Chapter 8 - Walla Walla Universitycurt.nelson/engr228/lecture/chap8.pdf · Engr228 -Chapter 8,...

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.


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.


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


• This is known as the characteristic equation.


LCRCRCs

LCRCRCs

121

21

121

21

2

2

2

1

-÷øö

çèæ--=

-÷øö

çèæ+-=

Using the quadratic formula, we get


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 += -


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


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


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)


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)


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


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)


)()( 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


7H10.5Ω+v(t)-

iR(t)

1/42f

iC(t)iL(t)


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-=



29722102 ==B

The solution is: tVetv t 2sin297)( 2-=

t

v(t)

tVetv t 2sin297)( 2-=


Solving:

Underdamped Example

Find iL for t > 0.

iL = e−1.2t (2.027 cos 4.75t + 2.561 sin 4.75t) A


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)

+


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


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.


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.

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