 = 120/13,200 = 1/110 = 1/n
P.P. 13.16
VS
+
–
Z1
+
v1
ZL/n2
–
As in Example 13.16, n2 = ZL/Z1 = 100/(2.5x103) = 1/25, n = 1/5 = 0.2
By voltage division, v1 = vs/2 (since Z1 = ZL/n2), therefore v1 = 30/2 = 15 volts, and
v2 = nv1 = (1/5)(15) = 3 volts
P.P. 13.17
(a)
s = 12x60 + 350 + 4,500 = 5.57 kW
(b)
iP = s/vP 5570/2400 = 2.321 A
February 5, 2006
CHAPTER 14
P.P.14.1
H(ω) =
Vo
jωL
=
Vs R + jωL
jω ω 0
jωL R
=
1 + jωL R 1 + jω ω0
R
where ω0 = .
L
H (ω) =
H = H(ω) =
At ω = 0 ,
As ω → ∞ ,
At ω = ω0 ,
H = 0,
H = 1,
1
H=
,
2
ω ω0
1 + (ω ω0 ) 2
φ = ∠H (ω) =
⎛ω⎞
π
− tan -1 ⎜ ⎟
2
⎝ ω0 ⎠
φ = 90°
φ = 0°
φ = 90° − 45° = 45°
Thus, the sketches of H and φ are shown below.
H
1
0.7071
0
ω0 = R/L
ω
φ
90°
45°
0
ω0 = R/L
ω
P.P.14.2
The desired transfer function is the input impedance.
Vo (s) ⎛
1 ⎞
⎟ || (3 + 2s)
Z i (s) =
= ⎜5 +
I o (s) ⎝ s 10 ⎠
Z i (s) =
(5 + 10 s)(3 + 2s) 5 (s + 2)(s + 1.5)
=
s 2 + 4s + 5
5 + 10 s + 3 + 2s
The poles are at
p1, 2 =
- 4 ± 16 − 20
= -2± j
2
The zeros are at
z1 = - 2 ,
P.P.14.3
H (ω) =
z 2 = - 1.5
1 + jω 2
( jω)(1 + jω 10)
H db = 20 log10 1 + jω 2 − 20 log10 jω − 20 log10 1 + jω 10
φ = -90° + tan -1 (ω 2) − tan -1 (ω 10)
The magnitude and the phase plots are shown in Fig. 14.14.
P.P.14.4
H (ω) =
50 400 jω
(1 + jω 4)(1 + jω 10) 2
H db = -20 log10 8 + 20 log10 jω − 20 log10 1 + jω 4 − 40 log10 1 + jω 10
φ = 90° − tan -1 (ω 4) − 2 tan -1 (ω 10)
The magnitude and the phase plots are shown in Fig. 14.16.
P.P.14.5
H (ω) =
10 400
⎛ jω8 ⎛ jω ⎞ 2 ⎞
( jω)⎜⎜1 +
+ ⎜ ⎟ ⎟⎟
⎝ 20 ⎠ ⎠
40
⎝
H db = -20 log10 40 − 20 log10 jω − 20 log10 1 + jω 5 − ω2 400
⎛ 0.2 ω ⎞
⎟
φ = -90° − tan -1 ⎜
⎝1 − ω2 400 ⎠
The magnitude and the phase plots are shown in Fig. 14.18.
P.P.14.6
A zero at ω = 0.5 ,
1 + jω 0.5
1
A pole at ω = 1 ,
1 + jω 1
1
Two poles at ω = 10 ,
(1 + jω 10) 2
Hence,
1 + j ω 0 .5
(1 0.5)(0.5 + jω)
2 =
(1 + jω 1)(1 + jω 10)
(1 100)(1 + jω)(10 + jω) 2
200 (s + 0.5)
H (ω) =
(s + 1)(s + 10) 2
H (ω) =
P.P.14.7
(a)
Q=
ω0 L
R
ω0 =
1
⎯
⎯→ ω0 =
⎯
⎯→ C =
LC
C = 0.625 μF
(b)
QR
(50)(4)
3
=
-3 = 8 × 10 rad / s
L
25 × 10
1
ω L
2
0
=
1
(64 × 10 )(25 × 10 -3 )
6
ω0 8 × 10 3
B=
=
= 160 rad / s
Q
50
Since Q > 10 ,
B
= 8000 − 80 = 7920 rad / s
2
B
ω 2 = ω0 + = 8000 + 80 = 8080 rad / s
2
ω1 = ω0 −
(c)
At ω = ω0 ,
Vin2 100 2
P=
=
= 1.25 kW
2R
8
At ω = ω1 ,
P = 0.5 ⋅
Vin2
= 0.625 kW
2R
At ω = ω2 ,
P = 0.5 ⋅
Vin2
= 0.625 kW
2R
P.P.14.8
ω0 =
1
LC
=
1
(20 × 10 )(5 × 10 )
-3
-9
Q=
R
100 × 10 3
=
= 50
ω0 L (10 5 )(20 × 10 -3 )
B=
ω0 10 5
=
= 2 krad / s
Q
50
= 10 5 = 100 krad / s
Since Q > 10 ,
B
= 100,000 − 1,000 = 99 krad / s
2
B
ω 2 = ω0 + = 100,000 + 1,000 = 101 krad / s
2
ω1 = ω0 −
P.P.14.9
1
50
= jω +
jω0.2
5 + j10ω
10 (1 − j2ω)
Z = jω +
1 + 4ω 2
Z = jω1 + 10 ||
Im(Z) = 0 ⎯
⎯→ ω −
20ω
=0
1 + 4ω 2
20ω
⎯
⎯→ 1 + 4ω 2 = 20
1 + 4ω 2
19
ω=
= 2.179 rad / s
2
ω=
Vo
R 2 || sL
=
,
s = jω
Vi R 1 + R 2 || sL
sR 2 L
H(s) =
R 1 R 2 + sR 1 L + sR 2 L
jωR 2 L
H (ω) =
R 1 R 2 + jωL (R 1 + R 2 )
H ( 0) = 0
jR 2 L
R2
H(ω) = lim
=
ω→ ∞ R R
R1 + R 2
1 2 ω + jL ( R 1 + R 2 )
i.e. a highpass filter.
P.P.14.10
H (s) =
The corner frequency occurs when H(ωc ) =
1
2
⋅ H(∞) .
⎛ R 2 ⎞⎛
⎞
jωL
⎟⎜
⎟
H (ω) = ⎜
⎝ R 1 + R 2 ⎠⎝ jωL + R 1 R 2 (R 1 + R 2 ) ⎠
⎛ R 2 ⎞⎛ jω ⎞
⎟⎜
⎟,
H (ω) = ⎜
⎝ R 1 + R 2 ⎠⎝ jω + k ⎠
where k =
At the corner frequency,
jωc
R2
R2
1
⋅
=
⋅
2 R 1 + R 2 R 1 + R 2 jω c + k
1
2
Hence,
=
ωc
⎯
⎯→ ωc = k =
ωc2 + k 2
R1R 2
(R 1 + R 2 ) L
⎛ R 2 ⎞⎛ jω ⎞
⎟
⎟⎜
H (ω) = ⎜
⎝ R 1 + R 2 ⎠⎝ jω + ωc ⎠
and the corner frequency is
(100)(100)
ωc =
= 25 krad / s
(100 + 100)(2 × 10 -3 )
P.P.14.11
B = 2π (20.3 − 20.1) × 10 3 = 400π
Assuming high Q,
ω1 + ω2 (2π)(40.4 × 10 3 )
ω0 =
=
= 40.4π × 10 3 rad / s
2
2
Q=
ω0 40.4π × 10 3
=
= 101
B
400π
R
B=
L
P.P.14.12
R 20 × 10 3
⎯
⎯→ L = =
= 15.916 H
B
400π
Q=
1
ω0 CR
C=
1
= 3.9 pF
(40.4π × 10 )(101)(20 × 10 3 )
⎯
⎯→ C =
1
ω0 QR
3
Given H (∞) = 5 and f c = 2 kHz
1
ω c = 2π f c =
R i Ci
R 1R 2
(R 1 + R 2 ) L
1
1
=
3
2πf c C i (2π )(2 × 10 )(0.1 × 10 -3 )
R i = 795.8 ≅ 800 Ω
Ri =
H(∞) =
P.P.14.13
- Rf
= -5 ⎯
⎯→ R f = 5R i = 3,978 ≅ 4 kΩ
Ri
Q = 10 ,
B=
ω0 = 20 krad / s
ω0
= 2 krad / s
Q
B
= 19 krad / s
2
B
ω 2 = ω0 + = 21 krad / s
2
ω1 = ω0 −
Since ω1 =
1
,
C2R
1
1
=
= 5.263 nF
3
ω1 R (19 × 10 )(10 × 10 3 )
1
1
C1 =
=
= 4.762 nF
3
ω2 R (21 × 10 )(10 × 10 3 )
Rf
K=
=5 ⎯
⎯→ R f = 5R i = 50 kΩ
Ri
C2 =
P.P.14.14
ω′c 2π × 10 4
Kf =
=
= 2π × 10 4
ωc
1
C
C′ =
KmKf
C
1
10 4
⎯
⎯→ K m =
=
=
C′ K f (15 × 10 -9 )(2π × 10 4 ) 3π
10 4
(1) = 1.061 kΩ
3π
Km
10 4
2
L′ =
L=
⋅
= 33.77 mH
Kf
3π 2π × 10 4
R′ = K mR =
Therefore,
R 1′ = R ′2 = 1.061 kΩ
C1′ = C′2 = 15 nF
L ′ = 33.77 mH
P.P.14.15
The schematic is shown in Fig. (a).
