Natural 8 and Step Responses of RLC
Circuits
Assessment Problems
AP 8.1 [a] 1 1 = , (2RC)2 LC 1 , [b] = 5000 = 2RC s1,2 = -5000 [c] 1 = 20,000, LC therefore C = 500 nF C = 1 F
therefore 25 106 - therefore
(103 )(106 ) = (-5000 j5000) rad/s 20 C = 125 nF
s1,2 = -40
(40)2 - 202 103 , s2 = -74.64 krad/s
s1 = -5.36 krad/s, AP 8.2 iL = = = = = 1 50 10-3 20
t 0
[-14e-5000x + 26e-20,000x ] dx + 30 10-3
t 0
-14e-5000x -5000
+
26e-20,000t -20,000
t 0
+ 30 10-3
56 10-3 (e-5000t - 1) - 26 10-3 (e-20,000t - 1) + 30 10-3 [56e-5000t - 56 - 26e-20,000t + 26 + 30] mA 56e-5000t - 26e-20,000t mA, t0
AP 8.3 From the given values of R, L, and C, s1 = -10 krad/s and s2 = -40 krad/s. [a] v(0- ) = v(0+ ) = 0, therefore 81 iR (0+ ) = 0
82
CHAPTER 8. Natural and Step Responses of RLC Circuits [b] iC (0+ ) = -(iL (0+ ) + iR (0+ )) = -(-4 + 0) = 4 A [c] C 4 dvc (0+ ) dvc (0+ ) = ic (0+ ) = 4, = = 4 108 V/s therefore dt dt C t 0+ [d] v = [A1 e-10,000t + A2 e-40,000t ] V, v(0+ ) = A1 + A2 , dv(0+ ) = -10,000A1 - 40,000A2 dt -A1 - 4A2 = 40,000; t0 A1 = 40,000/3 V
Therefore A1 + A2 = 0, [e] A2 = -40,000/3 V
[f] v = [40,000/3][e-10,000t - e-40,000t ] V, AP 8.4 [a] 1 = 8000, therefore 2RC 10 V = 160 mA [b] iR (0+ ) = 62.5 R = 62.5
iC (0+ ) = -(iL (0+ ) + iR (0+ )) = -80 - 160 = -240 mA = C Therefore -240 m dv(0+ ) = = -240 kV/s dt C dvc (0+ ) = d B2 - B1 dt
dv(0+ ) dt
[c] B1 = v(0+ ) = 10 V,
Therefore 6000B2 - 8000B1 = -240,000, [d] iL = -(iR + iC ); iR = v/R; iC = C dv dt
B2 = (-80/3) V
v = e-8000t [10 cos 6000t -
80 sin 6000t] V 3 1280 sin 6000t] mA 3
Therefore iR = e-8000t [160 cos 6000t - iC = e-8000t [-240 cos 6000t + iL = 10e-8000t [8 cos 6000t + AP 8.5 [a] 106 1 2 1 = , = 2RC LC 4 [b] 0.5CV02 = 12.5 10-3 ,
2 0.5LI0
460 sin 6000t] mA 3 t0 R = 100
82 sin 6000t] mA, 3 therefore
therefore
1 = 500, 2RC V0 = 50 V
[c]
= 12.5 10 ,
-3
I0 = 250 mA
Problems [d] D2 = v(0+ ) = 50, iR (0+ ) = dv(0+ ) = D1 - D2 dt
83
50 = 500 mA 100
Therefore iC (0+ ) = -(500 + 250) = -750 mA 10-3 dv(0+ ) = -750 = -75,000 V/s dt C 1 = 500, = Therefore D1 - D2 = -75,000; 2RC Therefore [e] v = [50e-500t - 50,000te-500t ] V iR = v = [0.5e-500t - 500te-500t ] A, R t 0+
D1 = -50,000 V/s
AP 8.6 [a] iR (0+ ) = [b] [c] [d] [e]
40 V0 = = 0.08 A R 500 iC (0+ ) = I - iR (0+ ) - iL (0+ ) = -1 - 0.08 - 0.5 = -1.58 A Vo 40 diL (0+ ) = = = 62.5 A/s dt L 0.64 1 1 = = 1000; = 1,562,500; s1,2 = -1000 j750 rad/s 2RC LC if = I = -1 A iL = if + B1 e-t cos d t + B2 e-t sin d t, iL (0+ ) = 0.5 = if + B1 , therefore B1 = 1.5 A therefore B2 = (25/12) A t0 diL (0+ ) = 62.5 = -B1 + d B2 , dt
Therefore iL (t) = -1 + e-1000t [1.5 cos 750t + (25/12) sin 750t] A, [f] v(t) = LdiL = 40e-1000t [cos 750t - (154/3) sin 750t]V dt 15 k (80) = 50 V 15 k + 9 k t0
AP 8.7 [a] i(0+ ) = 0, since there is no source connected to L for t < 0. [b] vc (0+ ) = vC (0- ) = [c] 50 + 80i(0+ ) + L
di(0+ ) di(0+ ) = 100, = 10,000 A/s dt dt 1 [d] = 8000; = 100 106 ; s1,2 = -8000 j6000 rad/s LC if = 0, i(0+ ) = 0 [e] i = if + e-t [B1 cos d t + B2 sin d t]; Therefore B1 = 0; Therefore B2 = 1.67 A; di(0+ ) = 10,000 = -B1 + d B2 dt i = 1.67e-8000t sin 6000t A, t0
84
CHAPTER 8. Natural and Step Responses of RLC Circuits vf = 100 V 50 = 100 + B1
AP 8.8 vc (t) = vf + e-t [B1 cos d t + B2 sin d t], vc (0+ ) = 50 V; B1 = -50 V; Therefore B2 = dvc (0+ ) = 0; dt 0 = -B1 + d B2 therefore
8000 B1 = (-50) = -66.67 V d 6000 t0
Therefore vc (t) = 100 - e-8000t [50 cos 6000t + 66.67 sin 6000t] V,
Problems
P 8.1 [a] = 1 1 = = 250 2RC 2(1000)(2 10-6 ) 1 1 = = 40,000 LC (12.5)(2 10-6 ) 2502 - 40,000 = -250 150
2 o =
s1,2 = -250 s1 = -100 rad/s s2 = -400 rad/s [b] overdamped
[c] Note -- we want d = 120 rad/s: d =
2 o - 2
2 2 .. 2 = o - d = 40,000 - (120)2 = 25,600
= 160 1 = 160; 2RC [d] s1 , s2 = -160 [e] = 40,000 = .. R = 1 = 1562.5 2(160)(2 10-6 )
1602 - 40,000 = -160 j120 rad/s .. R = 1 = 1250 2(200)(2 10-6 )
1 ; 2RC
Problems P 8.2 [a] - + - -
2 2 - o = -250 2 2 - o = -1000
85
Adding the above equations, = 625 rad/s 1 1 = = 625 2RC 2R(0.1 10-6 ) R = 8 k
2 2 2 - o = 750 2 4(2 - o ) = 562,500
- 2 = -1250
.. o = 500 rad/s
2 o = 25 104 =
1 LC 1 10-6 ) = 40 H t 0+ t 0+ t0
.. L = [b] iR =
(25
104 )(0.1
v(t) = -1e-250t + 4e-1000t mA, R
iC = C
dv(t) = 0.2e-250t - 3.2e-1000t mA, dt
iL = -(iR + iC ) = 0.8e-250t - 0.8e-1000t mA, P 8.3 [a] iR (0) = 15 = 75mA 200
iL (0) = -45mA iC (0) = -iL (0) - iR (0) = 45 - 75 = -30 mA [b] = 1 1 = = 12,500 2RC 2(200)(0.2 10-6 ) 1 1 = = 108 -3 )(0.2 10-6 ) LC (50 10 = -12,500 1.5625 108 - 108 = -12,500 7500 s2 = -20,000 rad/s
2 o =
s1,2
s1 = -5000 rad/s;
v = A1 e-5000t + A2 e-20,000t v(0) = A1 + A2 = 15
86
CHAPTER 8. Natural and Step Responses of RLC Circuits -30 10-3 dv (0) = -5000A1 - 20,000A2 = = -15 104 V/s dt 0.2 10-6 Solving, A1 = 10; A2 = 5 t0 v = 10e-5000t + 5e-20,000t V, [c] iC = = = C dv dt
0.2 10-6 [-50,000e-5000t - 100,000e-20,000t ] -10e-5000t - 20e-20,000t mA
