Chapter 4 – Second Order Circuit Transient
Chapter 4
Second Order Circuit Transient
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•
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Natural & Force Responses of Second Order Circuit
RLC Circuit Transient
Oscillatory, Underdamp, Critically Damped and Overdamped
Second Order Circuit Transient Parameters
2.4 Natural response of series RLC Circuit
Consider the series RLC circuit shown in Fig. 2.17. The circuit is being
excited by the energy initially stored in the capacitor and inductor. The initial
capacitor voltage v and inductor current i are given at t = 0 are
i (0) = I 0 and v(0) = V0
(2.43)
Applying KVL around the loop in Fig. 2.17,
di 1 t
0 = Ri + L + ∫ i dt
dt C − ∞
(2.44a)
i
d 2i R di
⇒
+
+
=0
dt 2 L dt LC
(2.44b)
This is a second-order differential equation. To solve such equation, we need
to use the initial conditions of (2.43). At t = 0, the first derivative of i(0) can be
obtained from (2.44a) by substituting (2.43) into (2.44a). Thus,
di (0)
1
= − ( RI 0 + V0 )
dt
L
(2.45)
With the two initial conditions in (2.43) and (2.45), we can now solve (2.44b).
Our experience in the proceeding section on first-order circuits suggests that the
solution is of exponential form. So we let
i = Ae st
-1-
(2.46)
Chapter 4 – Second Order Circuit Transient
Fig. 2.17 Source-free series RLC circuit
Substituting (2.46) into 2.44(b) yields
R
1
s+
)=0
L
LC
Ae st ( s 2 +
(2.47)
The equation inside the parentheses is known as characteristic equation of the
differential equation (2.44b). Since Aest cannot be zero, the roots of the
characteristic equation are
2
R
1
 R
s1 = −
+   −
= −α + α 2 − ω 0 2
LC
2L
 2L 
(2.48a)
2
R
1
 R
s2 = −
−   −
= −α − α 2 − ω 0 2
LC
2L
 2L 
where
α=
(2.48b)
1
R
and ω 0 =
.
LC
2L
The roots s1 and s2 are called natural frequency and measured in neper per
second (Np/s); ω0 is known as the undamped natural frequency which is
expressed in radians per second (rad/s); and α is the damping factor, expressed
in nepers per second. The two values in (2.48) indicate that there are two
possible solutions for i, each of which of the form of the assumed solution in
(2.46). Since (2.44b) is a linear equation, the complete solution requires a linear
combination of two possible solutions. Thus,
i = A1e s1t + A2e s 2 t
(2.49)
where A1 and A2 are determined from the initial values i(0) and di(0)/dt in (2.43).
-2-
Chapter 4 – Second Order Circuit Transient
From (2.48), there are three types of solutions:
1. If α > ω0, we have the overdamped case.
2. If α = ω0, we have the critically damped case.
3. If α < ω0, we have the underdamped case.
Overdamped case (α > ω0)
When C > 4L/R2, both s1 and s2 are negative and real. The solution is
i (t ) = A1e s1t + A2e s 2 t
(2.50)
which decays and approaches zero as t increases. Fig. 2.18(a) illustrates a typical
overdamped response.
Critcally damped case (α = ω0)
When α = ω0 , C = 4R/L2, (2.44b) becomes
di
d 2i
d  di

 di

2
+
2
α
+
α
t
=
0
or
+
α
i
+
α


 +α i = 0
2
dt
dt  dt
dt

 dt

Let f =
(2.51)
di
+ α i , then (2.51) becomes
dt
df
+ αf = 0
dt
which is a first-order O.D.E. with the solution f = A1e −α t , where A1 is a
constant. So, (2.51) then becomes
di
+ α i = A1e − αt or
dt
e αt
di
+ e αtα i = A1
dt
(2.52)
This can be written as
d αt
(e i ) = A1 ⇒ i (t ) = ( A1t + A2 ) e − αt
dt
where A2 is another constant. Fig. 2.18(b) shows the result.
