Engr228 - Chapter 8, Nilsson 10e 1
Chapter 8
Engr228
Circuit Analysis
Dr Curtis Nelson
Chapter 8 Objectives
• Be able to determine the natural and the step response of parallel RLC circuits;
• Be able to determine the natural and the step response of series RLC circuits.
Engr228 - Chapter 8, Nilsson 10e 2
RL and RC Circuit Review
• Transient, natural, or homogeneous response:– Fades over time;– Resists change.
• Forced, steady-state, particular response:– Follows the input;– Independent of time passed.
• The total response will be of the form:
tt
ekk-
+ 21
The RLC Circuit
• RLC circuits contain both an inductor and a capacitor;• These circuits have a wide range of applications including
oscillators, frequency filters, flight simulation, modeling automobile suspensions, and more;
• The response of RLC circuits with DC sources and switches will consist of a natural response and a forced response:
v(t) = vf(t)+vn(t)
The complete response must satisfy both the initial conditions and the final conditions of the forced response.
Engr228 - Chapter 8, Nilsson 10e 3
Source-Free Parallel RLC Circuits
We will first study the natural response of second-order circuit by looking at a source-free parallel RLC circuit:
LR+v(t)-
iR(t)C
iC(t)iL(t)
0)(1)(1)(
0)()(1)(0
2
2
=++
=++
=++
ò
tvLdt
tdvRdt
tvdC
dttdvCdttv
LRtv
iii CLR
Second-orderDifferential equation
This second-order differential equation can be solved by assuming the form of a solution:
0)(1)(1)(2
2
=++ tvLdt
tdvRdt
tvdC
stAetv =)(
011
0)11(
011
2
2
2
=++
=++
=++
Ls
RCs
Ls
RCsAe
AeL
AseR
eCAs
st
ststst
which means
Source-Free Parallel RLC Circuits
• This is known as the characteristic equation.
Engr228 - Chapter 8, Nilsson 10e 4
LCRCRCs
LCRCRCs
121
21
121
21
2
2
2
1
-÷øö
çèæ--=
-÷øö
çèæ+-=
Using the quadratic formula, we get
Source-Free Parallel RLC Circuits
Define resonant frequency:LC1
0 =w
Define damping factor:RC21
=a
Then: 20
22,1 waa -±-=s
Second-Order Differential Equation Solution
LCRCRCs 1
21
21 2
2,1 -÷øö
çèæ±-=
20
22,1 waa -±-=s
We will now divide the circuit response into three cases according to the sign of the term under the radical.
α > ω0 (overdamped): !(#) = &'()*+ + &-().+
α = ω0 (critically damped):
α < ω0 (underdamped):
stst eAteAtv 21)( +=
)sincos()( 21 tBtBetv ddt wwa += -
Engr228 - Chapter 8, Nilsson 10e 5
Types of Circuit Responses
Given initial conditions:vc(0) = 0, iL(0) = -10A
Find v(t) in the circuit at the right. Ignore the current arrows.
5.321
==RC
a 610 ==
LCw
α > ω0 therefore this is an overdamped case
20
22,1 waa -±-=s s1 = -1, s2 = -6
Overdamped Example
Engr228 - Chapter 8, Nilsson 10e 6
The solution is of the form: tt eAeAtv 621)( -- +=
Use initial conditions to find A1 and A2
From vc(0) = 0 at t = 0:
210
20
10)0( AAeAeAv +=+==
From KCL taken at t = 0:
( )( ) 4206
06421)10(0
0)()10()0(0
21
0
621
0
=--
=--+-+
=+-+
=++
=
--
=
AA
eAeAR
dttdvC
Rv
iii
t
tt
t
CLR
Overdamped Case - continued
Solving the two equations we get A1 = 84 and A2 = -84The solution is:
Veeeetv tttt )(848484)( 66 ---- -=-=
t
v(t)
)(84)( 6tt eetv -- -=
Overdamped Case - continued
Engr228 - Chapter 8, Nilsson 10e 7
Example: Overdamped RLC Circuit
vC(t) = 80e−50,000t − 20e−200,000t V for t > 0
Find vC(t) for t > 0.
7H8.573Ω+v(t)-
iR(t)
1/42f
iC(t)iL(t)
Given initial conditions:vc(0) = 0, iL(0) = -10A
Find v(t) in the circuitat the right.
45.2121
0 ====LCRC
wa
Critically damped when α = ω0 s1 = s2 = -2.45
The complete solution is of the form:
stst eAteAtv 21)( +=
Critically Damped Case (α = ω0)
Engr228 - Chapter 8, Nilsson 10e 8
Use initial conditions to find A1 and A2
From vc(0) = 0 at t = 0:
20
20
1 )0(0)0( AeAeAv =+==
Find A1 from KCL at t = 0:
( )
0)(42110
0)45.2(421)10(0
0)()10()0(0
1
0
45.21
45.21
0
=+-
=+-+-+
=+-+
=++
=
--
=
A
eAetAR
dttdvC
Rv
iii
t
tt
t
CLR
Therefore A2 = 0 and the solution is reduced to tteAtv 45.21)( -=
Critically Damped Case - continued
Solving the equation: A1 = 420The solution is:
Vtetv t45.2420)( -=
t
v(t)
ttetv 45.2420)( -=
Critically Damped Case - continued
Engr228 - Chapter 8, Nilsson 10e 9
Critically Damped Example
Find R1 such that the circuit is critically damped for t > 0 and R2such that v(0)=2V.
