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Unformatted text preview: Chapter 9  Complete Response of Circuits with Two Energy
Storage Elements
Exercises
Apply KVL to right mesh: Ex. 9.31 di ( t )
+ v ( t ) + 1( i ( t ) −is ( t ) ) = 0
dt
di ( t )
⇒ v ( t ) = −2
− ( i ( t )−is ( t ) )
dt
2 The capacitor current and voltage are related by
i (t ) = 1 dis ( t ) 1 di ( t ) d 2i ( t )
1 dv ( t ) 1 d di ( t )
=
−2
−i ( t )+ is ( t ) =
−
− dt
dt 2
2 dt
2 dt 2 dt 2 dt ∴ d 2i ( t ) 1 di ( t )
1 dis ( t )
+
+ i( t ) =
2
dt
2 dt
2 dt
The inductor voltage is related to the inductor
current by
di ( t )
v (t ) = 1
dt Ex. 9.32 Apply KCL at the top node:
is ( t ) =
Using the operator s = v (t )
1 dv ( t )
+ i (t ) +
1
2 dt d
we have
dt
v (t ) = s i (t ) v (t ) 1 + sv ( t ) ⇒ is ( t ) = v ( t ) +
1
s
2
is ( t ) = v ( t ) + i ( t ) + sv ( t ) 2
Therefore 2s is ( t ) = 2 s v ( t ) + 2 v ( t ) + s 2 v ( t ) ⇒ d 2v ( t )
dv ( t )
d i (t )
+2
+ 2 v (t ) = 2 s
2
dt
dt
dt 91 Ex. 9.33 Using the operator s = d
, apply KVL to the
dt left mesh:
i1 ( t ) + s ( i1 ( t ) −i 2 ( t ) ) = vs ( t ) Apply KVL to the right mesh:
1
2 i 2 ( t ) + 2 i 2 ( t ) + s ( i 2 ( t ) −i1 ( t ) ) = 0
s
1
2
i1 ( t ) = 2 i 2 ( t ) + 2 i 2 ( t ) + i 2 ( t )
s
s
Combining these equations gives:
3s i2 ( t ) + 4si2 ( t ) + 2i2 ( t ) = s vs ( t )
2 2 or Ex. 9.41 d 2i2 ( t )
di2 ( t )
d 2 vs ( t )
3
+4
+ 2i2 ( t ) =
dt 2
dt
dt 2 Using the operator s =
top node:
i s (t ) = d
, apply KCL at the
dt v (t )
1
+ i (t ) + s v (t )
4
4 Apply KVL to the rightmost mesh:
v (t ) − ( s i (t ) + 6 i (t )) = 0 Combining these equations gives:
s 2 i ( t ) + 7 s i ( t ) + 10 i ( t ) = 4 i s ( t ) The characteristic equation is: s 2 + 7 s + 10 = 0 .
The natural frequencies are: s = −2 and s = −5 . Ex. 9.42 Assume zero initial conditions . Write mesh
d
equations using the operator s =
:
dt
1
s i1 ( t ) − i 2 ( t ) + 7 + 10 i1 ( t ) − 10 = 0 2
and
1 v ( t ) − 7 − s i1 ( t ) − i 2 ( t ) = 0
2 92 Now 0.005 s v ( t ) = i 2 ( t ) ⇒ v ( t ) = 200 200 i2 (t )
s i2 (t )
s so the second mesh equation becomes: 1 − 7 − s i1 ( t ) − i 2 ( t ) = 0
2 Writing the mesh equation in matrix form:
s 10 + 2 −1 s
2 1 s i1 ( t ) 3 2 =
1
200 i 2 ( t ) 7 s+ 2
s
− Obtain the characteristic equation by calculating a determinant:
10 +
− 1
s
2
= s 2 + 20s + 400 = 0 ⇒ s1, 2 = −10 ± j 17.3
1
200
s+
2
s s
2 − 1
s
2 Ex. 9.51 After t = 0 , we have a parallel RLC circuit with
1
1
7
1
1
2
=
=
=
=6
α=
and ω o =
2 RC
2(6)(1 / 42) 2
LC
(7)(1 / 42)
2 7 − 6 = −1, − 6 2
dv ( t )
∴ vn (t ) = A1e −t + A2 e−6t . We need vn (0) and n
dt
7
∴ s 1 , s 2 = −α ± α −ω = − ±
2
2 2
o to evaluate A1 & A2 . t =0 At t = 0+ we have:
iC ( 0+ ) = −10 A ⇒ dv ( t )
10
=
= 420 V s
1
dt t =0+
42 Then
vn (0+ ) = 0 = A1 + A2 A1 = −84 , A2 = 84
dvn
=−420= − A1 − 6 A2 dt t =0+ 93 Finally
∴ vn (t ) = −84e −t + 84e −6t V 1
1
= 0 ⇒ s 2 + 40s + 100 = 0
s+
RC
LC
Therefore
s 1 , s 2 = −2.68 , −37.3 Ex. 9.52 s2 + vn ( t ) = A1 e −2.68t + A2 e−37.3t , v(0) = 0 = A1 + A2
v (0 + )
1 dv(0+ )
+
KCL at t = 0 yields
+ i (0 ) +
= 0 so
1
40 dt
dv(0+ )
= − 40 v(0+ ) − 40 i (0+ ) = − 40(0) − 40(1) = − 2.7 A1 − 37.3 A2
dt
Therefore: A1 = −1.16 , A2 = 1.16 ⇒ v(t ) = vn (t ) = −1.16e −2.68t + 1.16e−37.3t
+ Ex. 9.61 For parallel RLC circuits: α = 1
1
1
1
2
=
= 50, ω o =
=
= 2500
−3
LC (0.4)(10−3 )
2 RC 2(10)(10 ) The roots of the characteristic equations are:∴ s1, 2 = −50 ± (50) 2 − 2500 = −50, −50 The natural response is vn (t ) = A1 e−50t + A2 t e−50t .
At t = 0+ we have:
−v(0+ )
ic (0 ) =
= − . 8V
10Ω
dv( t )
i (0+ )
∴
=c
= −800 V / s
dt
c
+
t =0
+ 94 So vn (0+ ) = 8 = A1 ⇒ vn (t ) = 8e−50t + A2 t e−50t
dv(0+ )
= −800 = − 400 + A2 ⇒ A2 = − 400
dt
∴ vn (t ) = 8 e 50t − 400 t e−50t V 1
1
=
= 8000
2 RC 2(62.5)(10−6 )
1
1
ω o2 =
=
= 1 08
−6
LC (.01)(10 ) α= Ex. 9.71 2
∴ s = − α ± α 2 −ω o = − 8000 ± (8000) 2 −108 = − 8000 ± j 6000 ∴ vn (t ) = e −8000t [ A1 cos 6000 t + A2 sin 6000 t ]
at t = 0+ 10
+ ic (0+ ) = 0
62.5
⇒ ic (0+ ) = −.24 A 0.08 + ∴ dv(0+ ) ic (0+ )
=
= −2.4 ×10+5 V/s
dt
C dvn (0+ )
vn (0 ) = 10 = A1 and
= −2.4 × 105 = 6000 A2 − 8000 A1 ⇒ A2 = −26.7
dt
+ ∴ vn (t ) = e−8000t [10 cos 6000 t − 26.7 sin 6000 t ] V Ex. 9.81
d 2v ( t )
dv ( t )
+5
+ 6 v ( t ) = vs ( t ) so the characteristic equation is
2
dt
dt
s 2 + 5 s + 6 = 0 . The roots are s 1 , s 2 = −2, − 3 . The differential equation is d 2v ( t )
dv ( t )
(a)
+5
+ 6 v ( t ) = 8 . Try v f ( t ) = B . Substituting into the differential equation gives
2
dt
dt
6 B = 8 ∴ v f (t ) = 8 / 6 V . 95 (b) d 2v ( t )
dt 2 +5 dv ( t )
dt + 6 v ( t ) = 3 e − 4 t . Try v f ( t ) = B e− 4 t . Substituting into the differential 3
3
equation gives (−4) 2 B + 5(−4) B + 6 B = 3 ⇒ B = . ∴ v f ( t ) = e−4t .
2
2
2
d v (t )
dv ( t )
(c)
+5
+ 6 v ( t ) = 2 e − 2 t . Try v f ( t ) = B t e− 2 t because –2 is a natural frequency.
2
dt
dt
Substituting into the differential equation gives
( 4t − 4) B + 5 B (1 − 2t ) + 6 Bt = 2 ⇒ B = 2. ∴ v f ( t ) = 2 t e −2t . Ex. 9.82
d 2i ( t ) di ( t ) + 20 i ( t ) = 36 + 12 t . Try i f ( t ) = A + B t . Substituting into the differential
dt 2
dt
equation gives 0 + 9 B + 20( A + Bt ) = 36 + 12t ⇒ B = 0.6 and A = 1.53.