Underdamped case (α < ω0)
-3-
(2.53)
Chapter 4 – Second Order Circuit Transient
For α < ω0, C < 4L/R2. The roots can be written as
k1 = −α + − (ω 02 − α 2 ) = −α + jω d
(2.54a)
k 2 = −α − − (ω 02 − α 2 ) = −α − jω d
(2.54b)
where ω d = ω 02 − α 2 is called the damped natural frequency. The natural
response is
i (t ) = A1e − (α − jω d )t + A2 e −(α + jω d )t
=e
− αt
( A1e
jω d t
+ A2e
− jω d t
)
(2.55)
Using Euler’s identities,
e jθ = cosθ + j sin θ ,
e − jθ = cos θ − j sin θ
we get
i (t ) = e − αt [ A1 (cos ω d t + j sin ω d t ) + A2 (cos ω d t − j sin ω d t )]
⇒ i (t ) = e − αt [( A1 + A2 ) cos ω d t + j ( A1 − A2 ) sin ω d t ]
(2.56)
With the presence of sine and cosine functions, it is clear that the natural
response for this case is exponentially damped and oscillatory in nature. The
response has a time constant of 1/α and a period of T = 2π/ωd. Fig. 2.18(c)
depicts a typical underdamped response.
-4-
Chapter 4 – Second Order Circuit Transient
Fig. 2.18 Natural responses of a series RLC circuit
Three peculiar properties of an RLC network can be concluded.
1. The damping effect is due to the presence of resistance R. The damping
factor α determines the rate at which the response is damped. If R = 0, then α
= 0, and we have an LC circuit with 1
LC as undamped natural frequency.
Since α < ω0 in this case, the response is not only undamped but also
oscillatory.
2. Oscillatory response is possible due to the presence of the two types of
storage elements. Having both L and C allows the flow of energy back and
forth between them.
3. It is difficult to tell the difference between the overdamped and critically
damped responses. With the same initial conditions, the overdamped case has
the longest settling time because it takes the longest time to dissipate the
stored initial energy. If the fastest response without oscillation or ringing, the
critically damped circuit is the right choice.
Example 6 For the circuit in Fig. 2.19, find: (a) i (0 + ) and v(0 + ) , (b)
di (0 + ) dt and dv(0 + ) dt , and (c) i (∞) and v(∞) .
-5-
Chapter 4 – Second Order Circuit Transient
i(t)
t=0
v
0.4F
6Ω
12V
4Ω
2H
Fig. 2.19 Example 6
Solution i(0-) = 12/6 = 2A, v(0-) = 12V
At t = 0+, i(0+) = i(0-) = 2A, v(0+) = v(0-) = 12V
(b)
For t > 0,
vL = Ldi/dt or
di/dt = vL/L
Applying KVL at t = 0+, we obtain,
vL(0+) – v(0+) + 10i(0+) = 0
vL(0+) – 12 + 20 = 0,
or
vL(0+) = -8
Hence, di(0+)/dt = -8/2 = -4 A/s
Similarly,
iC = Cdv/dt,
or
dv/dt = iC/C
iC(0+) = -i(0+) = -2
dv(0+)/dt = -2/0.4 = -5 V/s
(c)
As t approaches infinity, the circuit reaches steady state.
i(∞) = 0 A, v(∞) = 0 V
Example 7 Refer to the circuit shown in Fig. 2.20. Calculate (a) iL (0 + ) ,
vC (0 + ) and vR (0 + ) , (b) diL (0 + ) dt , dvC (0 + ) dt and dvR (0 + ) dt , and (c)
iL (∞) , vC (∞) and vR (∞) .
-6-
Chapter 4 – Second Order Circuit Transient
i(t)
t=0
0.25F
vC
10Ω
2A
0.125H
10V
Fig. 2.20 Example 7
Solution At t = 0-, switch is opened. Consider the circuit shown in Figure (a).
iL(0-) = 0, and vR(0-) = 0. But, -vR(0-) + vC(0-) + 10 = 0, or vC(0-) = -10V.