Answer: R1 = 31.63 kΩ, R2=0.4Ω
20
22,1 waa -±-=s
For the underdamped case, the term inside the bracket will be negative and s will be a complex number.
Define 220 aww -=d
Then djs wa ±-=2,1
)()(
)(
21
)(2
)(1
tjtjt
tjtj
dd
dd
eAeAetveAeAtv
wwa
wawa
--
--+-
+=
+=
Underdamped Case (α < ω0)
Engr228 - Chapter 8, Nilsson 10e 10
)()( 21tjtjt dd eAeAetv wwa -- +=
Using Euler’s Identity qqq sincos je j +=
)sincos()(
sin)(cos)(()(
)sincossincos()(
21
2121
2211
tBtBetvtAAjtAAetv
tjAtAtjAtAetv
ddt
ddt
ddddt
ww
ww
wwww
a
a
a
+=
-++=
-++=
-
-
-
)sincos()( 21 tBtBetv ddt wwa += -
Underdamped Case - continued
)
Looking at the magnitude:
A pendulum is an example of an underdamped second-order mechanical system.
t
displacement(t)
Mechanical Analogue
Engr228 - Chapter 8, Nilsson 10e 11
7H10.5Ω+v(t)-
iR(t)
1/42f
iC(t)iL(t)
Given initial conditions:vc(0) = 0, iL(0) = -10A
Find v(t) in the circuitat the right.
221
==RC
a 610 ==
LCw
α < ω0 therefore, this is an underdamped case
2220 =-= awwd
v(t) is of the form: )2sin2cos()( 212 tBtBetv t += -
Underdamped Case - Example
Use initial conditions to find B1 and B2
From vc(0) = 0 at t = 0:
Find B2 from KCL at t = 0:
( )0)2(
421
10
02sin22cos2421
)10(0
0)(
)10()0(
0
2
0
22
22
0
=+-
=-+-+
=+-+
=++
=
--
=
B
teBteBR
dttdvC
Rv
iii
t
tt
t
CLR
Therefore B1 = 0 and the solution is reduced to121
0 )0sin0cos()0( BBBev =+=
)2sin()( 22 tBetv t-=
Underdamped Case - continued
Engr228 - Chapter 8, Nilsson 10e 12
29722102 ==B
The solution is: tVetv t 2sin297)( 2-=
t
v(t)
tVetv t 2sin297)( 2-=
Underdamped Case - continued
Solving:
Underdamped Example
Find iL for t > 0.
iL = e−1.2t (2.027 cos 4.75t + 2.561 sin 4.75t) A
Engr228 - Chapter 8, Nilsson 10e 13
Summary of Transient Responses
Source - Free Series RLC Circuit
0)(1)()(
0)(1)()(
0
2
2
=++
=++
=++
ò
tiCdt
tdiRdttidL
dttiCdt
tdiLRti
vvv CLR
LR +
vL(t)-
+ vR(t) -
C
i(t)
-vC(t)
+
Engr228 - Chapter 8, Nilsson 10e 14
Comparing Series and Parallel RLC Circuits
0)(1)()(2
2
=++ tiCdt
tdiRdttidL
Series RLCParallel RLC
0)(1)(1)(2
2
=++ tvLdt
tdvRdt
tvdC
tsts eAeAtv 2121)( +=
LCRCRCs 1
21
21 2
2,1 -÷øö
çèæ±-=
20
22,1 waa -±-=s
RC21
=aLC1
0 =w
tsts eAeAti 2121)( +=
LCLR
LRs 1
22
2
2,1 -÷øö
çèæ±-=
20
22,1 waa -±-=s
LR2
=aLC1
0 =w
Series RLC Circuit Solution
If:α > ω0 (overdamped):
α = ω0 (critically damped):
α < ω0 (underdamped):
w 0 = 1LC
a =R2L
tsts eAeAti 2121)( +=
( )21)( AtAeti t += -a
( ) ( )( )tBtBeti ddt wwa sincos)( 21 += -
s1 = -a + a 2 -w02
s2 = -a - a 2 -w02
Engr228 - Chapter 8, Nilsson 10e 15
Textbook Problem 8.49 - Nilsson 10th
vo(t) = 16 - 16e−400tcos300t − 21.33e−400tsin300t V
The circuit contains no initial energy. Find vo(t) for t ≥ 0.
Summary: Solving RLC Circuits
1. Identify the series or parallel RLC circuit;2. Find α and ω0;3. Determine whether the circuit is overdamped, critically
damped, or underdamped;4. Assume a solution (natural response + forced response):
fstst VeAteA ++ 21
ftsts VeAeA ++ 21
21
fddt VtBtBe ++- )sincos( 21 wwa
Overdamped
Critically damped
Underdamped
5. Find A, B, and Vf using initial and final conditions.
Engr228 - Chapter 8, Nilsson 10e 16
Chapter 8 Summary
• Showed how to determine the natural and the step response of parallel RLC circuits;
• Showed how to determine the natural and the step response of series RLC circuits.