+9 ∴ i f ( t ) = 1.53 + 0.6 t A Ex. 9.91 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions vC ( ∞ ) = R2
R1 + R 2 1 Next, represent the circuit by a 2nd order differential equation:
KCL at the top node of R2 gives:
KVL around the outside loop gives: vC ( t )
R2
vs ( t ) = L +C d
vC ( t ) = iL ( t )
dt d
iL ( t ) + R1 iL ( t ) + vC ( t )
dt Use the substitution method to get 96 vs ( t ) = L v (t ) d vC ( t )
d
d
+ C vC ( t ) + R1 C
+ C vC ( t ) + vC ( t ) R2 dt R 2
dt
dt = LC L
d R1 d2
+ R1 C vC ( t ) + 1 +
vC ( t ) + vt R2 dt R2 C ( ) dt 2 (a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω
Use the steady state response as the forced response:
R2
1
1= V
v f = vC ( ∞ ) =
2
R1 + R 2
The characteristic equation is
R1 1+ 1
R1 R2 2
2
s +
+
s+
= s + 6s + 8 = ( s + 2 ) ( s + 4 ) = 0 R 2 C L LC so the natural response is
vn = A1 e −2 t + A2 e −4 t V
The complete response is
vc ( t ) =
iL ( t ) =
+ vC ( t )
1.309 + 1
+ A1 e −2 t + A2 e−4 t V
2 d
vC ( t ) = −1.236 A1 e −2 t − 3.236 A2 e−4 t + 0.3819
dt At t = 0 () 0 = vc 0+ = A1 + A2 + 0.5 ( ) = −1.236 A − 3.236 A 0 = iL 0 + 1 2 + 0.3819 Solving these equations gives A1 = −1 and A2 = 0.5, so
vc ( t ) = 1 −2 t 1 −4 t
−e + e V
2
2 (b) C = 1 F, L = 1 H, R1 = 3 Ω, R2 = 1 Ω
Use the steady state response as the forced response:
R2
1
v f = vC ( ∞ ) =
1= V
R1 + R 2
4
The characteristic equation is 97 R1 1+ 1
R1 R2 2
2
2
s +
+
s+
= s + 4s + 4 = ( s + 2 ) = 0 R 2 C L LC so the natural response is
vn = ( A1 + A2 t ) e −2 t V The complete response is
vc ( t ) =
iL ( t ) = vC ( t ) +
At t = 0+ 1
+ ( A1 + A2 t ) e −2 t V
4 d
1
vC ( t ) = +
dt
4 () (( A 0 = vc 0+ = A1 + () 0 = iL 0+ = 2 ) − A1 ) − A2 t e −2 t 1
4 1
+ A2 − A1
4 Solving these equations gives A1 = −0.25 and A2 = −0.5, so
vc ( t ) = 1 1 1 −2 t
− + t e V
4 4 2 (c) C = 0.125 F, L = 0.5 H, R1 = 1 Ω, R2 = 4 Ω
Use the steady state response as the forced response:
R2
4
v f = vC ( ∞ ) =
1= V
R1 + R 2
5
The characteristic equation is
R1 1+ 1
R1 R2 2
2
s +
+
s+
= s + 4 s + 20 = ( s + 2 − j 4 ) ( s + 2 + j 4 ) = 0 R 2 C L LC so the natural response is
vn = e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
The complete response is iL ( t ) = vc ( t ) = 0.8 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
vC ( t )
4 + A 2 −2 t
A1
1d
vC ( t ) = 0.2 +
e cos 4 t − e−2 t sin 4 t
8 dt
2
2 98 At t = 0+ () 0 = vc 0+ = 0.8 + A1 () 0 = iL 0+ = 0.2 + A2
2 Solving these equations gives A1 = 0.8 and A2 = 0.4, so
vc ( t ) = 0.8 − e −2 t ( 0.8 cos 4 t + 0.4 sin 4 t ) V Ex 9.92 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
vC ( ∞ ) = R2
R1 + R 2 1, iL ( ∞ ) = 1
R1 + R 2 and vo ( ∞ ) = R2
R1 + R 2 1 Next, represent the circuit by a 2nd order differential equation:
KVL around the righthand mesh gives:
KCL at the top node of the capacitor gives: d
iL ( t ) + R 2 iL ( t )
dt
vs ( t ) − vC ( t )
d
− C vC ( t ) = iL ( t )
R1
dt
vC ( t ) = L Use the substitution method to get
vs ( t ) = R1 C d d
d L iL ( t ) + R 2 iL ( t ) + L iL ( t ) + R 2 iL ( t ) + R1 iL ( t )
dt dt dt d2
d
= R1 LC 2 iL ( t ) + ( L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t )
dt
dt Using iL ( t ) = vo ( t )
gives
R2 L
d R1 + R 2 d2
vs ( t ) =
LC 2 vo ( t ) + + R1 C vo ( t ) + vt R2 dt R2 o ( ) R2
dt R1 99 (a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω
Use the steady state response as the forced response:
R2
1
1= V
v f = vo ( ∞ ) =
2
R1 + R 2
The characteristic equation is
R2 1+ 1
R2 R1 2
2
s +
+
s+
= s + 6s + 8 = ( s + 2 ) ( s + 4 ) = 0 R1 C L LC so the natural response is
vn = A1 e −2 t + A2 e −4 t V
The complete response is
1
+ A1 e −2 t + A2 e −4 t V
2
A1 −2 t
A2 −4 t
v (t )
1
iL ( t ) = o
e+
eV
=
+
1.309 2.618 1.309
1.309
vo ( t ) = vC ( t ) = 1.309 iL ( t ) +
At t = 0+ 1d
1
iL ( t ) = + 0.6180 A1 e −2 t + 0.2361 A2 e −4 t
4 dt
2 () 0 = iL 0+ = () 0 = vC 0+ = A1
A2
1
+
+
2.618 1.309 1.309 1
+ 0.6180 A1 + 0.2361 A2
2 Solving these equations gives A1 = −1 and A2 = 0.5, so
vo ( t ) = 1 −2 t 1 −4 t
−e + e V
2
2 (b) C = 1 F, L = 1 H, R1 = 1 Ω, R2 = 3 Ω
Use the steadystate response as the forced response:
R2
3
v f = vo ( ∞ ) =
1= V
R1 + R 2
4
The characteristic equation is
R2 1+ 1
R2 R1 2
2
s2 + +
s+
= s + 4s + 4 = ( s + 2 ) = 0 R1 C L LC 910 so the natural response is vn = ( A1 + A2 t ) e −2 t V The complete response is
vo ( t ) =
iL ( t ) = vo ( t )
3 vC ( t ) = 3 iL ( t ) +
At t = 0+ 3
+ ( A1 + A2 t ) e −2 t V
4
= 1 A1 A2 −2 t
t e V
+ +
4 3
3 3 A1 A2 A2 −2 t
d
iL ( t ) = + +
t e
+
4 3
3 3 dt () 0 = iL 0+ = () 0 = vC 0+ A1 + 1
4 3
3 A1 A2
=+ +
43
3 Solving these equations gives A1 = 0.75 and A2 = 1.5, so
vo ( t ) = 3 3 3 −2 t
− + t e V
4 4 2 (c) C = 0.125 F, L = 0.5 H, R1 = 4 Ω, R2 = 1 Ω
Use the steady state response as the forced response:
R2
1
v f = vo ( ∞ ) =
1= V
R1 + R 2
5
The characteristic equation is
R2 1+ 1
R2 R1 2
2
s +
+
s+
= s + 4 s + 20 = ( s + 2 − j 4 ) ( s + 2 + j 4 ) = 0 R1 C L LC so the natural response is
vn = e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
The complete response is
vo ( t ) = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
iL ( t ) = vo ( t )
1 = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V 911 vC ( t ) = iL ( t ) +
At t = 0+ 1d
iL ( t ) = 0.2 + 2 A2 e−2 t cos 4 t − 2 A1 e−2 t sin 4 t
2 dt ()
( 0 ) = 0.2 + 2 A 0 = iL 0+ = 0.2 + A1
0 = vC + 2 Solving these equations gives A1 = 0.8 and A2 = 0.4, so
vc ( t ) = 0.2 − e −2 t ( 0.2 cos 4 t + 0.1sin 4 t ) V Ex. 9.101 At t = 0+ no initial stored energy ⇒ v1 (0+ ) = v2 (0+ ) = i (0+ ) = 0 3 di (0+ )
di (0+ )
+0=0 ⇒
=0
10 dt
dt
0
dv1 (0+ )
+
⇒
=0
KCL at A : + i1 (0 ) + 0 = 0
1
dt
5 dv2 (0+ )
= 10 ⇒
KCL at B : − 0 + i2 (0+ ) − 10 = 0 ⇒ i2 (0+ ) =
6
dt KVL : − 0 + For t > 0: dv2 (0+ )
= 12 V s
dt v1 1 d v1
+
+i = 0
1 12 dt
5 d v2
= 10
KCL at B : − i +
6 dt
3 di
− v1 +
+ v2 = 0
KVL:
10 dt
Eliminating i yields
1 d v1 5 d v2
v1 +
+
− 10 = 0
12 dt 6 dt
3 5 d 2 v2 −v1 + + v2 = 0
10 6 dt 2 KCL at A : Next
912 v1 = v2 + 1 d 2 v2
4 dt 2 ⇒ d v1 d 2 v2 1 d 3 v2
=
+
dt
dt 2
4 dt 3 Now, eliminating v1
v2 + 1 d 2 v2 1 d v2 1 d 3 v2 5 d v2
+
+
= 10
+
4 dt 2 12 dt 4 dt 3 6 dt Finally, the circuit is represented by the differential equation:
d 3 v2
d2 v
dv
+ 12 2 2 + 44 2 + 48v2 = 480
3
dt
dt
dt
The characteristic equation is s 3 + 12s 2 + 44s + 48 = 0 . It’s roots are s1, 2,3 = −2, −4, −6 . The natural response is
vn = A1e −2t + A2 e −4t + A3e −6t
Try v f = B as the forced response. Substitute into the differential equation and equate
coefficients to get B = 10. Then
v2 (t ) = vn (t ) + v f (t ) = A1e −2t + A2 e −4t + A3e−6t + 10
We have seen that v2 (0+ ) = 0 and dv2 (0+ )
d 2 v2 (0+ )
= 12 V/s . Also
= 4[v1 (0+ ) − v2 (0+ )] = 0 .
dt
dt 2 Then
v2 (0+ ) = 0 = A1 + A2 + A3 + 10
dv2 (0+ )
= 12 = −2 A1 − 4 A2 − 6 A3
dt
d 2 v2 (0+ )
= 0 = 4 A1 + 16 A2 + 36 A3
dt 2 Solving these equations yields A1 = −15, A2 = 6, A3 = −1 so
v2 (t ) = ( −15 e −2t + 6 e −4t − e −6t + 10 ) V Ex. 9.111
1
1
s2 +
s+
=0
RC
LC In our case L = 0.1, C = 0.1 so we have s 2 + 10
s + 100 = 0
R 913 a)
R = 0.4 Ω ⇒ s 2 + 25s + 100 = 0
s1,2 = −5, − 20 b)
R = 1 Ω ⇒ s 2 + 10 s + 100 = 0
s1,2 = − 5 ± j 5 3 914 Problems
Section 93: Differential Equations for Circuits with Two Energy Storage Elements
P9.31 KCL: iL ( t ) = v (t )
dv ( t )
+C
R2
dt KVL: vs ( t ) = R1iL ( t ) + L diL ( t )
+ v (t )
dt dv ( t ) L dv ( t )
d 2v ( t ) v (t )
vs ( t ) = R1 +C
+ LC
+ v (t )
+
dt R2 dt
dt 2 R2
d 2v ( t )
R L dv ( t )
= 1 + 1 v ( t ) + R1C + + [ LC ]
vs ( t )
R2 dt
dt 2 R2 In this circuit R1 = 2 Ω, R 2 = 100 Ω, L = 1 mH, C = 10 µ F so
2
dv ( t )
−8 d v ( t )
vs ( t ) = 1.02v ( t ) + .00003
+ 10
dt
dt 2
dv ( t ) d 2 v ( t )
8
8
10 vs ( t ) = 1.02 × 10 v ( t ) + 3000
+
dt
dt 2 P9.32 Using the operator s =
KCL: is ( t ) = d
we have
dt v (t )
+ iL ( t ) + Csv ( t )
R1 KVL: v ( t ) = R2iL ( t ) + LsiL ( t ) Solving by usingCramer's rule for iL ( t ) : iL ( t ) = is ( t )
R2 Ls
+
+ R2Cs + LCs 2 + 1
R1 R1 R2 L 2
1 + iL ( t ) + + R2C siL ( t ) + [ LC ] s iL ( t ) = is ( t )
R1 R1 915 In this circuit R1 = 100 Ω, R 2 = 10 Ω, L = 1 mH, C = 10 µ F so
1.1iL ( t ) + .00011siL ( t ) + 10−8 s 2iL ( t ) = is ( t ) diL ( t ) d 2iL ( t )
+
= 108 is ( t )
1.1× 10 iL ( t ) + 11000
2
dt
dt
8 P9.33 After the switch closes, a source
transformation gives: KCL:
iL ( t ) + C dvc ( t ) vs ( t ) + vc ( t )
+
=0
dt
R2 KVL: diL ( t )
− vc ( t ) − vs ( t ) = 0
dt
di ( t )
vc ( t ) = R1is ( t ) + R1iL ( t ) + L L
− vs ( t )
dt
Differentiating
R1is ( t ) + R1iL ( t ) + L d vc ( t )
d is ( t )
d iL ( t )
d 2iL ( t ) d vs ( t )
= R1
+ R1
+L
−
dt
dt
dt
dt 2
dt
Then d is ( t )
d iL ( t )
d 2iL ( t ) d vs ( t ) vs ( t )
+ R1
+L
−
iL ( t ) + C R1
+
dt
dt
dt 2
dt R2 di ( t ) 1
+ R1is ( t ) + R1iL ( t ) + L L
− vs ( t ) = 0
R2 dt Solving for i L ( t ) : d 2i L ( t ) R1
− R1
R1 di s ( t ) 1 dv s ( t )
1 di L ( t ) R1
1
+ +
+
+
i s (t ) −
+ i L (t ) =
2
dt
LC R 2
L dt
L dt L R 2C dt LR 2 C LC 916 Section 94: Solution of the Second Order Differential Equation  The Natural Response
P9.41 From Problem P 9.32 the characteristic equation is:
1.1×108 +11000 s + s 2 = 0 ⇒ s1 , s2 = −11000± (11000) 2 − 4(1.1×108 )
= −5500± j8930
2 P9.42 KVL: 40 ( is ( t ) − iL ( t ) ) = (100 × 10−3 ) diL ( t )
+ vc ( t )
dt The current in the inductor is equal to the current in
the capacitor so
1 dv ( t )
iL ( t ) = ×10−3 c
3 dt
2 40 dv ( t ) 100 d vC ( t )
vC ( t ) = 40 is ( t ) − × 10−3 C − × 10−6 2
3 dt
3 dt
d 2 vC ( t )
dv ( t )
+ 400 C + 30000 vC ( t ) = 40 is ( t )
2
dt
dt
2
s + 400s + 30000 = 0 ⇒ ( s + 100)( s + 300) = 0 ⇒ s1 = −100, s2 = −300 P9.43 v ( t ) − vs ( t )
dv ( t )
+ iL ( t ) + (10 × 10−6 )
=0
1
dt
di ( t )
KVL: v ( t ) = 2iL ( t ) + (1× 10−3 ) L
dt
KCL: 0 = 2 iL ( t ) + (1× 10−3 )
vs ( t ) = 3iL ( t ) + 0.00102 diL ( t )
dt
diL ( t )
dt − vs ( t ) + iL ( t ) + (10 × 10−6 ) ( 2 ) + 10−8 d 2iL ( t )
dt 2 ⇒ d 2iL ( t )
dt diL ( t )
dt + (10 × 10−6 ) (10−3 ) + 102000 diL ( t )
dt d 2iL ( t )
dt + 3 × 108 iL ( t ) = 108 vs ( t ) s 2 + 102000s + 3 ×108 = 0 ⇒ s1 = 3031, s2 = −98969 917 Section 9.5: Natural Response of the Unforced Parallel RLC Circuit
P9.51 dv( 0 )
d
= −3000 V/s . Using the operator s = , the node
dt
dt
− vs ( t )
v( t ) v( t )
L +
= 0 or L C s 2 + s + 1 v ( t ) = v s ( t )
equation is Csv ( t ) +
R
sL
R 1
1
s+
= 0 ⇒ s 2 + 500 s + 40, 000 = 0
The characteristic equation is: s 2 +
RC
LC The initial conditions are v ( 0 ) = 6 V, The natural frequencies are: s1,2 = −250 ± 2502 − 40, 000 = −100, −400
The natural response is of the form v ( t ) = Ae−100t + Be−400t . We will use the initial conditions
to evaluate the constants A and B.