(a)
Use the AC Sweep option of the Analysis Setup. Choose a Linear sweep type with the
following Sweep Parameters : Total Pts = 100, Start Freq = 1, and End Freq = 1K.
After saving and simulating the circuit, we obtain the magnitude and phase plots are
shown in Figs. (b) and (c).
(b)
(c)
P.P.14.16
The schematic is shown in Fig. (a).
(a)
Use the AC Sweep option of the Analysis Setup. Choose a Decade sweep type with
these Sweep Parameters : Pts/Decade = 20, Start Freq = 1K, and End Freq = 100K.
Save and simulate the circuit.
For the magnitude plot, choose DB( ) from the Analog Operators and Functions list.
Then, select the voltage V(R1:1) and OK. Another option would be to type DB(V(R1:1))
as the Trace Expression. For the phase plot, choose P( ) from the Analog Operators
and Functions list. Then, select the voltage V(R1:1) and OK. Another option would be
to type VP(R1:1) as the Trace Expression. The resulting magnitude and phase plots
are shown in Figs. (b) and (c).
(b)
(c)
P.P.14.17
or
ω0 = 2πf 0 =
C=
1
LC
1
4π f 02 L
2
For the high end of the band, f 0 = 108 MHz
1
C1 =
= 0.543 pF
2
2
4π (108 × 1012 )(4 × 10 -6 )
For the low end of the band, f 0 = 88 MHz
1
C2 =
= 0.818 pF
2
2
4π (88 × 1012 )(4 × 10 -6 )
Therefore, C must be adjustable and be in the range 0.543 pF to 0.818 pF .
P.P.14.18
For BP6 , f 0 = 1336 Hz and it passes frequencies in the range 1209 Hz < f < 1477 Hz .
B = 2π (1477 − 1209 ) = 1683 .9
P.P.14.19
L=
R
600
=
= 0.356 H
B 1683.9
C=
1
1
=
= 39.83 nF
2
2
4π f 0 L 4π (1336) 2 (0.356)
2
C = 10 μF
2π f c =
L=
1
R 1C
and
R1 = R 2 = 8 Ω
⎯
⎯→ f c =
1
1
=
= 1.989 kHz
2πR 1C (2π )(8)(10 × 10 -6 )
R2
8
=
= 0.64 mH
2πf c (2π)(1.989 × 10 3 )
February 5, 2006
CHAPTER 15
P.P.15.1
L[ t u ( t )] =
∫
∞
0
t e -st dt
Using integration by parts,
∫ u dv = uv − ∫ v du
Let
u=t
⎯
⎯→ du = dt .
e -st dt = dv ⎯
⎯→ v =
-t
L[ t u ( t )] = e -st
s
∞
0
+∫
∞
0
- 1 -st
e
s
1 -st
e -st
e dt = 0 + 2
s
s
∞
0
=
1
s2
Also,
L[ e at u ( t )] =
∫
∞
0
e at e -st dt =
P.P.15.2
L[ cos(ωt )] = ∫-∞
P.P.15.3
If
- 1 -( s − a ) t
e
s−a
∞
0
=
1
s−a
1 jωt
( e + e - jωt ) e -st dt
2
1 ∞
1 ∞
L[ cos(ωt )] = ∫0 e -(s- jω) t dt + ∫0 e -(s+ jω) t dt
2
2
s
1⎛ 1
1 ⎞
⎟= 2
L[ cos(ωt )] = ⎜
+
2 ⎝ s − jω s + jω ⎠ s + ω 2
∞
f ( t ) = cos(2 t ) + e -3t ,
s
1
s 2 + 3s + s 2 + 4
F(s) = 2
+
=
s + 4 s + 3 (s 2 + 4)(s + 3)
2s 2 + 3s + 4
F(s) =
(s + 3)(s 2 + 4)
Given f ( t ) = t 2 cos(3t )
s
From P.P.15.2,
L[ cos(3t )] = 2
s +9
P.P.15.4
Using Eq. 15.34,
F(s) = L[ t 2 cos(3t )] = ( - 1)
2
d2 ⎛ s ⎞
⎜
⎟
ds 2 ⎝ s 2 + 9 ⎠
[
]
[
d2
d2
-1
-1
-2
2
(
)
F(s) = 2 s s + 9 = 2 (1) ( s 2 + 9) − (s)( 2s) ( s 2 + 9)
ds
ds
-2
-2
-3
2
F(s) = ( - 2s) ( s + 9 ) − ( 4s) ( s 2 + 9 ) + ( 4s 2 ) ( 2s) ( s 2 + 9 )
2s 3 − 54s
-2
-3
F(s) = ( - 6s) ( s 2 + 9 ) + ( 8s 3 )( s 2 + 9 ) = 2
( s + 9) 3
2s ( s 2 − 27 )
F(s) =
( s 2 + 9) 3
P.P.15.5
]
h ( t ) = 10 [ u ( t ) − u ( t − 2)] + 5 [ u ( t − 2) − u ( t − 4)]
⎛ 1 e -2s ⎞ ⎛ e -2s e -4s ⎞
⎟ + 5⎜
⎟
H(s) = 10 ⎜ −
−
s ⎠ ⎝ s
s ⎠
⎝s
H (s ) =
P.P.15.6
T=5
f1 ( t ) = u ( t ) − u ( t − 2)
1
F1 (s) = (1 − e -2s )
s
F(s) =
P.P.15.7
5
( 2 − e -2s − e -4s )
s
F1 (s)
1 − e -2s
=
1 − e -Ts s (1 − e -5s )
s 3 + 2s + 6
2
s →∞
s → ∞ (s + 2s + 1)(s + 3)
2 6
1+ 2 + 3
s
s
=1
g(0) = lim
s →∞ ⎛
2 1 ⎞⎛ 3 ⎞
⎜1 + + 2 ⎟⎜1 + ⎟
⎝ s s ⎠⎝ s ⎠
g (0) = lim sF(s) = lim
Since all poles s = 0, - 1, - 1, - 3 lie in the left-hand s-plane, we can apply the final-value
theorem.
s 3 + 2s + 6
g (∞) = lim sF(s) = lim
2
s→0
s → 0 (s + 1) (s + 3)
6
=2
g(∞) = lim 2
s → 0 (1) (3)
P.P.15.8
F(s) = 1 +
4
5s
− 2
s + 3 s + 16
⎡ 4 ⎤ -1 ⎡ 5s ⎤
f ( t ) = L-1 [ 1 ] + L-1 ⎢
−L ⎢ 2
⎣ s + 3 ⎥⎦
⎣ s + 16 ⎥⎦
f ( t ) = δ( t ) + (4e -3t − 5 cos(4t ))u ( t ), t ≥ 0
F(s) =
P.P.15.9
A
B
C
+
+
s +1 s + 3 s + 4
6 (s + 2)
(s + 3)(s + 4)
6 (s + 2)
B = F(s) (s + 3) s= -3 =
(s + 1)(s + 4)
6 (s + 2)
C = F(s) (s + 4) s= -4 =
(s + 1)(s + 3)
A = F(s) (s + 1) s= -1 =
F(s) =
6
=1
(2)(3)
(6)(-1)
=
=3
(-2)(1)
(6)(-2)
=
= -4
(-3)(-1)
=
s = -1
s = -3
s = -4
1
3
4
+
−
s +1 s + 3 s + 4
f ( t ) = (e -t + 3e -3t − 4e -4t )u ( t ), t ≥ 0
P.P.15.10
G (s) =
s 3 + 2s + 6
A
B
C
D
= +
+
2
2 +
s (s + 1) (s + 3) s s + 1 (s + 1)
s+3
Multiplying both sides by s (s + 1) 2 (s + 3) gives
s 3 + 2s + 6 = A (s + 3)(s 2 + 2s + 1) + Bs (s + 1)(s + 3) + Cs (s + 3) + Ds (s + 1) 2
= A (s3 + 5s 2 + 7s + 3) + B (s 3 + 4s 2 + 3s) + C (s 2 + 3s) + D (s3 + 2s 2 + s)
Equating coefficients :
s0 :
6 = 3A ⎯
⎯→ A = 2
(1)
s1 :
2 = 7 A + 3B + 3C + D ⎯
⎯→ 3B + 3C + D = -12
(2)
2
0 = 5A + 4 B + C + 2 D ⎯
⎯→ 4B + C + 2D = -10
(3)
3
1= A+ B+ D ⎯
⎯→ B + D = -1
(4)
s :
s :
Solving (2), (3), and (4) gives
- 13
A = 2,
,
B=
4
G (s) =
C=
32
94
2 13 4
−
−
2 +
s s + 1 (s + 1)
s+3
-3
,
2
D=
9
4
g ( t ) = ( 2 − 3.25 e-t − 1.5 t e-t + 2.25 e-3t )u(t ), t ≥ 0
P.P.15.11
G (s) =
10
A
Bs + C
=
+ 2
(s + 1)(s + 4s + 13) s + 1 s + 4s + 13
2
Multiplying both sides by (s + 1)(s 2 + 4s + 13) gives
10 = A (s 2 + 4s + 13) + B (s 2 + s) + C (s + 1)
Equating coefficients :
s2 :
0= A+B ⎯
⎯→ A = -B
(1)
1
0 = 4A + B + C ⎯
⎯→ C = -3A
(2)
0
10 = 13A + C ⎯
⎯→ 10 = 10A
(3)
s :
s :
Solving (1), (2), and (3) gives
A = 1,
B = -1 ,
G (s) =
C = -3
1
s+3
1
s+2
1
−
=
−
−
2
2
s + 1 (s + 2) + 9 s + 1 (s + 2) + 9 (s + 2) 2 + 9
1
g ( t ) = (e - t − e - 2t cos(3t ) − e - 2t sin(3t )), t ≥ 0
3
P.P.15.12
2
For 0 < t < 1 , consider Fig. (a).
