February 5, 2006
CHAPTER 1
P.P.1.1
A proton has 1.602 x 10-19 C. Hence, 2 million 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 =
P.P.1.3
q=
1
∫ idt = ∫ 2dt + ∫ 0
= 2 + 14/3 =
P.P.1.4
(a)
2
1
7.358 mA
2
2t dt = 2 t
2
1
+ (2 / 3) t 0
3
2 1
6.667 C
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 =
(a)
v = 2 i = 10 cos (60 π t)
5V
p = v i = 50 cos2 (60 π t) At t = 5 ms, ms, p = 50 cos2 (60 π 5x10-3) = 50 cos2 (0.3 π) = (b)
v = 10 + 5
∫
t 0
17.27 watts
idt = 10 +
∫
t 0
25 cos 60 π t dt = 10+
25 60π
sin 60
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
π
t
P.P.1.6
p = v i = 15 x 120 = 1800 watts; w = p x t therefore, t = w/p = (30x103)/1800 =
P.P.1.7
16.667 seconds
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 P.P.1.10
cents/kWh
This assigned practice problem is to apply the detailed problem solving technique to some of the more difficult problems of Chapter 1.
P.P.1.6
p = v i = 15 x 120 = 1800 watts; w = p x t therefore, t = w/p = (30x103)/1800 =
P.P.1.7
16.667 seconds
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 P.P.1.10
cents/kWh
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 10cos(t)V/2cos(t)mA = 5 k
P.P.2.4
5 branches and 3 nodes. nodes. The 1 ohm and 2 ohm ohm resistors are in parallel. 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 v2 = -2i = –6V
P.P.2.6
Applying KVL to the loop we get: -35 + 10i + 2v x + 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
Applying KCL, 6 = i0 + [i0 /4] + [v0 /8], but i0 = v0/2
P.P.2.7
Which leads to: 6 = (v0/2) + (v0/8) + (v0/8) thus, v 0 = 8V and i0 = 4A P.P.2.8
2
+ -
5V
i1
+ V1 -
4 + V3 -
i2 + V2 -
Loop 1
i3
+
Loop 2
8
3V
At the top node,
i1 = i2 + i3
(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 or
V2 = 2 V
V1 = 5- V2 = 3V, V3 = 2+3 = 5V, i1 = (5-2)/2 = 1.5A, i2 = 250 mA, i3 = 1.25A 2
P.P.2.9
R eq
6
1
3
4
5
4
3
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. 2
Req
3
3
6
1
Thus, R eq = 1 + 2 + [6x6]/[6+6] = 6
20
P.P.2.10
8
Req
5
20
18
1 9 2
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: 8
4
6
1
20 2
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: 8
4
6
2
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: R eq = 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
P.P.2.12
15V
-
+ v
i1
i2 +
6
+ -
+ v1
10
15V
40
v2
+ -
+
4 8
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 i1
1k
+ 3k
v1
-
i2
+ 10mA
4k 5k
20k
v2
-
4k
10mA
Using current division, i 1 = 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 R a = [R 1 R 2 + R 2 R 3 + R 3 R 1]/ R 1 = [10x20 + 20x40 + 40x10]/10 = 140 ohms R b = [R 1 R 2 + R 2 R 3 + R 3 R 1]/ R 2 = 1400/20 = 70 ohms R c = [R 1 R 2 + R 2 R 3 + R 3 R 1]/ R 3 = 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: 13
i
a
24
100V + -
a
13 20
30
10
24
10
a’
b’
6
50
10
n
b
15 b
c’
R a’n = 20x30/[20 + 30 + 50] = 6 ohms, R b’n = 20x50/100 = 10 ohms R c’n = 30x50/100 = 15 ohms. Thus, R ab = 13 + (24 + 6)||(10 + 10) + 15 = 28 + 30x20/(30 + 20) = 40 ohms. i = 100/ R ab = 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 (a)
We use equation (2.61) -3 -3 R 1 = 50x10 / (1-10 ) = 0.05/999 = 50 mΩ (shunt)
(b)
R 2 = 50x10 /(100x10 – 10 ) = 50/99 = 505 mΩ (shunt)
(c)
R 3 = 50x10 /(10x10 -10 ) = 50/9 = 5.556 Ω (shunt)
P.P.2.17
-3
-3
-3
-3
-3
-3
February 5, 2006
CHAPTER 3
P.P.3.1
1A
6
i1
1
i1
i2
1A
2
4A i3
2
4A
7
At node 1, 1 = i1 + i2
1=
or 6 = 4v1 - v2
(1)
v1
− v2
+
v1
= 4+
v2
6
−0 2
At node 2, i1
v1
= 4 + i3
− v2 6
−0 7
or 168 = 7v1 - 13v2 (2) Solving (1) and (2) gives v1 = –2 V, v2 = –14 V 2
i1
P.P.3.2
4ix v1
i2
i2 3
10 A
v2
v3 ix 4
i3 6
At node 1, 10 = i1 + i2 =
v1 − v 3
+
− v2
v1
2 or 60 = 5v1 - 2v2 - 3v3
3 (1)
At node 2,
i2
v1
+ 4i x = i x
− v2 3
+3
v2 4
=0
or 4v1 + 5v2 = 0
(2)
At node 3,
v1
i1 = i3 + 4ix
− v3 2
=
v3
−0 6
+4
v2 4
or -3v1 + 6v2 + 4v3 = 0
(3)
Solving (1) to (3) gives v1 = 80 V, v2 = –64 V, v3 = 156 V
P.P.3.3 4
v
7V
v1
- +
+
3
+
2
v
-
6
-
At the supernode in Fig. (a),
7−v 4
v
v1
3
2
= +
+
or 21 = 7v + 8v1
v1 6 (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 2
= 1.4
i1 = 1.4 A P.P.3.4 v1
v2
3V
v3
+ -
-
+
+
+
v1
v2
v3
-
-
-
(a)
(b)
From Fig. (a),
v1
+
v2
+
v3
=0
6v1 + 3v2 + 4v3 = 0
(1)
- v1 + 10 + v2 = 0
v1 = v2 + 10
(2)
- v2 - 5i + v3 = 0
v3 = v2 + 5i
(3)
2
4
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⎤ ⎡ i 1 ⎤ ⎢ − 1 5 − 9⎥ ⎢i ⎥ ⎥ ⎢ 2⎥ ⎢ ⎢⎣− 2 − 4 9 ⎥⎦ ⎢⎣i 3 ⎥⎦ 3
−1 Δ = −2 3
−1 10 0
Δ1 =
0 10 0
−1 − 2 5 −9 − 4 9 = 135 - 8 - 18 - 20 - 108 - 9 = - 28 −1 − 2 5 −9 −1 − 2 5 −9 − 4 9 = 450 − 360 = 90 −1 − 2 5 −9
0
−2 −9
0
9
3
10
−1
0
−2 −9
3
−1 Δ2 = − 2
=
⎡10⎤ ⎢0⎥ ⎢ ⎥ ⎢⎣ 0 ⎥⎦
10
= 180 + 90 = 270
(3)
−1
10
−1 5 Δ3 = − 2 − 4 3 −1 −1 5
0
3
i1 =
= 40 + 100 = 140
0 10 0
Δ Δ 270 140 Δ1 90 = −9.643 , i3 = 3 = = −5A = = −3.214, i2 = 2 = Δ − 28 Δ − 28 Δ − 28
i0 = i3 = –5A P.P.3.7 2
i3
2
i1 6V
i1
i3
2
2
4
+
-
4
+
3A
-
3A
i2
8
i2 8
1
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)
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, G 13 = - 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, 0 ⎤ ⎡ 1.3 − 0.2 − 1 ⎢− 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 ⎦
R 11 = 50 + 40 + 80 = 170, R 22 = 40 + 30 + 10 = 80, R 33 = 30 + 20 = 50, R 44 = 10 + 80 = 90, R 55 = 20 + 60 = 80, R 12 = -40, R 13 = 0, R 14 = -80, R 15 = 0, R 21 = -40, R 23 = -30, R 24 = -10, R 25 = 0, R 31 = 0, R 32 = -30, R 34 = 0, R 35 = -20, R 41 = -80, R 42 = -10, R 43 = 0, R 45 = 0, R 51 = 0, R 52 = 0, R 53 = -20, R 54 = 0, v1 = 24, v2 = 0, v3 = -12, v4 = 10, v5 = -10
Hence the mesh-current equations are
⎡ 170 − 40 0 − 80 0 ⎤ ⎢− 40 80 − 30 − 10 0 ⎥ ⎢ ⎥ ⎢ 0 − 30 50 0 − 20⎥ ⎢ ⎥ − − 80 10 0 90 0 ⎢ ⎥ ⎢⎣ 0 − 20 0 0 80 ⎥⎦ P.P.3.10
⎡ i1 ⎤ ⎢i ⎥ ⎢ 2⎥ ⎢i 3 ⎥ ⎢ ⎥ ⎢i 4 ⎥ ⎢⎣i 5 ⎥⎦
=
⎡ 24 ⎤ ⎢ 0 ⎥ ⎢ ⎥ ⎢− 12⎥ ⎢ ⎥ ⎢ 10 ⎥ ⎢⎣− 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, 3
-5 + 10 x 10 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 10,000 + 20,000
= 142.38μA
V0 = 200 IE = 20,000 IB = 2.876 V From (2), -6
VCE = 12 - V0 - 500 IC = 9.124 - 500 x 100 x 142.38 x 10 VCE = 1.984 V {often, this is rounded to 2.0 volts}
P.P.3.13
20 k
i1 i0
iC 30 k
iB
+ 20 k
+
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 =
1 − 0.7
B
30k
= 10μA, iC = βiB = 0.8 mA
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 A
(2)
February 5, 2006
CHAPTER 4
P.P.4.1 6
i2 i1
+ 2
iS
By current division, i 2
2
=
is
2+6+4 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
=
1 6
4
vo
is
P.P.4.2 12
v1
+
+
VS = 10 V
Let v0 = 1. Then i =
1 8
5
−
and v 1
giving vs = 2.5V. If vs = 10V, then v0 = 4V
=
1 8
(12
+
8)
=
2 .5
vo
8
P.P.4.3
Let v0 = v1 + v2, where v1 and v2 are contributions to the 20-V and 8-A sources respectively. 3
5
i
+ v1
+
2
−
(a)
3
i2
i1
5
+ v2
8A
2
(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
v1
+
10 V
20
(a)
v2
2A
4
(b)
To obtain v1, consider Fig. (a).
0.1v1
+
10 − v1 20
=
v1
v1 = 2.5
4
For v2, consider Fig. (b). 2 + 0.1v2 +
0 − v2 20
=
vx = v1 + v2 = 12.5V
0.1v1
4
−
v2 4
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.
2 6
2
8
6
8 4A
i1 16V
+
i2
−
(a)
(b) 6
2
8 i3 12V
+ −
(c)
For i1, consider Fig. (a), i 1
=
16 6+2+8
= 1A
For i2, consider Fig. (b). By current division, i 2
For i3, consider Fig. (c), i 3
=
2 2 + 14
( 4 ) = 0 .5
− 12
= −0.75A 16 Thus, i = i1 + i2 + i3 = 1 + 0.5 - 0.75 = 750mA
P.P.4.6
=
Combining the 6-Ω and 3-Ω resistors in parallel gives 6 3 =
6x3
= 2Ω . 9 Adding the 1-Ω and 4-Ω resistors in series gives 1 + 4 = 5 Ω. Transforming the left current source in parallel with the 2-Ω resistor gives the equivalent circuit as shown in Fig. (a).
2
5V −
+
io
+
10V
−
7
5
3A
7
5
3A
(a)
io 7.5A
2
(b) io 10.5A
(10/7)
7
(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 io
=
7 10 7
+7
(10.5) = 1.78 A
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 15
(4 − 0.4i x )
ix = 1.176A
ix 4A
10
0.4ix
5
(a)
ix 4 – 0.4ix A
10
5
(b) P.P.4.8
To find R Th, consider the circuit in Fig. (a). 6
6
4
RTh
(a) 6 + 2A
6
2A
(b)
R Th
=
( 6 + 6) 4 =
12 x 4 18
=3
4
VTh
To find VTh, we use source transformations as shown in Fig. (b) and (c). 6
6 + 4
+
24 V
VTh
−
(c)
Using current division in Fig. (c), VTh
i=
=
4 4 + 12
VTh R Th
+1
( 24) = 6V
=
6 3 +1
=
1.5A
P.P.4.9 To find VTh, consider the circuit in Fig. (a).
5
6V
+ −
3
Ix
a
+
i2 i1
VTh
4 1.5Ix i2
i1 o
b (a)
0.5Ix
5
3
Ix
1.5Ix
i
a
+
4
(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 R Th, 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 m 2.25 i
P.P.4.10
Ix = -2
Since there are no independent sources, VTh = 0
4vx
10
+
−
+ vx
+ 5
15
vo
io
(a) 4vx
10
+
−
+ vx
15 +
5
vo
i
+ –
15io
(b)
To find R Th, 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 R Th = vo/(io) = –7.5
10i = -15io
P.P.4.11 3
3
6
RN
(a)
3
5A
3
4A
(b)
From Fig. (a), R N = (3 + 3) 6 = 3
From Fig. (b), I N =
1 2
(5 + 4) = 4.5A
IN
P.P.4.12
2vx i
+
−
+
+ 6
vx
2
ix
vx
+
1V
−
(a)
2vx
+
−
+ 6
2
10 A
Isc
vx
(b)
To get R N consider the circuit in Fig. (a). Applying KVL, 6 i x But vx = 1, 6ix = 3 ix = 0.5 v i = i x + x = 0.5 + 0.5 = 1 2 1 R N = R Th = = 1 i
−
2v x
−
1
=
0
To find I N, consider the circuit in Fig. (b). Because the 2Ω resistor is shorted, v x = 0 and the dependent source is inactive. Hence, I N = isc = 10A. P.P.4.13
Fig. (a).
We first need to find R Th and VTh. To find R Th, we consider the circuit in
vx
+
v0 4
−
+
i
2
vx
4
−
2 1
1
+ −
+
1V
+ −
9V
VTh
io +
3vx
3vx
(a)
(b)
Applying KCL at the top node gives
1 − vo 4
+
3v x
−
vo
1
=
vo 2
But vx = -vo. Hence
1 − vo 4
− 4v o =
1 − vo
1−
vo
vo = 1/(19)
2 1
19 = 9 4 4 38 R Th = 1/i = 38/(9) = 4.222Ω i=
=
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 R L = R Th = 4.222 Pmax
=
2 v Th
4R L
=
49 4(4.222)
=
2.901W
+
P.P.4.14
We will use PSpice to find Voc and Isc which then can be used to
find VTh and R th.