v ( 0) = 6 = A + B ⇒
dv ( 0 )
= −3000 = −100 A − 400 B dt A = −2 and B = 8 Therefore, the natural response is
v ( t ) = −2e−100t + 8e−400t t >0 P9.52
The initial conditions are v ( 0 ) = 2 V, i ( 0 ) = 0 . 1
1
s+
= 0 ⇒ s 2 + 4s + 3 = 0
RC
LC
The natural frequencies are: s1 , s2 = −1, − 3
The natural response is of the form v ( t ) = Ae − t + Be −3t . We will use the initial conditions to
evaluate the constants A and B.
dv ( t )
= − Ae − t − 3Be −3t . At t = 0 this becomes
Differentiating the natural response gives
dt
dv ( t ) v ( t )
dv ( t )
v (t ) i (t )
dv ( 0 )
+
+ i ( t ) = 0 or
=−
−
.
= − A − 3B . Applying KCL gives C
dt
R
dt
RC
C
dt
dv ( 0 )
v ( 0) i ( 0)
=−
At t = 0 this becomes
−
. Consequently
dt
RC
C
The characteristic equation is: s 2 + −1A − 3B = − v ( 0) i ( 0)
2
−
=−
−0 = − 8
RC
C
14
918 Also, v ( 0 ) = 2 = A + B . Therefore A = −1 and B = 3. The natural response is
v ( t ) = −e − t + 3e −3t V P9.53
di1 ( t )
di ( t )
−3 2
=0
dt
dt
di ( t )
di ( t )
KVL : − 3 1
+3 2
+ 2i2 ( t ) = 0
dt
dt
KVL : i1 + 5 (1)
( 2) d
, the KVL equations are
dt (1+5s )i1 + ( −3s )i2 = 0 3s
2
i1 = 0 ⇒ (1 + 5s ) ( 3s + 2 ) − ( 3s ) i1 = 0 ⇒ (1 + 5s ) i1 − ( 3s ) 3s + 2
( −3s ) i1 + ( 3s + 2 ) i2 = 0 1
The characteristic equation is (1+5s ) ( 3s + 2 ) − 9 s 2 = 6 s 2 + 13s + 2 = 0 ⇒ s1, 2 = − , −2
6
Using the operator s = −t −t The currents are i1 ( t ) = Ae 6 + Be−2t and i2 ( t ) = Ce 6 + De −2t , where the constants A, B, C
and D must be evaluated using the initial conditions. Using the given initial values of the
currents gives
i1 ( 0 ) = 11 = A + B and i 2 ( 0 ) = 11 = C + D
Let t = 0 in the KCL equations (1) and (2) to get
di1 ( 0 )
dt =− di2 ( 0 )
A
C
33
143
= − − 2 B and
=−
=− −D
2
6
6
6
dt So A = 3, B = 8, C = −1 and D = 12. Finally,
i1 (t ) = 3e− t / 6 + 8e−2t A and i 2 (t ) = −e −t / 6 + 12e −2t A 919 Section 9.6: Natural Response of the Critically Damped Unforced Parallel RLC Circuit
P9.61
After t = 0 Using KVL: 100 ic ( t ) + 0.025 dic ( t )
+ vc ( t ) = 0
dt The capacitor current and voltage are related by:
ic ( t ) = 10−5 ∴ d 2 vc ( t )
dt 2 + 4000 dvc ( t )
dt dvc ( t )
dt + 4 × 106 vc ( t ) = 0 The characteristic equation is: s 2 + 4000 s + 4 × 106 = 0
The natural frequencies are: s1,2 = − 2000, − 2000
The natural response is of the form: vc ( t ) = A1e−2000t + A2 t e−2000t
(The capacitor current is continuous at t = 0
in this circuit because it is equal to the
inductor current.) Before t = 0 the circuit is at steady state vc ( 0+ ) = 3 = A1
dvc ( 0+ ) = 0 = −2000 A1 + A2 ⇒ A2 = 6000
dt
∴ vc ( t ) = ( 3 + 6000t ) e−2000t V for t ≥ 0 P9.62
After t = 0 Using KCL: ∫ t −∞ 1 dvc ( t )
=0
4 dt
d 2 vc ( t )
dv ( t )
⇒
+4 c
+ 4vc ( t ) = 0
dt
dt vc (τ ) dτ + vc ( t ) + The characteristic equation is: s 2 + 4 s + 4 = 0 920 The natural frequencies are: s1,2 = −2, −2
The natural response is of the form: vc ( t ) = A1e−2t + A2 t e−2 t
Before t = 0 the circuit is at steady state i L ( 0+ ) = i L ( 0 − ) and vC ( 0+ ) = vC ( 0− ) i C ( 0 + ) = −i L ( 0 + ) = − 2 A
dv ( 0+ )
dt vc ( 0 + )=0= A1 and dvc ( 0+ )
dt = −8 = A2 = i C ( 0+ )
14 = −8 V ⇒ vc ( t ) = −8 t e−2 t V P9.63
Assume that the circuit is at steady state before t = 0. The initial conditions are vc ( 0− ) = 104 V & iL ( 0− ) = 0 A
After t = 0 − vc ( t ) + .01 KVL: diL ( t )
+ 106 iL ( t ) = 0
dt (1) KCL: d 2iL ( t )
dvC ( t )
di ( t ) = −C .01
+ 106 L iL ( t ) = −C
2
dt
dt
dt ∴ 0.01 C ( 2) d 2iL ( t )
di ( t )
+ 106 C L
+ iL ( t ) = 0
2
dt
dt The characteristic equation is: ( 0.01 C ) s 2 + (106 C ) s + 1 = 0 The natural frequencies are: s1,2 = −106 C ± (10 C )
6 2 − 4 ( 0.01C ) 2 ( 0.01C ) For criticallydamped response: 1012 C 2 − .04C = 0 ⇒ C = 0.04 pF so s1,2 = −5 ×107 , −5 × 107 . 921 The natural response is of the form: iL ( t ) = A1e −5×10 t + A2 t e −5×10
7 diL +
( 0 ) = 100
dt
di ( 0 )
= 106
So iL ( 0 ) = 0 = A1 and L
dt
Now from (1) ⇒ 7 t vc ( 0+ ) − 106 iL ( 0+ ) = 106 A s
= A2 ∴ iL ( t ) = 106 t e−5×10 t A
7 Now v ( t ) = 106 iL ( t ) = 1012 t e−5×10 t V
7 P9.64 The characteristic equation can be shown to be: s 2 +
The natural frequencies are: s1,2 = −250, − 250 1
1
= s 2 + 500 s + 62.5 × 103 = 0
s+
RC
LC The natural response is of the form: v ( t ) = Ae −250t + B t e−250t
dv ( 0 )
= −3000 = −250 A + B ⇒ B = −1500
dt
− 1500 t e−250t V v ( 0 ) = 6 = A and
∴ v ( t ) = 6e−250t P9.65
After t=0, using KVL yields: di ( t )
t
+ Ri ( t ) + 2 + 4 ∫0 i (τ ) dτ = 6
(1)
dt
v( t )
Take the derivative with respect to t:
d 2i ( t ) di ( t )
+R
+ 4i ( t ) = 0
dt 2
dt
The characteristic equation is s 2 + Rs + 4 = 0
Let R = 4 for critical damping ⇒ ( s + 2) 2 =0 So the natural response is i ( t ) = A t e −2t + B e −2t
i ( 0 ) = 0 ⇒ B = 0 and
∴ i ( t ) = 4 t e −2t A di ( 0 )
= 4 − R ( i( 0 ) ) = 4 − R ( 0 ) = 4 = A
dt 922 Section 97: Natural Response of an Underdamped Unforced Parallel RLC Circuit
P9.71
After t = 0 KCL:
vc ( t )
dv ( t )
+ iL ( t ) + 5 ×10−6 c
=0
dt
250
KVL:
vc ( t ) = 0.8 (1) diL ( t )
dt ( 2) d 2vc ( t )
dv ( t )
+ 800 c + 2.5 ×105 v ( t ) = 0 ⇒ s 2 + 800 s + 250, 000 = 0, s1,2 = −400 ± j 300
c
dt
dt 2
The natural response is of the form v c (t ) Before t = 0 the circuit is at steady state: = e −400t A1 cos 300t + A 2sin 300t −6
A
500
vc ( 0+ ) = vc ( 0− ) = 250 −6 iL ( 0+ ) = iL ( 0− ) = ( 500 )+6 = 3 V From equation (1) :
dvc ( 0+ )
dt = − 2 × 105 iL ( 0+ ) − 800vc ( 0+ ) = 0 vc ( 0+ ) = 3 = A1
dvc ( 0+ )
dt = 0 = −400 A1 + 300 A2 ⇒ A2 = 4 ∴ vc ( t ) = e −400t [3 cos 300t + 4 sin 300t ] V 923 P9.72
Before t = 0 v ( 0+ ) = v ( 0− ) = 0 V
i ( 0+ ) = i ( 0− ) = 2 A After t = 0
KCL :
t
v (t ) 1 d v (t )
+
+ 2 ∫ v (τ ) dτ + i ( 0 ) = 0
0
1
4 dt Using the operator s = d
we have
dt 1 2
v (t ) + s v (t ) + v (t ) + i (0) = 0 ⇒ 4
s (s 2 + 4s + 8) v ( t ) = 0 The characteristic equation and natural frequencies are: s 2 + 4s + 8 = 0
The natural response is of the form: v ( t ) = e
v ( 0 ) = 0 = B1 and −2 t B1 cos 2t + B 2 sin 2t dv ( 0 )
= 4 −i ( 0 ) − v ( 0 ) = −4 [ 2] = −8 = 2 B2 or B2 = −4 dt so v ( t ) = −4e −2t sin 2t V P9.73
After t = 0 1 dvc ( t ) vc ( t )
+
+ iL ( t ) = 0
4 dt
2
4 diL ( t )
KVL : vc ( t ) =
+ 8 iL ( t )
dt
KCL : Characteristic Equation: ⇒ s = −2 ± j 2 d 2i L ( t )
dt 2 +4 di L ( t )
dt (1)
( 2) + 5 i L ( t ) = 0 ⇒ s 2 + 4 s + 5 = 0 ⇒ s1,2 = −2 ± i Natural Response: i L ( t ) = e −2t A1 cos t + A2 sin t 924 Before t = 0 vc ( 0− ) 48 +
−
= 7 4 8 + 2 ⇒ vc ( 0 ) = vc ( 0 ) = 8 V 2 8
iL ( 0+ ) = iL ( 0− ) = − = −4 A
2 diL ( 0+ )
dt = vc ( 0+ )
4 − 2iL ( 0+ ) = 8
A
− 2 ( −4 ) = 10
4
s iL ( 0+ ) = −4 = A1
diL ( 0+ )
dt ∴ iL ( t ) = e −2t [ −4 cos t + 2 sin t ] A = 10 = − 2 A1 + A2 ⇒ A2 = 2 P9.74
The plot shows an underdamped response, i.e. v ( t ) = e − α t [ k1 cos ω t + k2 sin ω t ] + k3 . Examining the plot shows v ( ∞ ) = 0 ⇒ k3 = 0, v ( 0 ) = 0 ⇒ k1 = 0 .