y( t ) =
x1(t - λ)
x2(λ)
1
∫ (1)(1) dλ = t
t
0
t-1
0 t
(a)
1
2
λ
2
For 1 < t < 2 , consider Fig. (b).
y( t ) =
∫
(1)(1) dλ + ∫1 (1)(2) dλ = λ
t −1
1
t
t
t −1
+ 2λ
1
t
1
y( t ) = 1 − t + 1 + 2 ( t − 1) = t
0
t-1 1
t
2
λ
1 t-1 2
t
λ
(b)
For 2 < t < 3 , consider Fig. (c).
y( t ) =
∫
2
t −1
(1)( 2) dλ = 2 λ
2
2
t −1
1
y( t ) = 2 ( 2 − t + 1) = 6 − 2 t
0
For t > 3 , there is no overlap so y( t ) = 0 .
Thus,
(c)
y(t)
2
⎧ t
0<t<2
⎪
y( t ) = ⎨ 6 − 2t 2 < t < 3
⎪ 0
otherwise
⎩
0
1
The result of the convolution is shown in Fig. (d).
2
3
t
(d)
P.P.15.13
3e-λ
g(t-λ)
t-1
0 t
1
λ
(a)
∫ (1) 3 e
t
0
-λ
t-1 1
(b)
For 0 < t < 1 , consider Fig. (a).
y( t ) =
0
dλ = -3 e -λ
For t > 1 , consider Fig. (b).
t
0
= 3 (1 − e - t )
t
λ
y( t ) =
∫
t
t −1
(1) 3 e -λ dλ = -3 e -λ
t
t −1
= 3 e - t (e − 1)
Thus,
⎧ 3 (1 − e-t )
0≤t≤1
⎪⎪ -t
y( t ) = ⎨3 e (e − 1)
t≥1
⎪
0
elsewhere
⎪⎩
The circuit in the s-domain is shown below.
P.P.15.14
1
+
Vs
+
−
2/s
Vo
−
Vo =
2s
V
1+ 2 s s
H(s) =
Vo
2
=
Vs s + 2
⎯
⎯→ h ( t ) = 2 e -2t
v o (t) = h(t) ∗ v s (t) =
∫
t
0
h ( λ) v s ( t − λ) d λ
t
= ∫ 2 e- 2λ 10 e- (t − λ ) dλ
0
t
= 20 e- t ∫ e- 2 λ eλ dλ = 20 e- t (-e- λ ) 0t
0
= 20 (e − e -2t )u(t ) V
-t
Taking the Laplace transform of each term gives
[ s 2 V(s) − sv(0) − v′(0) ] + 4 [ sV(s) − v(0) ] + 4 V(s) = 1
s +1
2
1
s + 6s + 6
=
(s 2 + 4s + 4) V(s) = s + 5 +
s +1
s +1
2
s + 6s + 6
A
B
C
V (s) =
+
+
2 =
(s + 1)(s + 2)
s + 1 s + 2 (s + 2) 2
s 2 + 6s + 6 = A (s 2 + 4s + 4) + B (s 2 + 3s + 2) + C (s + 1)
P.P.15.15
Equating coefficients :
s2 :
A = 1− B
6 = 4A + 3B + C ⎯
⎯→ 6 = A + 3 + C or C = 3 − A
0
6 = 4A + 2B + C ⎯
⎯→ 6 = 6 − B or
s :
s :
Thus,
1= A+ B ⎯
⎯→ B = 1 − A or
1
A = 1,
B = 0,
V(s) =
1
2
+
s + 1 (s + 2) 2
B=0
C=2
and
Therefore,
v( t ) = (e -t + 2 t e -2t ) u(t )
Note, there were no units give for v(t).
Taking the Laplace transform of each term gives
2
2
sY (s) − y(0) + 3Y (s) + Y (s) =
s
s+3
[ s 2 + 3s + 2] Y(s) = 2s
s+3
2s
A
B
C
Y(s) =
=
+
+
(s + 1)(s + 2)(s + 3) s + 1 s + 2 s + 3
P.P.15.16
A = Y (s) (s + 1) s= -1 = -1
B = Y (s) (s + 2) s= -2 = 4
C = Y (s) (s + 3) s = -3 = -3
Y (s) =
-1
4
3
+
−
s +1 s + 2 s + 3
y( t ) = (-e -t + 4 e -2t − 3 e -3t ) u(t )
February 5, 2006
CHAPTER 16
P.P.16.1
Consider the circuit shown below.
s
Io
4/s
2/s
4
+
Vo(s)
−
Using current division,
4
2
8
s
⋅ =
Io =
2
4
s s (s + 4s + 4)
+s+4
s
Vo (s) = 4 I o =
32
s (s + 2) 2
32
A
B
C
+
+
2 =
s (s + 2)
s s + 2 (s + 2) 2
32 = A (s 2 + 4s + 4) + B (s 2 + 2s) + Cs
Equating coefficients :
s0 :
32 = 4A ⎯
⎯→ A = 8
1
0 = 4A + 2B + C
s :
2
s :
0= A+B ⎯
⎯→ B = -A = -8
Hence, 0 = 4A + 2B + C ⎯
⎯→ C = -16
Vo (s) =
8
8
16
−
−
s s + 2 (s + 2) 2
v o ( t ) = 8 (1 − e -2t − 2t e -2t ) u(t ) V
The circuit in the s-domain is shown below.
P.P.16.2
1
1/(s + 2)
+
−
Vo(s)
2s
i(0)/s
2
At node o,
Vo Vo i(0)
1
− Vo =
+
+
where i(0) = 0A
s+2
2s
2
s
1
⎛ 1 1⎞
= Vo ⎜1 + + ⎟
s+2
⎝ 2 2s ⎠
2s
2s / 3
A
B
Vo =
=
=
+
(s + 2)(3s + 1) (s + 2)(s + 1 3) s + 2 s + 1 3
Solving for A and B we get,
A = [2(–2)/3]/(–2+1/3) = 4/5, B = [2(–1/3)/3]/[(–1/3)+2] = –2/15
Vo =
45
2 15
−
s + 2 s +1 3
Hence,
2
⎛4
⎞
v o ( t ) = ⎜ e - 2 t − e - t 3 ⎟ u ( t )V
15
⎝5
⎠
P.P.16.3
v(0) = V0 is incorporated as shown below.
V(s)
+
I0/s
R
1/sC
V
CV0
−
We apply KCL to the top node.
I0
⎛
V
1⎞
+ CV0 = + sCV = ⎜ sC + ⎟ V
⎝
s
R
R⎠
I0
CV0
+
s (sC + 1 R ) sC + 1 R
V0
I0 C
V=
+
s + 1 RC s (s + 1 RC)
V0
A
B
V=
+ +
s + 1 RC s s + 1 RC
I0 C
I0 C
= I0R ,
B=
= - I0R
where A =
1 RC
- 1 RC
V=
V(s) =
V0
I0R
I0R
+
−
s + 1 RC
s
s + 1 RC
v( t ) = ((V0 − I 0 R ) e - t τ + I 0 R ), t > 0, where τ = RC
P.P.16.4
We solve this problem the same as we did in Example 16.4 up to the point
where we find V1. Once we have V1, all we need to do is to divide V1 by 5s to
and add in the contribution from i(0)/s to find IL.
IL = V1/5s – i(0)/s = 7/(s(s+1)) – 6/(s(s+2)) – 1/s
= 7/s – 7/(s+1) – 3/s + 3/(s+2) – 1/s = 3/s – 7/(s+1) + 3/(s+2)
Which leads to iL(t) = (3 – 7e–t + 3e–2t)u(t)A
We can use the same solution as found in Example 16.5 to find iL.
P.P.16.5
All we need to do is divide each voltage by 5s and then add in the contribution
from i(0). Start by letting iL = i1 + i2 + i3.