Clearly Isc = 12 A
Clearly VTh = Ioc = 5.333 volts . R Th = 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-k Ω resistor is replaced by 2-k Ω resistor. The plot of the power absorbed by R L is shown in the figure below. From the plot, it is clear that the maximum power occurs when R L = 2k Ω and it is 125 W.
VTh = 9V, R Th
P.P.4.16
=
(v oc − VL )
R L VL
=
(9 − 1)
20 8
=
2.5Ω
2.5 + 9V
VL
=
10 10 + 2.5
(9) = 7.2V
+ −
VL
10
P.P.4.17
R 1 = R 3 = 1k Ω, R 2 = 3.2k Ω R R x = 3 R 2 = R 2 = 3.2k R 1
P.P.4.18
We first find R Th and VTh. To get R Th, consider the circuit in Fig. (a). R Th
=
20 30 + 60 40 =
20 x 30 50
+
60 x 40 100
= 12 + 24 = 36Ω
20
30
20
30
a
+
v2
a
+ VTh
RTh
+ v 1
b
60
40
60
b
40
10 V
+
(a)
−
(b)
To find VTh, we use Fig. (b). Using voltage division, v1
But
−
v1
=
+
60 100
v2
(16) = 9.6,
+ v Th =
v2
20
=
50
0
(16) = 6.4
vTh = v1 - v2 = 9.6 - 6.4 = 32V IG
=
VTh R Th
+ R m
=
3.2 3.6 + 1.4
=
64mA
February 5, 2006
CHAPTER 5
The equivalent circuit is shown below:
P.P.5.1
2M
vd vs
+
+
-
40 k
1
+
5k
2
At node 2,
v s − v1 2x10 6
=
Av d − v 0 50
v1 5x10 3
+
+ Avd
v0
-
-
-
At node 1,
50
+
20 k
v1
i0
+
v1 − v 0 40x10 3
v1 − v 0 40x10 3
=
v1 =
v S + 50v 0 451
(1)
v0 20x133
But vd = v1 - vS. 5
[2 x 10 (v1 - vS) - v0] 4000/(5) + v1 - v0 = 2v0 1600 x 10 (vS - v1) + 803v0 ≅ 0 5
Substituting v1 in (1) into (2) gives 8
1.5914523 x 10 vS - 17737556v0 = 0 v0 vS
=
1.5964523x10 8 17737556
= 9.00041
If vS = 1 V, v0 = 9.00041 V, v1 = 1.0000455 -5
vd = vS - v1 = - 4.545 x 10 Av d − v 0 = 657 A Avd = - 9.0909, i0 = 50
(2)
P.P.5.2 20 k
i V1
10 k
VS
+
+
V2
-
+ V0
-
At node 1,
v S − v1 10
=
v1 − v 0 20
But v1 = v2 = 0,
vS 10
i0 =
=−
v0
v0
20
vS
0 − v0 20x103
=−
v0 20x10 3
When vs = 2V, v0 = -4, i0 =
P.P.5.3
v0 = −
i=
P.P.5.4
R 2 R 1
0 − v0 15k
(a) iS =
= –2
vi =
4 x10 −3 20
− 15 5
= 200 A
(40mV) = –120 mV
= 8 A
0 − v0
v0
R
iS
= −R
iS
R1
0V
V1
R2
R3
20 k
+
+
V2
V0
-
(b)
0 − v1
At node 2, iS =
At node 1,
v1 = -iSR 1
R 1
0 − v1 R 1
=
⎛ 1
-v1 ⎜⎜
⎝ R 1
v1 − 0 R 2
+
1 R 2
+
+
v1 − v 0 R 3
1 ⎞
− v0 ⎟⎟ = R 3 ⎠ R 3
⎛ 1
v0 = -iSR 1R 3 ⎜⎜
(1)
⎝ R 1
+
1 R 2
+
1 ⎞
⎟
R 3 ⎠⎟
⎛ R R ⎞ = − R 1 ⎜⎜1 + 3 + 3 ⎟⎟ iS ⎝ R 1 R 2 ⎠
v0
P.P.5.5
By voltage division v1 =
8 4+8
(3) = 2V
where v1 is the voltage at the top end of the 8k Ω resistor. Using the formula for noninverting amplifier,
⎛ ⎝
v0 = ⎜1 +
5 ⎞
⎟( 2) = 7 V
2 ⎠
This is a summer.
P.P.5.6
8 8 ⎡8 ⎤ v 0 = − ⎢ (1.5) + ( 2) + (1.2) ⎥ = –3.8 V 10 6 ⎣ 20 ⎦
i0 =
R 2 R 1 R 2 R 1
8
+
v0 4
=−
3.8 8
−
3.8 4
= –1.425 mA
If the gain is 4, then
P.P.5.7
But
v0
=
=4
R 2 = 4R 1
R 4
R 4 = 4R 3
R 3
If we select R1 = R3 = 10k , then R2 = R4 = 40k P.P.5.8
v0 =
R 2 ⎛ 2R 3 ⎞ ⎜1 + ⎟ (v2 - v1) R 1 ⎜⎝ R 4 ⎠⎟
R 3 = 0, R 4 = ∞, R 2 = 40k Ω, R 1 = 20k Ω v0 = i0 = P.P.5.9
40 20 v0
(8.01 − 8) = 0.02
0.02
=
10k 10x10 3
= 2 A
Due to the voltage follower va = 4V
For the noninverting amplifier,
⎛ ⎝
v0 = ⎜1 +
i0 =
v b 4
6 ⎞
⎟ va = (1 + 1.5) (4) = 10V
4 ⎠
mA
-
a +
+
vS
+
+
b
-
4k
v0 6k
-
But v b = va = 4 i0 = P.P.5.10
4 4
= 1mA
As a voltage follower, va = v1 = 2V
where va is the voltage at the right end of the 20 k Ω resistor. As an inverter, v b = −
50 10
v 2 = −7.5V
Where v b is the voltage at the right end of the 50k Ω resistor. As a summer 60 ⎤ ⎡ 60 v b v0 = − ⎢ v a + 30 ⎥⎦ ⎣ 20
= [6 - 15] = 9V The schematic is shown below. When it is saved and run, the results are P.P.5.11 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
P.P.5.12
or
-V0 =
R f R 1
V1 +
R f R 2
V2 +
R f
V3
R 3
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
V0
= 1.25, then V1 = 1, V2 = 0, V3 = 1, i.e.
[V1V2V3] = [101] (d)
V0
= 1.75, then V1 = 1, V2 = 1, V3 = 1, i.e.
[V1V2V3] = [111]
P.P.5.13
Av = 1 +
2R R G
R G =
R G =
2R Av −1
2x 25x10 3 142 − 1
= 354.6
February 5, 2006
CHAPTER 6
q
v=
P.P.6.1
w
=
=
= 40V
3x10 −6 1 Cv 2 = x 3x10 − 6 x1600 = 2.4mJ 2 2
C 1
i( t ) = C
P.P.6.2
120x10 −6
dv dt
= 10 x10 −6
d dt
(50 sin 2000 t )
= cos2000t A v=
P.P.6.3
=−
1
∫
t
C 0 500
idt
=
10 −3
t
∫ 50 sin 120πt dt V
0.1x10 −3
cos 120πt 0t
0
50
(1 − cos 120πt )V 120π 12π 50 (1 − cos 0.12π) = 93.14mV v(t = 1ms) = 12π 50 (1 − cos 0.6π) = 1.7361 V v(t = 5ms) = 12π
i(t) =
P.P.6.4
v
=
1
C∫
idt
=
−3
∫ idt ⋅10 1
−3
= ∫ idt
50 t dt = 25t x 10 C∫ 1 v = ∫ 100dt + v( 2) = (100 t − 0.2 + 0.1) C
For 0< t <2, v = For 2< t <6,
⎡50t, 0 〈 t 〈 2 ⎢100, 2 〈 t 〈 6 ⎣
1 10
=
t 0
t
2
= (100t - 0.1)V At t = 2ms, v = 100mV At t = 5ms, v = (500 - 100)mV = 400 mV
2
3
Under dc conditions, the capacitors act like open-circuits as shown below:
P.P.6.5
v2
+ 1k
i
−
3k +
+
10V
i=
10 1+ 3 + 6
6k
v1
−
= 1mA
= (3k + 6k )i = 9V v 2 = (3k )i = 3V v1
w1
=
w2
=
1 2 1 2
P.P.6.6
P.P.6.7
C1 v12
=
C 2 v 22
=
1
(10x10 − )(9)
2 1
2
6
= 405
2
(20 x10 − )(3) 6
2
=
J
90 J
Combining 60 and 120μF in series =
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 = = 40 F 180 60x 30 60μF in series with 30μF = = 20μF 90 20μF in parallel with 20μF = 40μF +
v2
40 60V
From the Figure, v1
= v2 =
60 2
+
−
− F + v1
40
F
= 30V
Hence v2 = 10V1, v4 = 20V Note that q 3 = q 4 = 60 x 10μC. Thus v1 = v2 = 30V, v3 = 10V, v4 = 20V
P.P.6.8
v=L
di dt
= 10 −3
d dt
( 20 cos 100 t ) ⋅ 10 −3
= –2 sin 100t mV w
=
1 2
Li 2
1
=
2
x10 −3 (400 cos 2 100 t ) ⋅ 10 −6
2
= 200 cos 100t J
P.P.6.9
i=
1 L
∫
t
v( t )dt + i( t 0 ) =
t0
1 2
t
∫ 10(1 − t )dt + 2 0
⎛ t 2 ⎞ = 5⎜⎜ t − ⎟⎟ + 2 ⎝ 2 ⎠ At t = 4, i = 5(4 - 8) + 2 = –18A
⎡ ⎣
p = vi = 10(1 - t) ⎢5t −
5
t2
+ 2⎤⎥ ⎦
2 3 = 20 + 30t - 75t + 25t 2
w
4
= ∫0 p dt = [20t + 15t2 - 25t3 +
25t 4 4]
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
+ 3k 4A
=
3
( 4) = 3A 1+ 3 vC = 1iC = 3V 1 1 ⎛ 1 ⎞ wL = Li 2L = ⎜ ⎟(3) 2 2 2 ⎝ 4 ⎠ 1 1 wC = Cv 2C = ( 2)(3) 2 2 2 iL
= 1.125J = 9J
1k
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) – i 1(0) = 1.4 - 0.6 = 800 mA di −2 t −2 t (b) v1 = 6 1 = 6(0.6)(−2)e = −7.2e dt 1 t 1 ( −7.2) − 2 t t i2 = v1dt + i 2 (0) = e 0 + 0.8 3 0 3 ( −2) –2t = -0.4 + 1.2e A –2t i = i1 + i2 = 0.4 + 1.8e A (c) From (b), –2t v1 = –7.2e V di -2t v2 = 8 = 8( −2)(1.8)e − 2 t = –28.8e V dt -2t v = v1 + v2 = –36e V
∫
P.P.6.13
3
-6
RC = 25 x 10 x 10 x 10 = 0.25 1 t 1 t vo = − v i ( t ) + v o ( 0) = − 10dt mV + 0 RC o 0.25 o = 40t mV
∫
P.P.6.14
P.P.6.15
∫
3
-6
-2
RC = 10 x 10 x 2 x 10 = 2 x 10 dv d v o = −RC i = −2x10 −2 (3t ) dt dt vo = -60mV dv o2 dt
2
= 4 cos10t − 3
dv o dt
− 2v o
Using this we obtain the analog computer as shown below. We may let RC = 1s.
2V
− C
C
R 2
d vo /dt
2
t=0
+
R
R
−
R/2
−
+
−
+
-dvo /dt
+
vo R
R
− +
R/3 dvo /dt
R R
− cos10t
+
−
R/4
+
-cos10t
2
d vo /dt
2
February 5, 2006
CHAPTER 7
The circuit in Fig. (a) is equivalent to the one shown in Fig. (b).
P.P.7.1
8
io
+
+ 12
6
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 -0.25t V vx =
4 v c = 10 e -0.25t V 4+8
⎯→ v x = v o + v c ⎯ 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). 6 + 24 V
+ −
vc(0)
12
4
(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). t=0
6
+ −
24 V
1/6 F
3
(b)
τ = R eq C = (3)(1 / 6) = 1 / 2 s v( t ) = v c (0) e - t τ = 8 e-2t 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. 3
io
+
vx
− 1
vo = 1 V
+ −
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
5
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. 3 i
+
vx
− 1
1/6 H
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 -53t i = - i1 = Be -53t i ( 0) = 5 = B i( t ) = 5e -53t A Therefore, i ( t ) = 5e -53t A and
v x ( t ) = -3i(t) = - 15e -53t V
5
where v x = 3i1 (3)
(4)
For t < 0, the equivalent circuit is shown in Fig. (a).