Therefore, v ( t ) = k2 e − α t sin ω t .
Again examining the plot we see that the maximum voltage is approximately 260 mV the time is
approximately 5 ms and that the minimum voltage is approximately −200 mV the time is
approximately 7.5 ms. The time between adjacent maximums is approximately 5 ms so
2π
ω≈
= 1257 rad/s . Then
5 ×10−3
−α ( 0.005 )
0.26 = k2 e
sin (1257 (.005 ) )
(1)
−0.2 = k2 e −α ( 0.0075 ) sin (1257 (.0075 ) ) ( 2) To find α we divide (1) by (2) to get ( 6.29 rad ) ⇒ sin ( 9.43 rad ) α ( 0.0025 ) sin − 1.3 = e e0.0025α = 1.95 ⇒ α = 267 From (1) we get k2 = 544 . Then
v ( t ) = 544e −267t sin 1257t V 925 P9.75
After t = 0 The characteristic equation is:
s2 + 1
1
s+
= 0 or s 2 + 2 s + 5 = 0
RC
LC The natural frequencies are: s1,2 = −1 ± j 2
The natural response is of the form:
v(t ) = e − t B1 cos 2t + B 2 sin 2t v ( 0+ ) 2
) = − 5 − i ( 0 ) = − 5 − 110 = − 1 V
2s
dv ( 0 )
3 1
= 10 − = − B + 2 B ⇒ B = − v(0 ) = 2 = B1 . From KCL, ic ( 0
+ + + so + dt 2 3
Finally, v ( t ) = 2e − t cos 2t − e− t sin 2t V
2 1 2 2 2 t≥0 926 Section 98: Forced Response of an RLC Circuit
P9.81 After t = 0 KCL : is ( t ) = v (t )
R KVL : v ( t ) = L
is ( t ) = 2 d 2iL ( t )
dt 2 dv ( t ) diL ( t ) dt dt d 2iL ( t )
L diL ( t )
+ iL ( t ) + LC
R dt
dt 2 d 2iL ( t )
dt + iL ( t ) + C + 1 diL ( t ) 1
1
+
iL ( t ) =
is ( t )
RC dt
LC
LC + ( 650 ) diL ( t )
dt + (105 ) iL ( t ) = (105 ) is ( t ) (a) Try a forced response of the form i f ( t ) = A . Substituting into the differential equations gives
0+0+ A 1
1
=
⇒
−3
(.01) (1×10 ) (.01) (1×10−3 ) A = 1 . Therefore i f ( t ) = 1 A . (b) Try a forced response of the form i f ( t ) = A t + B . Substituting into the differential equations
gives 0 + A 65
1
+ ( A t + B)
= 0.5 t . Therefore A = 0.5 and
( 0.01) ( 0.001)
(100 ) ( 0.001) B = −3.25 × 10 −3 . Finally i f ( t ) = 5 t − 3.25 × 10−3 A . (c) Try a forced response of the form i f ( t ) = A e−250 t . It doesn’t work so try a forced response of
the form i f ( t ) = B t e−250 t . Substituting into the differential equation gives ( −250 )2 B e−250t − 500 B e−250t + 650 ( −250 ) B t e−250t + B e−250t + 105 B t e−250t = 2 e−250t . Equating coefficients gives ( 250 )
and 2 2
B + 650 ( −250 ) B + 105 B = 0 ⇒ ( 250 ) + 650 ( −250 ) + 105 B = 0 ⇒ −500 B + 650 B = 2 ⇒ [ 0] B = 0 B = 0.0133 Finally i f ( t ) = 0.0133 t e −250t A . 927 P9.82 After t = 0 d 2v ( t )
dt
d 2v ( t )
dt + v (t )
R dv ( t ) 1
+
v (t ) = s
L dt
LC
LC + 70 dv ( t )
dt + 12000v ( t ) = 12000 vs ( t ) (a) Try a forced response of the form v f ( t ) = A . Substituting into the differential equations gives
0 + 0 + 12000 A = 24000 ⇒ A = 2 . Therefore v f ( t ) = 2 V . (b) Try a forced response of the form v f ( t ) = A + B t . Substituting into the differential equations
gives 70 A + 12000 A t + 12000 B = 2400 t . Therefore A = 0.2 and B =
Finally v f ( t ) = ( −1.167 × 10−3 ) t + 0.2 V . −70 A
= −1.167 × 10 −3 .
12000 (c) Try a forced response of the form v f ( t ) = A e −30 t . Substituting into the differential equations
gives 900 Ae −30t − 2100 Ae −30 t + 12000 Ae −30 t = 12000 e −30t . Therefore A = v f ( t ) = 1.11e−250 t V . 12000
= 1.11 . Finally
10800 928 Section 99: Complete Response of an RLC Circuit
P9.91 First, find the steady state response for t < 0, when the switch is open. Both inputs are constant
so the capacitor will act like an open circuit at steady state, and the inductor will act like a short
circuit. After a source transformation at the left of the circuit: i L ( 0) =
and 11 − 4
= 2.33 mA
3000 v C ( 0) = 4 V After the switch closes
Apply KCL at node a:
vC
R +C d
vC + iL = 0
dt Apply KVL to the right mesh:
L d
d
i L + Vs − vC = 0 ⇒ vC = L i L + Vs
dt
dt After some algebra:
V
1d
1
d2
i+
iL +
iL = − s
2L
dt
R C dt
LC
R LC ⇒ d2
d
i + ( 500 ) i L + (1.6 × 105 ) i L = −320
2L
dt
dt The characteristic equation is
s 2 + 500 s + 1.6 × 105 = 0 ⇒ s1,2 = −250 ± j 312 rad/s 929 After the switch closes the steadystate inductor current is iL(∞) = 2 mA so
i L ( t ) = −0.002 + e −250 t ( A1 cos 312 t + A2 sin 312 t ) d
i L (t ) + 4
dt
= 6.25 e −250 t −250 ( A1 cos 312 t − A2 sin 312 t ) − 312 ( A1 sin 312 t − A2 cos 312 t ) + 4 v C ( t ) = 6.25 = 6.25 e −250 t ( 312 A2 − 250 A1 ) cos 312 t + ( 250 A2 + 312 A1 ) sin 312 t + 4 Let t = 0 and use the initial conditions:
iL ( 0+ ) = 0.00233 = −0.002 + A1 ⇒ 0.00433 = A1
vC ( 0+ ) = 4 = 6.25 ( 312 A2 − 250 A1 ) + 4 ⇒ A2 = 250
250
A2 =
( 0.00433) = 0.00347
312
312 Then
i L ( t ) = −0.002 + e −250 t ( 0.00433 cos 312 t + 0.00345 sin 312 t )
= −0.002 + 0.00555 e −250 t cos ( 312 t − 36.68° ) A
v C ( t ) = 4 + 13.9 e −250 t sin ( 312 t ) V
i (t ) = vC (t )
2000 = 2 + 6.95 e −250 t sin ( 312 t ) mA
(checked using LNAP on 7/22/03) P9.92
First, find the steady state response for t < 0. The input is constant so the capacitor will act like
an open circuit at steady state, and the inductor will act like a short circuit. i ( 0) = −1
= −0.2 A
1+ 4 and
v (0) = 4
( −1) = −0.8 V
1+ 4 930 For t > 0
Apply KCL at node a:
v − Vs
d
+C v+i = 0
R1
dt Apply KVL to the right mesh:
R2 i + L d
d
i − v = 0 ⇒ v = R2 i + L iL
dt
dt After some algebra:
L + R1 R 2C d
R1 + R 2
d2
Vs
i+
i+
i=
2
dt
R1 L C dt
R1 L C
R1 L C
The forced response will be a constant, if = B so 1 = ⇒ d2
d
i +5 i +5i =1
2
dt
dt d2
d
B + 5 B + 5 B ⇒ B = 0.2 A .
2
dt
dt To find the natural response, consider the characteristic equation:
0 = s 2 + 5 s + 5 = ( s + 3.62 ) ( s + 1.38 ) The natural response is
in = A1 e −3.62 t + A2 e−1.38 t
so
i ( t ) = A1 e −3.62 t + A2 e−1.38 t + 0.2
Then
d v ( t ) = 4 i ( t ) + 4 i ( t ) = −10.48 A1 e−3.62 t − 1.52 A2 e−1.38 t + 0.8
dt At t=0+ −0.2 = i ( 0 + ) = A1 + A2 + 0.2
−0.8 = v ( 0 + ) = −10.48 A1 − 1.52 A2 + 0.8 so A1 = 0.246 and A2 = −0.646. Finally
i ( t ) = 0.2 + 0.246 e −3.62 t − 0.646 e−1.38 t A 931 P9.93
First, find the steady state response for t < 0. The input is constant so the capacitors will act like
an open circuits at steady state. v1 ( 0 ) =
and 1000
(10 ) = 5 V
1000 + 1000
v2 ( 0 ) = 0 V For t > 0,
Node equations: v1 − 10 1
v −v
d
+ × 10−6 v1 + 1 2 = 0
1000 6
1000 dt
1
d
⇒ 2 v1 + × 10−3 v1 − 10 = v2
6 dt
v1 − v2 1
d
= × 10−6 v2
1000 16 dt
1
d
⇒ v1 − v2 = × 10−3 v2 16 dt
After some algebra:
d2
d
v + ( 2.8 × 104 ) v1 + ( 9.6 ×107 ) v1 = 9.6 × 108
21
dt
dt
The forced response will be a constant, vf = B so
d2
d
B + ( 2.8 × 104 ) B + ( 9.6 × 107 ) B = 9.6 × 108
2
dt
dt ⇒ B = 10 V . To find the natural response, consider the characteristic equation:
s 2 + ( 2.8 × 104 ) s + ( 9.6 × 107 ) = 0 ⇒ s1,2 = −4 × 103 , −2.4 × 104
The natural response is
3 vn = A1 e −4×10 t + A2 e−2.4×10 4t so
3 v1 ( t ) = A1 e−4×10 t + A2 e−2.4×10 4t + 10 At t = 0
932 5 = v1 ( 0 ) = A1 e −4×103 ( 0 ) + A2 e −2.4×104 ( 0 ) + 10 = A1 + A2 + 10 (1) Next
1
d
2 v1 + ×10−3 v1 − 10 = v2
6 dt ⇒ d
v1 = −12000v1 + 6000 v2 + 6 × 104
dt At t = 0
d
v1 ( 0 ) = −12000v1 ( 0 ) + 6000 v2 ( 0 ) + 6 × 104 = −12000 ( 5 ) + 6000 ( 0 ) + 6 × 104 = 0
dt
so
3
4
d
v1 ( t ) = A1 −4 × 103 e −4×10 t + A2 −2.4 × 104 e−2.4×10 t
dt ( ) ( ) At t = 0+
0= d
−4×103 ( 0 )
−2.4×104 ( 0 )
v1 ( 0 ) = A1 −4 × 103 e
+ A2 −2.4 × 104 e
= A1 −4 × 103 + A2 −2.4 × 104
dt ( ) ( ) ( ) ( ) so A1 = −6 and A2 = 1. Finally
v1 ( t ) = 10 + e −2.4 ×10 4t P9.94
For t > 0 3t − 6 e−4 ×10 V for t > 0 KCL at top node:
diL ( t ) 1 dv ( t )
− 5 cos t + iL ( t ) +
=0 0.5
dt
12 dt (1) KVL for right mesh:
0.5 diL ( t ) 1 dv ( t )
=
+ v (t )
12 dt
dt ( 2) Taking the derivative of these equations gives:
d 2iL ( t ) diL ( t ) 1 d 2 v ( t )
of (1) ⇒ 0.5
+
+
= −5 sin t
dt
12 dt 2
dt 2
dt
2
2
d of ( 2 ) ⇒ 0.5 d iL ( t ) = 1 d v ( t ) + dv ( t )
dt
12 dt 2
dt 2
dt
d (3) ( 4) 933 d 2iL ( t )
di ( t )
Solving for
in ( 4 ) and L
in ( 2 ) & plugging into ( 3) gives
2
dt
dt
d 2v ( t )
dv ( t )
+7
+ 12 v ( t ) = − 30 s i n t
dt 2
dt The characteristic equation is: s 2 + 7s+12 = 0 .