I1 = V1/5s – 0/s = 6/(s(s+1)) – 6/(s(s+2)) = 6/s – 6/(s+1) – 3/s + 3/(s+2)
i1 = (3 – 6e–t + 3e–2t)u(t)A
or
I2 = V2/5s – 1/s = 2/(s(s+1)) – 2/(s(s+2)) – 1/s = 2/s – 2/(s+1) – 1/s + 1/(s+2) –1/s
i2 = (–2e–t + e–2t)u(t)A
or
I3 = V3/5s – 0/s = –1/(s(s+1)) + 2/(s(s+2)) = –1/s + 1/(s+1) + 1/s – 1/(s+2)
i3 = (e–t – e–2t)u(t)A
or
This leads to iL(t) = i1 + i2 + i3 = (3 – 7e–t + 3e–2t)u(t)A
P.P.16.6
Ix
1/s
1Ω
+
+
−
5/s
Vo
+
−
2Ω
4Ix
−
(a) Take out the 2 Ω and find the Thevenin equivalent circuit.
VTh =
Ix
1/s
1Ω
+
5/s
+
−
VTh
−
+
−
4Ix
Using mesh analysis we get,
–5/s +1Ix +Ix/s + 4Ix = 0 or (1 + 1/s + 4)Ix = 5/s or Ix = 5/(5s+1)
VTh = 5/s – 5/(5s+1) = (25s+5–5s)/(s(5s+1)
= 5(4s+1)/(s(5s+1) = 4(s+0.25)/(s(s+0.2)
Ix
5/s
+
−
1/s
1Ω
+
−
Isc
4Ix
Ix = (5/s)/1 = 5/s Isc = 5/s + 4(5/s)/(1/s) = 5/s + 20 = (20s+5)/s = 20(s+0.25)/s
ZTh = VTh/Isc = {4(s+0.25)/(s(s+0.2))}/{20(s+0.25)/s} = 1/(5(s+0.2))
1
5(s + 0.2)
4(s + 0.25)
s(s + 0.2)
+
+
−
Vo
2Ω
−
4(s + 0.25)
s(s + 0.2)
4(s + 0.25) 10(4s + 1)
Vo =
2 =
or
1
s(s + 0.3)
s(10s + 3)
+2
5(s + 0.2)
(b)
Initial value: vo(0+) = Lim sVo = 4V
s→∞
Final value:
vo(∞) = Lim sVo = 4(0+0.25)/(0+0.3) = 3.333V
s→0
(c)
Partial fraction expansion leads to Vo = 3.333/s + 0.6667/(s+0.3)
Taking the inverse Laplace transform we get,
vo(t) = (3.333 + 0.6667e–0.3t)u(t)V
P.P.16.7
If x ( t ) = e -3t u ( t ) , then X(s) =
Y(s) = H(s) X(s) =
1
.
s+3
2s
A
B
=
+
(s + 3)(s + 6) s + 3 s + 6
A = Y (s) (s + 3) s= -3 = -2
B = Y (s) (s + 6) s= -6 = 4
Y (s) =
-2
4
+
s+3 s+6
y( t ) = (-2 e -3t + 4 e -6t )u ( t )
H(s) =
2 (s + 6 − 6)
2s
12
=
= 2−
(s + 6)
s+6
s+6
h ( t ) = 2 δ(t ) − 12 e -6t u(t )
By current division,
2 + 1 2s
I1 =
I
s + 4 + 2 + 1 2s 0
P.P.16.8
H(s) =
I1
2 + 1 2s
4s + 1
=
= 2
I 0 s + 4 + 2 + 1 2s 2s + 12s + 1
P.P.16.9
(a)
2s
Vo
1 || 2 s
1+ 2 s
2
=
=
=
2s
Vi 1 + 1 || 2 s
s+4
1+
1+ 2 s
H(s) =
Vo
2
=
Vi s + 4
(b)
h ( t ) = 2 e -4t u(t )
(c)
Vo (s) = H (s) Vi (s) =
2
A
B
= +
s (s + 4) s s + 4
1
,
2
1 ⎛1
1 ⎞
⎟
Vo (s) = ⎜ −
2⎝s s + 4⎠
A = s Vo (s) s= 0 =
-1
2
1
(1 − e -4t ) u(t ) V
2
v o (t) =
(d)
B = (s + 4) Vo (s) s = -4 =
v i ( t ) = 8 cos( 2 t ) ⎯
⎯→ Vi (s) =
Vo (s) = H(s) Vi (s) =
8s
s +4
2
16s
A
Bs + C
=
+ 2
2
(s + 4)(s + 4) (s + 4) (s + 4)
A = (s + 4) Vo (s) s= -4 =
- 16
5
Multiplying both sides by (s + 4)(s 2 + 4) gives
16s = A (s + 4) + B (s 2 + 4s) + C (s + 4)
Equating coefficients :
s2 :
0= A+B ⎯
⎯→ B = -A =
s1 :
16 = 4B + C ⎯
⎯→ C =
16
5
0 = 4 A + 4C ⎯
⎯→ C = -A
s0 :
Vo (s) =
16
5
(1)
(2)
(3)
16 ⎛ − 1
s + 1 ⎞ 16 ⎛ − 1
s
1
2 ⎞
⎜⎜
⎟⎟ = ⎜⎜
⎟
+
+
+ ⋅
2
2
2
5 ⎝ s + 4 s + 4 ⎠ 5 ⎝ s + 4 s + 4 2 s + 4 ⎟⎠
[
]
v o ( t ) = 3.2 − e - 4t + cos( 2t ) + 0.5 sin( 2t ) u(t ) V
P.P. 16.10 Consider the circuit below.
R1
iR
i
L
+
+
vs
vL
-
C
+
_
R2
v
-
iR = i + C
vo = R2i
But
dv
dt
(1)
iR =
Hence,
vs − v
R1
vs − v
dv
= i+C
R1
dt
or
•
v=−
v
i
v
− + s
R1C C R1C
(2)
Also,
-
v + vL + vo =0
vL = L
di
= v − vo
dt
But vo = iR2 . Hence
•
+
vo
v iR2
−
L L
Putting (1) to (3) into the standard form
i = v / L − vo / L =
(3)
⎡ 1
⎡ • ⎤ ⎢−
⎢ v ⎥ = ⎢ R1C
⎢•⎥ ⎢ I
⎣i ⎦ ⎢
⎣ L
1⎤
⎡ 1 ⎤
C ⎥ ⎡v ⎤ ⎢
⎥ ⎢ ⎥ + R 1 C ⎥ vs
⎥
R2 ⎥ ⎣ i ⎦ ⎢
⎢⎣ 0 ⎥⎦
− ⎥
L⎦
⎡v ⎤
vo = [ 0 R2 ] ⎢ ⎥
⎣i ⎦
−
If we let R1= 1, R2 = 2, C = ½, L = 1/5, then
⎡ −2 −2 ⎤
⎡2⎤
A=⎢
,
B = ⎢ ⎥ , C = [ 0 2]
⎥
⎣ 5 −10 ⎦
⎣0⎦
2 ⎤
⎡s + 2
sI − A = ⎢
⎥
⎣ −5 s + 10 ⎦
⎡ s + 10 −2 ⎤
⎢ 5
s + 2 ⎥⎦
⎣
−1
( sI − A) = 2
s + 12s + 30
H (s) = C(sI − A) −1 B =
=
=
[0
⎡s + 10 − 2 ⎤ ⎡2⎤
2]⎢
s + 2⎥⎦ ⎢⎣0⎥⎦
⎣ 5
s 2 + 12s + 30
20
s 2 12s + 30
20
s 2 + 12s + 30
P.P. 16.11 Consider the circuit below.
i
1
i1
R1
vo
+
v
-
L
2
io
C
R2
i2
At node 1,
•
v
+ C v+ i
R1
•
1
1
i
or
v=−
v− i+ 1
R1C
C
C
This is one state equation.