P.P.7.4
i(t)
12
8 5
5A
12
8 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
For t < 0, the switch is closed. The inductor acts like a short so the equivalent circuit is shown in Fig. (a). P.P.7.5
3 i i
io
1H
io 6A
4
2 4
(a)
i=
4 (6) = 4 A , 4+ 2
2
(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 τ= = R th = (4 + 2) || 3 = 2 Ω , R th 2 3 (-i) - 1 - 4 -2t 8 = i= io = e and v o = -2i o = e -2t 6+3 3 3 3
Thus, t<0 2A t<0 ⎧ 4A ⎧ = i ⎨ o - 2t - 2t ⎩4 e A t > 0 ⎩- (4 3 ) e A t > 0
i= ⎨
P.P.7.6
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 − 10t
4 2
t 2
= 40 − 10 t
=0
Thus,
⎧ 0 t<0 ⎪⎪ 10t 04 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 − 2 t ) 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 P.P.7.8
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.7.9
(a)
(b)
∫
∞
-∞
∫
10
0
( t 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
For t < 0, the capacitor acts like an open circuit. v(0 ) = v(0 + ) = v(0) = 10
P.P.7.10
−
6 2 (10) − (50) = -5 2+6 6+2 3 3 1 1 τ = R th C = × = R th = 2 || 6 = Ω , 2 2 3 2 v(∞) =
For t > 0,
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v( t ) = -5 + (10 + 5) e -2t v( t ) = - 5 15 e -2t V v(0.5) = -5 + 15 e -1 = 518.2 mV
At t = 0.5,
For t < 0, only the left portion of the circuit is operational at steady state. i ( 0) = 0 v(0 ) = v(0 + ) = v(0) = 20 ,
P.P.7.11
−
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
i
10
+ v
0.2 F
+ −
30 V
5 (30) = 10 15 10 Ω, R th = 5 || 10 = 3 v(∞) =
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 ) i( t ) = i( t ) =
P.P.7.12
- v( t ) = -2 ( 1 + e -1.5 t ) 5 0
t
0
e -1.5t A
t
0
-2 1
20 V
v( t ) =
10 1
e - 1.5t V
0
t
0
Applying source transformation, the circuit is equivalent to the one below. i
1.5 H
10
t=0
5
+ −
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(∞) =
t
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
20 i(t)
6A
5H
15
Since switch S1 is open at t = 0 − , i(0 − ) = 0 . Also, since i cannot jump, i(0) = i(0 − ) = 0 . 90 =2A i(∞) = 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 = 3.6 A i(∞) = 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
Thus, i ( t ) =
t
2 (1 e -9t ) A 3.6 1.6 e -5( t
2)
0 A
0 t
t
2 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 R f 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. ⎯→ v o = -v 0 − v o = v ⎯ v o = - 4 e -2t V , t
0
Let v1 be the potential at the inverting terminal. v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v ( 0) = 0 where τ = RC = 100 × 10 3 × 10 -6 = 0.1 , P.P.7.15
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 -10t mV v o = v1 − v = -v v o = 40 e -10t
1 mV
This is a noninverting amplifier so that the output of the op amp is ⎛ R f ⎞ ⎟ vi v a = ⎜1 + ⎝ R 1 ⎠
P.P.7.16
⎛ R f ⎞ ⎛ 40 ⎞ ⎟ v i = ⎜1 + ⎟ 2 u ( t ) = 6 u ( t ) v th = v a = ⎜1 + ⎝ 20 ⎠ ⎝ R 1 ⎠ 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 k Ω 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
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). P.P.7.17
(a)
(b)
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). P.P.7.18
(a)
(b)
P.P.7.19
v(0) = 0 . When the switch is closed, we have the circuit shown below. 10 k
R
a
+ 9V
80 F
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 ,
τ = R th C
v( t ) = v(∞) ( 1 − e - t τ ) Since v(0) = 0 , v( t ) 9 ( 1 − e -t τ ) mA i( t ) = = R + 4 R + 4 Assuming R is in k Ω, 9 ( 1 − e -t 0 τ ) × 10 -3 R + 14 R + 14 = 1 − e -t 0 τ (0.12) 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 t0 = × ln ⎜ ⎝ 7.32 ⎠ 14
4k
When R = 6 k Ω, 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 Δ q 0.16 ΔI= = = 200 A Δ t 0.8 × 10 -3 Δw 6.4 = = 8 kW p = Δ t 0.8 × 10 -3 Δ q 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 , = 550 mA i(∞) = 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 t = 5τ = R 5 2 ⎛ ⎞ 1 2 1 12 W = LI = ( 20 × 10 -3 ) ⎜ ⎟ = 57.6 mJ ⎝ 5 ⎠ 2 2 ⎛ 12 5 ⎞ di ⎟ = 24 kV V = L = 20 × 10 -3 ⎜ ⎝ 2 × 10 -6 ⎠ dt
February 5, 2006
CHAPTER 8
P.P.8.1
(a)
At t = 0-, we have the equivalent circuit shown in Figure (a). 10
i
+ 2
v
vL
a +
24V
+ −
2
vC
(a)
+ i
50mF
+ −
(b)
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).
5
5
a iR
2A
iL 3A
+ vC
+
b
vR
−
10 F
+ vL
2H 3A
(a)
(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.
5 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, s1,2 =
(b)
Since α <
ωo,
−α±
α
2
ωo
= 1
2
− ωo = −1 ±
LC =1
5x 2 x10 −2 = 10
1 − 100 = -1
j9.95.
we clearly have an underdamped response.
For t < 0, the inductor is connected to the voltage source although it acts P.P.8.4 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
α
LC
=1
1x
1 9
=
3
= R/(2L) = 5/(2x1) = 2.5
Since α < ωo, we have an underdamped case. s1,2 =
−α±
α
2
i(t) = e
2
− ωo = −2.5 ±
6.25 − 9 = -2.5 ± j1.6583
-2.5t
[A1cos1.6583t + A2sin1.6583t]
We now determine A1 and A2 . i(0) = 5 = A1 -2.5t
di/dt = -2.5{e [A1cos1.6583t + A2 sin1.6583t]} -2.5t + 1.6583e [-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 -2.5t
Thus, i(t) = e
P.P.8.5
α
-3
= 1/(2RC) = 1/(2x2x25x10 ) = 10 ω0 = 1
since α =
ωo,
[5cos1.6583t – 7.538sin1.6583t] A
LC
=1
0.4x 25x10 −3
= 10
we have a critically damped response. Therefore, -10t
v(t) = [(A1 + A2t)e
] -10t
v(0) = 0 = A1 + A2x0 = A1, which leads to v(t) = [A2 te
].
-3
dv(0)/dt = -(v(0) + Ri(0))/(RC) = -2x3/(2x25x10 ) = -120 dv/dt = [(A2 - 10A2t)e At t = 0,
-10t
] –10t
-120 = A2 therefore, v(t) = (–120t)e
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
-3
= 1/(2RC) = 1/(2x20x4x10 ) = 6.25 ωo = 1
Since α >
ωo,
LC
10 x 4 x10 −3 = 5
=1
this is an overdamped response.
s1,2 =
−α±
α
2
− ωo = −6.25 ±
(6.25) 2 -2.5t
Thus, v(t) = A1e
−
25 = -2.5 and –10 -10t
+ A2 e
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
-10t
-10A2e
At t = 0, -500 = -2.5A1 -10A2 = 7.5A1 since A1 = -A2 A1 = -66.67,
–10t
v(t) = 66.67(e
Thus,
P.P.8.7
A2 = 66.67 –e
–2.5t
) V
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 -2t
Thus, v(t) = vf + [(A1cos3.464t + A2 sin3.464t)e ] 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 -2t
But, dv/dt = 3.464[(-A1sin3.464t + A2 cos3.464t)e ] -2t -2[(A1cos3.464t + A2sin3.464t)e ] dv(0)/dt = 0 – 2A1 + 3.464A2 , which leads to A2 = -4/3.464 = -1.1547 Thus,
-2t
v(t) = {10 + [(–2cos3.464t – 1.1547sin3.464t)e ]} V
i = C(dv/dt), vR = Ri = RC(dv/dt) = (1/4)dv/dt -2t
= (1/4)[(4 – 4)cos3.464t + (2x1.1547 + 2x3.464)sin3.464t]e -2t
vR (t) = {[2.31sin3.464t]e } 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 + A1 cost + A2 sint 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
iC 4
2A (1/20)F
i
10 +
2H
4 2A
v
(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
4 i
+ v
(1/20)F
2H (c)
Applying KVL gives
-v – 10iC + 4i + 2di/dt = 0
Applying KCL to the top node,
(6)
i – iC = 0
Namely,
i = iC = -Cdv/dt = -(1/20)dv/dt
(7)
Combining (6) and (7),
-v – (10/20)dv/dt – (4/20)dv/dt – (2/20)d v/dt = 0.
2
or
2
2
2
(d v/dt ) + 7(dv/dt) + 10 = 0 2
The characteristic equation is s + 7s + 10 = 0 = (s + 2)(s + 5) -2t
This means that vn = (Ae Thus, v = vf + vn
-2t
v = 8 + (Ae
-5t
+ Be ) and vf = v(∞) = 8. -5t
+ Be )
(8)
v(0) = 0 = 8 + A +B, or A + B = -8 -2t
(9)
-5t
dv/dt = (-2Ae -5Be ) dv(0)/dt = 40 = -2A – 5B 2A + 5B = -40
(10)
From (9) and (10), A = 0 and B = -8. -5t
Thus, v(t) = 8(1 – e ) 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 ) A for all t > 0.
-5t
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).
1
1
v1
1
v2
1 +
5V
+ −
0.5F
(1/3)F
5V
+
+
v1
−
(a)
v2
(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).
1
1
v1
0.5F
(c)
v2 (1/3)F
Applying KCL at node 1 gives (v1/1) + (1/2)(dv1/dt) + (v1 – v2)/1 = 0 or
v2 = 2v1 + (1/2)dv1/dt
(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, 2
v2 = 2v2 +(2/3)(dv2/dt) + (1/2)(dv2/dt) + (1/6)d v2/dt or,
2
2
2
(d v2/dt ) + (7dv2/dt) + 6v2 = 0 2
Now we have, s + 7s + 6 = 0 = (s + 1)(s + 6) -t
-6t
Thus, v2n = (Ae + Be ) and v2f = v2(∞) = 5V. -t
-6t
v2 = v2n + v2f = 5 + (Ae + Be ) v2(0) = 0 which implies that A + B = -5 -t
(6)
-6t
dv2/dt = (-Ae - 6Be ) dv2(0) = 0 = -A – 6B
(7)
From (6) and (7), A = -6 and B = 1. -t
-6t
Thus, v2(t) = (5 – 6e + e ) V From (5),
v1 = v2 + (1/3)dv2/dt -t
-6t
Thus, v1(t) = (5 – 4e – e ) V
Now we can find,
–t
–6t
vo = v1 – v2 = (2e – 2e )V, t > 0
Let v1 equal the voltage at non-inverting terminal of the op amp. P.P.8.11 Then vo is equal to the output of the op amp. At the non-inverting terminal, (vs – vo)/R 1 = C1dv1/dt
(1)
At the output terminal of the op amp, (v1 – vo)/R 2 = C2dvo/dt
(2)
We now eliminate v1 from (2), v1 = vo + R 2C2dvo/dt
(3)
From (1)
(4)
vs = v1 + R 1C1dv1/dt
Substituting (3) into (4) gives 2
2
vs = vo + R 2C2dvo/dt + R 1C1dvo/dt + R 1C1R 2C2d vo/dt or
2
2
d vo/dt + [(1/(R 1C1)) + (1/(R 2C2))]dvo/dt + vo/(R 1R 2C1C2) = vs/(R 1R 2C1C2)
With the given parameters, 4
4
-6
-6
-2
(R 1R 2C1C2) = 10 x10 x20x10 x100x10 = 2x10 1/(R 1R 2C1C2) = 5 -4
-6
-6
[(1/(R 1C1)) + (1/(R 2C2))] = 10 [(1/20x10 ) + (1/200x10 )] = 6 2
Hence, we now have s + 6s + 5 = 0 = (s +1)(s + 5) -t
-5t
-t
-5t
Therefore von = Ae + Be , and vof = 4V Thus,
vo = 4 + Ae + Be
(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+)]/R 2C2 = 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 -t
-5t
vo(t) = (4 – 5e + e ) V, t > 0
P.P.8.12 We follow the same procedure as in Example 8.12. The schematic is shown in Figure (a). (a). The current marker is inserted inserted to display the inductor current. 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) (0) = 8 volt volts. s.
(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 redrawn as shown in Figure (b). (b). 3H 50mA
3F 4H 10
50mV
+ −
0.1
(a)
4 F
3H 0.1 50mV
+
4F
−
(b)
P.P.8.15
The dual circuit is obtained in Figure (a) and redrawn in Figure (b).
5
0.2F
4H
0.2 4F
0.2 H
2A
1/3
2V
+
+
3
−
20A
−
(a)
20 V
1/3
4F
0.2 H
2V
0.2
+
20A
−
(b)
P.P.8.16
Since 12 = 4i + vL + vC
or vC = 12 12 – 4i - vL
-250t
-(vC – 12) = 4i + vL = e
-250t
vC(t) = [12 – 12e
(12cosωdt + 0.2684sinωdt – 268sinωd t) -250t
cos11.18t + 267.7e
sin11.18t] V
We follow the same procedure as in Example 8.17. The schematic P.P.8.17 is as shown in Figure (a) with two voltage markers to display both input and output voltages. When the schematic is saved saved and run, the result is as displayed in Figure (b). (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
frequency (f) =
1 = 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° i1 leads i2 by 155 Thus, 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
(b)
Since sin(A) = cos(A – 90°),
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( t + 210 ) V
(b)
I = j (5 – j12) = 12 + j5 = 13 ∠22.62°
The sinusoid is i(t) = 13 cos( t + 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( t – 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 ω = 5 2jω V +5 V + V = 20∠-30°, 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 -6 I = jωC V = (j100)(50x10 )(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 Z1 = impedance of the 2-mF capacitor in series with the 20-Ω resistor Let 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) - j25 + 50 + j20 = 20 – j50 + 12.38 – j23.76 = 32.38 – j73.76
Zin = 20 − j50 + Zin Zin
In the frequency domain, the voltage source is Vs = 10∠75° the 0.5-H inductor is j ωL = j (10)(0.5) = j5 1 1 1 the -F capacitor is = = - j2 20 jωC j (10)(1 20)
P.P.9.11
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) Z2 = -j2 = 2 + j4 and 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 =
We need to find the equivalent impedance via a delta-to-wye transformation as shown below. P.P.9.12
c Zcn
Zan 30 0 V
n
Zbn
+
− b
a 5 10 -2
4 (-5 + j8) j4 (8 + j5) = = 0.32 + j3.76 8 + j6 j4 + 8 + j5 − j3 3 (5 − j8)(8 − j6) - j3 (8 + j5) = = = -0.24 – j2.82 100 8 + j6 j4 (- j3) 12 (8 − j6) = = = 0.96 – j0.72 100 8 + j6
Zan = Z bn Zcn
The total impedance from the source terminals is Z = Zcn + (Zan + 5 – j2) || ( Z bn + 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, I = V / Z =
30 ∠0° 4.714 ∠ − 3.802°
I = 6.364 3.802 A
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.