(a)
At t = 0+, since the inductor current and capacitor voltage cannot change
abruptly, the inductor current must still be equal to 0A, the capacitor has a
voltage equal to –10V. Since it is in series with the +10V source, together
they represent a direct short at t = 0+. This means that the entire 2A from
the current source flows through the capacitor and not the resistor.
Therefore, vR(0+) = 0 V.
(b)
At t = 0+, vL(0+) = 0, therefore LdiL(0+)/dt = vL(0+) = 0,
Thus,
diL/dt = 0A/s, iC(0+) = 2A.
This means that
dvC(0+)/dt = 2/C = 8 V/s.
Now for the value of dvR(0+)/dt. Since vR = vC + 10, then
dvR(0+)/dt = dvC(0+)/dt + 0 = 8 V/s.
(c) As t approaches infinity, we end up with the equivalent circuit shown in
Figure (b).
iL(∞) = 10(2)/(40 + 10) = 400 mA
vC(∞) = 2[10||40] –10 = 16 – 10 = 6V
vR(∞) = 2[10||40] = 16 V
-7-
Chapter 4 – Second Order Circuit Transient
40 Ω
40 Ω
+
+
vR
+
+
vC
10 Ω
2A
iL
vC
vR
+
10 Ω
10V
(a)
+
10V
(b)
Example 8 The voltage in an series RLC network is described by the
differential equation
d 2v
dv
+
4
+4=0
dt
dt 2
subject to the initial condition v(0) = 1 and dv(0)/dt = -1. Determine the
characteristic equation. Find v(t) for t > 0.
2
Solution s + 4s + 4
− 4 ± 42 − 4 × 4
= -2, repeated roots.
= 0, thus s1,2 =
2
v(t) = [(A + Bt) e-2t] ⇒ v(0) = 1 = A
dv/dt = [Be-2t] + [-2(A + Bt) e-2t]
dv(0)/dt = -1 = B – 2A = B – 2 or B = 1.
Therefore, v(t) = [(1 + t) e-2t] V
-8-
Chapter 4 – Second Order Circuit Transient
Example 9 Find i(t) for t > 0 in the circuit of Fig. 2.21.
60Ω
t =0
10Ω
i(t)
30V
1mF
40Ω
2.5H
Fig. 2.21
Solution At t = 0, i(0) = 0, vC(0) = 40x30/50 = 24V
For t > 0, we have a source-free RLC circuit.
α = R/(2L) = (40 + 60)/5 = 20
and ωo =
1
=
LC
1
10
−3
x 2.5
= 20
ωo = α leads to critical damping
i(t) = [(A + Bt) e-20t],
i(0) = 0 = A
di/dt = {[Be-20t] + [-20(Bt) e-20t]},
but di(0)/dt = -(1/L)[Ri(0) + vC(0)] = -(1/2.5)[0 + 24]
Hence,
B = -9.6 or
i(t) =
[-9.6t e-20t] A
2.5 Natural response of parallel RLC circuit
Consider the parallel RLC network shown in Fig. 2.22. Assuming that the
initial values of inductor current i and capacitor voltage v are
i (0) = I 0 and v(0) = V0
-9-
(2.57)
Chapter 4 – Second Order Circuit Transient
Fig. 2.22 Source-free parallel RLC circuit
According to passive sign convention, the current is entering each element:
the current through each element is leaving from the top node. Applying KCL at
the top node gives
v 1 t
dv
+ ∫ vdt + C
= 0 (2.58a) ⇒
R L −∞
dt
d 2v
1 dv 1
+
+
v=0
2
RC dt LC
dt
(2.58b)
Similar to the (2.47), the characteristic equation is obtained as
s2 +
1
1
s+
= 0 (2.59)
RC
LC
The roots of (2.59) are
2
s1, 2
where
1
1
 1 
=−
± 
= −α ± α 2 − ω 02 (2.60)
 −
2 RC
LC
 2 RC 
α=
1
and ω 0 =
2 RC
1
LC
(2.61)
Overdamped case (α > ω0)
When L > 4R2C, the roots of the characteristic equation are real and negative.