At node 2,
io = i + i2
(2)
Applying KVL around the loop containing C, L, and R2, we get
i1 =
(1)
•
−v + L i + io R2 = 0
or
•
v R2
− io
L L
Substituting (2) into (3) gives
•
v R
R
(4)
i = − 2 i − 2 i2
L L
L
vo = v
(5)
From (1), (3), (4), and (5), we obtain the state model as
1⎤
⎡ 1
⎡1
⎤
− ⎥
0 ⎥
⎡ • ⎤ ⎢−
⎢C
v
v
⎡ i1 ⎤
⎡
⎤
R
C
C
⎢ ⎥=⎢ 1
⎥
+
⎢
⎥
⎢
⎥
⎢•⎥ ⎢ 1
R2 ⎥ ⎢⎣i2 ⎥⎦
R2 ⎥ ⎣ i ⎦ ⎢
i
0
−
− ⎥
⎣ ⎦ ⎢
⎢⎣
L ⎥⎦
L⎦
⎣ L
i=
⎡vo ⎤ ⎡1 0⎤ ⎡v ⎤ ⎡0 0⎤ ⎡ i1 ⎤
⎢ i ⎥ = ⎢ 0 1 ⎥ ⎢ i ⎥ + ⎢ 0 1 ⎥ ⎢i ⎥
⎦⎣ ⎦ ⎣
⎦⎣ 2⎦
⎣ o⎦ ⎣
Substituting R1 = 1, R2 =2, C = ½, L = ¼ yields
⎡•⎤
⎢ v ⎥ = ⎡ −2 −2 ⎤ ⎡v ⎤ + ⎡ 2 0 ⎤ ⎡ i1 ⎤
⎢ • ⎥ ⎣⎢ 4 −8⎦⎥ ⎣⎢ i ⎦⎥ ⎣⎢ 0 −8⎦⎥ ⎢⎣i2 ⎥⎦
⎣i ⎦
⎡vo ⎤ ⎡1 0⎤ ⎡v ⎤ ⎡0 0⎤ ⎡ i1 ⎤
⎢ i ⎥ = ⎢ 0 1 ⎥ ⎢ i ⎥ + ⎢ 0 1 ⎥ ⎢i ⎥
⎦⎣ ⎦ ⎣
⎦⎣ 2⎦
⎣ o⎦ ⎣
(3)
P.P. 16.12
Let
so that
x1 = y
(1)
•
•
•
•
•
••
x1 = y (2)
Let
Finally, let
x2 = x1 = y
(3)
x3 = x2 = y
(4)
then
•
•••
••
•
x3 = y = −6 y − 11 y − 6 y + z
= −6 x3 − 11x2 − 6 x1 + z
(5)
From (1) to (5), we obtain,
⎡•⎤
⎢ x1 ⎥ ⎡ 0
1
0 ⎤ ⎡ x1 ⎤ ⎡0⎤
⎢•⎥ ⎢
0
1 ⎥⎥ ⎢⎢ x2 ⎥⎥ + ⎢⎢0⎥⎥ z (t )
⎢ x2 ⎥ = ⎢ 0
⎢ • ⎥ ⎢ −6 −11 −6 ⎥ ⎢ x ⎥ ⎢1 ⎥
⎦⎣ 3⎦ ⎣ ⎦
⎢ x3 ⎥ ⎣
⎣ ⎦
⎡ x1 ⎤
y (t ) = [1 0 0] ⎢⎢ x2 ⎥⎥
⎢⎣ x3 ⎥⎦
P.P.16.13
The circuit in the s-domain is equivalent to the one shown below.
βVo
+
Z
Z
Vo
−
- Vo = (βVo ) Z ⎯
⎯→ - 1 = βZ ,
R
Z = R || 1 sC =
1 + sRC
βR
or
Thus, - 1 =
- (1 + sRC ) = βR
1 + sRC
where
For stability,
βR > -1
β>
or
-1
R
From another viewpoint,
Vo = -(βVo ) Z ⎯
⎯→ (1 + βZ) Vo = 0
⎛
βR ⎞
⎜1 +
⎟V = 0
⎝ 1 + sRC ⎠ o
(sRC + βR + 1) Vo = 0
⎛ βR + 1 ⎞
⎜s +
⎟V = 0
⎝
RC ⎠ o
βR + 1
must be positive, i.e.
RC
-1
β>
βR + 1 > 0
or
R
For stability
P.P.16.14
(a)
(b)
Following Example 15.24, the circuit is stable when
10 + α > 0
α > -10
or
For oscillation,
10 + α = 0
or
α = -10
P.P.16.15
Vo
=
Vi
R
L
=
1
R
1
R + sL +
s2 + s ⋅ +
sC
L LC
R
s⋅
Comparing this with the given transfer function,
R
=4
L
and
If we select R = 2 , then
2
L = = 0.5 H
4
1
= 20
LC
and
C=
1
1
=
= 0 .1 F
20L 10
Consider the circuit shown below.
P.P.16.16
Y3
Y4
Y1
Clearly,
V1
+
−
Vin
Y2
V2
−
+
Vo
V2 = 0
At node 1,
or
(Vin − V1 ) Y1 = (V1 − Vo ) Y3 + (V1 − 0) Y2
Vin Y1 = V1 (Y1 + Y2 + Y3 ) − Vo Y3
(1)
At node 2,
(V1 − 0) Y2 = (0 − Vo ) Y4
or
V1 =
- Y4
V
Y2 o
Substituting (2) into (1),
- Y4
Vin Y1 =
V (Y + Y2 + Y3 ) − Vo Y3
Y2 o 1
Vo
- Y1 Y2
=
or
Vin Y4 (Y1 + Y2 + Y3 ) + Y2 Y3
If we select Y1 =
1
1
, Y2 = sC1 , Y3 = sC 2 , and Y4 =
, then
R1
R2
(2)
-s⋅
C1
R1
Vo
=
⎞
Vin
1 ⎛ 1
⎜
+ sC1 + sC 2 ⎟ + s 2 C1C 2
R 2 ⎝ R1
⎠
1
Vo
R 1C 2
=
Vin
1 ⎛1
1 ⎞
1
⎜ +
⎟+
s2 + s ⋅
R 2 ⎝ C 1 C 2 ⎠ R 1 R 2 C1 C 2
-s⋅
Comparing this with the given transfer function shows that
1 ⎛1
1 ⎞
1
1
⎜ +
⎟ = 6,
= 2,
= 10
R 2 ⎝ C1 C 2 ⎠
R 1C 2
R 1 R 2 C1 C 2
If R 1 = 10 kΩ , then
1
C2 =
= 0.5 mF
2 × 10 3
1
=5 ⎯
⎯→
R 2 C1
1
= 5C1
R2
⎛ C1 ⎞
C2
1 ⎛1
1 ⎞
⎜ +
⎟=6 ⎯
⎟=6 ⎯
⎯→ 5⎜1 +
⎯→ C1 =
= 0.1 mF
R 2 ⎝ C1 C 2 ⎠
5
⎝ C2 ⎠
R2 =
1
1
=
= 2 kΩ
5C1 (5)(0.1 × 10 -3 )
Therefore,
C1 = 0.1 mF ,
C 2 = 0.5 mF ,
R 2 = 2 kΩ
February 5, 2006
CHAPTER 17
T = 2, ωo = 2π/T = π
P.P.17.1
f(t) = 1,
–1,
0<t<1
1<t<2
ao =
2
1 T
1 1
f ( t )dt = ⎡ ∫ (1)dt + ∫ (−1)dt ⎤ = 0.5(1 – 1) = 0
∫
⎥⎦
1
T 0
2 ⎢⎣ 0
an =
2
2 T
2 1
f ( t ) cos nω o dt = ⎡ ∫ 1 cos nπtdt + ∫ (−1) cos nπtdt ⎤
∫
⎥⎦
1
T 0
2 ⎢⎣ 0
=
bn =
=
1
[sin nπt ]10 − 1 [sin nπt ]12 = 0
nπ
nπ
2
2 T
2⎡ 1
ω
=
π
+
f
(
t
)
sin
n
dt
1
sin
n
tdt
(−1) sin nπtdt ⎤
o
∫
∫
∫
⎢
⎥⎦
0
0
1
T
2⎣
−1
[cos nπt ]10 + 1 [cos nπt ]12 = 2 [1 − cos nπ]
nπ
nπ
nπ
bn = 4/(nπ), for n = odd
= 0,
for n = even
f(t) =
4 ∞ 1
∑ sin nπt ,
π k =1 n
n = 2k – 1
T = 1, ωo = 2π/T = 2π, f(t) = t, 0 < t < 1.
P.P.17.2
ao =
2
1 T
⎡ 1 ( t )dt ⎤ = t
f
(
t
)
dt
=
⎢⎣ ∫0
⎥⎦
2
T ∫0
an =
2 T
2 1
f ( t ) cos nω o dt = ⎡ ∫ t cos nπtdt ⎤
∫
⎥⎦
T 0
1 ⎢⎣ 0
1
0
= 0.5
1
⎡ 1
= 2⎢
[cos 2nπt ] + t [sin 2nπt ]⎤⎥
2
2 nπ
⎣ (2nπ)
⎦0
=
bn =
2
[[cos 2nπ1] − 1] = 0
4n 2 π 2
2 T
2 1
f ( t ) sin nω o dt = ⎡ ∫ t sin 2nπtdt ⎤
∫
⎥⎦
0
T
1 ⎢⎣ 0
t
⎡ 1
[cos 2nπt ]⎤⎥ = − 2 [cos 2nπ] = –1/(nπ)
= 2 ⎢ 2 2 [sin 2nπt ] −
2nπ
⎣ 4n π
⎦ 0 2nπ
1
f(t) = 0.5 −
1 ∞ 1
∑ sin 2nπt
π n =1 n
P.P.17.3
f(t) =
1,
–1,
–π < t < 0
0<t<π
f(t) is an odd function,
ao = 0 = an
T = 2π, ωo = 2π/T = 1
bn =
4 T/2
4 ⎡ π
⎤
f
(
t
)
sin
n
dt
( −1) sin ntdt ⎥
ω
=
o
∫
∫
⎢
0
0
T
2π ⎣
⎦
π
2
⎡2
⎤
[cos nπ − 1]
= ⎢ [cos nt ]⎥ =
⎣ nπ
⎦ 0 nπ
= –4/(nπ),
0,
n = odd
n = even
f(t) =
−4 ∞ 1
∑ sin nt ,
π k =1 n
n = 2k – 1
f(t) = t/π, 0 < t < π, T = 2π, ωo = 1
P.P.17.4
This is an even function, bn = 0.