To show that the circuit in Fig. (a) meets the requirement, consider the equivalent circuit in Fig. (b). P.P.9.13
Z = -j10 || (10 – j10) =
10
V1
10
+ Vi
- j10 (10 − j10) - j (10 − j10) = = 2 – j6 Ω 10 − j20 1 − j2 10 +
+ - 10
-j10
(a)
Vo
Vi = 10 V
+
V1
− (b)
Z = 2 j6
V1 =
2 − j6 10 (10) = (1 − j) 10 + 2 − j6 3
⎛ - j ⎞⎛ 10 ⎞ 10 - j10 ⎟⎜ ⎟ (1 − j) = - j V1 = ⎜ 3 10 − j10 ⎝ 1 − j ⎠⎝ 3 ⎠ 10 Vo = ∠ - 90° 3 This implies that the RC circuit provides a 90 ° lagging phase shift. 10 The output voltage is = 3.333 V 3 Vo =
P.P.9.14
the 1-mH inductor is j ωL = j (2π)(5 × 10 3 )(1 × 10 -3 ) = j31.42 the 2-mH inductor is j ωL = j (2π)(5 × 10 3 )(2 × 10 -3 ) = j62.83 Consider the circuit shown below. 31.42
V1
j62.83
+ Vi
+ 10
50
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 Vo = V1 = = 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 k Ω Z1 = 4.8 k Ω Z2 = 10 + j ωL = 10 + j ( 2π )(6 × 10 6 ) (0.25 × 10 -6 ) = 10 + j9.425 Zx =
12k (10 + j9.425) = 25+ j23.5625 Ω 4.8k
R x = 25, Xx = 23.5625 = ωLx Xx 23.5625 = = 0.625 μH Lx = 2πf 2π (6 × 10 6 ) i.e. a 25-
resistor in series with a 0.625- H inductor.
February 5, 2006
CHAPTER 10
P.P.10.1
⎯→ 10 ∠0°, ω = 2 10 sin( 2t ) ⎯
⎯→ jωL = j4 2 H ⎯ 1 ⎯→ = - j2.5 0.2 F ⎯ jωC Hence, the circuit in the frequency domain is as shown below. -j2.5
V1
4
V2
+ 10 0 A
At node 1,
At node 2,
10 =
2
V1
+
V1
Vx
4
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
=
V1
3Vx
− V2
2 - j2.5 100 = (5 + j4) V1 − j4V2 V2
+
− V2
(1)
+
(2)
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 °
⎡2.5 + j1.5 j4 ⎤ ⎡ 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° = = 3.376∠165.7° V2 = 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 -2
8
For mesh 1,
6
I1
(8 − j2 + j4) I1 − j4 I 2 = 0 (8 + j2) I1 = j4 I 2
4
I2
+
−
10 30 V
(1)
For mesh 2,
(6 + j4) I 2 − j4 I 1 − 6 I 3 + 10∠30° = 0
For mesh 3,
I3
= -2
Thus, the equation for mesh 2 becomes (6 + j4) I 2 − j4 I 1 = -12 − 10∠30° From (1),
I2
=
(2)
8 + j2 I = (0.5 − j2) I1 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 Hence,
P.P.10.4
Io
= - I1 =
Io
=
20.66 + j5 21.256∠13.6° = 11 − j14 17.8∠ - 51.84°
1.194 65.44 A
Meshes 2 and 3 form a supermesh as shown in the circuit below. 10 -j4
j8 I2
50 0 V
+
−
I1 I3 5
-j6
− 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
For mesh 1,
= I2 + 2
(3)
Eliminating I 3 from (1) and (2) (15 − j4) I 1 + (-5 + j4) I 2 = 60 (-5 + j4) I 1 + (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
- 10 + j12
5 - j2
= I1 =
Δ1 = Δ
= 298 − j20 = 298.67 ∠ - 3.84°
Thus,
Io
P.P.10.5
Let I o = I 'o + I "o , where I 'o and I "o are due to the voltage source and
5.074 5.94 A
current source respectively. For I 'o consider the circuit in Fig. (a). -2 Io 8
6
'
I1
4
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° = 0.08 + j0.556 I 'o = I 1 = 11 − j14
For I "o consider the circuit in Fig. (b). 2 0 A
-2
6
"
Io 8
4
(b)
Let
j24 = 1.846 + j2.769 Ω 6 + j4 Z2 (2)(1.846 + j2.769) = 0.4164 + j0.53 I "o = (2) = Z1 + Z 2 9.846 + j0.77 Z1
= 8 − j2 Ω ,
Z2
= 6 || j4 =
= I 'o + I "o = 0.4961 + j1.086 I o = 1.1939 65.45 A
Therefore,
Io
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 ∠0°, ω = 5 30 sin(5t ) ⎯ 1 1 0.2 F ⎯ ⎯→ = = - j jωC j (5)(0.2) ⎯→ jωL = j (5)(1) = j5 1 H ⎯ The circuit in the frequency domain is shown in Fig. (a). 8 + 30 0 V
+
−
Vo
'
-
(a)
5
- j || j5 = -j1.25
Note that
By voltage division, - j1.25 (30) = 4.631∠ - 81.12 ° 8 − j1.25 v 'o = 4.631sin(5t − 81.12°)
Vo'
Thus,
=
For v "o , we remove the voltage source.
⎯→ 2 ∠0°, ω = 10 2 cos(10 t ) ⎯ 1 1 0.2 F ⎯ ⎯→ = = - j0.5 jωC j (10)(0.2) ⎯→ jωL = j (10)(1) = j10 1 H ⎯ The corresponding circuit in the frequency domain is shown in Fig (b). I
+ 8
j10
Vo
"
-j0.5
2 0
(b)
Let
Z1 = - j0.5 ,
Z2
= 8 || j10 =
j80 = 4.878 + j3.9 8 + j10
By current division, I= Vo"
Z2 Z1 + Z 2
(2)
= I (-j0.5) =
Z2 Z1 + Z 2
(2)(-j0.5) =
- j (4.877 + j3.9) 4.878 + j3.4
Thus,
6.245∠ - 51.36° = 1.051∠ - 86.24° 5.94 ∠34.88° v "o = 1.051 cos(10t − 86.24°)
Therefore,
v o = v 'o + v "o
Vo"
=
v o = 4.631 sin(5t – 81.12 ) + 1.051 cos(10t – 86.24 ) V
If we transform the current source to a voltage source, we obtain the circuit shown in Fig. (a). P.P.10.7
4
-3
2 Io
VS
1
+
j5
−
-2
(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 Z = 4 − j3 + 2 + j = 6 − j2 . Then, Is = s = = 1.5 + j3 . Let Z 6 − j2 Io 6
1
IS
j5 -j2
-j2
(b)
Note that
Z || j5 =
(6 − j2)( j5) 10 = (1 + j) . 6 + j3 3
By current division, Io
=
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° 3.288 99.46 A
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
P.P.10.8
Z th
=
12.4 – j3.2
By voltage division,
- j4 (- j4)(30∠20°) (30∠20°) = 6 + j2 − j4 6 − j2 ( 4 ∠ - 90°)(30 ∠20°) Vth = 6.324 ∠ - 18.43° Vth
=
Vth
=
18.97 -51.57 V
To find Vth , consider the circuit in Fig. (a).
P.P.10.9
8 + j4 +
8 + j4 +
Vo
Vo
5 0 V2
V1 4 – j2
At node 2,
a
0.2Vo
(a)
At node 1,
VS a
4 – j2
b
5 + 0.2Vo +
V1 − V2
8 + j4
= 0,
Hence, the equation for node 2 becomes
+
0.2Vo
(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
Is
−
b
(1) where Vo = V1 − V2 .
1 0
5 + 0.2 (V1 − V2 ) +
= V2 −
V1
− V2 =0 8 + j4
V1
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 = 7.35∠72.9° V2 = 2 + j
0 = -50 − ( 2 + j) V2 +
Vth
= V2 =
7.35 72.9 V
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
Is
= -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 = Is Is 2.6 + j0.8 2.72∠17.10° Vs
=1
Z th
=
and
4.473 –7.64
– Vo =
P.P.10.10
To find Z N , consider the circuit in Fig. (a).
4
j2
4
j2
I3 8
1
-3
8
1
-3
a
a ZN
20 0
+
I1
−
b
(a)
(b)
Z N
= (4 + j2) || (9 − j3) =
Z N
=
IN
I2
-j4
b
(4 + j2)(9 − j3) 13 − j
3.176 + j0.706
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
Also,
I1
For mesh 3,
(13 − j) I 3 − 8 I 1 − (1 − j3) I 2 = 0
= I 2 + j4
Solving for I 2 , we obtain I N
= I2 =
I N
=
(2)
50 − j62 79.65∠ - 51.11° = 9 − j3 9.487 ∠ - 18.43°
8.396 -32.68 A
Using the Norton equivalent, we can find I o as in Fig. (c). Io IN
ZN
(c)
(1)
10 – j5
(3)
By current division,
3.176 + j0.706 (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° Io
=
Io
=
Z N
I N
=
1.971 -2.10 A
P.P.10.11
1 1 = = -j20 k Ω jωC1 j (5 × 10 3 )(10 × 10 -9 ) 1 1 ⎯→ = = -j10 k Ω 20 nF ⎯ jωC 2 j (5 × 10 3 )(20 × 10 -9 )
⎯→ 10 nF ⎯
Consider the circuit in the frequency domain as shown below.
- 20 k 10 k
20 k V1
2 0 V
V2
+
Io Vo
−
+
- 10 k
−
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 V1
− Vo
=
Vo
(1)
−0
20 - j10 V1 = (1 + j2) Vo
(2)
Substituting (2) into (1) gives 4 = j6Vo
or
Vo
2 3
= ∠ - 90°
Hence,
v o ( t ) = 0.667 cos(5000 t − 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(5000 t − 90°) μA i o ( t ) = 66.67 sin(5000t) A
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 1 + jωRC = R 2 + jωRC R + 1 + jωRC
where ωRC = (1000)(10 × 10 3 )(1 × 10 -6 ) = 10 1/ G =
1 + j10 10.05∠84.29° = 2 + j10 10.2∠78.69°
G = 1.0147 –5.6
P.P.10.13
The schematic is shown below.
⎯→ f = 477.465 Hz . Setup/Analysis/AC Sweep as Since ω = 2πf = 3000 rad / s ⎯ 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($N_0005) -1.546E+02
From the output file, we obtain Vo = 0.2682 ∠-154.6 ° V
and
Io
Therefore, v o ( t ) = 0.2682 cos(3000t – 154.6 ) V i o ( t ) = 0.544 cos(3000t – 55.12 ) mA
= 0.544∠-55.12 ° mA
P.P.10.14
The schematic is shown below.
We select ω = 1 rad/s and f = 0.15915 Hz. We use this to obtain the values of capacitances, where C = 1 ωX c , and inductances, where L = X L ω . Note that IAC does not allow for an AC PHASE component; thus, we have used VAC in conjunction with G to create an AC current source with a magnitude and a phase. To obtain the desired output use Setup/Analysis/AC Sweep as Linear for 1 point starting and ending at a frequency of 0.15915 Hz. When the schematic is saved and run, the output file includes Frequency 1.592E-01
IM(V_PRINT1) 2.584E+00
IP(V_PRINT1) 1.580E+02
Frequency 1.592E-01
VM($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
Ix
=
2.584 158 A
⎛ R 2 ⎞ ⎛ 10 × 10 6 ⎞ ⎟ C = ⎜1 + ⎟( × -9 ) = C eq = ⎜1 + 3 10 10 ⎝ 10 × 10 ⎠ ⎝ R 1 ⎠
10 F
C = C1 = C 2 = 1 nF If R = R 1 = R 2 = 2.5 k Ω and 1 1 = = 63.66 kHz f o = 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 3
8 45 V
I=
+ −
I
j1
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. 8
40 V
+ −
j4
I1
I2
-j2
+ −
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) I 1 = 0 - j I 1 + j I 2 = j10
(2)
For mesh 2,
In matrix form, ⎡ 4 − j - j⎤⎡ I 1 ⎤ ⎡ 20 ⎤ ⎢ - j j ⎥⎢ I ⎥ = ⎢ j10 ⎥ ⎣ ⎦⎣ 2 ⎦ ⎣ ⎦
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
10
8
Zth
5
(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). -4
j10 I
8
2A
5
(b)
By current division, I=
8 − j4 (2) 8 − j4 + j10 + 5
+ Vth
VTh = 5 I =
(10)(8 − j4) = 6.25∠ - 51.34° 13 + j6
Z L = Z *Th = 3.415 j0.7317
Pmax = P.P.11.6
Let
VTh
2
8 R L
(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
VTh =
Z2 (9)(1 − j3) (120∠60°) = (120∠60°) Z1 + Z 2 89 + j33
VTh = 35.98∠ - 31.91°
R L = Z Th = 30 The current through the load is VTh 35.98∠ - 31.91° = = 0.6764 ∠ - 4.4° I= 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 2 i dt ( 4 t ) dt ( 8 4 t ) dt = + − ∫ ∫ ∫ 1 T 0 2 0 2 16 1 2 t dt ( 4 − 4 t + t 2 ) dt = + ∫ ∫ 0 1 2 ⎡ 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 = ∫ ∫ π 0 T 0 64 π 1 2 Vrms = [1 − cos(2 t )] dt = 32 π ∫0 2 2 Vrms =
Vrms = 5.657 V 2 Vrms 32 = = 5.333 W P= R 6
P.P.11.9
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 V 150 ∠10° = = 2.08∠ - 23.7° A I= 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 Z=
P.P.11.12 (a)
(b)
(c)
If Z = 250∠ - 75° ,
pf = cos( -75°) = 0.2588
⎯→ S = Q = S sin θ ⎯
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 I1
V
+ −
(30–j10)
+ Vo
I2 (60+j20)
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
30 − j10
= 3.2 + j2.4
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° S1 = 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 ,
pf = 0.75 = cos θ1 ⎯ ⎯→ θ1 = -41.41° P ⎯→ S1 = 1 = 2666.67 P1 = S1 cos θ1 ⎯ 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 θ2 Q 2 = S 2 sin θ2 = 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 ⎯→ S = = = 265.8 kVA Q = S sin θ ⎯ sin θ sin( 31.79°) P = S cos θ = 225.93 kW
For pf = 1 = cos θ1 ⎯ ⎯→ θ1 = 0° Since P remains the same,
⎯→ S1 = P = P1 = S1 cos θ1 ⎯
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 C= = 30.69 mF (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 V bn by 120° and V bn leads Vcn by 120°.