Thus, the response is
v(t ) = A1e s1t + A2e s 2 t
(2.62)
Critcally damped case (α = ω0)
For α = ω0 , L = 4RC2, the roots are real and equal so that the response is
v(t ) = ( A1t + A2 ) e − αt
-10-
(2.63)
Chapter 4 – Second Order Circuit Transient
Underdamped case (α < ω0)
For α < ω0, L < 4RC2. The roots are complex and can be expressed as
s1, 2 = −α ± jω d (2.64)
where ω d = ω 02 − α 2 .
The response is
⇒ v(t ) = e − αt [( A1 + A2 ) cos ω d t + j ( A1 − A2 ) sin ω d t ]
(2.65)
The constants A1 and A2 can be determined from the initial conditions. v(0)
and dv(0)/dt are needed. The first term can be directly obtained from (2.57). The
second term can be founded by combining (2.57) and (2.58a), as
V + RI 0
v ( 0)
dv(0)
dv(0)
+ i ( 0) + C
=0 ⇒
=− 0
R
dt
dt
RC
(2.66)
Example 10 For the network in Fig. 2.23, what value of C is needed to make
the response underdamped with unity damping factor (α = 1)?
0.5H
10Ω
C
10mF
Fig. 2.23
Solution Let Co = C + 0.01. For a parallel RLC circuit,
α = 1/(2RCo), ωo = 1/ LCo
α = 1 = 1/(2RCo), we then have Co = 1/(2R) = 1/20 = 50 mF
ωo = 1/ 0.5 x0.5 = 6.32 > α (underdamped)
∴ Co = C + 10 mF = 50 mF
C = 40 mF
-11-
Chapter 4 – Second Order Circuit Transient
Example 11 Find v(t) for t > 0 in the circuit in Fig. 2.24.
5Ω
t =0
25V
0.1H
i(t)
1mF
Fig. 2.24
Solution For t < 0, the switch is on.
v(0) = 0,
i(0) = 25/5 = 5A
For t > 0, the switch is open and we have a source-free parallel RLC circuit.
α = 1/(2RC) = 1/(2x5x10-3) = 100 and ωo = 1/ LC = 1 / 0.1 × 10 − 3 = 100
Since ωo is equal to α, the circuit is critically damped. The response of v(t) can
be written as
v(t) = [(A1 + A2t) e-100t]
Since v(0) = 0, substituting t = 0 into v(t) gives A1 = 0.
By using (2.66), we have
dv(0)/dt = -[v(0) + Ri(0)]/(RC) = -[0 + 5x5]/(5x10-3) = -5000
But,
dv/dt = [(A2 + (-100)A2t)e-100t]
Therefore, dv(0)/dt = -5000 = A2 – 0 ⇒ v(t) = -5000t e-100t V
2.6 Step response of a series RLC circuit
The step response is obtained by the sudden application of a dc source.
Consider the series RLC circuit shown in Fig. 2.25. Applying KVL around the
loop for t > 0,
Ri + L
di
+ v = VS
dt
Since i = C dv/dt, (2.67) can be rearranged as
-12-
(2.67)
Chapter 4 – Second Order Circuit Transient
VS
v
d 2v R dv
+
+
=
dt 2 L dt LC LC
(2.68)
The solution to (2.68) has two components: steady-state (natural) response
and transient (forced) response; that is
v(t ) = vT (t ) + vSS (t )
(2.69)
Fig. 2.25 Step response of series RLC circuit
The transient response is the solution when we set VS = 0 in (2.68) and is the
same as the one obtained in the last two sections. The transient response vT (t )
for the overdamped, underdamped, and critically damped cases are:
v(t ) = A1e s1t + A2e s 2 t
(Overdamped)
v(t ) = ( A1t + A2 ) e − αt
(Critically damped)
v(t ) = e − αt [( A1 + A2 ) cos ω d t + j ( A1 − A2 ) sin ω d t ]
(Underdamped)
The steady-state response is the final value of v(t). In the circuit in Fig. 2.25, the
final value of the capacitor voltage is the same as the source voltage VS. Hence,
vSS (t ) = v(∞) = VS
(2.70)
Thus, the complete solution for the overdamped, underdamped, and critically
damped cases are:
-13-
Chapter 4 – Second Order Circuit Transient
v(t ) = VS + A1e s1t + A2e s 2 t
(Overdamped)
v(t ) = VS + ( A1t + A2 ) e − αt
(Critically damped)
v(t ) = VS + e − αt [( A1 + A2 ) cos ω d t + j ( A1 − A2 ) sin ω d t ]
(Underdamped)
The values of the constants A1 and A2 are obtained from the initial conditions:
v(0) and dv(0)/dt.