iR = 50e-5000t + 25e-20,000t mA iL = -iC - iR = -40e-5000t - 5e-20,000t mA, P 8.4 1 1 = = 8000 2RC 2(312.5)(0.2 10-6 ) 1 1 = = 108 -3 )(0.2 10-6 ) LC (50 10 s1,2 = -8000 80002 - 108 = -8000 j6000 rad/s t0
.. response is underdamped v(t) = B1 e-8000t cos 6000t + B2 e-8000t sin 6000t v(0+ ) = 15 V = B1 ; iR (0+ ) = 15 = 48 mA 312.5
iC (0+ ) = [-iL (0+ ) + iR (0+ )] = -[-45 + 48] = -3 mA -3 10-3 dv(0+ ) = = -15,000 V/s dt 0.2 10-6 dv(0) = -8000B1 + 6000B2 = -15,000 dt 6000B2 = 8000(15) - 15,000; .. B2 = 17.5 V t0
v(t) = 15e-8000t cos 6000t + 17.5e-8000t sin 6000t V,
Problems P 8.5 = 1 1 = 104 = -6 ) 2RC 2(250)(0.2 10
2 .. 2 = o
87
2 = 108 ;
Critical damping: v = D1 te-t + D2 e-t iR (0+ ) = 15 = 60 mA 250
iC (0+ ) = -[iL (0+ ) + iR (0+ )] = -[-45 + 60] = -15 mA v(0) = D2 = 15 dv = D1 [t(-e-t ) + e-t ] - D2 e-t dt dv iC (0) -15 10-3 (0) = D1 - D2 = = -75,000 = dt C 0.2 10-6 D1 = D2 - 75,000 = (104 )(15) - 75,000 = 75,000 v = (75,000t + 15)e-10,000t V, P 8.6 = 1000/2 = 500 R= 1 1 = = 400 2C 2(500)(2.5 10-6 ) t0
v(0+ ) = 3(1 + 1) = 6 V iR (0+ ) = 6 = 15 mA 400
dv = -300e-100t - 2700e-900t dt dv(0+ ) = -300 - 2700 = -3000 V/s dt iC (0+ ) = 2.5 10-6 (-3000) = -7.5 mA iL (0+ ) = -[iR (0+ ) + iC (0+ )] = -[15 - 7.5] = -7.5 mA
88 P 8.7
CHAPTER 8. Natural and Step Responses of RLC Circuits [a] = 20,000; d =
2 o - 2
d = 15,000
2 2 .. o = d + 2 = 225 106 + 400 106 = 625 106
1 = 625 106 LC 1 = 40 mH L= 6 )(40 10-9 ) (625 10 [b] = 1 2RC 1 1 = = 625 2C 2(20,000)(40 10-9 )
.. R =
[c] Vo = v(0) = 100 V [d] Io = iL (0) = -iR (0) - iC (0) iR (0) = 100 Vo = = 160 mA R 625 dv iC (0) = C (0) dt dv = 100{e-20,000t [-15,000 sin 15,000t - 30,000 cos 15,000t]- dt 20,000e-20,000t [cos 15,000t - 2 sin 15,000t] dv (0) = 100{1(-30,000) - 20,000} = -500 104 dt dv C (0) = -500 104 (40 10-9 ) = -200 mA dt . Io = 200 - 160 = 40 mA . [e] dv dt = = C 100e-20,000t [25,000 sin 15,000t - 50,000 cos 15,000t] 25 105 e-20,000t [sin 15,000t - 2 cos 15,000t]
dv = 0.1e-20,000t (sin 15,000t - 2 cos 15,000t) dt
iC (t) = 0.1e-20,000t (sin 15,000t - 2 cos 15,000t) A iR (t) = 0.16e-20,000t (cos 15,000t - 2 sin 15,000t) A iL (t) = -iR (t) - iC (t) = e-20,000t (40 cos 15,000t + 220 sin 15,000t) mA, t0
Problems P 8.8 [a] 2 = 1000; = 500 rad/s o = 400 rad/s
89
2 2 2 - o = 600;
C= L=
1 1 = = 4 F 2R 2(500)(250) 1
2 o C
=
(400)2 (4
1 = 1.5625 H 10-6 )
iC (0+ ) = A1 + A2 = 45 mA diC diL diR + + =0 dt dt dt diL (0) diR (0) diC (0) =- - dt dt dt 15 diL (0) = = 9.6 A/s dt 1.5625 1 dv(0) diR (0) 1 iC (0) 45 10-3 = = = 45 A/s = dt R dt R C (250)(4 10-6 ) diC (0) .. = -9.6 - 45 = -54.6 A/s dt .. 200A1 + 800A2 = 54.6 A1 + A2 = 0.045 Solving, A1 = -31 mA; A2 = 76 mA t 0+
.. iC = -31e-200t + 76e-800t mA, [b] By hypothesis v = A3 e-200t + A4 e-800t , v(0) = A3 + A4 = 15 45 10-3 dv(0) = = 11,250 V/s dt 4 10-6 -200A3 - 800A4 = 11,250; v = 38.75e-200t - 23.75e-800t V, [c] iR (t) = t0
.. A3 = 38.75 V; t0 t 0+
A4 = -23.75 V
v = 155e-200t - 95e-800t mA, 250 [d] iL = -iR - iC iL = -124e-200t + 19e-800t mA, t0
810
CHAPTER 8. Natural and Step Responses of RLC Circuits 1 2RC
2
P 8.9
[a]
=
1 = (500)2 LC
.. C =
1 = 1 F (500)2 (4)
1 = 500 2RC .. R = 1 = 1 k 2(500)(10-6 )
v(0) = D2 = 8 V iR (0) = 8 = 8mA 1000
iC (0) = -8 + 10 = 2 mA 2 10-3 dv (0) = D1 - 500D2 = = 2000 V/s dt 10-6 .. D1 = 2000 + 500(8) = 6000 V/s [b] v = 6000te-500t + 8e-500t V, t0
dv = [-3 106 t + 2000]e-500t dt iC = C P 8.10 [a] = dv = (-3000t + 2)e-500t mA, dt t 0+
1 = 0.5 rad/s 2RC 1 = 25.25 LC 25.25 - (0.5)2 = 5 rad/s
2 o =
d =
.. v = B1 e-t/2 cos 5t + B2 e-t/2 sin 5t v(0) = B1 = 0; iR (0+ ) = 0 A; v = B2 e-t/2 sin 5t iC (0+ ) = 4 A; dv + 4 (0 ) = = 50 V/s dt 0.08
50 = -B1 + d B2 = -0.5(0) + 5B2 .. B2 = 10 .. v = 10e-t/2 sin 5t V, t0
Problems [b] dv = -5e-t/2 sin 5t + 10e-t/2 (5 cos 5t) dt dv = 0 when 10 cos 5t = sin 5t or dt .. 5t1 = 1.47, t1 = 294.23 ms 5t2 = 1.47 + , 5t3 = 1.47 + 2, [c] t3 - t1 = 1256.6 ms; t2 = 922.54 ms t3 = 1550.86 ms Td = tan 5t = 10
811
2 2 = 1256.6 ms = d 5 1256.6 Td = = 628.3 ms [d] t2 - t1 = 628.3 ms; 2 2 [e] v(t1 ) = 10e-(0.147115) sin 5(0.29423) = 8.59 V v(t2 ) = 10e-(0.46127) sin 5(0.92254) = -6.27 V v(t3 ) = 10e-(0.77543) sin 5(1.55086) = 4.58 V [f]
P 8.11
[a] = 0;
d = o =
25.25 = 5.02 rad/s v(0) = B1 = 0; v = B2 sin o t
v = B1 cos o t + B2 sin o t; C dv (0) = -iL (0) = 4 dt
50 = -B1 + d B2 = -0 + 25.25B2 .. B2 = 50/ 25.25 = 9.95 V v = 9.95 sin 5.02t V, [b] 2f = 5.02; f= t0 5.02 = 0.80 Hz 2
812
CHAPTER 8. Natural and Step Responses of RLC Circuits [c] 9.95 V
P 8.12
2 [a] o =
1 1 = 25 106 = LC (12.5)(3.2 10-9 ) 1 = 31.25 k 2(5000)(3.2 10-9 )
o = 5000 rad/s 1 = 5000; 2RC R=
[b] v(t) = D1 te-5000t + D2 e-5000t v(0) = 100 V = D2 dv = (D1 t + 100)(-5000e-5000t ) + D1 e-5000t dt dv iC (0) (0) = -500 103 + D1 = dt C iC (0) = -iR (0) - iL (0) iR (0) = 100 = 3.2 mA 31,500