Hence,
Van
= 110∠(30° + 120°) =
Vcn
= 110∠(30° − 120°) = 110
Vab
=
Van
Vab
=
(103.92 + j60) + j120
Vab
= 207.8
P.P.12.2 (a)
–90 V ˚
= 120∠30° − 120∠ - 90°
V bn
−
110 150 V
60 V ˚
Alternatively, using the fact that Vab leads Van by 30° and has a magnitude of Vab
=
3 times that of Van ,
3 (120) ∠(30° + 30°) = 207.85∠60°
Following the abc sequence, V bc = 207.8 –60 V ˚
Vca
(b)
Ia
=
= 207.8
180 V ˚
Van Z
Z = (0.4 + j0.3) + (24 + j19) + (0.6 + j0.7) Z = 25 + j20 = 32 ∠38.66°
Ia
=
120∠30° 32 ∠38.66°
=
3.75
- 8.66 A
Following the abc sequence, I b = I a ∠ - 120° = 3.75 Ic
=
- 128.66 A
I a ∠ - 240° = 3.75 111.34 A ˚
P.P.12.3 The phase currents are VAB I AB = ZΔ
=
180∠ - 20° 20 ∠40°
I BC
=
I AB ∠ - 120° = 9
I CA
=
I AB ∠120° = 9 60
=
9
- 60 A
- 180 A
The line currents are
P.P.12.5
=
I AB 3 ∠ - 30° = 9 3 ∠ - 90° = 15.59
I b
=
I a ∠ - 120° = 15.59 150 A
Ic
=
I a ∠120° = 15.59 30 A
- 90 A
˚
In a delta load, the phase current leads the line current by 30° and has a
P.P.12.4
magnitude
Ia
1 3
times that of the line current. Hence, Ia
∠30° =
I AB
=
ZΔ
= 18 + j12 =
3
22.5 3
∠65° =
21.63∠33.69° Ω (13∠65°)(21.63∠33.69°)
VAB
=
I AB Z Δ
VAB
=
281.2 98.69 V
ZY
=
13 65 A
= 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 Van
=
I b
=
I a ∠ - 120° = 7.21
Ic
=
I a ∠120° = 7.21 53.66 A
ZY
=
138.56∠ - 15°
Ia
19.21∠51.34°
=
7.21
- 66.34 A
- 186.34 A
P.P.12.6 For the source,
S = 3 V p I p*
=
(3)(120∠30°)(3.75∠8.66°)
S = −1350∠38.66° = –1054.2 – j843.3 VA For the load, 2
S = 3 I p where
Z
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
IL
⎯ ⎯→
P cos θ
30 × 10 3
=
0.85
S
=
=
=
35.29 kVA
35.29 × 10 3
3 VL
=
3 ( 440)
46.31 A
Alternatively,
P p
=
P p
=
I p
P.P.12.8 (a)
=
30 × 103 3
= 10
kW ,
V p
=
440 3
V
V p I p cos θ
P p
=
(10 × 10 3 ) 3
V p cos θ
(440)(0.85)
= 46.31 A
For load 1, V p
=
I a1
=
S1
VL
=
840
3
=
Va Z p 2 Vrms
Z*
3
=
840∠0° 3
=
⋅
1 30 + j40
(840) 2 50∠ - 53.15°
=
9.7 ∠ - 53.13°
= 14.112 ∠53.13°
kVA
For load 2,
S2
P2
=
cos θ 2
=
48 0.8
=
Q2
=
S 2 sin θ2
S2
=
48 + j36 kVA
S = S1 + S 2
=
=
60 kVA
(60)(0.6) = 36 kVAR
56.47 j47.29 kVA
S = 73.65∠39.94° kVA
with pf = cos(39.94°) = 0.7667 (b)
Qc
=
P (tan θold − tan θ new )
Qc
=
(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 I AB = Z AB
I BC
=
I CA
=
VBC
=
=
200∠0° 10 − j5
200∠ - 120° 16
Z BC VCA Z CA
The line currents are I a = I AB
= 17.89∠26.56°
=
200∠120° 8 + j6
− I CA =
− I AB =
-6.25 − j10.825
20∠83.13°
(16 + j8) − (2.392 + j19.856)
= 13.608 − j11.856 =
I b
=
= 12.5∠ - 120° =
18.05
- 41.06 A
(-6.25 − j10.825) − (16 + j8)
=
I BC
=
-22.25 − j18.825 = 29.15 220.2 A
− I BC =
(2.392 + j19.856) − (-6.25 − j10.825)
Ic
=
I CA
Ic
=
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
=
I CA
=
220∠0° j10
=
220 ∠120°
The line currents are I a = I AB
10
− I CA =
22∠30°
=
22 ∠ - 120°
( j44) − (-11 − j19.05)
Ia
= 11 + j63.05 =
I b
=
I b
= 19.05 − j33 =
Ic
=
I CA
Ic
=
-30.05 − j30.05 = 42.5 225 A
I BC
− I AB =
− I BC =
64 80.1 A
(19.05 + j11) − ( j44) 38.1
- 60 A
(-11 − j19.05) − (19.05 + j11)
The real power is absorbed by the resistive load P = I CA
2
(10) = ( 22) 2 (10) = 4.84 kW
The schematic is shown below. First, use the AC Sweep option of the P.P.12.11 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
L = XL
ω = 10
and C = 1 ωX c
ω = 1 rad / s
=
0.1 .
, so that f = 1 2π = 0.1592 Hz . Hence,
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)
=
37.25 83.79 A
If point o is connected to point B, P2
=
IAB 15.55 –75.01 A ˚
0W
P1
=
Re (VAB I *a )
P1
=
(200)(18.05) cos(0° + 41.06°) = 2722 W
P3
=
Re ( VCB I *c )
where VCB P3 (b)
I cC
=
-VBC
=
=
200∠(-120° + 180°) = 200∠60°
( 200)(31.87) cos(60° − 74.27°) = 6177 W
Total power is PT = P1 + P2
+ P3 =
P.P.12.14
VL
=
208 V ,
(a)
PT
=
P1 + P2
=
-560 + 800 = 240 W
(b)
QT
3 ( P2
−
P1 ) =
(c)
tan θ =
=
QT PT
=
P1
2722 + 0 + 6177 = 8899 W
2355.6 240
=
-560 W ,
P2
=
800 W
3 (800 + 560) = 2.356 kVAR
=
9.815
⎯ ⎯→
θ=
84.18°
pf = cos θ = 0.1014
(d)
(lagging / inductive)
It is inductive because P2 > P1 For a Y-connected load, VL 208 I p = I L , V p = = 3 3 P p
=
V p I p cos θ
Z p
=
Z p
=
V p I p
=
120 6.575
⎯ ⎯→
I p
=
= 120
V
80 (120)(0.1014)
=
6.575 A
= 18.25
Z p ∠θ = 18.25 84.18
The impedance is inductive . P.P.12.15
ZΔ
30 − j40 = 50∠ - 53.13°
=
The equivalent Y-connected load is ZΔ ZY = = 16.67 ∠ - 53.13° 3 440 V p = = 254 V 3 V p 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
3 ( P2
−
P1 ) =
QT
=
QT
=
- 9.292 kVAR
3 (802.1 − 6167 )
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 ⎤ ⎡ I 1 ⎤ ⎡ 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 Since I1 enters the coil with reactance 2Ω and I2 enters the coil with P.P. 13.2 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)I 1 – 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
-j4
4
VS
+ –
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)I 2
Substituting (2) into (1),
I2
2
(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
If we reverse the direction of i 2 so that we replace I2 by –i2, we P.P. 13.6 have the circuit shown in Figure (a). j3
-j4
+ –
j3
i1
j6
i2
12
o
12 0
(a)
We now replace the coupled coil by the T-equivalent circuit and assume ω = 1. La = 5 – 3 = 2 H L b = 6 – 3 = 3 H Lc = 3 H Hence the equivalent circuit is shown in Figure (b). We apply mesh analysis.
-j4
j2
j3 j3
o
12 0
+ –
I1
I2
(b)
12
12 = i1(-j4 + j2 + j3) + j3i 2 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
8
+ v0 – 4
i1
1
2
1:2
2
i2 + – 60 0o
+
+
v1
v2
–
–
+ v3
10
–
Consider the circuit shown above. At node 1,
(60 – v1)/4 = i1 + (v1 – v3)/8
(1)
At node 2,
[(v1 – v3)/8] + [(v 2 – 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
We should note that the current and voltage of each winding of the P.P. 13.10 autotransformer in Figure (b) are the same for the two-winding transformer in Figure (a). 6A 0.5A
+
6A
10V
– +
6.5A
+
+
120V
10V
–
–
+
120V
120V
–
0.5A
(a)
–
+ 130V
–
(b)
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
The process is essentially the same as in Example 13.13. We are P.P. 13.13 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
Following the same basic steps in Example 13.14, we first assume P.P. 13.14 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 /n
2
–
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 Vs
=
j L R j L
jω ω0 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 ⎞ − tan -1 ⎜ ⎟ 2 ⎝ ω0 ⎠
π
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) = ⎜5 + Z i (s) = 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
1 + jω 0.5 1 A pole at ω = 1 , 1 + jω 1 1 Two poles at ω = 10 , (1 + jω 10) 2 A zero at ω = 0.5 ,
Hence, H (ω) =
1 + jω 0.5 (1 0.5)(0.5 + jω) 2 = (1 + jω 1)(1 + jω 10) (1 100)(1 + jω)(10 + jω) 2 200 (s
H (ω) =
(s
1)(s
0. 5 ) 10 ) 2
P.P.14.7
(a)
Q=
ω0 L R
ω0 =
⎯ ⎯→ ω0 =
QR (50)(4) = = 8 × 10 3 rad / s -3 L 25 × 10
1 1 1 ⎯ ⎯→ C = 2 = 6 ω0 L (64 × 10 )(25 × 10 -3 ) LC
C = 0.625 F (b)
ω0
8 × 10 3 = = 160 rad / s B= Q 50
Since Q > 10 , B = 8000 − 80 = 7920 rad / s 2 B ω2 = ω0 + = 8000 + 80 = 8080 rad / s 2
ω1 = ω0 −
(c)
At ω = ω0 ,
Vin2 100 2 = = 1.25 kW P= 2R 8
At ω = ω1 ,
Vin2 = 0.625 kW P = 0.5 ⋅ 2R
At ω = ω2 ,
Vin2 = 0.625 kW P = 0.5 ⋅ 2R
P.P.14.8
ω0 =
1 LC
=
1 (20 × 10 )(5 × 10 ) -3
-9
= 10 5 = 100 krad / s
R 100 × 10 3 = = 50 Q= ω0 L (10 5 )(20 × 10 -3 )
ω0
10 5 B= = = 2 krad / s Q 50
Since Q > 10 , B = 100,000 − 1,000 = 99 krad / s 2 B ω2 = 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 + 4ω2 Z = jω1 + 10 ||
Im(Z) = 0 ⎯ ⎯→ ω −
20ω =0 1 + 4ω2
20ω ⎯ ⎯→ 1 + 4ω2 = 20 2 1 + 4ω 19 ω= = 2.179 rad / s 2
ω=
R 2 || sL s = jω , Vi R 1 + R 2 || sL sR 2 L H(s) = R 1R 2 + sR 1L + sR 2 L jωR 2 L H(ω) = R 1R 2 + jωL (R 1 + R 2 ) H (0) = 0 jR 2 L R 2 = H(ω) = lim ω→∞ R R ω + jL ( R + R ) R 1 + R 2 1 2 1 2 i.e. a highpass filter. P.P.14.10
H(s) =
Vo
=
The corner frequency occurs when H (ωc ) =
1 ⋅ H(∞) . 2
⎛ R 2 ⎞⎛ ⎞ jωL ⎜ ⎟⎜ ⎟ ⎝ R 1 + R 2 ⎠⎝ jωL + R 1R 2 (R 1 + R 2 ) ⎠ ⎛ R 2 ⎞⎛ jω ⎞ R 1 R 2 ⎟⎜ ⎟, where k = H (ω) = ⎜ (R 1 + R 2 ) L ⎝ R 1 + R 2 ⎠⎝ jω + k ⎠ H (ω) =
At the corner frequency, R 2 R 2 jωc 1 ⋅ = ⋅ 2 R 1 + R 2 R 1 + R 2 jωc + k
ω 1 = 2c 2 2 ωc + k Hence,
H (ω) =
⎯ ⎯→ ωc = k =
R2 R1
R 2 j
R1R 2 (R 1
R2 )L
j c
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,
ω0 =
ω1 + ω2 2
(2π)(40.4 × 10 3 ) = = 40.4π × 103 rad / s 2
ω0
40.4π × 10 3 = = 101 Q= B 400π R R 20 × 10 3 ⎯ ⎯→ L = = = 15.916 H B= L B 400π 1 1 ⎯ ⎯→ C = ω0 CR ω0 QR 1 = 3.9 pF C= (40.4π × 10 3 )(101)(20 × 10 3 ) Q=
P.P.14.12
Given H (∞) = 5 and f c = 2 kHz 1 ωc = 2πf c = R i C i
1 1 = 2πf c C i ( 2π)(2 × 10 3 )(0.1 × 10 -3 ) R i = 795.8 ≅ 800 R i =
H(∞) = P.P.14.13
- R f = -5 ⎯ ⎯→ R f = 5R i = 3,978 ≅ 4 k R i
ω0 = 20 krad / s
Q = 10 ,
B=
ω0 Q
= 2 krad / s
B = 19 krad / s 2 B ω2 = ω0 + = 21 krad / s 2
ω1 = ω0 −
Since ω1 =
1 , C 2 R 1 1 = = 5.263 nF 3 ω1R (19 × 10 )(10 × 10 3 ) 1 1 = = 4.762 nF C1 = ω2 R (21 × 10 3 )(10 × 10 3 ) R f = 5 ⎯ ⎯→ R f = 5R i = 50 k K = R i C2 =
P.P.14.14
ω′c 2π × 10 4 = = 2π × 10 4 K f = ωc 1 C C 1 10 4 ⎯ ⎯→ K m = = = C′ = K m K f C′ K f (15 × 10 -9 )(2π × 10 4 ) 3π 10 4 R ′ = K m R = (1) = 1.061 k Ω 3π K m 10 4 2 ⋅ = 33.77 mH L′ = L= K f 3π 2π × 10 4
Therefore,
R 1′ = R ′2 = 1.061 k 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 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 magnitude plot, choose choose DB( ) from the the Analog Operators and Functions list. Then, select the voltage V(R1:1) and OK. Another option option would be to type DB(V(R1:1)) as the Trace Expression. For the the phase plot, choose choose P( ) from from the Analog Operators and Functions list. Then, select the voltage voltage V(R1:1) and OK. Another option option would 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 = 0.543 pF C1 = 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 = 2 = 39.83 nF 2 4π f 0 L 4π (1336) 2 (0.356) 2
C = 10 μF
2πf c =
L=
and
R 1 = R 2 = 8 Ω
1 1 1 ⎯ ⎯→ f c = = = 1.989 kHz R 1C 2πR 1C ( 2π )(8)(10 × 10 -6 )
R 2 8 = = 0.64 mH 2πf c (2π)(1.989 × 10 3 )
February 5, 2006
CHAPTER 15
[ t u ( t )] = ∫0
∞
L
P.P.15.1
t e -st dt
Using integration by parts, u dv = uv − v du
Let
⎯→ u = t ⎯
du = dt .