Example 12 A branch voltage in a series RLC circuit is described by
d 2v
dv
+ 4 + 8v = 24
dt
dt
If the initial conditions are v(0) = dv(0)/dt = 0. Find v(t).
Solution
s2 + 4s + 8 = 0 leads to s =
− 4 ± 16 − 32
= −2 ± j 2
2
∴ v(t) = VS + (A1cos2t + A2sin2t)e-2t
Since dv(0)/dt = 0, d2v(0)/dt2 also equals to 0.
∴ 8VS = 24 ⇒ VS = 3
v(0) = 0 = 3 + A1 ⇒ A1 = -3
dv/dt = -2(A1cos2t + A2sin2t)e-2t + (-2A1sin2t + 2A2cos2t)e-2t
0 = dv(0)/dt = -2A1 +2A2 or A2 = A1 = -3
∴ v(t) = [3 – 3(cos2t + sin2t)e-2t] volts
-14-
Chapter 4 – Second Order Circuit Transient
Example 13 Consider the circuit in Fig. 2.26. Find vL(0) and vC(0).
40Ω
t=0
2A
0.5H
10Ω
1F
vL
vC 50V
Fig. 2.26
Solution For t = 0-, we have the equivalent circuit in Figure (a). For t = 0+, the
equivalent circuit is shown in Figure (b). By KVL,
v(0+) = v(0-) = 40,
i(0+) = i(0-) = 1
By KCL, 2 = i(0+) + i1 = 1 + i1 which leads to i1 = 1. By KVL, -vL + 40i1 + v(0+)
= 0 which leads to vL(0+) = 40 x 1 + 40 = 80
vL(0+) = 80 V,
40 Ω
vC(0+) = 40 V
10 Ω
i1 40 Ω
+
i
+
+
50V
v
+
vL
−
(a)
10 Ω
v
0.5H
50V
+
−
(b)
Fig. 2.26
2.7 Step response of a parallel RLC circuit
Consider the parallel RLC circuit shown in Fig. 2.27. Applying KCL at the
top node for t > 0
v
dv
+i+C
= IS
R
dt
Since v = L di/dt, (2.71) can be rearranged as
-15-
(2.71)
Chapter 4 – Second Order Circuit Transient
IS
i
d 2i
1 di
+
+
=
dt 2 RC dt LC LC
(2.72)
Fig. 2.27 Parallel RLC circuit with an applied current
The characteristic equation of (2.72) is the same as (2.44b). The complete
solution consists of the steady-state response and transient response; that is
i (t ) = iT (t ) + iSS (t )
(2.73)
In Fig. 2.27, the final value of the current through the inductor is the same as the
source current IS. Thus, the complete solution of i(t) can be obtained as
i (t ) = I S + A1e s1t + A2e s 2 t
(Overdamped)
i (t ) = I S + ( A1t + A2 ) e − αt
(Critically damped)
i (t ) = I S + e − αt [( A1 + A2 ) cos ω d t + j ( A1 − A2 ) sin ω d t ] (Underdamped)
Example 14 For the circuit in Fig. 2.28, find v(t) for t > 0.
2A
1H
t =0
0.04F
v
2Ω
4Ω
50u(t)V
Fig. 2.28
Solution i(0-) = 0, v(0-) = -2 x 6 = -12V
-16-
Chapter 4 – Second Order Circuit Transient
For t > 0, we have a series RLC circuit with a step input.