.. iC (0) = -(3.2 + 6.4) = -9.6 mA .. 9.6 10-3 dv (0) = - = -3 106 -9 dt 3.2 10
.. -500 103 + D1 = -3 106 D1 = -25 105 V/s .. v(t) = (-25 105 t + 100)e-5000t V, [c] iC (t) = 0 when dv (t) = 0 dt t0
dv = (-25 105 t + 100)(-5000)e-5000t + e-5000t (-25 105 ) dt = (125 108 t - 30 105 )e-5000t dv = 0 when 125 108 t1 = 3 106 ; dt .. t1 = 240 s
v(240s) = e-1.2 [(-25 105 )(240 10-6 ) + 100] = -150.6 V
Problems [d] iL (240s) = -iR (240s) = -150.6 = -4.82 mA 31,250
813
1 C (240s) = (3.2 10-9 )(-150.6)2 = 36.29 J 2 1 L (240s) = (12.5)(-4.82 10-3 )2 = 145.2 J 2 (240s) = C + L = 181.49 J 1 1 (0) = (12.5)(6.4 10-3 )2 + (3.2 10-9 )(100)2 = 272 J 2 2 % remaining = P 8.13 [a] = 181.49 (100) = 66.72% 272 o = 103 , therefore overdamped
1 = 1250, 2RC
s1 = -500,
s2 = -2000
therefore v = A1 e-500t + A2 e-2000t v(0+ ) = 0 = A1 + A2 ; Therefore A1 = +980 , 15 dv(0+ ) iC (0+ ) = 98,000 V/s = dt C
- 500A1 - 2000A2 = 98,000 A2 = -980 15 t0
v(t) = [b]
980 -500t [e - e-2000t ] V, 15
Example 8.4:
vmax 74.1 V =
at 1.4 ms
814
CHAPTER 8. Natural and Step Responses of RLC Circuits Example 8.5: Problem 8.13: vmax 36.1 V = at 1.0 ms
vmax 30.9 at 0.92 ms =
P 8.14
From the form of the solution we have v(0) = A1 + A2 dv(0+ ) = -(A1 + A2 ) + jd (A1 - A2 ) dt We know both v(0) and dv(0+ )/dt will be real numbers. To facilitate the algebra we let these numbers be K1 and K2 , respectively. Then our two simultaneous equations are K1 = A1 + A2 K2 = (- + jd )A1 + (- - jd )A2 The characteristic determinate is = 1 1 = -j2d
(- + jd ) (- - jd )
The numerator determinates are N1 = K1 1 = -( + jd )K1 - K2
K2 (- - jd ) 1
and
N2 =
K1
(- + jd ) K2 A1 =
= K2 + ( - jd )K1
It follows that
d K1 - j(K1 + K2 ) N1 = 2d
and
A2 =
d K1 + j(K1 + K2 ) N2 = 2d A1 = A 2
We see from these expressions that
Problems P 8.15 By definition, B1 = A1 + A2 . From the solution to Problem 8.14 we have A1 + A2 = 2d K1 = K1 2d
815
But K1 is v(0), therefore, B1 = v(0), which is identical to Eq. (8.30). By definition, B2 = j(A1 - A2 ). From Problem 8.14 we have B2 = j(A1 - A2 ) = It follows that K2 = -K1 + d B2 , Thus we have dv + (0 ) = -B1 + d B2 , dt which is identical to Eq. (8.31). P 8.16 t<0: Vo = 15 V, Io = -60 mA but K2 = dv(0+ ) dt and K1 = B1 j[-2j(K1 + K2 )] K1 + K2 = 2d d
t > 0:
iR (0) =
15 = 150 mA; 100
iL (0) = -60 mA
iC (0) = -150 - (-60) = -90 mA = 1 1 = = 5000 rad/s 2RC 2(100)(10-6 )
816
CHAPTER 8. Natural and Step Responses of RLC Circuits
2 o =
1 1 = 16 106 = -3 )(10-6 ) LC (62.5 10 25 106 - 16 106 = -5000 3000 s2 = -8000 rad/s
s1,2 = -5000
s1 = -2000 rad/s;
.. vo = A1 e-2000t + A2 e-8000t A1 + A2 = vo (0) = 15 dvo -90 10-3 = -90,000 (0) = -2000A1 - 8000A2 = dt 10-6 Solving, A1 = 5 V, A2 = 10 V t0
.. vo = 5e-2000t + 10e-8000t V, P 8.17
2 o =
1 1 = = 16 106 LC (62.5 10-3 )(10-6 )
=
1 1 = = 2500 2RC 2(250)(10-6 ) 25002 - 16 106 = -2500 j3122.5rad/s
s1,2 = -2500
vo (t) = B1 e-2500t cos 3122.5t + B2 e-2500t sin 3122.5t vo (0) = B1 = 15 V iR (0) = 15 = 75 mA 200
iL (0) = -60 mA iC (0) = -iR (0) - iL (0) = -15 mA .. iC (0) = -15,000 V/s C
dvo (0) = -2500B1 + 3122.5B2 = -15,000 V/s dt .. 3122.5B2 = 2500(15) - 15,000 .. B2 = 7.21 V t0
vo (t) = 15e-2500t cos 3122.5t + 7.21e-2500t sin 3122.5t V,
Problems P 8.18
2 o =
817
1 1 = 16 106 = -3 )(10-6 ) LC (62.5 10
=
1 1 = = 4000 2RC 2(125)(10-6 )
2 .. 2 = o (critical damping)
vo (t) = D1 te-4000t + D2 e-4000t vo (0) = D2 = 15 V iR (0) = 15 = 120 mA 125
iL (0) = -60 mA iC (0) = -60 mA dvo (0) = -4000D2 + D1 dt -60 10-3 iC (0) = = -60,000 C 10-6 D1 - 4000D2 = -60,000; vo (t) = 15e-4000t V, P 8.19 t0 D1 = 0
vT = -2 104 i + 16 103 iT ;
i =
20 (-iT ) 100
= 4000it + 16,000iT = 20,000iT vT = 20 k iT
818
CHAPTER 8. Natural and Step Responses of RLC Circuits Vo = 3000 (50) = 30 V; 5000 Io = 0 30 = -1.5 mA 20,000
iC (0) = -iR (0) - iL (0) = -
-1.5 10-3 iC (0) = = -6000 C 0.25 10-6
2 o =
1 1 = = 105 LC (40)(0.25 10-6 )
=
1 1 = = 100 rad/s 3 )(0.25 10-6 ) 2RC 2(20 10 105 - 1002 = 300 rad/s
d =
vo = B1 e-100t cos 300t + B2 e-100t sin 300t vo (0) = B1 = 30 V dvo (0) = 300B2 - 100B1 = -6000 dt .. 300B2 = 100(30) - 6000; .. B2 = -10 V t0 t0
vo = 30e-100t cos 300t - 10e-100t sin 300t V, P 8.20 [a] v = L [b] iR = diL dt = 16[e-20,000t - e-80,000t ] V,
v = 40[e-20,000t - e-80,000t ] mA, t 0+ R [c] iC = I - iL - iR = [-8e-20,000t + 32e-80,000t ] mA, [a] v = L diL dt = 40e-32,000t sin 24,000t V, v - iL 625 t0
t 0+
P 8.21
[b] iC (t) = I - iR - iL = 24 10-3 -
= [24e-32,000t cos 24,000t - 32e-32,000t sin 24,000t] mA, P 8.22 v=L diL dt = 960,000te-40,000t V, t0
t 0+
Problems P 8.23 t<0: t > 0: iL = 9/3000 = 3 mA
819
6 k 3 k = 2 k
iL (0) = 3 mA,
2 o =
iL () = 9 mA o = 80 rad/s 2 = 104
1 1 = = 6400; LC (62.5)(2.5 10-6 )
=
1 1 = = 100; 2RC 2(2000)(2.5 10-6 )
2 2 - o = 104 - 6400 = 3600
s1,2 = -100 60 rad/s s1 = -40 rad/s; s2 = -160 rad/s
iL = If + A1 e-40t + A2 e-160t iL () = If = 9mA iL (0) = A1 + A2 + If = 3 mA .. A1 + A2 + 9 m = 3 m so A1 + A2 = -6 mA