⎯→ e -st dt = dv ⎯
[ t u ( t )] =
L
-t s
e
-st ∞ 0
-1
v=
+∫
s
∞1
e
s
0
e -st
-st
dt = 0 +
e -st s
∞ 0
2
1
=
s2
Also,
[ e at u ( t )] = ∫0 e at e -st ∞
L
P.P.15.2
-1
dt =
s−a
e -( s− a ) t
∞
1 [ cos(ωt )] = ∫-∞ ( e jωt + e - jωt ) e -st
s
a
∞
L
dt
2
[ cos(ωt )] =
L
1
=
0
1
∫ 2
∞
0
e -(s - jω) t dt +
1
∫ 2
∞
0
e -(s+ jω) t dt
⎛ 1 s 1 ⎞ ⎟= 2 + 2 ⎝ s − jω s + jω ⎠ s
1 [ cos(ωt )] = ⎜
L
P.P.15.3
If
2
f ( t ) = cos(2 t ) + e -3t , F(s) = F(s) =
s s2 + 4 2s 2 (s
+
1 s+3
3s
3)(s
2
=
s 2 + 3s + s 2 + 4 (s 2 + 4)(s + 3)
4 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) =
[t
L
2
cos(3t )] = ( - 1)
2
d 2 ⎛ s ⎞
⎜
⎟
ds 2 ⎝ s 2 + 9 ⎠
d 2
F(s) =
2
[ s(s
+ 9)
2
-1
]=
d 2 2
[ (1) (s
2
-1 -2 + 9) − (s)( 2s) ( s 2 + 9) ]
ds ds -2 -2 -3 F(s) = ( - 2s) ( s 2 + 9 ) − ( 4s) ( s 2 + 9 ) + ( 4s 2 ) ( 2s) ( s 2 + 9 ) F(s) = ( - 6s) ( s + 9 ) 2
2s s 2
F(s) =
P.P.15.5
-2
+ ( 8s )( s + 9) = 3
2
-3
2s 3 − 54s
( s 2 + 9) 3
27
s2
9
3
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
e - 2s
2
s
e -4s
T = 5 f 1 ( t ) = u ( t ) − u ( t − 2) 1 F1 (s) = (1 − e - 2s ) s F(s) =
P.P.15.7
5
F1 (s)
=
1 − e -Ts
1 e -2s s (1 e -5s )
g (0) = lim sF(s) = lim s →∞
s →∞
1+
2
s 3 + 2s + 6 (s 2 + 2s + 1)(s + 3)
+
6
s2 s3 =1 s →∞ ⎛ 2 1 ⎞⎛ 3 ⎞ ⎜1 + + 2 ⎟⎜1 + ⎟ ⎝ s s ⎠⎝ s ⎠
g (0) = 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 g (∞) = lim 2 =2 s → 0 (1) (3) P.P.15.8
F(s) = 1 +
4 s+3
−
5s s 2 + 16
f ( t ) =
-1
L
⎡ 4 ⎤ -1⎡ 5s ⎤ −L ⎢ 2 ⎣ s + 3 ⎥⎦ ⎣ s + 16 ⎥⎦
[ 1 ] + L-1⎢
-3t f ( t ) = δ( t ) + (4e − 5 cos(4 t ))u ( t ),
P.P.15.9
F(s) =
A s +1
B s+3
+
A = F(s) (s + 1)
s = -1
=
B = F(s) (s + 3)
s = -3
=
C = F(s) (s + 4)
s = -4
=
F(s) =
1 s +1
f ( t ) = (e
P.P.15.10
+
G (s) =
-t
+
3 s+3
−
t≥0
C s+4
6 (s + 2) (s + 3)(s + 4) 6 (s + 2)
s = -1
=
(s + 1)(s + 4) 6 (s + 2)
s = -3
=
(s + 1)(s + 3)
s = -4
=
6 ( 2)(3) (6)(-1)
=1
(-2)(1) (6)(-2)
=3
(-3)(-1)
= -4
4 s+4
+ 3e -3t − 4e -4t )u ( t ), t ≥ 0
s 3 + 2s + 6 s (s + 1) 2 (s + 3)
=
A s
+
B s +1
+
C (s + 1) 2
+
D 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)
s2 :
0 = 5A + 4B + C + 2D ⎯ ⎯→
4B + C + 2D = -10
(3)
s3 :
1 = A + B + D ⎯ ⎯→
Solving (2), (3), and (4) gives - 13 A = 2 , B = , 4 G (s) =
2 s
−
13 4 s +1
−
C =
32 (s + 1)
B + D = -1
2
+
9 4 s+3
-3 2
,
(4)
D =
9 4
g ( t ) = ( 2 3.25 e
1.5 t e
-t
2.25 e
10
G (s) =
P.P.15.11
-t
(s + 1)(s 2 + 4s + 13)
-3t
=
)u( t ),
A s +1
+
t
0
Bs + C s 2 + 4s + 13
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
s1 :
0 = 4A + B + C ⎯ ⎯→
s0 :
10 = 13A + C ⎯ ⎯→ 10 = 10A
Solving (1), (2), and (3) gives A = 1 , B = -1 , G (s) =
1 s +1
−
s+3 (s + 2) 2 + 9
(1)
C = -3A
(2) (3)
C = -3
=
g ( t ) = (e - t − e - 2t cos(3t ) −
1 s +1
−
s+2 (s + 2) 2 + 9
1 - 2t e sin(3t )), 3
−
1 (s + 2) 2 + 9
t≥0
P.P.15.12
2 For 0 < t < 1 , consider Fig. (a).
x1(t - λ)
x2(λ)
1
∫ (1)(1) d λ = t
y( t ) =
t
0
t-1
0
t (a)
1
2
λ
2
For 1 < t < 2 , consider Fig. (b). y( t ) =
∫
1
t −1
(1)(1) d λ +
∫ (1)(2) d λ = λ t
1
t t −1
+ 2λ
1
t 1
y( t ) = 1 − t + 1 + 2 ( t − 1) = t
0
t-1
1
t
2
λ
2
t
λ
(b)
For 2 < t < 3 , consider Fig. (c).
∫
y( t ) =
2
t −1
(1)( 2) d λ = 2 λ
2 2 t −1
1
y( t ) = 2 (2 − t + 1) = 6 − 2 t
0
1
t-1
For t > 3 , there is no overlap so y( t ) = 0 .
(c)
y(t)
Thus, t y( t ) = 6
2t 0
0
t
2
2
t
3
2
otherwise
0
1
2
The result of the convolution is shown in Fig. (d).
3
t
(d)
P.P.15.13
3e
g(t-λ)
t-1
0
t
-λ
1
λ
(a)
∫ (1) 3 e t
0
-λ
t-1 (b)
For 0 < t < 1 , consider Fig. (a). y( t ) =
0
d λ = -3 e -λ
For t > 1 , consider Fig. (b).
t 0
= 3 (1 − e - t )
1
t
λ
y( t ) =
∫
t
t −1
(1) 3 e -λ d λ = -3 e -λ
t t −1
= 3 e -t (e − 1)
Thus, 3 (1 e-t )
0
y( t ) = 3 e-t (e 1)
t t
0
1 1
elsewhere
The circuit in the s-domain is shown below.
P.P.15.14
1 + Vs
Vo =
2s 1+ 2 s
H(s) =
Vo Vs
+
−
2/s
Vo
Vs
=
2 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 ) d λ 0
t
= 20 e- t ∫ e- 2λ eλ d λ = 20 e- t (-e-λ ) 0t 0
= 20 (e -t
e -2t )u( t ) V
Taking the Laplace transform of each term gives 1 [ s 2 V(s) − sv(0) − v′(0) ] + 4 [ sV (s) − v(0) ] + 4 V(s) = 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) = = + + (s + 1)(s + 2) 2 s + 1 s + 2 (s + 2) 2
P.P.15.15
s 2 + 6s + 6 = A (s 2 + 4s + 4) + B (s 2 + 3s + 2) + C (s + 1) Equating coefficients :
s2 :
1 = A + B ⎯ ⎯→
s1 :
⎯→ 6 = 4A + 3B + C ⎯
6 = A +3+ C
s0 :
6 = 4A + 2B + C ⎯ ⎯→
6 = 6− B
Thus,
B = 0 ,
A = 1 ,
B = 1− A
or A = 1 − B or C = 3 − A
or B = 0
C = 2
and V(s) =
1 s +1
Therefore, v( t ) = (e -t
+
2 (s + 2) 2
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 2s [ s 2 + 3s + 2] Y(s) = 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 s +1
y( t ) = ( -e -t
+
4 s+2 4 e -2t
−
3 s+3 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
Using current division, 4 Io
=
2
s 4 s
⋅ = s
+s+4
Vo (s) = 4 I o 32 s (s + 2) 2
=
32 = A (s 2
= A s
8 s (s 2
+ 4s + 4)
32 s (s + 2) 2
+
B s+2
+
C ( s + 2) 2
+ 4s + 4) + B (s 2 + 2s) + Cs
Equating coefficients : s0 : s1 : s2 :
32 = 4A
⎯ ⎯→ A = 8 0 = 4A + 2B + C 0 = A + B ⎯ ⎯→ B = -A = -8
Hence, 0 = 4A + 2B + C Vo (s) =
8 s
−
8 s+2
v o ( t ) = 8 (1 e -2t
⎯ ⎯→ C = -16
−
16 (s + 2) 2 2t e -2t ) u(t ) V
4
+ Vo(s)
P.P.16.2
The circuit in the s-domain is shown below. 1
+
1/(s + 2)
At node o, 1
s+2 1
− Vo =
Vo(s)
2s
−
Vo 2s
+
Vo
+
i(0)
2 s 1 ⎞
i(0)/s
2
where i(0) = 0A
= Vo ⎛ ⎜1 + + ⎟ s+2 ⎝ 2 2s ⎠ Vo
=
1
2s (s + 2)(3s + 1)
=
2s / 3 (s + 2)(s + 1 3)
=
A s+2
+
B 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 s+2
−
2 15 s +1 3
Hence, v o (t) =
⎛ 4 e - 2 t − 2 e - t 3 ⎞ u ( t )V ⎜ ⎟ 15 ⎝ 5 ⎠
v(0) = V0 is incorporated as shown below.
P.P.16.3
V(s) + I0 /s
1/sC
R
V
CV0
We apply KCL to the top node. I0 V 1 ⎞ ⎛ + CV0 = + sCV = ⎜sC + ⎟ V ⎝ s R R ⎠ V= V= V= where A =
I0 s (sC + 1 R ) V0 s + 1 RC V0 s + 1 RC
I0 C 1 RC
V(s) =
+
+
CV0 sC + 1 R I0 C
s (s + 1 RC)
+
A s
+
B s + 1 RC
= I 0 R , V0
s + 1 RC
v( t ) = (( V0
+
B=
I 0 R s
−
I0 C - 1 RC
= - I 0 R
I 0 R s + 1 RC
− 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) –t
–2t
Which leads to iL(t) = (3 – 7e + 3e )u(t)A
P.P.16.5
We can use the same solution as found in Example 16.5 to find iL.
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 –t
or
–2t
i2 = (–2e + e )u(t)A
I3 = V3/5s – 0/s = –1/(s(s+1)) + 2/(s(s+2)) = –1/s + 1/(s+1) + 1/s – 1/(s+2) –t
or
–2t
i3 = (e – e )u(t)A –t
–2t
This leads to iL(t) = i1 + i2 + i3 = (3 – 7e + 3e )u(t)A
P.P.16.6
Ix
1/s
1 +
+
5/s
Vo
−
+
2
4Ix
(a) Take out the 2 Ω and find the Thevenin equivalent circuit. VTh = Ix
1/s
1 +
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
1
+
+
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 0.25)
+
s(s
0.2)
−
Vo =
0.25)
s(s
0.2)
1 5(s
(b)
+
4(s
4(s
0.2)
0.2)
Vo
2 = 2
4(s + 0.25) s(s + 0.3)
or
2
10( 4s + 1) s(10s + 3)
+
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 If x ( t ) = e -3t u ( t ) , then X (s) =
P.P.16.7
Y(s) = H(s) X (s) =
2s (s + 3)(s + 6)
A = Y (s) (s + 3)
s = -3
= -2
B = Y (s) (s + 6)
s = -6
=4
Y (s) =
-2 s+3
+
-3t y( t ) = (-2 e
H(s) =
2s (s + 6)
=
1 s+3
A s+3
+
B s+6
4 s+6
+ 4 e -6t )u ( t ) =
2 (s + 6 − 6) s+6
= 2−
12 s+6
h ( t ) = 2 (t ) 12 e -6t u(t ) P.P.16.8
I1
=
By current division, 2 + 1 2s I s + 4 + 2 + 1 2s 0
H(s) =
I1 I0
=
2 + 1 2s s + 4 + 2 + 1 2s
=
4s 2s 2
1 12s
1
P.P.16.9
2s (a)
Vo Vi
=
1 || 2 s 1 + 1 || 2 s
=
1+ 2 s 2s
1+
1+ 2 s
=
.