α = R/(2L) = 6/2 = 3,
ωo = 1/ LC = 1 / 0.04
s = − 3 ± 9 − 25 = −3 ± j 4
Thus, v(t) = Vf + [(Acos4t + Bsin4t)e-3t]
where Vf = final capacitor voltage = 50 V
v(t) = 50 + [(Acos4t + Bsin4t)e-3t]
v(0) = -12 = 50 + A
which gives A = -62
i(0) = 0 = C dv(0)/dt
dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t]
0 = dv(0)/dt = -3A + 4B or B = (3/4)A = -46.5
∴ v(t) = {50 + [(-62cos4t – 46.5sin4t)e-3t]} V
Example 14 Find vo(t) in the circuit of Fig. 2.29 for t > 0.
t=0
10Ω
3A
5Ω
1H
10mF
vo
Fig. 2.29
Solution At t = 0-, we obtain, iL(0) = 3x5/(10 + 5) = 1A
and
vo(0) = 0.
For t > 0, the 10-ohm resistor is short-circuited and we have a parallel RLC
circuit with a step input.
α = 1/(2RC) = (1)/(2x5x0.01) = 10
ωo = 1/ LC = 1/ 1x0.01 = 10
Since α = ωo, we have a critically damped response.
-17-
Chapter 4 – Second Order Circuit Transient
s1,2 = -10
Thus, i(t) = Is + [(A + Bt)e-10t],
Is = 3
i(0) = 1 = 3 + A or
A = -2
vo = Ldi/dt = [Be-10t] + [-10(A + Bt)e-10t]
vo(0) = 0 = B – 10A or
Thus,
B = -20
vo(t) = (200t e-10t) V
2.8 General Second-order Circuits
After studying the series and parallel 2nd order circuit, we are prepared to
apply the ideas to any 2nd order circuit. Given a 2nd order circuit, we determine
its step response x(t) by taking the following 4 steps:
(1) Determine the initial condition x(0) and dx(0)/dt and the final value x( ∞ ).
(2) Find the natural response xn(t) (or transient response xT(t)) by turning off
independent sources and applying KCL and KVL. Once a 2nd order
differential equation is obtained, determine its characteristic roots. We obtain
xn(t) (or xT(t)) with two unknown constants.
(3) Obtain the forced response (or steady state response) as
x f (t ) = x(∞) or xSS (t ) = x(∞)
(2.74)
(4) The total response is now found as the sum of natural and forced responses
x(t ) = xn (t ) + x f (t ) or x(t ) = xT (t ) + xSS (t )
(2.75)
Finally, we determine the constants associated with the natural response by
imposing the initial conditions x(0) and dx(0)/dt determined in step (1).
-18-
Chapter 4 – Second Order Circuit Transient
Example 15 Obtain the differential equation for vo in the circuit of Fig. 2.30.
i
L
R1
vs
R2
vo
C
Fig. 2.30
Solution Let i be the inductor current, we obtain
By applying KCL,
vs = R1i + L di dt + vo
(2.76)
i = vo R2 + C dvo dt
(2.77)
Substituting (2.77) into (2.76) yields,
dvo L dvo
d 2vo
R1
vs =
vo + R1C
+
+ LC 2 + vo
R2
dt R2 dt
dt
⇒ LC
d 2vo
L dvo
R1
+
(
R
C
+
)
+
(
1
+
)vo = vs
1
R2 dt
R2
dt 2
(2.78)
Example 16 Derive the 2nd differential equation for vo in the Fig. 2.31.