2 T/2
2 ⎡ π
1 t2
⎤
ao =
f ( t )dt =
( t / π)dt = 2 x
⎥⎦ π
T ∫0
2π ⎢⎣ ∫0
2
an =
=
π
0
= 0.5
4 T/2
4 ⎡ π t
⎤
f ( t ) cos nω o dt =
cos ntdt ⎥
∫
∫
⎢
T 0
2π ⎣ 0 π
⎦
π
2 ⎡
[( t / n ) sin nt ]0π − 1 ∫0 sin ntdt ⎤⎥
2 ⎢
n
π ⎣
⎦
π
− 2 −1
2
cos nt = 2 2 (cos nπ − 1)
=
2
nπ n
n π
0
= –4/(n2π2),
0,
f(t) =
n = odd
n = even
1 4
−
2 π2
∞
1
∑n
k =1
2
cos nt ,
n = 2k – 1
f(t) = t/π, 0 < t < π, ωo = 2π/T = 1
P.P.17.5
This is half-wave symmetric. For odd n,
an =
=
4 T/2
4 ⎡ π t
⎤
f ( t ) cos nω o dt =
cos ntdt ⎥
∫
∫
⎢
T 0
2π ⎣ 0 π
⎦
π
2 ⎡
[( t / n ) sin nt ]0π − 1 ∫0 sin ntdt ⎤⎥
2 ⎢
n
π ⎣
⎦
π
− 2 −1
2
cos nt = 2 2 (cos nπ − 1)
=
2
nπ n
n π
0
= –4/(n2π2),
0,
n = odd
n = even
bn =
4 T/2
4 ⎡ πt
⎤
f ( t ) sin nω o dt =
sin ntdt ⎥
∫
∫
⎢
0
0
T
2π ⎣ π
⎦
π
2
⎡ 2
⎤
= ⎢ 2 2 [sin nt − nt cos nt ]⎥ =
,
⎣n π
⎦ 0 nπ
Thus, f(t) =
n = odd
2 ∞ ⎛ −2
1
⎞
⎜ 2 cos nt + sin nt ⎟ ,
∑
π k =1 ⎝ n π
n
⎠
n = 2k – 1
P.P.17.6
∞
1
vs(t) = 0.5 – (1/π) ∑ sin 2πnt , ω = 2πn
n =1 n
vo(ω) = (1/(jωC))vs/(R + (1/jωC)) = vs/(1 + jωRC) = vs/(1 + j2ω), RC = 2
For the DC component (ω = 0, or n = 0),
vs = 0.5 and vo = 0.5
For the nth harmonic, vs = –(1/(nπ))∠90° or
vo = –(1/(nπ))∠–90°/ 1 + 4ω2 ∠tan–12ω
or vo = –1∠(–90 – tan–12ω)/(nπ 1 + 4ω2 )
∞
Hence,
vo(t) = 0.5 – (1/π) ∑
n =1
=
1
n 1 + 4ω
2
cos(2πnt – 90° – tan–12ω)
1 1 ∞ sin( 2πnt − tan −1 4πn
− ∑
2 π n =1
n 1 + 16π 2 n 2
P.P.17.7
2
v(t) = (1/3) + (1/π )
∞
⎛ 1
∑ ⎜⎝ n
n =1
= (1/3) + (1/π2)
2
∞
∑A
n =1
n
cos nt −
π
⎞
sin nt ⎟
n
⎠
cos(nt - φn)
where
1
n
An =
1
1
+ π2 = 2 1 + n 2π2
2
n
n
φn = tan–1(bn/an) = tan–1(–nπ)
v(t) = (1/3) +
1
π2
1
∑n
2
1 + n 2 π 2 cos[nt – tan-1(–nπ)]
Z = 2 + 1||(1/(jω)),
ω = n
= 2 + (1/(jω))/(1 + (1/(jω))) = 2 + (1/(1 + (jω))
ω = n
= (3 + 2jω)/(1 + jω),
= (3 + j2n)/(1 + jn)
I = V/Z = [(1 + jn)/(3 + j2n)] V
2Ω
I
V(ω)
By current division,
+
−
1/jω
1Ω
Io = (1/jω)I/[1 + (1/jω)] = I/(1 + jω) = V/(3 + j2n)
For the DC component (n = 0),
For the nth harmonic,
Io
V = 1/3 and Io = V/3 = 1/9
V = [1/(n2π2)] 1 + n 2 π 2 ∠–tan–1(–nπ)
Io = V/[ 9 + 4n 2 ∠–tan–1(2n/3)]
=
But,
1 + n 2 π 2 ∠[–tan–1(–nπ)–tan–1(2n/3)]/[n2π2 9 + 4n 2 ]
tan–1(–nπ) = –tan–1(nπ)
In the time domain,
io(t) = {(1/9) +
∑
1 + n2π2
cos[nt – tan–1(2n/3) + tan–1(nπ)]}A
n π 9 + 4n
Note, the summation is to be carried out from n=1 to ∞.
2
2
2
∞
P = VDC IDC + 0.5 ∑ Vn In cos(φn – θn)
P.P.17.8
n =0
= 80(0) + 0.5(120)(5) cos(10°) + 0.5(60)(2) cos(30°)
= 295.44 + 51.962
= 347.4 watts
I2rms = 82 + 0.5[302 + 202 + 152 + 102]
P.P.17.9
= 64 + 0.5x1625 = 876.5
= 29.61 A
P.P.17.10
f(t) = 1,
0<t<1
= 0,
1<t<2
T = 2,
ωo = 2π/T = π
T
1
0
0
Cn = (1/T) ∫ f ( t )e − jnωo t dt = 0.5[ ∫ 1e − jnπt dt + 0]
= 0.5[1/(–jnπ)]e–jnπt
But
1
0
= [j/(2nπ)](e–jnπ – 1)
e–jnπ = cos(nπ) – jsin(nπ) = cos(nπ) = (–1)n
Cn = [j/(2πn)][(–1)n – 1]
= 0,
n = even
= [–j/(nπ)],
n = odd ≠ 0
1
For n = 0,
Co = 0.5 ∫ 1dt = 0.5
Hence,
f(t) =
0
∞
1
j jnπt
e
− ∑
2 n = −∞ nπ
n≠0
n = odd
f(t) = t, –1 < t < 1, T = 2, ωo = 2π/T = π
P.P.17.11
Cn = (1/T) ∫
T/2
1
f ( t )e − jnωo t dt = 0.5 ∫ te − jnπt dt
−T / 2
−1
= 0.5[e–jnπt/(–jnπ)2](–jnπt – 1)| 1−1
= [–1/(2n2π2)][e–jnπ(–jnπ – 1) – ejnπ(jnπ – 1)]
= [–1/(2n2π2)][(cos nπ – j sin nπ)(–jnπ – 1)
– (cos nπ + j sin nπ)(jnπ – 1)]
= [j cos nπ]/(nπ)
Cn = j(–1)n/nπ, n ≠ 0
Co = (1/T) ∫
For n = 0,
T/2
−T / 2
∞
Thus,
f(t) =
∑
n = −∞
n ≠0
f ( t )dt = 0
j(−1) n jnπt
e
nπ
|Cn| = 1/(nπ), n ≠ 0, θn = (–1)n 90°, n ≠ 0
The amplitude and phase spectra are shown below.
0.32
0.8
–4
0.11
–3
|Cn|
0.32
0.16
–2
0.16
–1
0
1
2
0.11
3
0.8
4
n
90°
–3
–4
–1
–2
1
0
3
2
4
n
–90°
P.P.17.12
The schematic is shown below. The attributes of the voltage source is
entered as shown. After entering the final time (5 or 6T), the Print Step, the Step Ceiling,
and the Center Frequency in the transient dialog box, the circuit is saved. Once the
PSpice is run, the output contains the following Fourier coefficients.
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)
DC COMPONENT = 4.950490E-01
HARMONIC FREQUENCY FOURIER NORMALIZED
NORMALIZED
NO
(HZ) COMPONENT COMPONENT (DEG)
1
2
3
4
5
6
7
8
9
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
3.184E-01
1.593E-01
1.063E-01
7.978E-02
6.392E-02
5.336E-02
4.583E-02
4.020E-02
3.583E-02
1.000E+00
5.002E-01
3.338E-01
2.506E-01
2.008E-01
1.676E-01
1.440E-01
1.263E-01
1.126E-01
-1.782E+02
-1.764E+02
-1.747E+02
-1.729E+02
-1.711E+02
-1.693E+02
-1.675E+02
-1.657E+02
-1.640E+02
PHASE
PHASE (DEG)
0.000E+00
1.783E+00
3.564E+00
5.347E+00
7.128E+00
8.912E+00
1.069E+01
1.248E+01
1.426E+01
TOTAL HARMONIC DISTORTION = 7.363360E+01 PERCENT
P.P.17.13
The schematic is shown below. Since T = 1/f = 0.55 ms, in the
transient dialog box, we set Print Step = 0.01 ms, Final Time = 4 ms, Center Frequency
= 2,000 Hz, Number of Harmonics = 5, and Output Vars = V(R1:1). Once the circuit
is saved, we simulate it and obtain the following results.