The natural frequencies are s1,2 = −3, −4 .
The natural response is of the form vn (t ) = A1e −3t + A2 e −4t . Try a forced response of the form
v f ( t ) = B1 cos t + B 2 sin t . Substituting the forced response into the differential equation and equating like terms gives B1 = 21
33
and B 2 = − .
17
17 v ( t ) = vn (t ) + v f ( t ) = A1e−3t + A2 e−4t + 21
33
cos t − sin t
17
17 We will use the initial conditions to evaluate A1 and A2. We are given iL ( 0 ) = 0 and v ( 0 ) = 1 V .
Apply KVL to the outside loop to get
1 iC ( t ) + iL ( t ) + 1( iC ( t ) ) + v ( t ) − 5 cos t = 0 At t = 0+
iC ( 0 ) = 5cos ( 0 ) + iL ( 0 ) − v ( 0 ) 5 + 0 − 1
=
=2 A
2
2
dv ( 0 ) iC ( 0 )
2
=
=
= 24 V/s
dt
1 12 1 12 ⇒
+
33 dv(0 )
= 24 = −3 A1 − 4 A2 −
dt
17 v(0+ ) = 1 = A1 + A2 + 21
17 A1 = 25
429
A2 = −
17 Finally,
∴ v(t ) = 25e −3t − 429e −4t − 21 cos t + 33 sin t
V
17 934 P9.95
Use superposition. Find the response to inputs 2u(t) and –2u(t2) and then add the two responses.
First, consider the input 2u(t): For 0< t < 2 s Using the operator s = d
we have
dt KVL:
vc ( t ) + siL ( t ) + 4 iL ( t ) − 2 = 0 (1) KCL:
1
3
s vc ( t ) ⇒ vc ( t ) = iL ( t )
(2)
s
3
Plugging (2) into (1) yields the characteristic equation: ( s 2 + 4 s + 3) = 0 . The natural frequencies
are s1,2 = −1 , −3 . The inductor current can be expressed as
iL ( t ) = iL (t ) = in (t ) + i f (t ) = ( A1 e −t + A2 e−3t ) + 0 = A1 e− t + A2 e −3t .
Assume that the circuit is at steady state before t = 0. Then vc (0+ ) = 0 and iL (0+ ) = 0 .
Using KVL we see that (1) diL (0+ )
= 4 2 − iL (0+ ) − vC (0+ ) = 8 A/s . Then dt
iL (0) = 0 = A1 + A2 A1 = 4 , A2 = −4 .
diL (0)
= 8 = − A1 − 3 A2 dt Therefore iL (t ) = 4e − t − 4e−3t A . The response to 2u(t) is
0
t<0 v1 (t ) = 8 − 4 iL (t ) = −t
−3t
8 − 16 e + 16 e V t > 0 .
= 8 − 16 e − t + 16 e −3t u ( t ) V The response to –2u(t2) can be obtained from the response to 2u(t) by first replacing t by t2
everywhere is appears and the multiplying by –1. Therefore, the response to –2u(t2) is
v2 (t ) = −8 + 16e − (t − 2 ) − 16 e−3(t − 2 ) u ( t  2 ) V . By superposition, v(t ) = v1 (t ) + v2 (t ) . Therefore
v(t ) = 8 − 16e −t + 16e−3t u (t ) + −8 + 16e− (t − 2) − 16 e −3(t − 2) u (t − 2) V 935 P9.96 First, find the steady state response for t < 0, when the switch is closed. The input is constant so
the capacitor will act like an open circuit at steady state, and the inductor will act like a short
circuit. i ( 0) = −
and After the switch opens 5
= −1.25 A
4 v (0) = 5 V Apply KCL at node a:
v
d
+ 0.125 v = i
2
dt
Apply KVL to the right mesh:
−10 cos t + v + 4 d
i+4i =0
dt After some algebra:
d2
d
v + 5 v + 6 v = 20 cos t
2
dt
dt
The characteristic equation is
s 2 + 5 s + 6 = 0 ⇒ s1,2 = −2, − 3 Try
vf = A cos t + B sin t d2
d
A cos t + B sin t ) + 5 ( A cos t + B sin t ) + 6 ( A cos t + B sin t ) = 20 cos t
2(
dt
dt 936 ( − A cos t − B sin t ) + 5 ( − A sin t + B cos t ) + 6 ( A cos t + B sin t ) = 20 cos t
( − A + 5 B + 6 A) cos t + ( − B − 5 A + 6 B ) sin t = 20 cos t
Equating the coefficients of the sine and cosine terms yields A =2 and B =2. Then
vf = 2 cos t + 2 sin t
v ( t ) = 2 cos t + 2 sin t + A1 e −2 t + A2 e −3 t Next v (t )
d
+ 0.125 v ( t ) = i ( t ) ⇒
2
dt d
v (t ) = 8 i (t ) − 4 v (t )
dt d
V 5
v ( 0 ) = 8 i ( 0 ) − 4 v ( 0 ) = 8 − − 4 ( 5 ) = −30
dt
s 4
Let t = 0 and use the initial conditions:
5 = v ( 0 ) = 2 cos 0 + 2 sin 0 + A1 e −0 + A2 e −0 = 2 + A1 + A2 d
v ( t ) = −2 sin t + 2 cos t − 2 A1 e −2 t − 3 A2 e−3 t
dt
−30 = d
v ( 0 ) = −2 sin 0 + 2 cos 0 − 2 A1 e −0 − 3 A2 e −0 = 2 − 2 A1 − 3 A2
dt So A1 = −23 and A2 = 26 and
v ( t ) = 2 cos t + 2 sin t − 23 e −2 t + 26 e −3 t V for t > 0 937 P9.97 First, find the steadystate response for t < 0. The input is constant so the capacitor will act like
an open circuit at steady state, and the inductor will act like a short circuit.
i ( 0) = 0 A and After t = 0 Apply KCL at node a: v (0) = 0 V C d
v=i
dt Apply KVL to the right mesh:
d
8 i + v + 2 i + 4 (2 + i) = 0
dt
d
12 i + v + 2 i = −8
dt
2
1
d
d
4
After some algebra:
v + ( 6) v + v = −
2
dt
dt
C
2C
The forced response will be a constant, vf = B so
1
d2
d
4
B + ( 6) B + B = −
2
dt
dt
C
2C (a) ⇒ B = −8 V d2
d
v + ( 6 ) v + ( 9 ) v = −72 .
2
dt
dt
2
The characteristic equation is s + 6 s + 9 = 0 ⇒ s1,2 = −3, −3 When C = 1/18 F the differential equation is
Then v ( t ) = ( A1 + A2 t ) e −3t − 8 .
Using the initial conditions: 938 0 = v ( 0 ) = ( A1 + A2 ( 0 ) ) e0 − 8 ⇒
0 = i ( 0) = C
So (b) A1 = 8 d
v ( 0 ) = C −3 ( A1 + A2 ( 0 ) ) e0 + A2 e0 ⇒ dt A2 = 3 A1 = 24 v ( t ) = ( 8 + 24 t ) e −3t − 8 V for t > 0 d2
d
v + ( 6 ) v + ( 5 ) v = −40
2
dt
dt
2
The characteristic equation is s + 6 s + 5 = 0 ⇒ s1,2 = −1, −5 When C = 1/10 F the differential equation is
Then v ( t ) = A1 e − t + A2 e −5 t − 8 . Using the initial conditions:
0 = v ( 0 ) = A1 e0 + A2 e0 − 8 ⇒ A1 + A2 = 8 ⇒ A1 = 10 and A2 = −2
d
0
0
0 = v ( 0 ) = − A1 e − 5 A2 e ⇒ − A1 − 5 A2 = 0 dt So
v ( t ) = 10 e − t − 2 e −5 t − 8 V for t > 0
(c) d2
d
v + ( 6 ) v + (10 ) v = −80
2
dt
dt
2
The characteristic equation is s + 6 s + 10 = 0 ⇒ s1,2 = −3 ± j
When C = 1/20 F the differential equation is Then v ( t ) = e −3 t ( A1 cos t + A2 sin t ) − 8 . Using the initial conditions:
0 = v ( 0 ) = e0 ( A1 cos 0 + A2 sin 0 ) − 8 ⇒
0=
So A1 = 8 d
v ( 0 ) = −3 e0 ( A1 cos 0 + A2 sin 0 ) + e0 ( − A1 sin 0 + A2 cos 0 ) ⇒
dt A2 = 3 A1 = 24 v ( t ) = e −3 t ( 8 cos t + 24 sin t ) − 8 V for t > 0 939 P9.98 The circuit will be at steady state for t<0:
so
iL(0+) = iL(0−) = 0.5 A
and
vC(0+) = vC(0−) = 2 V. For t>0:
Apply KCL at node b to get:
1
1d
1 1d
= i (t ) +
v (t ) ⇒ i (t ) = −
v (t )
L
C
L
4
4 dt
4 4 dt C
Apply KVL at the rightmost mesh to get:
4i L( t) + 2 d
1 d i (t ) = 8 vc ( t ) + v ( t )
L
dt 4 dt
c Use the substitution method to get
d 1 1 d
1 1 d 1 d 4 −
vC ( t ) + 2 −
vC ( t ) = 8 vc ( t ) + v ( t )
dt 4 4 dt 4 4 dt 4 dt
c
or
d2
d
v t + 6 v (t ) + 2 v (t )
2 C( )
C
dt
dt C
d2
d
The forced response will be a constant, vC = B so 2 =
B + 6 B + 2B ⇒ B = 1 V .
dt
dt 2
2= To find the natural response, consider the characteristic equation:
0 = s 2 + 6 s + 2 = ( s + 5.65 ) ( s + 0.35 ) The natural response is
vn = A1 e −5.65 t + A2 e −0.35 t vC ( t ) = A1 e −5.65 t + A2 e −0.35 t so
+1 Then
iL ( t ) = 1 1d
1
−5.65 t
−0.35 t
+
+ 0.0875 A2 e
vC ( t ) = + 1.41A1 e
4 4 dt
4 940 At t=0+ 2 = vC ( 0 + ) = A1 + A + 1
2 1
1
= iL ( 0+ ) = + 1.41A1 + 0.0875 A
2
2
4
so A1 = 0.123 and A2 = 0.877. Finally
vC ( t ) = 0.123 e −5.65 t + 0.877 e −0.35 t +1 V P9.99 The inductor current and voltage are related be After t = 0 v (t ) = L di ( t )
dt (1) Apply KCL at the top node to get
C dv( t )
v( t )
+ i (t ) +
=5
dt
2 (2) d
, and substituting (1) into (2) yields ( s 2 + 4s + 29 ) i ( t ) = 5 .
dt
The characteristic equation is s 2 + 4 s + 29 = 0 . The characteristic roots are s1,2 = − 2 ± j 5 . Using the operator s = The natural response is of the form in ( t ) = e −2t [ A cos 5t + B sin 5t ] .
Try a forced response of the form i f ( t ) = A . Substituting into the differential equation gives A = 5 . Therefore i f ( t ) = 5 A .