diL (0) = 0 = -40A1 - 160A2 dt Solving, A1 = -8 mA, A2 = 2 mA t0
iL = 9 - 8e-40t + 2e-160t mA,
820 P 8.24
CHAPTER 8. Natural and Step Responses of RLC Circuits
2 o =
1 1 = 108 ; = -3 )(0.2 10-6 ) LC (50 10
o = 104 rad/s .. overdamped
=
1 1 = = 12,500 rad/s 2RC 2(200)(0.2 10-6 )
s1,2 = -12,500 s1 = -5000 rad/s; If = 60 mA
(12,500)2 - 108 = -12,500 7500 rad/s s2 = -20,000 rad/s
iL = 60 10-3 + A1 e-5000t + A2 e-20,000t .. -45 10-3 = 60 10-3 + A1 + A2 ; diL 15 = -5000A1 - 20,000A2 = = 300 dt 0.05 Solving, A1 = -120 mA; A2 = 15 mA t0 2 = 64 106 A1 + A2 = -105 10-3
iL = 60 - 120e-5000t + 15e-20,000t mA, P 8.25 =
1 1 = = 8000; 2RC 2(312.5)(0.2 10-6 ) underdamped
o = 104
s1,2 = -8000 j 80002 - 108 = -8000 j6000 rad/s iL = 60 10-3 + B1 e-8000t cos 6000t + B2 e-8000t sin 6000t -45 10-3 = 60 10-3 + B1 .. B1 = -105 mA
diL (0) = -8000B1 + 6000B2 = 300 dt .. B2 = -90 mA iL = 60 - 105e-8000t cos 6000t - 90e-8000t sin 6000t mA, t0
Problems P 8.26 = 1 1 = 104 = -6 ) 2RC 2(250)(0.2 10 critical damping
4 4 4 4t
821
2 2 = 108 = o
iL = If + D1 te-10 t + D2 e-10 t = 60 10-3 + D1 te-10 t + D2 e-10 iL (0) = -45 10-3 = 60 10-3 + D2 ; diL (0) = -104 D2 + D1 = 300 A/s dt .. D1 = 300 + 104 (-105 10-3 ) = -750 A/s iL = 60 - 750,000te-10 t - 105e-10 t mA, P 8.27 For t>0
4 4
.. D2 = -105 mA
t0
=
1 = 1000; 2RC
1 = 64 104 LC
s1,2 = -1000 600 rad/s s1 = -400 rad/s; s2 = -1600 rad/s
vo = Vf + A1 e-400t + A2 e-1600t Vf = 0; vo (0+ ) = 0; iC (0+ ) = 30 mA
.. A1 + A2 = 0 iC (0+ ) dvo (0+ ) = = 24,000 V/s dt 1.25 10-6 dvo (0+ ) = -400A1 - 1600A2 = 24,000 dt Solving, A1 = 20 V; A2 = -20 V t0
vo = 20e-400t - 20e-1600t V,
822 P 8.28
CHAPTER 8. Natural and Step Responses of RLC Circuits [a] From the solution to Prob. 8.27 s1 = -400 rad/s and s2 = -1600 rad/s, therefore io = If + A1 e-400t + A2 e-1600t If = 30 mA; io (0 ) = 0;
+
dio (0+ ) =0 dt -400A1 - 1600A2 = 0
.. 0 = 30 10-3 + A1 + A2 ; Solving A1 = -40 mA; A2 = 10 mA
.. io = 30 - 40e-400t + 10e-1600t mA, [b] dio = 16e-400t - 16e-1600t dt vo = L dio = 20e-400t - 20e-1600t V, dt
t0
t0
This agrees with the solution to Problem 8.27 P 8.29 = 1 1 = = 1000 2RC 2(400)(1.25 10-6 ) 1 1 = = 64 104 LC (1.25 10-6 )(1.25) 10002 - 64 104 = -1000 600 rad/s s2 = -1600 rad/s
2 o =
s1,2 = -1000
s1 = -400 rad/s; vo () = 0 = Vf
.. vo = A1 e-400t + A2 e-1600t vo (0) = 12 = A1 + A2 Note: .. iC (0+ ) = 0
dvo (0) = 0 = -400A1 - 1600A2 dt A1 = 16 V, A2 = -4 V t>0
Solving,
vo (t) = 16e-400t - 4e-1600t V,
Problems P 8.30 [a] io = If + A1 e-400t + A2 e-1600t If = 12 = 30mA; 400 io (0) = 0 .. A1 + A2 = -30 10-3
823
0 = 30 10-3 + A1 + A2 ,
12 dio (0) = = -400A1 - 1600A2 dt 1.25 Solving, A1 = -32 mA; A2 = 2 mA t0 io = 30 - 32e-400t + 2e-1600t mA, [b] dio = [12.8e-400t - 3.2e-1600t ] dt vo = L dio = 16e-400t - 4e-1600t V, dt t0
This agrees with the solution to Problem 8.29. P 8.31 iL (0- ) = iL (0+ ) = 37.5 mA For t > 0
iL (0- ) = iL (0+ ) = 37.5 mA = 1 = 100 rad/s; 2RC
2 o =
1 = 6400 LC
s1 = -40 rad/s vo () = 0 = Vf
s2 = -160 rad/s
vo = A1 e-40t + A2 e-160t iC (0+ ) = -37.5 + 37.5 + 0 = 0 .. dvo =0 dt
824
CHAPTER 8. Natural and Step Responses of RLC Circuits dvo (0) = -40A1 - 160A2 dt .. A1 + 4A2 = 0; .. A1 = 0; A1 + A2 = 0
A2 = 0
.. vo = 0 for t 0 Note: vo (0) = 0; vo () = 0; dvo (0) =0 dt
Hence the 37.5 mA current circulates between the current source and the ideal inductor in the equivalent circuit. In the original circuit the 7.5 V source sustains a current of 37.5 mA in the inductor. This is an example of a circuit going directly into steady state when the switch is closed. There is no transient period, or interval. P 8.32 t < 0: vo (0- ) = vo (0+ ) = 625 (25) = 20 V 781.25
iL (0- ) = iL (0+ ) = 0 t>0
-160 10-3 +
20 + iC (0+ ) + 0 = 0; 125
.. iC (0+ ) = 0
1 1 = 800 rad/s = 2RC 2(125)(5 10-6 )
2 o =
1 1 = = 64 104 -3 )(5 10-6 ) LC (312.5 10 critically damped
2 .. 2 = o
Problems [a] vo = Vf + D1 te-800t + D2 e-800t Vf = 0 dvo (0) = -800D2 + D1 = 0 dt vo (0+ ) = 20 = D2 D1 = 800D2 = 16,000 V/s .. vo = 16,000te-800t + 20e-800t V, [b] iL = If + D3 te-800t + D4 e-800t iL (0+ ) = 0; If = 160 mA; 20 diL (0+ ) = = 64 A/s dt 312.5 10-3 t 0+
825
.. 0 = 160 + D4 ; -800D4 + D3 = 64;
D4 = -160 mA; D3 = -64 A/s t0
.. iL = 160 - 64,000te-800t - 160e-800t mA P 8.33 [a] wL =
0
pdt =
0
vo iL dt
vo = 16,000te-800t + 20e-800t V iL = 0.16 - 64te-800t - 0.16e-800t A p = 3.2e-800t + 2560te-800t - 3840te-1600t -1,024,000t2 e-1600t - 3.2e-1600t W wL = 3.2
0
e-800t dt + 2560
0
0
te-800t dt - 3480
0
0
te-1600t dt
-1,024,000 e-800t = 3.2 -800
0
t2 e-1600t dt - 3.2
e-1600t dt
2560 -800t + e (-800t - 1) (800)2 0
3840 -1600t - e (-1600t - 1) (1600)2 0 -
1,024,000 -1600t e (16002 t2 + 3200t + 2) (-1600)3 0
0
e-1600t - 3.2 (-1600)
826