DC COMPONENT = -1.507149E-04
HARMONIC FREQUENCY FOURIER NORMALIZED
NORMALIZED
NO
(HZ) COMPONENT COMPONENT (DEG)
1
2
3
4
5
2.000E+03
4.000E+03
6.000E+03
8.000E+03
1.000E+04
1.455E-04
1.845E-06
1.401E-06
1.015E-06
8.345E-07
1.000E+00
1.268E-02
9.629E-03
6.974E-03
5.736E-03
9.006E+01
9.624E+01
9.318E+01
8.118E+01
5.922E+01
PHASE
PHASE (DEG)
0.000E+00
6.177E+00
3.120E+00
-8.880E+00
-3.084E+01
TOTAL HARMONIC DISTORTION = 1.830344E+00 PERCENT
From this, we use the amplitude and phase of the Fourier components to get
v(t) = {–150.72 + 145.5sin(4π103t + 90°) + 1.845sin(8π103t + 96.24°)
+ - - -}μV
It should be noted that these answers are not quite the same as in the book. This is
probably due to different versions of PSpice.
P.P.17.14
From Example 16.14,
2ωo
3ωo
4ωo
5ωo
6ωo
= 4π =
= 6π =
= 8π =
= 10π =
= 12π =
12.566 rad/s
18.84 rad/s
25.13 rad/s
31.41 rad/s
37.7 rad/s
Since the ideal bandpass filter passes only 15 < ω < 25, it means that only the 3rd, 4th, and
5th harmonics will be passed. Hence,
y(t) = (–1/3π)sin(3ωot) – (1/(4π))sin(4ωot) – (/(5π))sin(5ωot), ωo = 2π
February 5, 2006
CHAPTER 18
P.P.18.1
1< t < 2
⎡1,
(a) g(t) = u(t + 1) - u(t - 2) = ⎢
otherwise
⎣0,
2
1
G(ω) = ∫ 1 ⋅ e − jωt dt = − e − jωt 12
1
jω
=
e − jω − e − j2 ω
jω
(b) F(t) = 4δ(t + 2)
F(ω) =
∫
∞
∞
f ( t )e − jωt dt = ∫ 4δ( t + 2)e − jωt dt
−∞
jωt
= 4e
−∞
t =2
= 4e
j2ω
(c) F(t) = sin(ωot)
⎡ − e jωo t ⎤ 1
F e j ωo t − F e - jωo
F(ω) = F ⎢
⎥=
⎣ 2 j ⎦ j2
[( ) (
)]
= − jπ[δ(ω − ω o ) − δ(ω + ω o )]
P.P.18.2
P.P.18.3
∫
F(ω) =
∞
−∞
1
f ( t )e − jωt = ∫ (−1)e − jωt dt
0
− jωt
=
e
− jω
=
2(cos ω − 1)
jω
f (t) =
e at ,
0,
F(ω) =
∫
∞
−∞
0
−1
−
− j ωt
e
− jω
1
0
=
[
]
j
1 − e jω − e − jω + 1
ω
t<0
t>0
∞
f ( t )e jωt dt = ∫ e at e − jωt dt
−∞
Let x = -t, then dt = -dx
0
0
∞
∞
F(ω) = ∫ e −ax e jωx ( −dx ) = − ∫ e
=
1
1
e a − jω ) x =
a − jω
a − jω
− ( a − jω ) x
dx
P.P.18.4
(a) g(t) = u(t) - u(t - 1)
F(ω) = u(ω) - e-jωu(ω) = (1 - e-jω)u(ω)
= (1 - e-jω)(πδ(ω) + 1/(jω))
(b) f(t) = te-2tu(t)
Let g(t) = e-2tu(t)
f(t) = tg(t)
F(ω) =
(c)
G(ω) = 1/(2 + jω)
j
dG
= j(-1) (2 + jω)-2(j)
dω
1
(2 + jω) 2
f(t) is sketched below.
f(t)
5
t
2
f '(t) = –10δ(t – 2) – 20δ(t – 2)
f "(t) = 10δ(t) - 10δ(t - 2) - 20δ'(t - 2)
(jω)2F(ω) = 10(1 - e-jω2) - 20jωe-jω2
F(ω) =
P.P.18.5
(
) + 20je − jω2
10 e − jω2 − 1
ω
ω2
Given f(t), f '(t) and f "(t) are sketched below:
f(t)
2
–4
–3
–2
–1
0
1
2
3
4
t
f ’(t)
–4
–3
–2
–1
0
1
2
3
t
4
f “(t)
2
–4
2
–3
2
–2
–1
0
1
–4
2
2
3
4
t
–4
f "(t) = 2δ(t + 4) - 4δ(t + 3) + 2δ(t + 2) + 2δ(t - 2) - 4δ(t + 3) + 2δ(t - 4)
We take the Fourier transform of each term.
(jω)2F(ω) = 2(ej4ω + e-j4ω) - 4(ej3ω + e-j3ω) + 2(ej2ω + e-j2ω)
= 4 cos 4ω - 8 cos 3ω + 4 cos 2ω
F(ω) = [1/(ω2)](8 cos 3ω - 4 cos 4ω - 4 cos 2ω)
P.P.18.6
(b)
6(2 jω + 3)
( jω + 1)( jω + 4)( jω + 2)
2
3
5
=
+
−
jω + 1 j ω + 2 jω + 4
h(t) = (2e-t + 3e-2t - 5e-4t)u(t)
(a) H(ω) =
y(t) = u(t) + 2e-t cos 4t u(t)
= (1 + 2e-t cos 4t) u(t)
P.P.18.7
vi = 2 sgn (t)
H(ω) = 4/(4 + jω)
Vi(ω) = 4/(jω)
Vo (ω) = H(ω)Vi (ω) =
=
4
4
−
j ω 4 + jω
16
A
B
=
+
jω(4 + jω) jω 4 + jω
vo(t) = 2 sgn (t) - 4e-4tu(t) = 2[-1 + u(t)] - 4e-4tu(t)
= - 2 + 4 [1 - e-4t]u(t)
P.P.18.8
Is(ω) = 20π[δ(ω + 4) + δ(ω - 4)]
6 + jω2
3 + jω
H(ω) =
=
10 + 6 + j2ω 8 + jω
⎛ 3 + jω ⎞
⎟⎟(20π)[δ(ω + 4) + δ(ω − 4)]
I0(ω) = H(ω)Is(ω) = ⎜⎜
⎝ 8 + jω ⎠
20π ∞ ⎛ 3 + jω ⎞
⎜
⎟[δ(ω + 4) + δ(ω − 4)e jωt dω]
2π ∫−∞ ⎜⎝ 8 + jω ⎟⎠
⎡ 3 − j4 − j4 t 3 + j4 j4 t ⎤
e
e ⎥
= 10 ⎢
+
8 + j4
⎣ 8 − jω
⎦
i o ( t ) = F -1Io(ω) =
But
3 + j4
5∠53.13D
=
= 0.559∠26.57 D
D
8 + j4
80∠26.56
(
D
i o ( t ) = 5.59 e − j( 4 t + 26.57 ) + e j( 4 t + 26.57
io(t) = 11.18 cos (4t + 26.57°)A
P.P.18.9
∞
(a) W1Ω = ∫ 100 e
−4 t
−∞
D
)
)
∞
dt = 200∫ e − 4 t dt
0
since t is even.
W1Ω =
(b)
200e −4 t
−4
40
4 + ω2
1 ∞ 1600
= ∫
π 0 4 + ω2
∞
0
= 50J
H (ω) =
W1Ω
(
W1Ω =
)
2
dω =
1600 1 ⎛ ω
1
ω⎞
⋅ ⎜ 2
+ tan −1 ⎟ 0∞
2⎠
π 8⎝ω + 4 2
200 ⎛
π
⎞
⎜ 0 + − 0 − 0 ⎟ = 50J
4
π ⎝
⎠
P.P.18.10 F(ω) =
W2 Ω
for -4 < ω < 4,
1
1
2
F(ω) =
1 + jω
1 + ω2
2 ∞ dω
2
2 π
= ∫
= tan −1 ω ∞0 = ⋅ = 1
2
0
π 1+ ω
π
π 2
2 4 dω
= 2 tan −1 ω 04
2
∫
0
π 1+ ω
2 76
= ⋅
π = 0.844 = 84.4%
π 180
W=
i.e. 84.4% of the total energy.
P.P.18.11 If fc = 2 MHz, fm = 4 kHz
upper sideband = 2,000,000 + 4,000 = 2,004,000 Hz
Carrier = 2,000,000 Hz
Lower sideband = 2,000,000 -4,000 = 1,996,000 Hz
P.P.18.12 W = 12.5 kHz, fs = 2W = 25 kHz
1
1
= 40 μs
Ts = =
f s 25x10 3
February 5, 2006
P.P.19.11
V1 − 0
R1
0 − V2
I1 =
R2
I1 =
Also,
⎯
⎯→ V1 = I 1 R 1
⎯
⎯→ V2 = - I 1 R 2
Comparing these with
V1 = z 11 I 1 + z 12 I 2
V2 = z 21 I1 + z 22 I 2
shows that
z 11 = R 1 ,
z 21 = - R 2 ,
z 12 = z 21 = 0
Hence,
⎡ R1
[z ] = ⎢
⎣ - R2
0⎤
0 ⎥⎦
Since Δ z = z 11 z 22 − z 12 z 21 = 0 , [z ]-1 does not exist . Consequently, [y ] does not exist .