2 s+4
–0.3t
)u(t)V
H(s) =
Vo
=
Vi
2 s
4
(b)
h ( t ) = 2 e -4t u(t )
(c)
Vo (s) = H (s) Vi (s) =
A
s (s + 4)
=
1
= s Vo (s) s=0 = ,
s
+
B s+4
s = -4
=
-1 2
1 ⎛ 1
1 ⎞ ⎜ − ⎟ 2 ⎝ s s + 4 ⎠
1
v o (t) =
A
B = (s + 4) Vo (s)
2
Vo (s) =
(d)
2
2
(1
e -4t ) u(t ) V
v i ( t ) = 8 cos( 2 t )
⎯ ⎯→
Vo (s) = H(s) Vi (s) =
A = (s + 4) Vo (s)
s = -4
Vi (s) =
8s s2
16s (s + 4)(s 2
=
=
+ 4)
+4 A
+
(s + 4)
Bs + C (s 2
+ 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 : 0= A+B
s1 :
16 = 4B + C
⎯ ⎯→
(2)
s0 :
0 = 4 A + 4C
⎯ ⎯→ C = -A
(3)
Vo (s) =
⎯ ⎯→
B = -A =
16
s2 :
C=
16 5
5
16 ⎛
(1)
−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.
iR
R 1
i
L +
vL
-
+ C
+ _
vs
R 2
v -
i R
= i + C
vo
= R2i
But
dv dt
(1)
−v
=
vs
vs
−v
i R
R1
Hence, R1
= i + C
dv dt
or •
v=−
v R1C
i
vs
C
R1C
− +
(2)
Also, v L
v + v L + vo =0
=L
di dt
= v − vo
But vo = iR2 . Hence •
+ vo
i = v / L − vo / L =
v
−
iR2
L L Putting (1) to (3) into the standard form
(3)
⎡ 1 −1⎤ ⎡v• ⎤ ⎢ − ⎥ ⎡v ⎤ ⎡ 1 ⎤ R C C ⎢ ⎥=⎢ 1 ⎥ ⎢ ⎥ + ⎢ R 1C ⎥ vs • ⎥ ⎢ ⎥ ⎢ I R2 ⎥ ⎣ i ⎦ ⎢ i ⎢⎣ 0 ⎥⎦ − ⎥ ⎣ ⎦ ⎢ L⎦ ⎣ L ⎡v ⎤ vo = [ 0 R2 ] ⎢ ⎥ ⎣i ⎦ If we let R 1= 1, R 2 = 2, C = ½, L = 1/5, then ⎡ −2 −2 ⎤ ⎡2⎤ = A = ⎢ , B ⎥ ⎢0 ⎥ , C = [ 0 2] 5 10 − ⎣ ⎦ ⎣ ⎦ sI
⎡s + 2 − A = ⎢ ⎣ −5
⎤ ⎥ s + 10 ⎦ 2
⎡ s + 10 −2 ⎤ ⎢ 5 ⎥ s + 2⎦ ⎣ −1 ( sI − A) = 2 s + 12s + 30
H(s) = C(sI − A ) −1 B =
=
=
P.P. 16.11
[0
⎡s + 10 − 2 ⎤ ⎡2⎤ 2]⎢ ⎥⎢ ⎥ s + 2 ⎦ ⎣0 ⎦ ⎣ 5 s2
+ 12s + 30
20 s 2 12s + 30
20 s2
+ 12s + 30
Consider the circuit below. i 1
L
vo
2 io
i1
R 1
+ v -
C
R 2
i2
At node 1, i1
=
v R1
•
or
v=−
•
+ C v+ i 1
v−
R1C
1 C
i+
i1
C
(1)
This is one state equation. At node 2, io = i + i2
(2)
Applying KVL around the loop containing C, L, and R 2, we get •
−v + L i + io R2 = 0 or •
i=
v
−
R2
io L L Substituting (2) into (3) gives • v R R i = − 2 i − 2 i2 (4) L L L vo = v (5) From (1), (3), (4), and (5), we obtain the state model as ⎡ 1 −1⎤ ⎡1 ⎤ 0 ⎥ ⎡v• ⎤ ⎢− ⎥ ⎢ ⎡i1 ⎤ R1C C ⎡v ⎤ C
⎢ ⎥=⎢ ⎢•⎥ ⎢ ⎣i ⎦ ⎢ ⎣
1 L
⎥ ⎢ ⎥ + ⎢ ⎥ ⎢ ⎥ R2 ⎥ ⎣i2 ⎦ R2 ⎥ ⎣ i ⎦ ⎢ 0 − − ⎥ ⎢ ⎣ L ⎦⎥ L⎦
⎡vo ⎤ ⎡1 0⎤ ⎡v ⎤ ⎡0 0 ⎤ ⎡ i1 ⎤ ⎢ i ⎥ = ⎢ 0 1 ⎥ ⎢ i ⎥ + ⎢0 1 ⎥ ⎢ i ⎥ ⎦⎣ ⎦ ⎣ ⎦ ⎣ 2 ⎦ ⎣ o⎦ ⎣ Substituting R 1 = 1, R 2 =2, C = ½, L = ¼ yields
⎡v• ⎤ −2 −2 v ⎤ ⎡ ⎤ ⎡2 0 ⎤ ⎡ i1 ⎤ ⎢ ⎥ = ⎡ + ⎢ • ⎥ ⎢⎣ 4 −8⎥⎦ ⎢⎣ i ⎥⎦ ⎢⎣0 −8⎥⎦ ⎣⎢i2 ⎥⎦ ⎣i ⎦ ⎡vo ⎤ ⎡1 ⎢ i ⎥ = ⎢0 ⎣ o⎦ ⎣
0 ⎤ ⎡v ⎤
⎡0 + ⎥⎢ ⎥ ⎢ 1 ⎦ ⎣ i ⎦ ⎣0
0 ⎤ ⎡ i1 ⎤
⎥⎢ ⎥
1 ⎦ ⎣i2 ⎦
(3)
P.P. 16.12
Let so that
x1 = y •
x1
(1) •
= y (2)
•
Let Finally, let
•
x2
= x1 = y
x3
=
•
x2
(3)
••
= 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,
⎡ x• ⎤ ⎢ 1 ⎥ ⎡ 0 1 0 ⎤ ⎡ x1 ⎤ ⎡0⎤ ⎢•⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ x2 ⎥ = ⎢ 0 0 1 ⎥ ⎢ x2 ⎥ + ⎢0⎥ z (t ) ⎢ • ⎥ ⎣⎢ −6 −11 −6⎦⎥ ⎣⎢ x3 ⎦⎥ ⎣⎢1⎥⎦ ⎢ x3 ⎥ ⎣ ⎦
y(t ) = [1 0
P.P.16.13
⎡ x1 ⎤ ⎢ ⎥ 0] x2 ⎢ ⎥ ⎢⎣ x3 ⎥⎦
The circuit in the s-domain is equivalent to the one shown below. Vo
+ Z
Z
Vo
= (βVo ) Z ⎯ ⎯→
- Vo
where
R
Z = R || 1 sC =
Thus, - 1 =
- 1 = βZ , 1 + sRC
βR or 1 + sRC
- (1 + sRC ) = βR
For stability,
βR > -1
-1
or
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 For stability
βR + 1 RC
must be positive, i.e.
βR + 1 > 0
-1
or
R
P.P.16.14
(a)
Following Example 15.24, the circuit is stable when -10 10 + α > 0 or
(b)
For oscillation, 10 + α = 0
-10
or
P.P.16.15
Vo Vi
=
s⋅
R R + sL +
1 sC
= s2
+s⋅
R
L R L
+
1 LC
Comparing this with the given transfer function,
R L
= 4
1
and
LC
If we select R = 2 , then 2 L = = 0 .5 H 4 P.P.16.16
= 20
C=
and
1 20L
=
1 10
= 0. 1 F
Consider the circuit shown below. Y3 Y4 Y1
Y2
− +
Vin
Clearly,
V1
V2
+
Vo
−
=0
V2
At node 1,
− V1 ) Y1 = (V1 − Vo ) Y3 + (V1 − 0) Y2 Vin Y1 = V1 (Y1 + Y2 + Y3 ) − Vo Y3
( Vin or
(1)
At node 2, ( V1 − 0) Y2 or
=
V1
- Y4 Y2
= (0 − Vo ) Y4
Vo
(2)
Substituting (2) into (1), Vin Y1 or
If we select Y1
Vo Vin
=
= 1
R 1
=
- Y4 Y2
Vo ( Y1 + Y2
+ Y3 ) − Vo Y3
- Y1 Y2 Y4 ( Y1 + Y2 , Y2
= sC1 ,
+ Y3 ) + Y2 Y3 Y3
= sC 2 , and
Y4
=
1 R 2
, then
-s⋅
Vo
=
Vin
C1 R 1
⎛ 1 ⎞ ⎜ + sC1 + sC 2 ⎟ + s 2 C1C 2 R 2 ⎝ R 1 ⎠ 1
-s⋅
Vo
=
Vin
s2
+s⋅
1 R 1C 2
⎛ 1 1 ⎞ 1 ⎜ + ⎟+ R 2 ⎝ C1 C 2 ⎠ R 1 R 2 C1C 2 1
Comparing this with the given transfer function shows that 1 R 1C 2
If R 1
⎛ 1 1 ⎞ ⎜ + ⎟ = 6, R 2 ⎝ C1 C 2 ⎠ 1
= 2,
1 R 1 R 2 C1C 2
= 10
= 10 k Ω , then C2
=
1 R 2 C1
1 2 × 10 3
= 0.5 mF 1
= 5 ⎯ ⎯→
R 2
= 5C1
⎛ 1 1 ⎞ ⎜ + ⎟ = 6 ⎯ ⎯→ R 2 ⎝ C1 C 2 ⎠ 1
R 2
=
1 5C1
=
1 (5)(0.1 × 10 -3 )
Therefore, C1 = 0.1 mF ,
C2
⎛ ⎝
5 ⎜1 +
C1 ⎞
⎟ = 6 ⎯ ⎯→ C 2 ⎠
C1
= 2 k Ω
= 0.5 mF ,
R 2
= 2k
=
C2 5
= 0.1 mF
February 5, 2006
CHAPTER 17 T = 2, ωo = 2π/T = π
P.P.17.1
f(t) = 1, –1, ao =
1 T
an =
2 T
=
bn =
=
∫
T
∫
T
0
0
f ( t )dt =
0
1⎡ 1 (1)dt + 2 ⎢⎣ 0
∫
∫
2
1
( −1)dt ⎤ = 0.5(1 – 1) = 0
⎥⎦
2⎡ 1 1 cos nπtdt + 2 ⎢⎣ 0
∫
f ( t ) cos nω o dt =
∫ (−1) cos nπtdt ⎤⎦⎥ 2
1
1 1 [sin nπt ]10 − [sin nπt ]12 = 0 nπ nπ 2 T
∫
T
0
f ( t ) sin nωo dt =
2⎡ 1 1sin nπtdt + 2 ⎢⎣ 0
∫
∫ (−1) sin nπtdt ⎤⎦⎥ 2
1
1 2 −1 [cos nπt ]10 + [cos nπt ]12 = [1 − cos nπ] nπ nπ nπ
bn = 4/(nπ), for n = odd = 0, for n = even f(t) =
4
1 k 1 n
n = 2k – 1
sin n t ,
T = 1, ωo = 2π/T = 2π, f(t) = t, 0 < t < 1.
P.P.17.2
1 ao = T
an =
2 T
∫
T
∫
T
0
0
2
1 t f ( t )dt = ⎡ ∫ ( t )dt ⎤ = ⎢⎣ 0 ⎥⎦ 2
f ( t ) cos nω o dt =
1 0
= 0.5
2⎡ 1 t cos nπtdt ⎤ ⎦⎥ 1 ⎢⎣ 0
∫
1
⎡ 1 ⎤ t π + π cos 2 n t sin 2 n t = 2⎢ [ ] [ ] ⎥ 2 2nπ ⎣ (2nπ) ⎦0
=
bn =
2 4n 2 π 2 2 T
∫
T
0
[[cos 2nπ1] − 1] = 0
f ( t ) sin nωo dt =
2⎡ 1 t sin 2nπtdt ⎤ ⎦⎥ 1 ⎢⎣ 0
∫
1
−2 t ⎡ 1 [cos 2nπt ]⎤⎥ = [cos 2nπ] = –1/(nπ) = 2 ⎢ 2 2 [sin 2nπt ] − 2nπ ⎣ 4n π ⎦ 0 2nπ f(t) = 0.5
1
1 n 1
n
sin 2n t
P.P.17.3
f(t) =
1, –1,
– π < t < 0 0
f(t) is an odd function,
a o = 0 = an
T = 2π, ωo = 2π/T = 1 bn =
4 T/2
∫
T 0
f ( t ) sin nωo dt
=
⎡ π ( −1) sin ntdt ⎤ 2π ⎢⎣ ∫0 ⎦⎥ 4
π
2 ⎡2 ⎤ [cos nπ − 1] = ⎢ [cos nt ]⎥ = ⎣ nπ ⎦ 0 nπ
= –4/(nπ), 0,
n = odd n = even
−4 ∞ 1 f(t) = sin nt , n π k ∑ =1
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 ao = T an =
=
4 T
∫
T/2
∫
T/2
0
0
2 ⎡ f ( t )dt = 2π ⎢⎣
∫
π
0
f ( t ) cos nωo dt =
1 t2 ⎤ ( t / π)dt = 2 x ⎥⎦ π 2 4 ⎡ 2π ⎢⎣
π
t
0
π
∫
π 0
= 0.5
⎤ ⎦
cos ntdt ⎥
2 ⎡
1 π ⎤ π ( t / n ) sin nt sin ntdt ⎥ − [ ] 0 ∫ 2 ⎢ n 0 π ⎣ ⎦ π
− 2 −1 2 = cos nt (cos nπ − 1) = 2 2 π nπ 2 n n 0 = –4/(n2π2), 0, f(t) =
n = odd n = even 1
4
1
2
2
2 k 1 n
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
∫
T/2
0
f ( t ) cos nωo dt =
4 ⎡ 2π ⎢⎣
π
t
0
π
∫
2 ⎡
1 π ⎤ π ( t / n ) sin nt sin ntdt ⎥ − [ ] 0 ∫ 2 ⎢ n 0 π ⎣ ⎦ π
− 2 −1 2 = cos nt (cos nπ − 1) = 2 2 π nπ 2 n n 0 = –4/(n2π2), 0,
n = odd n = even
⎤ ⎦
cos ntdt ⎥
bn =
4 T
∫
T/2
0
4 ⎡ 2π ⎢⎣
f ( t ) sin nω o dt =
π
t
0
π
∫
⎤ ⎦
sin ntdt ⎥
π
2 ⎡ 2 ⎤ = ⎢ 2 2 [sin nt − nt cos nt ]⎥ = , ⎣n π ⎦ 0 nπ Thus, f(t) =
2
2 n
k 1
2
n = odd 1
cos nt
n
sin nt ,
n = 2k – 1
P.P.17.6
vs(t) = 0.5 – (1/π)
∞
1
∑ n sin 2πnt ,
ω = 2πn
n =1
vo(ω) = (1/(j ωC))vs/(R + (1/j ωC)) = vs/(1 + jωRC) = vs/(1 + j2ω), RC = 2 For the the DC component (ω = 0, or n = 0),
vs = 0.5 and vo = 0.5
For the nth harmonic, 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 n =1
=
1 2
1
1 1 + 4ω 2
sin( 2 nt n 1
cos(2πnt – 90° – tan –12ω) 1
tan
n 1 16
2
4 n
n2
P.P.17.7 2
v(t) = (1/3) + (1/ π )
∞
∑ ⎛ ⎝ ⎜ n
1
n =1
2
= (1/3) + (1/π )
2
∞
∑A n =1
n
cos nt −
π n
⎞ ⎠
sin nt ⎟
cos(nt - φn)
where
1 1 1 + π2 = 2 1 + n 2 π2 2 n n n
An =
φn = tan –1(bn/an) = tan –1(–nπ) v(t) = (1/3) +
1
π
2
∑ n1
2
1 + n 2 π 2 cos[nt – tan-1(–nπ)]
Z = 2 + 1||(1/(j ω)),
ω = n
= 2 + (1/(jω))/(1 + (1/(j ω))) = 2 + (1/(1 + (j ω)) = (3 + 2jω)/(1 + jω),
ω = n
= (3 + j2n)/(1 + jn) I = V/Z = [(1 + jn)/(3 + j2n)] V 2
I
V( )
Io
+
1/j
−
1
By current division, I o = (1/jω)I/[1 + (1/j ω)] = I/(1 + j ω) = V/(3 + j2n) For the DC component (n = 0), For the nth harmonic,
V = 1/3 and I o = 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)]/[n 2π2 9 + 4n 2 ] tan –1(–nπ) = –tan –1(nπ)
In the time domain, io(t) = {(1/9) +
1 n2 2
2
2
–1
2
–1
cos[nt – tan (2n/3) + tan (n )]}A
n 9 4n Note, the summation is to be carried out from n=1 to
.