CHAPTER 8. Natural and Step Responses of RLC Circuits All the upper limits evaluate to zero hence wL = 3.2 3840 (1,024,000)(2) 3.2 2560 - - - + = 4 mJ 2 2 3 800 800 1600 1600 1600 Note this value corresponds to the final energy stored in the inductor, i.e. 1 wL () = (312.5 10-3 )(0.16)2 = 4 mJ. 2 [b] v = 16,000te-800t + 20e-800t V iR = v = 128te-800t + 0.16e-800t A 125
0
pR = viR = 2,048,000t2 e-1600t + 5120te-1600t + 3.2e-1600t wR = pR dt
0
= 2,048,000 =
t2 e-1600t dt + 5120
0
te-1600t + dt 3.2
0
0
e-1600t dt
2,048,000e-1600t [16002 t2 + 3200t + 2] -16003 5120e-1600t (-1600t - 1) 16002
0
+
0
+
3.2e-1600t (-1600)
Since all the upper limits evaluate to zero we have 2,048,000(2) 5120 3.2 = 5 mJ + + 16003 16002 1600 [c] 160 = iR + iC + iL (mA) wR = iR + iL = 160 + 64,000te-800t mA .. iC = 160 - (iR + iL ) = -64,000te-800t mA = -64te-800t A pC = viC = [16,000te-800t + 20e-800t ][-64te-800t ] = -1,024,000t2 e-1600t - 1280e-1600t wC = -1,024,000 wC =
0
t2 e-1600t dt - 1280
0
te-1600t dt
0
-1,024,000e-1600t [16002 t2 + 3200t + 2] -16003 -1,024,000(2) 1280(1) - = -1 mJ 16003 16002
-
1280e-1600t (-1600t - 1) 16002
0
Since all upper limits evaluate to zero we have wC =
Problems
827
Note this 1 mJ corresponds to the initial energy stored in the capacitor, i.e., 1 wC (0) = (5 10-6 )(20)2 = 1 mJ. 2 Thus wC () = 0 mJ which agrees with the final value of v = 0. [d] is = 160 mA ps (del) = 160v mW = 0.16[16,000te-800t + 20e-800t ] = 3.2e-800t + 2560te-800t W ws = 3.2 = = [e] wL = 4 mJ wR = 5 mJ wC = 1 mJ wS = 8 mJ
0
e-800t dt +
0
0
2560te-800t dt
0
3.2e-800t -800
+
2560e-800t (-800t - 1) 8002
2560 3.2 + = 8 mJ 800 800 (absorbed) (absorbed) (delivered) (delivered)
wdel = wabs = 9 mJ. P 8.34 vC (0+ ) = 3.75 103 (150) = 50 V 11.25 103 iL () = 150 = 20 mA 7500
iL (0+ ) = 100 mA; =
1 1 = = 800 2RC 2(2500)(0.25 10-6 ) 1 1 = = 106 LC (4)(0.25 10-6 )
2 2 < o ;
2 o =
2 = 64 104 ;
..
underdamped
s1,2 = -800 j 8002 - 106 = -800 j600 rad/s
828 iL
CHAPTER 8. Natural and Step Responses of RLC Circuits = = If + B1 e-t cos d t + B2 e-t sin d t 20 + B1 e-800t cos 600t + B2 e-800t sin 600t B1 = 100 m - 20 m = 80 mA
iL (0) = 20 10-3 + B1 ;
50 diL (0) = 600B2 - 800B1 = = 12.5 dt 4 .. 600B2 = 800(80 10-3 ) + 12.5; B2 = 127.5 mA t0
.. iL = 20 + 80e-800t cos 600t + 127.5e-800t sin 600t mA, P 8.35 [a] 2 = 5000; = 2500 rad/s
2 o = 4 106 ;
2 2 - o = 1500;
o = 2000 rad/s
=
R = 2500; 2L 1 = 4 106 ; LC
R = 5000L L= 109 = 5H 4 106 (50)
2 o =
R = 25,000 [b] i(0) = 0 L di(0) = vc (0); dt 1 2 (50) 10-9 vc (0) = 90 10-6 2 vc (0) = 60 V
2 .. vc (0) = 3600;
60 di(0) = = 12 A/s dt 5 [c] i(t) = A1 e-1000t + A2 e-4000t i(0) = A1 + A2 = 0 di(0) = -1000A1 - 4000A2 = 12 dt Solving, .. A1 = 4 mA; A2 = -4 mA t0
i(t) = 4e-1000t - 4e-4000t mA
Problems [d] di(t) = -4e-1000t + 16e-4000t dt di = 0 when 16e-4000t = 4e-1000t dt or e3000t = 4 .. t = ln 4 s = 462.10 s 3000 t 0+
829
[e] imax = 4e-0.4621 - 4e-1.8484 = 1.89 mA di [f] vL (t) = 5 = [-20e-1000t + 80e-4000t ] V, dt P 8.36 = 2000 rad/s; d = 1500 rad/s
2 o = 625 104 ;
2 o - 2 = 225 104 ;
wo = 2500 rad/s
=
R = 2000; 2L
R = 4000L L= 1 (625 104 )(80 10-9 ) = 2H
1 = 625 104 ; LC .. R = 8 k i(0+ ) = B1 = 7.5 mA;
at t = 0+
60 + vL (0+ ) - 30 = 0; di(0+ ) -30 = = -15 A/s dt 2
..
vL (0+ ) = -30 V
+ . di(0 ) = 1500B2 - 2000B1 = -15 . dt
.. 1500B2 = 2000(7.5 10-3 ) - 15; .. i = 7.5e-2000t sin 1500t mA, t0
.. B2 = 0 A
830 P 8.37
CHAPTER 8. Natural and Step Responses of RLC Circuits From Prob. 8.36 we know vc will be of the form vc = B3 e-2000t cos 1500t + B4 e-2000t sin 1500t From Prob. 8.36 we have vc (0) = -30 V = B3 and iC (0) dvc (0) 7.5 10-3 = = 93.75 103 = dt C 80 10-9 dvc (0) = 1500B4 - 2000B3 = 93,750 dt .. 1500B4 = 2000(-30) + 93,750; B4 = 22.5 V t0
vc (t) = -30e-2000t cos 1500t + 22.5e-2000t sin 1500t V P 8.38
2 [a] o =
1 1 = = 25 106 -3 )(0.5 10-6 ) LC (80 10 R = o = 5000 rad/s 2L
=
.. R = (5000)(2)L = 800 [b] i(0) = iL (0) = 30 mA vc (0) = 800i(0) + 80 10-3 di(0) dt
20 - 800(30 10-3 ) di(0) = 80 10-3 dt .. di(0) = -50 A/s dt
[c] vC = D1 te-5000t + D2 e-5000t vC (0) = D2 = 20 V -iL (0) iC (0) dvC (0) = D1 - 5000D2 = = dt C C D1 - 100,000 = - 30 10-3 = -60,000 0.5 10-6 .. D1 = 40,000 V/s
vC = 40,000te-5000t + 20e-5000t V,
t0
Problems P 8.39 [a] For t > 0:
831
Since i(0- ) = i(0+ ) = 0 va (0+ ) = 72 V [b] va = 5000i + 1 0.1 10-6
t 0
i dx + 72
di dva = 5000 + 10 106 i dt dt di(0+ ) di(0+ ) dva (0+ ) = 5000 + 10 106 i(0+ ) = 5000 dt dt dt -L di(0+ ) = 72 dt
72 di(0+ ) =- = -28.8 A/s dt 2.5
+ . dva (0 ) = -144,000 V/s . dt
[c] =
R 12,500 = = 2500 rad/s 2L 2(2.5) 1 1 = = 4 106 LC (2.5)(0.1 10-6 ) = -2500 25002 - 4 106 = -2500 1500 rad/s
2 o =
s1,2
Overdamped: va = A1 e-1000t + A2 e-4000t va (0) = 72 = A1 + A2 dva (0) = -144,000 = -1000A1 - 4000A2 dt Solving, A1 = 48; A2 = 24 t 0+ va = 48e-1000t + 24e-4000t V,
832 P 8.40
CHAPTER 8. Natural and Step Responses of RLC Circuits
2 o =
1 1 = 25 = LC (10)(4 10-3 ) 2 = 16
=
R 80 = = 4; 2L 2(10) ..