The complete response is i(t ) = 5 + e−2t [ A cos 5t + B sin 5t ] where the constants A and B are
yet to be evaluated using the initial condition:
i (0) = 0 = A + 5 ⇒ A = −5
di (0)
di (0)
2A
0 = v ( 0) = L
⇒
= 0 = −2 A + 5 B ⇒ B =
= −2
dt
dt
5
Finally, i (t ) = 5 + e −2t [ −5 cos 5t − 2 sin 5t ] A . P9.910 941 Assume that the circuit is at steady before t = 0. 2
×9 = 6 A
2+1
1
v ( 0 + ) = v (0 − ) =
× 9× 1.5 = 4.5 V
2+1 i (0 + ) = i (0 − ) = Apply KCL at the top node of the current source
to get After t = 0: i ( t ) + 0.5 dv( t ) v( t )
+
= is ( t )
dt
1.5 (1) Apply KVL and KCL to get dv( t ) v( t ) 5di( t ) v ( t ) + 0.5
+
+ i (t ) 0.5 =
dt
dt
1.5 (2) Solving for i(t) in (1) and plugging into (2) yields
d 2 v( t ) 49 dv( t ) 4
di ( t )
2
+
+ v ( t ) = is ( t ) + 2 s
2
dt
dt
30 dt
5
5 where is ( t ) = 9 + 3e −2t A d
49
4
, the characteristic equation is s 2 + s + = 0 and the characteristic
30
5
dt
roots are s1,2 =−.817 ± j.365 . The natural response has the form Using the operator s = vn (t ) = e−0.817t A1 cos (0.365 t ) + A2 sin (0.365 t ) Try a forced response of the form v f (t ) = B0 + B1e −2 t . Substituting into the differential equations gives B0 = 4.5 and B1 = −7.04 . The complete response has the form
v(t ) = e−.817 t A1 cos(0.365 t ) + A2 sin (0.365 t ) + 4.5 − 7.04 e −2t Next, consider the initial conditions:
v (0) = 4.5 = A1 + 4.5 − 7.04 ⇒ A1 = 7.04 942 d v ( 0)
4
4
= 2 is (0)  2 i (0) − v(0) = 2(9 + 3) − 2(6) − (4.5) = 6
dt
3
3 6= d v( 0 )
= −0.817 A1 + 0.365 A2 + 14.08 ⇒ A2 = −6.38
dt So the voltage is given by
v(t ) = e−0.817 t 7.04 cos(0.365 t ) + A2 sin (0.365 t ) + 4.5 − 7.04 e−2t Next the current given by
i (t ) = is (t ) − Finally v(t )
d v(t )
− 0 .5
1.5
dt i (t ) = e −0.817 t [ 2.37 cos(0.365t ) + 7.14 sin(0.365t ) ] + 6 + 0.65e −2t A P9.911 First, find the steady state response for t < 0. The input is constant so the capacitor will act like
an open circuit at steady state, and the inductor will act like a short circuit. va ( 0− ) = −4 i ( 0− ) ( ) i ( 0− ) = 2va ( 0− ) = 2 −4 i ( 0− ) ⇒ i ( 0− ) = 0 A
and v ( 0− ) = 10 V 943 For t > 0
Apply KCL at node 2:
va
d
+ K va + C
v=0
R
dt KCL at node 1 and Ohm’s Law:
va = − R i
so d
1+ K R
v=
i
dt
C Apply KVL to the outside loop: L d
i + R i + v − Vs = 0
dt After some algebra:
d2
Rd
1+ K R
1+ K R
v+
v+
v=
Vs
2
dt
L dt
LC
LC ⇒ d2
d
v + 40 v + 144 v = 2304
2
dt
dt The forced response will be a constant, vf = B so
d2
d
B + ( 40 ) B + (144 ) B = 2304 ⇒ B = 16 V
2
dt
dt
The characteristic equation is s 2 + 40 s + 144 = 0 ⇒ s1,2 = −4, −36 . v ( t ) = A1 e− 4 t + A2 e−36 t + 16 . Then
Using the initial conditions: 10 = v ( 0+ ) = A1 e0 + A2 e0 + 16 ⇒
0= d
v ( 0+ ) = −4 A1 e0 − 36 A2 e0
dt ⇒
⇒ − 4 A1 − 36 A2 = 0 A1 + A2 = −6 A1 = −6.75 and A2 = 0.75 So
v ( t ) = 0.75 e −36 t − 6.75 e−4 t + 16 V for t > 0 (checked using LNAP on 7/22/03) 944 Section 910: State Variable Approach to Circuit Analysis
P9.101 At t = 0− the circuit is source free ∴ iL (0) = 0 and v(0) = 0.
Apply KCL at the top node to get After t = 0 iL ( t ) + 1 dv( t )
=4
5 dt (1) Apply KVL to the right mesh to get
v( t ) −(1)
Solving for i1 ( t ) in (1) and plugging into (2) ⇒ diL ( t )
− 6 iL ( t ) = 0
dt (2) d 2 v( t )
dv ( t )
+6
+ 5v ( t ) = 120 .
2
dt
dt The characteristic equation is s 2 + 6 s + 5= 0 . The natural frequencies are s1,2 =−1, −5 . The natural
response has the form vn (t ) = A1 e−t + A2 e−5t . Try v f ( t ) = B as the forced response.
Substituting into the differential equation gives B = 24 so v f ( t ) = 24 V. The complete response
has the form v (t ) = A1 e−t + A2 e−5t + 24 .
Now consider the initial conditions. From (1) dv(0)
= 20 − 5 iL (0) = 20 V . Then
s
dt v(0) = 0 = A1 + A2 + 24 ⇒
dv(0)
= 20 = − A1 −5 A2 dt A1= −25, A2 = 1 Finally v (t ) = − 25e −t + e −5t + 24 V . 945 P9.102
Before t = 0 there are no sources in the circuit so iL (0) = 0 and v(0) = 0 . After t = 0 we have: Apply KCL at the top node to get
iL ( t ) = 4 − 1 dv ( t )
10 dt (1) Apply KVL to the left mesh to get
v( t ) − diL ( t )
− 6iL ( t ) = 0
dt (2) Substituting iL ( t ) from (1) into (2) gives
d 2 v( t )
dv ( t )
+6
+ 10v ( t ) = 2 40
2
dt
dt The characteristic equation is s 2 + 6 s + 10 = 0 . The natural frequencies are s1,2 = −3 ± j . The
natural response has the form vn (t ) = e −3t A1 cos t + A2 sin t . Try v f ( t ) = B as the forced response. Substituting into the differential equation gives B = 24 so v f ( t ) = 24 V. The complete
response has the form v(t ) = e −3t A1 cos t + A2 sin t + 24 . Now consider the initial conditions. From (1) dv(0)
= 40 − 10 iL (0) = 40 V . Then
s
dt v (0) = 0 = A1 + 24 ⇒ A1 = −24 dv(0)
= 40 = −3 A1 + A2 = 72 + A2 ⇒ A2 = −32
dt
Finally, v (t ) = e −3t [ −24 cos t −32 sin t ] + 24 V 946 P9.103
Assume that the circuit is at steady state before t = 0 so iL (0) = −3 A and v(0) = 0 V . After t = 0 we have
KCL: i ( t ) + C
KVL: v( t ) = L i (t ) + C
d 2i ( t )
−6
1 di ( t )
1
+
+
i (t ) =
2
dt
R C dt
LC
LC ⇒ dv( t ) v( t )
+
+ 6=0
dt
R di ( t )
dt d di( t ) 1 di ( t ) L + L
+ 6=0
dt dt R dt d 2i ( t )
di ( t )
+ 100
+ 250i ( t ) = −1500
2
dt
dt The characteristic equation is s 2 + 100 s + 250 = 0 . The natural frequencies are
s1,2 = −2.57, − 97.4 . The natural response has the form in (t ) = A1 e−2.57 t + A2 e−97.4t . Try ( t ) = B as the forced response. Substituting into the differential equation gives B = −6 so
i f ( t ) = −6 A . The complete response has the form i(t ) = A1 e−2.57 t + A2 e−97.4t − 6 .
i f Now consider the initial conditions:
i (0) = A1 + A2 − 6 = −3 A1 = 3 .081 di (0)
= 0 = − 2.57 A1 −97.4 A2 A2 =−0.081
dt Finally:
i (t ) = 3. 081 e
v(t ) = .2 −2.57 t −.081e di ( t )
= −1.58e
dt −97.4 t −2.57 t −6 A +1.58e−97.4t V 947 P9.104 Apply KCL to the supernode corresponding to
the dependent voltage source to get
ix ( t ) − 2ix ( t ) − 0.01 dv ( t ) vx ( t )
+
=0
2
dt Apply KCL at node 1 to get
i ( t ) − 2ix ( t ) +
(Encircled numbers are node numbers.) vx ( t )
=0
2 Apply KVL to the topright mesh to get
vx ( t ) + v ( t ) − 0.1 di ( t )
=0
dt Apply KVL to the outside loop to get ix ( t ) = −2 vx ( t ) − v ( t ) .
Eliminate ix ( t ) to get Then eliminate vx ( t ) to get dv ( t )
5
=0
vx ( t ) + v ( t ) − 0.01
2
dt
9
i ( t ) + vx ( t ) + 2 v ( t ) = 0
2
d i (t )
vx ( t ) = −v ( t ) + 0.01
dt
dv ( t )
di ( t )
+ 0.25
=0
dt
dt
di ( t )
−2.5 v ( t ) + i ( t ) + 0.45
=0
dt
−1.5 v ( t ) − 0.01 Using the operator s = d
we have
dt
(−1.5 − .01s )v ( t ) + (.25s ) i ( t ) = 0
(−2.5)v( t ) + (1+.45s ) i ( y ) = 0 The characteristic equation is s 2 +13.33 s + 333.33 = 0 . The natural frequencies are s1 , s2 = −6.67 ± j 17 . The natural response has the form vn (t ) = [ A cos 17 t + B sin 17 t ] e−6.67 t . ( t ) = 0 . The complete response has the form
+ B sin 17 t ] e−6.67 t . The forced response is v v(t ) = [ A cos17 t f 948 The given initial conditions are i (0) = 0 and v (0) = 10 V. Then
v(0) =10 = A and dv(0)
=−111= − 6.67 A +17 B ⇒ B =−2.6
dt Finally i (t ) = [3.27 sin 17 t ] e−6.67 t A .
(Checked using LNAP on 7/22/03)
P9.105 Assume that the circuit is at steady state before t = 0 so v(0) = 10 V and iL (0) =
The switch is open when 0 < t < 0.5 s v ( 0) 10
=
A.
3
3 For this series RLC circuit we have: α= R
1
= 3 and ω 02 =
= 12
2L
LC −α ± α 2 −ω02 = s1,2 = −3 ± j 3 The natural response has the form vn (t ) = e −3t ( A cos 1.73 t + B sin 1.72 t ) . There is no source so v f ( t ) =0 . The complete response has the form v (t ) = e −3t ( A cos 1.73 t + B sin 1.72 t ) .
Next v(0) = 10 = A A= 10 dv(0) i ( 0 ) 10 3
⇒
=−
=−
= − 20 = −3 A +1.73 B B =5.77
dt
C
16 so v(t ) = e−3t (10 cos1.73 t + 5.77 sin 1.73 t ) V
i(t ) = e−3t ( 3.33 cos1.73 t − 5.77 sin 1.73 t ) A In particular,
1.73
1.73 v (0.5) = e −1.5 10 cos
+ 5.77 sin = 0.2231× ( 6.4864 + 4.3915 ) = 2.43 V
2
2 and
1.73
1.73 i (0.5) = e −1.5 3.33 cos
−5.77 sin = 0.2231× ( 2.1600 − 4.3915 ) = −0.50 A
2
2 949 The switch is closed when t > 0.5 s Apply KCL at the top node:
v( t ) −30
1 dv ( t )
+ iL ( t ) +
=0
6
6 dt
dv( t ) 1
⇒ iL ( t ) = 5 − v( t ) + 6
dt ⇒ 2
d iL ( t )
1 dv ( t ) d v ( t ) =− + dt
6 dt
dt 2 Apply KVL to the right mesh:
1 diL ( t )
v( t ) = 3 iL ( t ) +
2 dt
The circuit is represented by the differential equation
d 2v ( t )
dv ( t )
+7
+ 1 8 v ( t ) = 18 0
2
dt
dt
The characteristic equation is 0 = s 2 + 7 s + 18 . The natural frequencies are s1,2 = −3.7 ± j 2.4 .