P.P.19.12
This is a series connection of two two-ports.
For N a ,
For N b ,
z 12 a = z 21a = 20 ,
z 12 b = z 21b = 50 ,
Thus,
[z ] = [z a ] + [z b ]
z 11a = 20 − j15 ,
z 11b = 50 + j40 ,
z 22 a = 20 + j10
z 22 b = 50 − j20
⎡ 20 − j15
20 ⎤ ⎡50 + j40
50 ⎤
[z ] = ⎢
+⎢
⎥
20 + j10 ⎦ ⎣ 50
50 − j20 ⎥⎦
⎣ 20
⎡ 70 + j25
70 ⎤
[z ] = ⎢
70 − j10 ⎥⎦
⎣ 70
V2
z 12 Z L
=
Vs (z 11 + Z s )(z 22 + Z L ) − z 12 z 21
V2
(70)(40)
=
Vs (70 + j25 + 5)(70 − j10 + 40) − 4900
V2
2800
=
Vs 8250 − j750 + j2750 + 250 − 4900
V2
2800
=
= 0.6799∠ - 29.05°
Vs 3600 + j2000
P.P.19.13
We convert the upper T network N a to a Π network, as shown below.
25 S
-j5 S
j5 S
y 1 y 2 + y 2 y 3 + y 3 y 1 (-j5)(j5) + (j5)(1) + (1)(-j5)
=
= -j5
j5
y2
yb = 5 ,
y c = 25
ya =
For N a ,
y 12 a = -25 = y 21a ,
y 11a = 25 − j5 ,
y 22 a = 25 + j5
⎡ 25 − j5
- 25 ⎤
[y a ] = ⎢
25 + j5⎥⎦
⎣ - 25
For N b ,
y 12 b = j10 = y 21b ,
y 11b = 2 − j10 = y 22 b
⎡ 2 − j10
j10 ⎤
[y b ] = ⎢
2 − j10 ⎥⎦
⎣ j10
Since N a and N b are in parallel, [y ] = [y a ] + [y b ]
⎡ 27 − j15 - 25 + j10 ⎤
[y ] = ⎢
⎥S
⎣ - 25 + j10 27 − j5 ⎦
P.P.19.14
Convert the left Π network to a T network.
(20)(50)
(20)(30)
R1 =
= 6, R2 =
= 10 ,
100
20 + 30 + 50
R3 =
(30)(50)
= 15
100
Putting this network into the given network produces the network shown below. This
may be regarded as a cascaded connection of T two-port networks.
6Ω
15 Ω
60 Ω
40 Ω
10 Ω
20 Ω
Na
Nb
For N a ,
Aa = 1+
Ca =
6
= 1 .6 ,
10
1
= 0 .1 ,
10
⎛6⎞
B a = 15 + ⎜ ⎟ ( 25) = 30
⎝ 10 ⎠
15
Da = 1 +
= 2 .5
10
⎡1.6 30 ⎤
[Ta ] = ⎢
⎥
⎣0.1 2.5⎦
For N b ,
Ab = 1+
Cb =
40
= 3,
20
1
= 0.05 ,
20
⎛ 40 ⎞
B b = 60 + ⎜ ⎟ (80) = 220
⎝ 20 ⎠
60
Db = 1 +
=4
20
⎡ 3
220 ⎤
[Tb ] = ⎢
⎥
⎣ 0.05 4 ⎦
Hence,
220⎤
⎡1.6 30 ⎤ ⎡ 3
[T] = [Ta ][Tb ] = ⎢
⎥
⎢
⎥
⎣0.1 2.5⎦ ⎣0.05 4 ⎦
We can now use MATLAB to obtain T.
>> Ta=[1.6,30;0.1,2.5]
Ta =
1.6000 30.0000
0.1000 2.5000
>> Tb=[3,220;0.05,4]
Tb =
3.0000 220.0000
0.0500 4.0000
>> T=Ta*Tb
T=
6.3000 472.0000
0.4250 32.0000
472 Ω ⎤
⎡ 6.3
[T] = ⎢
32 ⎥⎦
⎣ 0.425 S
P.P.19.15
To obtain h11 and h 21 , simulate the schematic in Fig. (a) using PSpice.
(a)
Insert a 1-A dc current source to account for I 1 = 1 A . Also, include pseudocomponents
VIEWPOINT and IPROBE to display V1 and I 2 respectively. When the circuit is saved
and run, the values of V1 and I 2 are displayed on the pseudocomponents as shown in
Fig. (a). Thus,
I
V
h 21 = 2 = -0.6190
h11 = 1 = 4.238 Ω ,
1
1
To obtain h12 and h 22 , insert a 1-V dc voltage source at the output port to account for
V2 = 1 V . The pseudocomponents VIEWPOINT and IPROBE are included to display
V1 and I 2 respectively. After simulation, the schematic displays the results as shown in
Fig. (b).
V
I
h12 = 1 = -0.7143 ,
h 22 = 2 = -0.1429 S
1
1
(b)
Thus,
⎡ 4.238 Ω - 0.7143 ⎤
[h ] = ⎢
⎥
⎣ - 0.6190 - 0.1429 S ⎦
P.P.19.16
Insert a 1-A ac current source at the output terminals to account for
I 1 = 1 A . Include two VPRINT1 pseudocomponents to output V1 and V2 . For each
VPRINT1, set the attributes to AC = yes, PHASE = yes, and MAG = yes. In the AC
Sweep and Noise Analysis dialog box, set Total pt : 1, Start Freq : 60, and End Freq : 60.
The schematic is shown in Fig. (a).
(a)
Once the schematic is saved and run, the output results include :
FREQ
6.000E+01
VM($N_0002)
3.987E+00
VP($N_0002)
1.755E+02
FREQ
6.000E+01
VM($N_0003)
1.752E-02
VP($N_0003)
-2.651E+00
From this table,
z 11 =
V1
= 3.987 ∠175.5° ,
1
z 21 = 0.0175∠ - 2.65°
Similarly, insert a 1-A ac source at the output port with the two pseudocomponents in
place as in Fig. (a). The result is the schematic in Fig. (b).
(b)
When the schematic is saved and run, the output results include :
FREQ
6.000E+01
VM($N_0002)
1.000E-30
VP($N_0002)
0.000E+00
FREQ
6.000E+01
VM($N_0003)
2.651E-01
VP($N_0003)
9.190E+01
From this table,
z 12 =
V1
≅0
1
z 22 = 0.265∠91.9°
Thus,
⎤
⎡ 3.987 ∠175.5°
0
[z ] = ⎢
⎥Ω
⎣ 0.0175 ∠ - 2.65° 0.265 ∠91.9° ⎦
P.P.19.17
In this case, R s = 150 kΩ , R L = 3.75 kΩ .
h ie h oe − h re h fe = (6 × 10 3 )(8 × 10 -6 ) − (1.5 × 10 -4 )(200) = 18 × 10 -3
The gain for the transistor is given as,
- (200)(3750)
Av =
= Vo /Vb = –123.61
6000 + (18 × 10 - 3 )(3.75 × 10 3 )
To calculate the gain of the circuit we need to use,
–Vs + 150kIb + Vb = 0 or 0.002 = 150k(0.002/156k) – Vc/123.61
Vc = –9.506 mV which leads to the gain = –9.506/2 = –4.753
200
= 194.17
1 + (8 × 10 )(3.75 × 10 3 )
Ai =
-6
Z in = 150,000 + 6000 − (1.5 × 10 -4 )(194.17) ≅ 156 kΩ
150 × 10 3 + 6 × 10 3
(150 × 10 3 )(8 × 10 -6 ) − (1.5 × 10 -4 )(200)
156
=
kΩ = 128.08 kΩ
1.248 − 0.03
Z out =
P.P.19.18
Let D(s) = (s 3 + 4s) + (s 2 + 2)
Dividing both numerator and denominator by s 3 + 4s gives
2
3
s + 4s
H (s) =
s2 + 2
1+ 3
s + 4s
i.e.
y 21 =
-2
3
s + 4s
y 22 =
s2 + 2
s 3 + 4s
As a third order function, we can realize H (s) by the LC network shown in Fig. (a).
L3
L1
+
+
C2
V1
V2
−
1Ω
−
(a)
ZA =
1
s 3 + 4s
2s
= 2
=s+ 2
= s L3 + Z B
y 22
s +2
s +2
L3 = 1 H
ZB =
2s
s +2
2
L3
L1
ZB
C2
(b)
YB =
y22 = 1 / ZA
1
s2 + 2
1
1
=
= 0.5s + = s C 2 +
ZB
2s
s
YC
C 2 = 0.5 F
YC =
1
1
=
s L1 s
⎯
⎯→
L1 = 1 H
Hence,
L1 = 1 H ,
C 2 = 0 .5 F ,
L3 = 1 H