2 2 < o
underdamped
s1,2 = -4 j 9 = -4 j3 rad/s i = B1 e-4t cos 3t + B2 e-4t sin 3t i(0) = B1 = -240/100 = -2.4 A di (0) = 3B2 - 4B1 = 0 dt .. B2 = -3.2 A i = -2.4e-4t cos 3t - 3.2 sin 3t A, P 8.41 t < 0: t0
i(0) =
240 240 = = 6A 8 + 30 70 + 11 40 70 (6)(20) = 108 V 100
vo (0) = 240 - 8(6) -
Problems t > 0:
833
=
20 R = = 10, 2L 2(1)
2 = 100
2 o =
1 1 = 200 = LC (1)(5 10-3 ) underdamped 100 - 200 = -10 j10 rad/s
2 o > 2
s1,2 = -100
vo = B1 e-10t cos 10t + B2 e-10t sin 10t vo (0) = B1 = 108 V C dvo (0) = -6, dt dvo -6 = = -1200 V/s dt 5 10-3
dvo (0) = -10B1 + 10B2 = -1200 dt 10B2 = -1200 + 10B1 = -1200 + 1080; .. vo = 108e-10t cos 10t - 12e-10t sin 10t V, P 8.42 [a] t < 0: 80 = 100 mA; vo = 500io = (500)(0.01) = 50 V 800 t > 0: R 500 = = = 105 rad/s -3 ) 2L 2(2.5 10 io =
2 o =
B2 = -120/10 = -12 V t0
1 1 = = 100 108 LC (2.5 10-3 )(40 10-9 ) .. critically damped
2 2 = o
834
CHAPTER 8. Natural and Step Responses of RLC Circuits .. io (t) = D1 te-10 t + D2 e-10
5 5t
io (0) = D2 = 100 mA dio (0) = -D2 + D1 = 0 dt .. D1 = 105 (100 10-3 ) = 10,000
5 5
io (t) = 10,000te-10 t + 0.1e-10 t A, [b] vo (t) = D3 te-10 t + D4 e-10 vo (0) = D4 = 50 C dvo (0) = -0.1 dt
5 5t
t0
dvo -0.1 = -25 105 V/s = -D4 + D3 (0) = dt 40 10-9 .. D3 = 105 (50) - 25 105 = 25 105
5 5
vo (t) = 25 105 te-10 t + 50e-10 t V, P 8.43 = 8000 R = = 4000 rad/s 2L 2(1) 1 1 = = 20 106 LC (1)(50 10-9 )
t0
2 o =
s1,2 = -4000
40002 - 20 106 = -4000 j2000 rad/s
vo = Vf + B1 e-4000t cos 2000t + B2 e-4000t sin 2000t vo (0) = 0 = Vf + B1 vo () = 80 V; .. B1 = -80 V
dvo (0) = 0 = 2000B2 - 4000B1 dt .. 2000B2 = 4000(-80) .. B2 = -160 V t0
vo = 80 - 80e-4000t cos 2000t - 160e-4000t sin 2000t V,
Problems P 8.44 t < 0:
835
iL (0) =
-150 = -5 A 30
vC (0) = 18iL (0) = -90 V t > 0:
=
10 R = = 50 rad/s 2L 2(0.1) 1 1 = = 5000 LC (0.1)(2 10-3 ) .. underdamped 502 - 5000 = -50 j50
2 o =
o > 2
s1,2 = -50
vc = 60 + B1 e-50t cos 50t + B2 e-50t sin 50t vc (0) = -90 = 60 + B1 C dvc (0) = -5; dt .. B1 = -150
-5 dvc (0) = = -2500 dt 2 10-3 .. B2 = -200 t0
dvc (0) = -50B1 + 50B2 = -2500 dt
vc = 60 - 150e-50t cos 50t - 200e-50t sin 50t V,
836 P 8.45
CHAPTER 8. Natural and Step Responses of RLC Circuits iC (0) = 0; = vo (0) = 50 V
8000 R = = 25,000 rad/s 2L 2(160 10-3 ) 1 1 = = 625 106 -3 )(10 10-9 ) LC (160 10 critical damping
2 o =
2 .. 2 = o ;
vo (t) = Vf + D1 te-25,000t + D2 e-25,000t Vf = 250 V vo (0) = 250 + D2 = 50; D2 = -200 V
dvo (0) = -25,000D2 + D1 = 0 dt D1 = 25,000D2 - 5 106 V/s vo = 250 - 5 106 te-25,000t - 200e-25,000t V, P 8.46 [a] t < 0: t0
io (0- ) =
120 = 4 mA 30,000
vC (0- ) = 80 - (10,000)(0.004) = 40 V t = 0+ :
5 k 20 k = 4 k .. vo (0+ ) = -(0.004)(4000) + 40 = 40 - 16 = 24 V
Problems [b] vo (t) = vc - 4000io dvo + dvc + dio (0 ) = (0 ) - 4000 (0+ ) dt dt dt dvc + -4 10-3 (0 ) = = -2048 V/s dt (125/64) 10-6 -vL (0+ ) + vo (0+ ) + 40 = 0 vL = 64 V 64 dio + = 12.8 A/s (0 ) = dt 5 dvo + (0 ) = -2048 - 4000(12.8) = -53,248 V/s dt
2 [c] o =
837
1 1 = = 10.24 104 LC (5)[(125/64) 10-6 ] 4000 R = = 400 rad/s; 2L 2(5) overdamped 2 = 16 104
=
2 2 > o
s1,2 = -400 240 rad/s vo (t) = Vf + A1 e-160t + A2 e-640t Vf = vo () = -40 V -40 + A1 + A2 = 24 -160A1 - 640A2 = -53,248 Solving, A1 = -25.6; A2 = 89.6 t 0+ .. vo (t) = -40 - 25.6e-160t + 89.6e-640t V, P 8.47 [a] vc = Vf + [B1 cos d t + B2 sin d t] e-t dvc = [(d B2 - B1 ) cos d t - (B2 + d B1 ) sin d t]e-t dt Since the initial stored energy is zero, vc (0 ) = 0 and
+
dvc (0+ ) =0 dt and B2 = B1 d
It follows that
B1 = -Vf
838
CHAPTER 8. Natural and Step Responses of RLC Circuits When these values are substituted into the expression for [dvc /dt], we get dvc = dt But 2 + d Vf e-t sin d t d and
2 2 2 + d 2 + d = = o d d d 2 o V e-t sin d t d
Vf = V
Therefore [b]
dvc = dt
dvc = 0 when dt
sin d t = 0,
or
d t = n
where n = 0, 1, 2, 3, . . . Therefore t = [c] When tn = and n , d n d cos d tn = cos n = (-1)n
sin d t = sin n = 0
Therefore vc (tn ) = V [1 - (-1)n e-n/d ] [d] It follows from [c] that v(t1 ) = V + V e-(/d ) Therefore But = and vc (t3 ) = V + V e-(3/d )
e-(/d ) vc (t1 ) - V = -(3/ ) = e(2/d ) d vc (t3 ) - V e thus = 1 [vc (t1 ) - V ] ln Td [vc (t3 ) - V ] 2 3 - = ms 7 7 7
2 = t3 - t1 = Td , d ;
P 8.48
vc (t1 ) - V 1 ln Td vc (t3 ) - V
Td = t3 - t1 =
=
7000 63.84 ln 1000; 2 26.02
d =
2 = 7000 rad/s Td
2 2 o = d + 2 = 49 106 + 106 = 50 106
L=
1 = 200 mH; (50 106 )(0.1 10-6 )
R = 2L = 400
Problems P 8.49
839
[a] Let i be the current in the direction of the voltage drop vo (t). Then by hypothesis i = if + B1 e-t cos d t + B2 e-t sin d t if = i() = 0, i(0) = Vg = B1 R
Therefore i = B1 e-t cos d t + B2 e-t sin d t L di(0) = 0, dt therefore di(0) =0 dt
di = [(d B2 - B1 ) cos d t - (B2 + d B1 ) sin d t] e-t dt Therefore d B2 - B1 = 0; Therefore vo = L 2 Vg d Vg di =- L + sin d t e-t dt d R R LVg R Vg L R 2 + d sin d t e-t d
2 2 + d -t e sin d t d
B2 =
Vg B1 = d d R
=- =- vo = - [b]
Vg -t e sin d t V, RCd
t 0+
Vg dvo =- {d cos d t - sin d t}e-t dt d RC dvo = 0 when dt tan d t = d (smallest t)
Therefore d t = tan-1 (d /) t= P 8.50 d 1 tan-1 d -Vg -t e sin d t RCd
[a] From Problem 8.49 we have vo = =