The natural response has the form vn (t ) = e −3.5 t ( A cos 2.4 t + B sin 2.4 t ) . The forced response is v f ( t ) = 10 V . The complete response has the form v (t ) = e −3.5 t ( A cos 2.4 t + B sin 2.4 t ) + 10 . Next v (0.5) = e −3.5×0.5 ( A cos 1.2 + B sin 1.2 ) = 0.063 A + 0.162 B dv ( 0.5 )
= e−3.5×0.5 ( −3.5 A + 2.4 B ) cos1.2 − ( 3.5B + 2.4 A ) sin 1.2 dt
= e−3.5×0.5 ( −3.5 cos 1.2 − 2.4 sin 1.2 ) A + e−3.5×0.5 ( 2.4 cos1.2 − 3.5 sin 1.2 ) B
= −0.6091A − 0.4158B
Using the initial conditions yields
2.43=v(0.5) = 0.063 A+ 0.162 B A=−20.65 dv(0.5) i ( 0.5 ) −1 2
⇒
B = 23.03
=−
=−
=3= −0.6091A− 0.4158B dt
C
16 Finally
In summary v(t ) = e −3.5 t ( −20.65 cos 2.4 t + 23.03 sin 2.4 t ) + 10 e −3t (10 cos1.73 t + 5.77 sin 1.73 t ) V
0<t <0.5 v(t ) = −3.5 t
e ( −20.65 cos 2.4 t + 23.03 sin 2.4 t )+10 V 0.5<t 950 Section 911: Roots in the Complex Plane
P9.111
After t = 0 i1 + i 2 +
2 ×10−3 d i1
dt 2 ×10−3 d i1 dt
2000 −6 =0 = 3000 i 2 + 2 ×10−3 d i2
dt d
yields
dt 2000 + 2 × 10−3 s i1 6 2000
= −3
−3 2 × 10 s
3000 + 2 × 10 s i 2 0 Using the operator s = s 2 + 3.5 × 106 s + 1.5 ×1012 = 0
⇒ s1,2 = −5 × 105 , −3 ×106 P9.112
From P9.71
s2 + 1
1
1
1
= 0 ⇒ s2 +
=0
s+
s+
−6
RC
LC
( 250 ) ( 5 ×10 ) ( 0.8) ( 5 ×10−6 )
⇒ s 2 + 800 s + 250000 = 0 s1,2 = 400 ± j 300 P9.113 dv( t ) v( t )
1
KCL: i L ( t ) = × 10−6
+
4
dt
4000
di ( t )
KVL: vs ( t ) = 4 L + v( t )
dt d 2v ( t )
dv ( t )
dv( t ) v( t ) d 1
v ( t ) = 4 ×10−6
+
+ v ( t ) = 10−6
+ 10−3
+ v (t ) 2
s
dt 4
dt 4000 dt
dt
951 Characteristic equation: s 2 + 103 s + 106 = 0
Characteristic roots: s1, 2 = −500 ± j 866 P9.114
Before t = 0 the voltage source voltage is 0 V so vb (0+) = vb (0−) = 0 V and
i (0+) = i (0−) = 0 A . Apply KCL at node a to get va (0+ ) −36
v (0+ ) − vb (0+ )
− i (0 + ) + a
= 0 ⇒ va (0+ ) + 2 va (0+ ) = 36 ⇒ va (0) = 12 V
12
6 After t = 0 the node equations are:
− va ( t ) − vs ( t ) 1 t
v ( t ) − va ( t )
+ ∫ ( vb (τ ) − va (τ ) ) dτ + b
=0
0
12
6
L
C Using the operator s = d vb ( t ) vb ( t ) − va ( t ) 1 t
+
+ ∫ ( vb (τ ) − va (τ ) ) dτ = 0
6
dt
L0 d
we have
dt
v (t )
1 1
1 1 1
+ + va ( t ) + − − vb ( t ) = s 6
12
s 12 6 s 1 1 1 1
1 − − va ( t ) + s + + vb ( t ) = 0 6 s 18 6 s Using Cramer’s rule
952 ( s 2 +5s + 6) vb ( t ) = ( s + 6) vs ( t ) =( s + 6) ( 36 ) The characteristic equation is s 2 + 5s + 6 = 0 . The natural
frequencies are s1,2 = −2, −3 . The natural response has the
form vn (t ) = A1 e −2 t + A2 e −3 t . Try v f ( t ) = B as the forced
response. Substituting into the differential equation gives
B = 36 so v f ( t ) = 36 V. The complete response has the form
vb (t ) = A1 e −2 t + A2 e −3t + 36 . Next
vb (0+ ) = 36 + A1 + A2
dvb +
(0 ) = −2 A1 − 3 A2
dt
Apply KCL at node a to get At t = 0+ 1 dvb ( t ) vb ( t ) − va ( t )
+
+ i (t ) = 0
18 dt
6 1
1 d vb ( 0
(−2 A1 − 3 A2 ) =
18
18 dt + ) = v ( 0 )−v ( 0 ) − i
+ + a b 6 (0+ ) = 126−0 − 0 = 2 So
0 = vb (0+ ) =36+ A1 + A 2 ⇒
1
( −2 A 1 − 3 A 2 ) = 2 18 Finally vb ( t ) = 36 − 72e −2 t A1 = −72, A2 = 36 + 36e −3t V for t ≥ 0 953 PSpice Problems
SP 91 Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure
to edit the labels of the parts so, for example, there is only one R1.) 954 V(C1:2), V(C2:2) and V(C3:2) are the capacitor voltages, listed from top to bottom. 955 SP 92 Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure
to edit the labels of the parts so, for example, there is only one R1.) 956 V(R2:2), V(R4:2) and V(R6:2) are the output voltages, listed from top to bottom. 957 SP 93 958 SP 94 959 Verification Problems
VP 91 This problem is similar to the verification example in this chapter. First, check the steadystate
inductor current
v ( t ) 25
i (t ) = s
=
= 250 mA
100 100
This agrees with the value of 250.035 mA shown on the plot. Next, the plot shows an
underdamped response. That requires
12 ×10−3 = L < 4 R 2C = 4(100) 2 (2 × 10−6 ) = 8 × 10−2
This inequality is satisfied, which also agrees with the plot.
The damped resonant frequency is given by ω=
d 1
LC 1
− 2 RC 2 2 1
1
=
−
= 5.95 × 103
−6 −6
−3
2(100)(2 × 10 ) ( 2 ×10 ) (12 ×10 ) The plot indicates a maxima at 550.6µs and a minima at 1078.7µs. The period of the damped
oscillation is
T d = 2 (1078.7 µ s − 550.6 µ s) = 1056.2 µ s
Finally, check that
5.95 ×103 = ω d = 2π
2π
=
= 5.949 × 103
−6
1056.2 ×10
Td The value of ω d determined from the plot agrees with the value obtained from the circuit.
The plot is correct. VP 92 This problem is similar to the verification example in this chapter. First, check the steadystate
inductor current.
v ( t ) 15
i (t ) = s
=
= 150 mA
100 100 960 This agrees with the value of 149.952 mA shown on the plot.
Next, the plot shows an underdamped response. This requires
8 ×10−3 = L < 4 R 2C = 4 (100)2 (0.2 × 10−6 ) = 8 × 10−3
This inequality is not satisfied. The values in the circuit would produce a critically damped, not
underdamped, response.
This plot is not correct. Design Problems
DP 91 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
vC ( ∞ ) =
The specifications require that vC ( ∞ ) = 1
so
2
R2 1
=
2 R1 + R 2 R2
R1 + R 2 1 ⇒ R1 = R 2 Next, represent the circuit by a 2nd order differential equation:
KCL at the top node of R2 gives:
KVL around the outside loop gives: vC ( t )
R2
vs ( t ) = L +C d
vC ( t ) = iL ( t )
dt d
iL ( t ) + R1 iL ( t ) + vC ( t )
dt Use the substitution method to get 961 vs ( t ) = L v (t ) d vC ( t )
d
d
+ C vC ( t ) + R1 C
+ C vC ( t ) + vC ( t ) R2 dt R 2
dt
dt = LC L
d R1 d2
+ R1 C vC ( t ) + 1 +
vC ( t ) + vt R2 dt R2 C ( ) dt 2 The characteristic equation is
R1 1+
1
R1 R2
+ s+
s2 + R 2 C L LC = s 2 + 6s + 8 = ( s + 2 ) ( s + 4 ) = 0 Equating coefficients of like powers of s:
1
R2 C + R1
L 1+
= 6 and R1
R2 =8 LC Using R1 = R 2 = R gives
1
R
+ =6 ⇒
RC L 1
=4
LC These equations do not have a unique solution. Try C = 1 F. Then L = 1
H and
4 1
3
1
+ 4 R = 6 ⇒ R 2 − R + = 0 ⇒ R = 1.309 Ω or R = 0.191 Ω
R
2
4
Pick R = 1.309 Ω. Then
vc ( t ) =
iL ( t ) =
At t = 0+ vC ( t )
1.309 + 1
+ A1 e −2 t + A2 e−4 t V
2 d
vC ( t ) = −1.236 A1 e−2 t − 3.236 A2 e−4 t + 0.3819
dt () 0 = vc 0+ = A1 + A2 + 0.5 ( ) = −1.236 A − 3.236 A 0 = iL 0 + 1 2 + 0.3819 Solving these equations gives A1 = 1 and A2 = 0.5, so
vc ( t ) = 1 −2 t 1 −4 t
−e + e V
2
2 962 DP 92 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
vC ( ∞ ) =
The specifications require that vC ( ∞ ) = 1
so
4
R2 1
=
4 R1 + R 2 R2
R1 + R 2 1 ⇒ 3 R 2 = R1 Next, represent the circuit by a 2nd order differential equation:
KCL at the top node of R2 gives:
KVL around the outside loop gives: vC ( t )
R2
vs ( t ) = L +C d
vC ( t ) = iL ( t )
dt d
iL ( t ) + R1 iL ( t ) + vC ( t )
dt Use the substitution method to get
vs ( t ) = L v (t ) d vC ( t )
d
d
+ C vC ( t ) + R1 C
+ C vC ( t ) + vC ( t ) R2 dt R 2
dt
dt L
d R1 d2
v t+
+ R1 C vC ( t ) + 1 +
vt
2 C( ) R2 dt R2 C ( ) dt The characteristic equation is
R1 1+ 1
R1 R2 2
2
2
+ s+
= s + 4s + 4 = ( s + 2 ) = 0
s + R 2 C L LC Equating coefficients of like powers of s:
R
1+ 1
R1
R2
1
+
= 4 and
=4
R2 C L
LC
= LC 963 Using R 2 = R and R1 = 3R gives
1
3R
+
=4 ⇒
RC L 1
=1
LC These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and
1
4
1
1
+ 3 R = 4 ⇒ R 2 − R + = 0 ⇒ R = 1 Ω or R = Ω
R
3
3
3
Pick R = 1 Ω. Then R1 = 3 Ω and R 2 = 1 Ω .