C1
R2
vs
R1
C2
vo
Fig. 2.31
Solution Let i1 and i2 be the currents flowing through C1 and C2 respectively,
and let v1 be the voltage across C1.
i1 = C1
dv
dv1
and i2 = C2 o
dt
dt
-19-
(2.79)
Chapter 4 – Second Order Circuit Transient
0 = R2i2 + R1 (i2 − i1 ) + vo
By KVL,
(2.80)
Substituting (2.80) into (2.79) obtains
dvo
dv
dv
+ R1C2 o − R1C1 1
dt
dt
dt
Applying KVL to the outer loop produces,
dv
vs = v1 + R2i2 + vo = v1 + R2C2 o + vo which leads to
dt
dv
v1 = vs − vo − R2C2 o
dt
Substituting (2.82) into (2.81) leads to,
0 = R2C2
d 2vo
dv
dv
R1R2C1C2 2 + ( R1C1 + R2C2 + R1C2 ) o = R1C1 s
dt
dt
dt
(2.81)
(2.82)
(2.83)
Example 17 For the circuit in Fig. 2.32, find i and v for t > 0.
i
4Ω
4u(t)A
1H
v
6Ω
0.25F
Fig. 2.32
Solution For t > 0, we obtain the natural response by considering the circuit
below.
iL
a
1H
4Ω
At node a,
But,
0.25F
vc
6Ω
vc
dv
+ 0.25 c + iL = 0
4
dt
di
vc = L + 6iL
dt
Combining (2.84) and (2.85),
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v
(2.84)
(2.85)
Chapter 4 – Second Order Circuit Transient
diL
d 2i L
+
7
+ 10iL = 0
dt
dt 2
(2.86)
s 2 + 7 s + 10 = 0 ⇒ s1, 2 = −2,−5
(2.87)
iL (t ) = iL (∞) + [ Ae −2t + Be −5t ]
(2.88)
The characteristic equation is
Thus,
where i L (∞) represents the final inductor current = 4(4)/(4 + 6) = 1.6
iL (t ) = 1.6 + [ Ae −2t + Be −5t ]
and
iL(0) = 1.6 + [A+B] or
-1.6 = A+B
(2.89)
diL
= −2 Ae − 2t − 5 Be − 5t
dt
diL (0)
= −2 A − 5 B = 0 ⇒ A = −2.5 B
dt
(2.90)
Combining (2.89) and (2.90) obtains
A = − 8 3 and B = 16 15
So, the inductor current is
iL (t ) = 1.6 + [−(8 3)e −2t + (16 15) e −5t ]
(2.91)
The voltage across the 6-Ω resistor can be obtained as
v(t ) = 6iL (t ) = 9.6 + [−16e −2t + 6.4e −5t ]V
(2.92)
The voltage across the 0.25F capacitor is
vc =
Thus,
diL
16
16
+ 6iL = e − 2t − e − 5t + 9.6 − 16e − 2t + 6.4e − 5t
dt
3
3
32
vc = 9.6 − e − 2t + 1.0667e − 5t
3
{
}
i (t ) = vc 4 = 2.4 − 2.667e − 2t + 0.2667e − 5t A
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(2.93)
Chapter 4 – Second Order Circuit Transient
Exercise
1. The step response of a series RLC circuit is given by
di
d 2i
+
2
+ 5 = 10
dt
dt 2
Given i(0) = 2 and di(0)/dt = 4, solve for i(t).
2. In the following circuit, find v(t) and i(t) for t > 0. Assume v(0) = 0V and i(0)
= 1A.
t=0
i
2Ω
4A
1H
0.5F
v
Assignment
1. A branch voltage in a series RLC circuit is described by
d 2v
dv
+
4
+ 8v = 24
dt
dt 2
If the initial conditions are v(0) = dv(0)/dt = 0, find v(t).
2. Find i(t) for t > 0 in the circuit in Fig. 2.33.
i(t)
12u(t)V
8mH
5µF
2kΩ
Fig. 2.33
3. No energy is stored in the 100mH inductor or the 0.4µF capacitor when the
switch in the circuit shown in Fig. 2.34 is closed. Find vc(t) for t ≥ 0 .
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Chapter 4 – Second Order Circuit Transient
t=0
280Ω
0.1H
48V
vc
0.4µF
Fig. 2.34
4. If the switch in Fig. 2.35 has been closed for a long time before t = 0,
determine: (a) the characteristic equation of the circuit, (b) ix and vR for t > 0.
t=0
16V
ix
12Ω
1/36 µF
8Ω
vR
1H
Fig. 2.35
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