R 4800 = 37,500 rad/s = 2L 2(64 10-3 ) 1 1 = = 3906.25 106 LC (64 10-3 )(4 10-9 )
2 o =
840
CHAPTER 8. Natural and Step Responses of RLC Circuits d =
2 o - 2 = 50 krad/s
-Vg -(-72) = 75 = RCd (4800)(4 10-9 )(50 103 ) .. vo = 75e-37,500t sin 50,000t V, [b] From Problem 8.49 td = d 1 tan-1 d = 50,000 1 tan-1 50,000 37,500 t0
td = 18.55 s [c] vmax = 75e-0.0375(18.55) sin[(0.05)(18.55)] = 29.93 V [d] R = 480 ; = 3750 rad/s d = 62,387.4 rad/s vo = 601.08e-3750t sin 62,387.4t V, td = 24.22 s vmax = 547.92 V P 8.51 [a] 1 d2 vo = vg dt2 R1 C1 R2 C2 1 10-6 = = 250 R1 C1 R2 C2 (100)(400)(0.5)(0.2) 10-6 10-6 d2 vo = 250vg .. dt2 0 t 0.5- : vg = 80 mV d2 vo = 20 dt2 Let
g(t) g(0)
t0
g(t) =
dvo , dt
t
then
dg = 20 or dt
dg = 20 dt
dx = 20
dy
0
g(t) - g(0) = 20t, g(t) = dvo = 20t dt
g(0) =
dvo (0) = 0 dt
Problems dvo = 20t dt
vo (t) vo (0)
841
dx = 20
t
x dx;
0
vo (t) - vo (0) = 10t2 ,
vo (0) = 0
vo (t) = 10t2 V,
0 t 0.5-
1 dvo1 =- vg = -20vg = -1.6 dt R1 C1 dvo1 = -1.6 dt
vo1 (t) vo1 (0)
dx = -1.6
t
dy
0
vo1 (t) - vo1 (0) = -1.6t, vo1 (t) = -1.6t V, 0.5+ t tsat : d2 vo = -10, dt2 dg(t) = -10; dt
g(t) g(0.5+ )
vo1 (0) = 0
0 t 0.5-
let
g(t) =
dvo dt
dg(t) = -10 dt
t
dx = -10
dy
0.5
g(t) - g(0.5+ ) = -10(t - 0.5) = -10t + 5 g(0.5+ ) = C dvo (0.5+ ) dt
0 - vo1 (0.5+ ) dvo (0.5+ ) = dt 400 103
vo1 (0.5+ ) = vo (0.5- ) = -1.6(0.5) = -0.80 V .. C 0.80 dvo (0.5+ ) = = 2 A dt 0.4 106
2 10-6 dvo + (0.5 ) = = 10 V/s dt 0.2 10-6 .. g(t) = -10t + 5 + 10 = -10t + 15 = .. dvo = -10t dt + 15 dt
vo (t) vo (0.5+ )
dvo dt
dx =
t 0.5+
-10y dy +
t 0.5+
15 dy
842
CHAPTER 8. Natural and Step Responses of RLC Circuits vo (t) - vo (0.5+ ) = -5y 2
t 0.5
+ 15y
t 0.5
vo (t) = vo (0.5+ ) - 5t2 + 1.25 + 15t - 7.5 vo (0.5+ ) = vo (0.5- ) = 2.5 V .. vo (t) = -5t2 + 15t - 3.75 V, dvo1 = -20(-0.04) = 0.8, dt dvo1 = 0.8 dt;
vo1 (t) vo1 (0.5+ )
0.5+ t tsat
0.5+ t tsat dx = 0.8
t 0.5+
dy
vo1 (t) - vo1 (0.5+ ) = 0.8t - 0.4; .. vo1 (t) = 0.8t - 1.2 V, Summary: 0 t 0.5- s : 0.5+ s t tsat : vo1 = -1.6t V,
vo1 (0.5+ ) = vo1 (0.5- ) = -0.8 V
0.5+ t tsat
vo = 10t2 V vo = -5t2 + 15t - 3.75 V
vo1 = 0.8t - 1.2 V,
[b] -12.5 = -5t2 + 15tsat - 3.75 sat .. 5t2 - 15tsat - 8.75 = 0 sat Solving, tsat = 3.5 sec
vo1 (tsat ) = 0.8(3.5) - 1.2 = 1.6 V P 8.52 1 = (106 )(0.5 10-6 ) = 0.50 s 1 = 2; 1 .. 2 = (5 106 )(0.2 10-6 ) = 1 s; .. 1 =1 2
dvo d2 vo + 2vo = 20 +3 2 dt dt
s2 + 3s + 2 = 0 (s + 1)(s + 2) = 0; s1 = -1, s2 = -2 20 = 10 V 2
vo = Vf + A1 e-t + A2 e-2t ; vo = 10 + A1 e-t + A2 e-2t
Vf =
Problems vo (0) = 0 = 10 + A1 + A2 ; .. A1 = -20, A2 = 10 V 0 t 0.5 s 0 t 0.5 s dvo (0) = 0 = -A1 - 2A2 dt
843
vo (t) = 10 - 20e-t + 10e-2t V, dvo1 + 2vo1 = -1.6; dt
.. vo1 = -0.8 + 0.8e-2t V,
vo (0.5) = 10 - 20e-0.5 + 10e-1 = 1.55 V vo1 (0.5) = -0.8 + 0.8e-1 = -0.51 V At t = 0.5 s
iC =
0 + 0.51 1.55 - 0 - = 0.954 A 400 103 5 106 0.954 dvo = = 4.773 V/s dt 0.2
C
dvo = 0.954 A; dt
t 0.5 s d2 vo dvo + 2vo = -10 +3 2 dt dt vo () = -5 .. vo = -5 + A1 e-(t-0.5) + A2 e-2(t-0.5) 1.55 = -5 + A1 + A2
844
CHAPTER 8. Natural and Step Responses of RLC Circuits dvo (0.5) = 4.773 = -A1 - 2A2 dt .. A1 + A2 = 6.55; Solving, A1 = 17.87 V; A2 = -11.32 V t 0.5 s -A1 - 2A2 = 4.773
.. vo = -5 + 17.87e-(t-0.5) - 11.32e-2(t-0.5) V, dvo1 + 2vo1 = 0.8 dt
.. vo1 = 0.4 + (-0.51 - 0.4)e-2(t-0.5) = 0.4 - 0.91e-2(t-0.5) V, P 8.53
t 0.5 s
At t = 0 the voltage across each capacitor is zero. It follows that since the operational amplifiers are ideal, the current in the 500 k is zero. Therefore there cannot be an instantaneous change in the current in the 1 F capacitor. Since the capacitor current equals C(dvo /dt), the derivative must be zero. [a] From Example 8.13 therefore d2 vo =2 dt2 g(t) = dvo dt g(0) = dvo (0) dt
P 8.54
dg(t) = 2, dt
g(t) - g(0) = 2t;
g(t) = 2t + g(0);
iR =
5 dvo (0) 10-3 = 1 A = iC = -C 500 dt
-1 10-6 dvo (0) = = -1 = g(0) dt 1 10-6 dvo = 2t - 1 dt dvo = 2t dt - dt vo - vo (0) = t2 - t; vo = t2 - t + 8, vo (0) = 8 V 0 t tsat
Problems [b] t2 - t + 8 = 9 t2 - t - 1 = 0 t = (1/2) ( 5/2) 1.62 s, = tsat 1.62 s =
845
(Negative value has no physical significance.) P 8.55 Part (1) -- Example 8.14, with R1 and R2 removed: [a] Ra = 100 k; 1 d2 vo = 2 dt Ra C1 vg = 250 10-3 ; [b] Since vo (0) = 0 = vo = 6 V, [c] or C1 = 0.1 F; 1 vg ; Rb C2 therefore Rb = 25 k; 1 = 100 Ra C1 d2 vo = 1000 dt2 vo = 500t2 V C2 = 1 F 1 = 40 Rb C2
dvo (0) , our solution is dt The second op-amp will saturate when tsat = 6/500 0.1095 s =
dvo1 1 vg = -25 =- dt Ra C1 [d] Since vo1 (0) = 0, vo1 = -25t V At t = 0.1095 s, vo1 -2.74 V =
Therefore the second amplifier saturates before the first amplifier saturates. Our expressions are valid for 0 t 0.1095 s. Once the second op-amp saturates, our linear model is no longer valid. Part (2) -- Example 8.14 with vo1 (0) = -2 V and vo (0) = 4 V: [a] Initial conditions will not change the differential equation; hence the equation is the same as Example 8.14. [b] vo = 5 + A1 e-10t + A2 e-20t (from Example 8.14) vo (0) = 4 = 5 + A1 + A2
846
CHAPTER 8. Natural and Step Responses of RLC Circuits 2 4 + iC (0+ ) - =0 100 25 iC (0+ ) = dvo (0+ ) 4 mA = C 100 dt