∞
∞
P = VDC IDC + 0.5
P.P.17.8
∑
Vn In cos(φn – θn)
n =0
= 80(0) + 0.5(120)(5 0.5(120)(5)) cos(10°) + 0.5(60)(2 0.5(60)(2)) cos(30 °) = 295.44 + 51.962 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
= 0,
1
T = 2, ωo = 2π/T = π Cn = (1/T)
∫
T
0
1
∫
f ( t )e − jnωo t dt = 0.5[ 1e − jnπt dt + 0] 0
= 0.5[1/(–jn π)]e –jnπt
–jnπ π = [j/(2n )](e – 1) 0
1
But e –jnπ = cos(nπ) – jsin(n π) = 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 2
j n n 0 n odd
n
jn t
e
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/(2n 2π2)][e –jnπ(–jnπ – 1) – e jnπ(jnπ – 1)] = [–1/(2n2π2)][(cos nπ – j sin n π)(–jnπ – 1) – (cos nπ + j sin n π)(jnπ – 1)] = [j cos nπ]/(nπ) Cn = j(–1)n/nπ, n ≠ 0
∫
T/2
f ( t )dt = 0
For n = 0,
Co = (1/T)
Thus,
j(−1) n jnπt e f(t) = nπ n = −∞
−T / 2
∞
∑ n≠0
|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
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. P.P.17.12
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)
DC COMPONENT = 4.950490E-01 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (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
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
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. P.P.17.13
DC COMPONENT = -1.507149E-04 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 2 3 4 5
2.000E+03 1.455E-04 1.000E+00 4.000E+03 1.845E-06 1.268E-02 6.000E+03 1.401E-06 9.629E-03 8.000E+03 1.015E-06 6.974E-03 1.000E+04 8.345E-07 5.736E-03
9.006E+01 9.624E+01 9.318E+01 8.118E+01 5.922E+01
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(4 103t + 90 ) + 1.845sin(8 103t + 96.24 ) + - - -} V
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 = 12.566 rad/s = 6π = 18.84 rad/s = 8π = 25.13 rad/s = 10π = 31.41 rad/s = 12π = 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) = (–1/3 )sin(3
ot)
– (1/(4 ))sin(4
ot)
– (/(5 ))sin(5
ot),
o
= 2
February 5, 2006
CHAPTER 18
P.P.18.1
1< t < 2
⎡1, (a) g(t) = u(t + 1) - u(t - 2) = ⎢ ⎣0,
∫
G(ω) =
2
1
j
e
=
j t 1 ⋅ e − ω dt = −
1 jω
otherwise
e− ω
j t 2 1
j 2
e j
(b) F(t) = 4δ(t + 2)
∫
F(ω) =
∞
f ( t )e − jωt dt =
−∞ jωt
= 4e
t =2
∫
∞
−∞
4δ( t + 2)e − jωt dt
= 4e j2
(c) F(t) = sin(ωot)
⎡ − e jω t ⎤ 1 jω t - jω F(ω) = F ⎢ ⎥ = [F(e ) − F(e ⎣ 2 j ⎦ j2 o
o
j
=
P.P.18.2
−∞
=
f ( t ) =
f ( t )e − jωt =
e − jωt
− jω
0
−1
−
2 cos
=
P.P.18.3
∞
∫
F(ω) =
o
o
o
1
∫ (−1)e
− jωt
0
e − jωt
1 0
− jω
=
j
ω
dt
[1 − e ω − e − ω + 1] j
1
j
e at ,
t<0
0,
t >0
F(ω) =
∫
∞
−∞
f ( t )e jωt dt =
∫
∞
e at e − jωt dt
−∞
Let x = -t, then dt = -dx F(ω) =
=
0
∫e
− ax jωx
∞
e
1 a − jω
( −dx ) = −
e a − jω) x =
)]
∫
0
∞
− ( a − jω ) x
e
1 a − jω
dx
j
P.P.18.4
(a) g(t) = u(t) - u(t - 1) -j -j F(ω) = u(ω) - e ωu(ω) = (1 - e ω)u(ω) -j = (1 - e )( ( ) + 1/(j )) -2t
(b) f(t) = te u(t) -2t Let g(t) = e u(t) f(t) = tg(t) F(ω) =
(c)
G(ω) = 1/(2 + j ω) j
dG d ω
= j(-1) (2 + j ω) (j) -2
1 ( 2 + jω) 2
f(t) is sketched below. f(t) 5
t
2
f '(t) = –10δ(t – 2) – 20δ(t – 2) f "(t) = 10 δ(t) - 10δ(t - 2) - 20 δ'(t - 2) 2 -jω2 -jω2 (jω) F(ω) = 10(1 - e ) - 20jωe F(ω) =
P.P.18.5
j 2 10 e − ω − 1
ω
2
+
j 2 20 je − ω
ω
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
2
–4
–3
–2
2
–1
0
1
2
2
3
–4
4
t
–4
f "(t) = 2δ(t + 4) - 4 δ(t + 3) + 2 δ(t + 2) + 2 δ(t - 2) - 4 δ(t + 3) + 2 δ(t - 4) We take the Fourier transform of each term. (jω) F(ω) = 2(e ω + e ω) - 4(e ω + e ω) + 2(e ω + e = 4 cos 4ω - 8 cos 3ω + 4 cos 2ω 2 F(ω) = [1/( )](8 cos 3 - 4 cos 4 - 4 cos 2 ) 2
j4
P.P.18.6
-j4
(a) H(ω) =
=
j3
-j3
j2
-j2ω
)
6( 2 jω + 3) ( jω + 1)( jω + 4)( jω + 2) 2 3 5
+
−
jω + 1 jω + 2 jω + 4 -t -2t -4t h(t) = (2e + 3e - 5e )u(t) (b)
-t
y(t) = u(t) + 2e cos 4t u(t) -t = (1 + 2e cos 4t) u(t)
P.P.18.7
Vi(ω) = 4/(jω)
vi = 2 sgn (t) H(ω) = 4/(4 + jω)
Vo (ω) = H(ω)Vi (ω) =
=
4 jω
−
4 4 + jω
16 jω( 4 + jω)
=
A jω
+
B 4 + jω
-4t
-4t
vo(t) = 2 sgn (t) - 4e u(t) = 2[-1 + u(t)] - 4e u(t) -4t = - 2 + 4 [1 - e ]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)] ⎝ 8 + jω ⎠ 20π ∞ ⎛ 3 + jω ⎞ -1 ⎜⎜ ⎟⎟[δ(ω + 4) + δ(ω − 4)e jωt d ω] io ( t ) = F Io(ω) = ∫ 2π −∞ ⎝ 8 + jω ⎠
I0(ω) = H(ω)Is(ω) = ⎜⎜
⎡ 3 − j4 − j4 t 3 + j4 j4 t ⎤ = 10⎢ + e e ⎥ − ω + 8 j 8 j 4 ⎣ ⎦ But 3 + j4 8 + j4
5∠53.13
=
80∠26.56
(
= 0.559∠26.57
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
∫
dt = 200
∞
0
e −4 t dt
since t is even. 200e −4 t
W1Ω =
(b)
H (ω) =
−4
∞ 0
= 50J
40 4 + ω2 1 ∞ 1600
1600 1 ⎛ ω 1 ω ⎞ ⋅ ⎜ 2 + tan −1 ⎟ 2 ⎠ π 8 ⎝ ω + 4 2
W1Ω =
π ∫ (4 + ω )
W1Ω =
π 200 ⎛ ⎞ ⎜ 0 + − 0 − 0 ⎟ = 50J π ⎝ 4 ⎠
2 2
0
P.P.18.10 F(ω) = W2 Ω =
1 2
=
2
d ω
π ∫0 1 + ω 2
π∫
4
d ω
1+ ω 76
0
2
∞
1
2
F(ω) =
1 + jω
for -4 < ω < 4, W=
d ω =
2
=
2
π
1 + ω2
tan −1 ω ∞0 =
= 2 tan −1 ω 04
⋅ π = 0.844 = 84.4% π 180
2 π
⋅ =1 π 2
∞ 0
i.e. 84.4% of the total energy. P.P.18.11 If f c = 2 MHz, f m = 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, f s = 2W = 25 kHz 1 1 Ts = = = 40 s f s 25x10 3
February 5, 2006
P.P.19.11 I1 =
Also,
I1 =
V1 − 0
R 1 0 − V2 R 2
⎯ ⎯→ V1 = I1 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, [z ] =
R 1
0
- R 2
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 ,
z 12a = z 21a = 20 ,
z 11a = 20 − j15 ,
z 22a = 20 + j10
For N b ,
z 12 b = z 21 b = 50 ,
z 11 b = 50 + j40 ,
z 22 b = 50 − j20
Thus,
[z ] = [z a ] + [z b ]
⎡ 20 − j15
⎤ ⎡50 + j40 50 ⎤ +⎢ ⎥ 20 + j10 ⎦ ⎣ 50 50 − j20 ⎥⎦ ⎣ 20 ⎡ 70 + j25 70 ⎤ [z ] = ⎢ 70 − j10 ⎥⎦ ⎣ 70 [z ] = ⎢
V2 Vs V2 Vs V2 Vs
=
20
z 12 Z L
(z 11 + Z s )(z 22 + Z L ) − z 12 z 21
=
(70)(40) (70 + j25 + 5)(70 − j10 + 40) − 4900
=
2800 8250 − j750 + j2750 + 250 − 4900
V2 Vs P.P.19.13
=
2800 = 0.6799 3600 + j2000
- 29.05
We convert the upper T network N a to a Π network, as shown below. 25 S
-j5 S
ya =
j5 S
y1 y 2 + y 2 y 3 + y 3 y1 y2
y b = 5 ,
=
(-j5)(j5) + (j5)(1) + (1)(-j5) = -j5 j5
y c = 25
For N a , y 12a = -25 = y 21a ,
⎡ 25 − j5
y 11a = 25 − j5 ,
y 22a = 25 + j5
- 25 ⎤
[y a ] = ⎢
25 + j5 ⎥⎦
⎣ - 25
For N b , y 12 b = j10 = y 21 b ,
⎡ 2 − j10
[y b ] = ⎢
⎣ j10
y 11 b = 2 − j10 = y 22 b
j10 ⎤
2 − j10 ⎥⎦
Since N a and N b are in parallel, [y ] = [y a ] + [y b ] [y ] =
P.P.19.14
27 j15
- 25 j10
- 25 j10
27 j5
S
Convert the left Π network to a T network. (20)(50) (20)(30) R 1 = = 6 , R 2 = = 10 , 100 20 + 30 + 50
R 3 =
(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. 6
15
40
60
10
20
Na
Nb
For N a , Aa = 1 + Ca =
1 10
⎛ 6 ⎞ ⎟ ( 25) = 30 ⎝ 10 ⎠
6 = 1.6 , 10
B a = 15 + ⎜
= 0.1 ,
Da = 1 +
15 10
= 2 .5
⎡1.6 30 ⎤ [Ta ] = ⎢ ⎥ ⎣0.1 2.5⎦ For N b , A b = 1 + C b =
1 20
40 20
= 0.05 ,
⎡ 3
[T b ] = ⎢
= 3,
⎣ 0.05
⎛ 40 ⎞ ⎟ (80) = 220 ⎝ 20 ⎠
B b = 60 + ⎜ D b = 1 +
60 20
220 ⎤ 4 ⎥⎦
Hence, 220⎤ ⎡1.6 30 ⎤ ⎡ 3 [T] = [Ta ][T b ] = ⎢ ⎥⎢ ⎥ ⎣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
=4
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 h 11 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, V1 I2 h11 = h 21 = = 4.238 Ω , = -0.6190 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). V1 I2 h12 = h 22 = = -0.7143 , = -0.1429 S 1 1
(b)
Thus, [h ] =
4.238 - 0.6190
- 0.7143 - 0.1429 S
Insert a 1-A ac current source at the output terminals to account for P.P.19.16 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($N_0003) -2.651E+00
From this table, z 11 =
V1
1
= 3.987∠175.5° ,
z 21 = 0.0175∠ - 2.65°
Similarly, insert a 1-A ac source at the output port with the two pseudocomponents in place as in Fig. (a). The result is the schematic in Fig. (b).
(b)
When the schematic is saved and run, the output results include : FREQ 6.000E+01
VM($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
1
z 22 = 0.265∠91.9°
≅0
Thus, [z ] =
3.987 175.5 0.0175
- 2.65
0 0.265 91.9
P.P.19.17
In this case, R s = 150 k Ω , R L = 3.75 k Ω . 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 -3 3 6000 + (18 × 10 )(3.75 × 10 ) 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 Ai =
200 1 + (8 × 10 -6 )(3.75 × 10 3 )
= 194.17
Z in = 150,000 + 6000 − (1.5 × 10 -4 )(194.17) ≅ 156 k Ω Z out
P.P.19.18
150 × 10 3 + 6 × 10 3 = (150 × 10 3 )(8 × 10 -6 ) − (1.5 × 10 -4 )(200) 156 k Ω = 128.08 k = 1.248 − 0.03
Let D(s) = (s 3 + 4s) + (s 2 + 2)
Dividing both numerator and denominator by s 3 + 4s gives 2 H(s) =
s 3 + 4s s2 + 2
1+
i.e.
y 21 =
s 3 + 4s
-2 s 3 + 4s
y 22
s2 + 2 = 3 s + 4s
As a third order function, we can realize H (s) by the LC network shown in Fig. (a) .