vc ( t ) =
iL ( t ) = vC ( t ) +
At t = 0+ 1
+ ( A1 + A2 t ) e −2 t V
4 d
1
vC ( t ) = +
dt
4 () (( A 0 = vc 0+ = A1 + () 0 = iL 0+ = 2 ) − A1 ) − A2 t e −2 t 1
4 1
+ A2 − A1
4 Solving these equations gives A1 = 0.25 and A2 = 0.5, so
vc ( t ) = 1 1 1 −2 t
− + t e V
4 4 2 DP 93 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
R2
vC ( ∞ ) =
1
R1 + R 2
4
The specifications require that vC ( ∞ ) = so
5 964 R2
4
=
5 R1 + R 2 ⇒ 4 R1 = R 2 Next, represent the circuit by a 2nd order differential equation:
vC ( t ) KCL at the top node of R2 gives: R2
vs ( t ) = L KVL around the outside loop gives: +C d
vC ( t ) = iL ( t )
dt d
iL ( t ) + R1 iL ( t ) + vC ( t )
dt Use the substitution method to get
vs ( t ) = L v (t ) d vC ( t )
d
d
+ C vC ( t ) + R1 C
+ C vC ( t ) + vC ( t ) R2 dt R 2
dt
dt = LC L
d R1 d2
v t+
+ R1 C vC ( t ) + 1 +
vt
2 C( ) R2 dt R2 C ( ) dt The characteristic equation is
R1 1+
1
R1 R2
+ s+
s2 + R 2 C L LC = s 2 + 4 s + 20 = ( s + 2 − j 4 ) ( s + 2 + j 4 ) = 0 Equating coefficients of like powers of s:
R1
1
+
= 4 and
R2 C L 1+ R1
R2 LC = 20 Using R1 = R and R 2 = 4 R gives
1
R
+ = 4 and
4R C L 1
= 16
LC
1
1
These equations do not have a unique solution. Try C = F . Then L = H and
8
2
2
+ 2 R = 4 ⇒ R2 − 2R + 1 = 0 ⇒ R = 1 Ω
R
Then R1 = 1 Ω and R 2 = 4 Ω . Next
vc ( t ) = 0.8 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V 965 iL ( t ) =
+ vC ( t )
4 + A 2 −2 t
A1
1d
vC ( t ) = 0.2 +
e cos 4 t − e−2 t sin 4 t
8 dt
2
2 At t = 0 () 0 = vc 0+ = 0.8 + A1 () 0 = iL 0+ = 0.2 + A2
2 Solving these equations gives A1 = −0.8 and A2 = −0.4, so
vc ( t ) = 0.8 − e −2 t ( 0.8 cos 4 t + 0.4 sin 4 t ) V DP 94 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
vC ( ∞ ) =
The specifications require that vC ( ∞ ) = 1
so
2
R2 1
=
2 R1 + R 2 R2
R1 + R 2 1 ⇒ R1 = R 2 Next, represent the circuit by a 2nd order differential equation:
KCL at the top node of R2 gives:
KVL around the outside loop gives: vC ( t )
R2
vs ( t ) = L +C d
vC ( t ) = iL ( t )
dt d
iL ( t ) + R1 iL ( t ) + vC ( t )
dt Use the substitution method to get 966 v (t ) d vC ( t )
d
d
+ C vC ( t ) + R1 C
+ C vC ( t ) + vC ( t ) R2 dt R 2
dt
dt vs ( t ) = L = LC L
d R1 d2
vC ( t ) + + R1 C vC ( t ) + 1 +
vt R2 dt R2 C ( ) dt 2 The characteristic equation is
R1 1+
1
R1 R2
+ s+
s2 + R 2 C L LC = s 2 + 4 s + 20 = ( s + 2 − j 4 ) ( s + 2 + j 4 ) = 0 Equating coefficients of like powers of s:
R1
1
+
= 4 and
R2 C L 1+ R1
R2 LC = 20 Using R1 = R 2 = R gives
1
R
+ = 4 and
RC L
Substituting L = 1
= 10
LC 1
into the first equation gives
10 C ( RC ) 2 0.4 ± 0.42 − 4 ( 0.1)
4
1
− ( RC ) + = 0 ⇒ RC =
10
10
2 Since RC cannot have a complex value, the specification cannot be satisfied. DP 95 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions 967 vC ( ∞ ) = R2
R1 + R 2 1, iL ( ∞ ) = The specifications require that vo ( ∞ ) = 1
so
2
R2 1
R1 + R 2 1
=
2 R1 + R 2 and vo ( ∞ ) = R2
R1 + R 2 1 ⇒ R1 = R 2 Next, represent the circuit by a 2nd order differential equation:
KVL around the righthand mesh gives:
KCL at the top node of the capacitor gives: d
iL ( t ) + R 2 iL ( t )
dt
vs ( t ) − vC ( t )
d
− C vC ( t ) = iL ( t )
R1
dt
vC ( t ) = L Use the substitution method to get
vs ( t ) = R1 C
= R1 LC Using iL ( t ) = vo ( t )
gives
R2
vs ( t ) = d d
d L iL ( t ) + R 2 iL ( t ) + L iL ( t ) + R 2 iL ( t ) + R1 iL ( t )
dt dt dt d2
d
i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t )
2 L( ) (
dt
dt R1
R2 LC L
d R1 + R 2 d2
+ R1 C vo ( t ) + v t+
vt
2 o( ) R2 dt R2 o ( ) dt The characteristic equation is
R2 1+
1
R2 R1
+
s2 + s+ R1 C L LC Equating coefficients of like powers of s: = s 2 + 6s + 8 = ( s + 2 ) ( s + 4 ) = 0 R2
1
+
= 6 and
R1 C L 1+ R2
R1 LC =8 Using R1 = R 2 = R gives
1
R
+ =6 ⇒
RC L 1
=4
LC These equations do not have a unique solution. Try C = 1 F. Then L = 1
H and
4 968 1
3
1
+ 4 R = 6 ⇒ R 2 − R + = 0 ⇒ R = 1.309 Ω or R = 0.191 Ω
R
2
4
Pick R = 1.309 Ω. Then
1
+ A1 e −2 t + A2 e −4 t V
2
A1 −2 t
A2 −4 t
v (t )
1
iL ( t ) = o
e+
eV
=
+
1.309 2.618 1.309
1.309
vo ( t ) = vC ( t ) = 1.309 iL ( t ) +
At t = 0+ 1d
1
iL ( t ) = + 0.6167 A1 e−2 t + 0.2361 A2 e−4 t
4 dt
2 () 0 = iL 0+ = () 0 = vC 0+ = A1
A2
1
+
+
2.618 1.309 1.309 1
+ 0.6167 A1 + 0.2361 A2
2 Solving these equations gives A1 = 1 and A2 = 0.5, so
vo ( t ) = 1 −2 t 1 −4 t
−e + e V
2
2 DP 96 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
vC ( ∞ ) = R2
R1 + R 2 1, iL ( ∞ ) = The specifications require that vo ( ∞ ) = 1
R1 + R 2 and vo ( ∞ ) = R2
R1 + R 2 1 3
so
4 R2
3
=
4 R1 + R 2 ⇒ 3R1 = R 2 969 Next, represent the circuit by a 2nd order differential equation:
d
iL ( t ) + R 2 iL ( t )
dt
vs ( t ) − vC ( t )
d
− C vC ( t ) = iL ( t )
R1
dt
vC ( t ) = L KVL around the righthand mesh gives:
KCL at the top node of the capacitor gives:
Use the substitution method to get
vs ( t ) = R1 C
= R1 LC Using iL ( t ) = vo ( t )
gives
R2
vs ( t ) = d d
d L iL ( t ) + R 2 iL ( t ) + L iL ( t ) + R 2 iL ( t ) + R1 iL ( t )
dt dt dt d2
d
i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t )
2 L( ) (
dt
dt R1
R2 LC L
d R1 + R 2 d2
+ R1 C vo ( t ) + vo ( t ) + vt R2 dt R2 o ( ) dt 2 The characteristic equation is
R2 1+
1
R2 R1
+
s2 + s+ R1 C L LC Equating coefficients of like powers of s: = s 2 + 4 s + 4 = ( s + 2 )2 = 0 R2
1
+
= 4 and
R1 C L 1+ R2
R1 LC =4 Using R1 = R and R 2 = 3R gives
1
3R
+
= 4 and
RC L 1
=1
LC These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and
1
4
1
1
+ 3 R = 4 ⇒ R 2 − R + = 0 ⇒ R = 1 Ω or R = Ω
R
3
3
3
Pick R = 1 Ω. Then R1 = 1 Ω and R 2 = 3 Ω .
vo ( t ) = 3
+ ( A1 + A2 t ) e −2 t V
4 970 iL ( t ) = vo ( t ) vC ( t ) = 3 iL ( t ) + = 3 1 A1 A2 −2 t
t e V
+ +
4 3
3 3 A1 A2 A2 −2 t
d
iL ( t ) = + +
t e
+
4 3
3 3 dt At t = 0+
0 = iL ( 0 + ) = A1 + 1
4 3
3 A1 A2
0 = vC ( 0 + ) = + +
43
3
Solving these equations gives A1 = −0.75 and A2 = −1.5, so
vo ( t ) = 3 3 3 −2 t
− + t e V
4 4 2 DP 97 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
vC ( ∞ ) = R2
R1 + R 2 1, iL ( ∞ ) = 1
The specifications require that vo ( ∞ ) = so
5
R2
1
=
5 R1 + R 2 1
R1 + R 2 and vo ( ∞ ) = R2
R1 + R 2 1 ⇒ R1 = 4 R 2 Next, represent the circuit by a 2nd order differential equation:
KVL around the righthand mesh gives:
KCL at the top node of the capacitor gives: d
iL ( t ) + R 2 iL ( t )
dt
vs ( t ) − vC ( t )
d
− C vC ( t ) = iL ( t )
R1
dt
vC ( t ) = L 971 Use the substitution method to get
vs ( t ) = R1 C
= R1 LC Using iL ( t ) = vo ( t )
gives
R2
vs ( t ) = d d
d L iL ( t ) + R 2 iL ( t ) + L iL ( t ) + R 2 iL ( t ) + R1 iL ( t )
dt dt dt d2
d
i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t )
2 L( ) (
dt
dt R1
R2 LC L
d R1 + R 2 d2
+ R1 C vo ( t ) + v t+
vt
2 o( ) R2 dt R2 o ( ) dt The characteristic equation is
R2 1+
1
R2 R1
+
s2 + s+ R1 C L LC = s 2 + 4 s + 20 = ( s + 2 − j 4 ) ( s + 2 + j 4 ) = 0 Equating coefficients of like powers of s:
R2
1
+
= 4 and
R1 C L 1+ R2
R1 LC = 20 Using R 2 = R and R1 = 4 R gives
1
R
+ = 4 and
4R C L
These equations do not have a unique solution. Try C = 1
= 16
LC
1
1
F . Then L = H and
8
2 2
+ 2 R = 4 ⇒ R2 − 2R + 2 = 0 ⇒ R = 1 Ω
R
Then R1 = 4 Ω and R 2 = 1 Ω . Next
vo ( t ) = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
iL ( t ) = vo ( t ) vC ( t ) = iL ( t ) +
At t = 0+ 1 = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V 1d
iL ( t ) = 0.2 + 2 A2 e−2 t cos 4 t − 2 A1 e−2 t sin 4 t
2 dt 972 ()
( 0 ) = 0.2 + 2 A 0 = iL 0+ = 0.2 + A1
0 = vC + 2 Solving these equations gives A1 = −0.8 and A2 = −0.4, so
vc ( t ) = 0.2 − e −2 t ( 0.2 cos 4 t + 0.1sin 4 t ) V DP 98 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
vC ( ∞ ) = R2
R1 + R 2 1, iL ( ∞ ) = The specifications require that vC ( ∞ ) = 1
R1 + R 2 and vo ( ∞ ) = R2
R1 + R 2 1 1
so
2
R2 1
=
⇒ R1 = R 2
2 R1 + R 2
Next, represent the circuit by a 2nd order differential equation:
KVL around the righthand mesh gives:
KCL at the top node of the capacitor gives: d
iL ( t ) + R 2 iL ( t )
dt
vs ( t ) − vC ( t )
d
− C vC ( t ) = iL ( t )
R1
dt
vC ( t ) = L Use the substitution method to get
vs ( t ) = R1 C
= R1 LC Using iL ( t ) = vo ( t )
gives
R2 d d
d L iL ( t ) + R 2 iL ( t ) + L iL ( t ) + R 2 iL ( t ) + R1 iL ( t )
dt dt dt d2
d
i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t )
2 L( ) (
dt
dt 973 vs ( t ) = R1
R2 LC L
d R1 + R 2 d2
+ R1 C vo ( t ) + v t+
vt
2 o( ) R2 dt R2 o ( ) dt The characteristic equation is
R2 1+
1
R2 R1
+
s2 + s+ R1 C L LC = s 2 + 4 s + 20 = ( s + 2 − j 4 ) ( s + 2 + j 4 ) = 0 Equating coefficients of like powers of s:
R2
1
+
= 4 and
R1 C L 1+ R2
R1 LC = 20 Using R1 = R 2 = R gives
1
R
+ =4 ⇒
RC L
Substituting L = 1
= 10
LC 1
into the first equation gives
10 C ( RC ) 2 0.4 ± 0.42 − 4 ( 0.1)
4
1
− ( RC ) + = 0 ⇒ RC =
10
10
2 Since RC cannot have a complex value, the specification cannot be satisfied. 974 DP 99
Let’s simulate the three copies of the circuit simultaneously. Each copy uses a different value of
the inductance. The PSpice transient response shows that when L = 1 H the inductor current has its maximum at
approximately t=0.5 s. Consequently, we choose L = 1 H. 975 ...
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This note was uploaded on 06/10/2011 for the course PHYSICS physics taught by Professor Physics during the Spring '11 term at HKU.
 Spring '11
 physics
 Physics, Current, Energy

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