0.04 10-3 dvo (0+ ) = = 40 V/s dt 10-6 dvo = -10A1 e-10t - 20A2 e-20t dt dvo + (0 ) = -10A1 - 20A2 = 40 dt Therefore -A1 - 2A2 = 4 and Thus, A1 = 2 and A2 = -3 vo = 5 + 2e-10t - 3e-20t V, [c] Same as Example 8.14: dvo1 + 20vo1 = -25 dt [d] From Example 8.14: vo1 () = -1.25 V; Therefore vo1 = -1.25 + (-2 + 1.25)e-20t = -1.25 - 0.75e-20t V, P 8.56 [a] t0 v1 (0) = -2 V (given) A1 + A2 = -1
t0
2C
dva va - vg va + + =0 dt R R va vg dva + = ; dt RC 2RC va dvb + = 0, dt RC 0 - va d(0 - vb ) +C =0 R dt dvb dt
(1) Therefore (2) Therefore
va = -RC
Problems dvb d(vb - vo ) 2vb +C +C =0 R dt dt (3) Therefore dvb vb 1 dvo + = dt RC 2 dt d2 vb dva = -RC 2 dt dt and va = -RC dvb dt
847
From (2) we have
When these are substituted into (1) we get (4) - RC vg d2 vb dvb = - 2 dt dt 2RC
Now differentiate (3) to get (5) 1 d2 vo 1 dvb d2 vb = + dt2 RC dt 2 dt2 vg 1 dvb d2 vb =- 2 2 + 2 dt RC dt 2R C
But from (4) we have (6)
Now substitute (6) into (5) vg d2 vo =- 2 2 2 dt R C [b] When R1 C1 = R2 C2 = RC : d2 vo vg = 2 2 2 dt R C The two equations are the same except for a reversal in algebraic sign.
[c] Two integrations of the input signal with one operational amplifier. P 8.57 [a] f (t) = = [b] inertial force + frictional force + spring force M [d2 x/dt2 ] + D[dx/dt] + Kx dx K - x dt M then d2 x dy 2 vB dy = dy = - 1 dx R1 C1 dt
D f d2 x - = 2 dt M M Given vA = d2 x , dt2
t 0 t 0
1 vB = - R1 C1 vC = - vD = - 1 R2 C2
1 x R1 R2 C1 C2
R3 R3 dx vB = R4 R4 R1 C1 dt
848
CHAPTER 8. Natural and Step Responses of RLC Circuits vE = vF = R5 + R6 R5 + R6 1 vC = x R6 R6 R1 R2 C1 C2 -R8 f (t), R7 vA = -(vD + vE + vF )
Therefore
R8 R3 dx R5 + R6 d2 x - = f (t) - x dt2 R7 R4 R1 C1 dt R6 R1 R2 C1 C2 R7 , R8 D= R3 R7 R8 R4 R1 C1 and K= R7 (R5 + R6 ) R8 R6 R1 R2 C1 C2
Therefore M =
Box Number Function 1 2 3 4 5 6 P 8.58 inverting and scaling inverting and scaling integrating and scaling integrating and scaling inverting and scaling noninverting and scaling
[a] Given that the current response is underdamped we know i will be of the form i = If + [B1 cos d t + B2 sin d t]e-t where and = d = R 2L
2 o - 2 =
1 - 2 LC
The capacitor will force the final value of i to be zero, therefore If = 0. By hypothesis i(0+ ) = Vdc /R therefore B1 = Vdc /R. At t = 0+ the voltage across the primary winding is zero hence di(0+ )/dt = 0. From our equation for i we have di = [(d B2 - B1 ) cos d t - (d B1 + B2 ) sin d t]e-t dt Hence di(0+ ) = d B2 - B1 = 0 dt Thus Vdc B2 = B1 = d d R It follows directly that Vdc cos d t + sin d t e-t i= R d
Problems [b] Since d B1 - B1 = 0 it follows that di = -(d B1 + B2 )e-t sin d t dt But B2 = 2 Vdc d R and d B1 = d Vdc R
849
Therefore d B1 + B2 = But
2 Vdc d + 2 d Vdc 2 Vdc + = R d R R d
2 2 d + 2 = o =
1 LC
Hence d B1 + B2 = Now since v1 = -L Vdc d RLC di dt we get
v1 = L
Vdc Vdc -t e-t sin d t = - e sin d t d RLC d RC di dt sin d t e-t d Vdc -t e sin d t sin d t e-t + d d RC Vdc -t Vdc - e sin d t d RC d 1 d 1 - e-t sin d t RC
[c] vc = Vdc - iR - L
iR = Vdc cos d t +
vc = Vdc - Vdc cos d t +
= Vdc - Vdc e-t cos d t + = Vdc 1 - e-t cos d t +
= Vdc [1 - e-t cos d t + Ke-t sin d t] P 8.59 vsp = Vdc 1 - a e-t sin d t d RC
-aVdc d -t dvsp = [e sin d t] dt d RC dt = = -aVdc [-e-t sin d t + d cos d te-t ] d RC aVdc e-t [ sin d t - d cos d t] d RC
850
CHAPTER 8. Natural and Step Responses of RLC Circuits dvsp = 0 when dt or tan d t = d ; sin d t = d cos d t d t = tan-1 d
.. tmax =
d 1 tan-1 d
Note that because tan is periodic, i.e., tan = tan( n), where n is an integer, there are an infinite number of solutions for t where dvsp /dt = 0, that is t= tan-1 (d /) n d
Because of e-t in the expression for vsp and knowing t 0 we know vsp will be maximum when t has its smallest positive value. Hence tmax = P 8.60 tan-1 (d /) . d
[a] vc = Vdc [1 - e-t cos d t + Ke-t sin d t] dvc d = Vdc [1 + e-t (K sin d t - cos d t)] dt dt = Vdc {(-e-t )(K sin d t - cos d t)+ e-t [d K cos d t + d sin d t]} = Vdc e-t [(d - K) sin d t + ( + d K) cos d t] dvc = 0 when dt or tan d t = (d - K) sin d t = -( + d K) cos d t + d K K - d + d K K - d
.. d t n = tan-1 tc = =
+ d K 1 tan-1 n d K - d 4 103 R = = 666.67 rad/s 2L 6 109 - (666.67)2 = 28,859.81 rad/s 1.2
d =
Problems K= tc = 1 d 1 - = 21.63 RC
851
1 1 tan-1 (-43.29) + n = {-1.55 + n} d d
The smallest positive value of t occurs when n = 1, therefore tc max = 55.23 s [b] vc (tc max ) = 12[1 - e-tc max cos d tc max + Ke-tc max sin d tc max ] = 262.42 V [c] From the text example the voltage across the spark plug reaches its maximum value in 53.63 s. If the spark plug does not fire the capacitor voltage peaks in 55.23 s. When vsp is maximum the voltage across the capacitor is 262.15 V. If the spark plug does not fire the capacitor voltage reaches 262.42 V. P 8.61 1 1 [a] w = L[i(0+ )]2 = (5)(16) 10-3 = 40 mJ 2 2 3 R 3 10 = 300 rad/s [b] = = 2L 10 d = 109 - (300)2 = 28,282.68 rad/s 1.25
1 106 4 106 = = RC 0.75 3 tmax = d 1 tan-1 d = 55.16 s
vsp (tmax ) = 12 -
12(50)(4 106 ) -tmax e sin d tmax = -27,808.04 V 3(28,282.68)
[c] vc (tmax ) = 12[1 - e-tmax cos d tmax + Ke-tmax sin d tmax ] K= 1 1 - = 47.13 d RC
vc (tmax ) = 568.15 V