Natural and Step Response of Series & Parallel RLC ... · PDF fileNatural and Step Response of...

Natural and Step Response of Series & Parallel RLC Circuits (Second-order Circuits)

Objectives: Determine the response form of the circuit Natural response parallel RLC circuits Natural response series RLC circuits Step response of parallel and series RLC circuits

Natural Response of Parallel RLC Circuits

The problem – given initial energy stored in the inductor and/or capacitor, find v(t) for t ≥ 0.

It is convenient to calculate v(t) for this circuit because

A. The voltage must be continuous for all time

B. The voltage is the same for all three components

C. Once we have the voltage, it is pretty easy to calculate the branch current

D. All of the above



0)(1

)(1)(

0)(1

)(1)(

0)(

)(1)(

2

2

2

2

0

0

dt

tdv

RCtv

LCdt

tvdC

dt

tdv

Rtv

Ldt

tvdC

R

tvIdxxv

Ldt

tdvC

t

:form standard in place to by sides both Divide

:integral the remove to sides both ateDifferenti

:KCL



0)(1

)(1)(

2

2

dt

tdv

RCtv

LCdt

tvd:equation Describing

This equation is Second order Homogeneous Ordinary differential equation With constant coefficients

Once again we want to pick a possible solution to this differential equation. This must be a function whose first AND second derivatives have the same form as the original function, so a possible candidate is

A. Ksin t

B. Keat

C. Kt2



0)(1

)(1)(

2

2

dt

tdv

RCtv

LCdt

tvd:equation Describing

The circuit has two initial conditions that must be satisfied, so the solution for v(t) must have two constants. Use

0)]1()1([)]1()1([

0)(1

)(1

)(

)(

21

212121

21

22

2

211

2

1

2122112

2

21

2

1

21

tsts

tstststststs

tsts

eALCsRCseALCsRCs

eAeALC

eAseAsRC

eAseAs

eAeAtv :SubstituteV;



0)1()1(

)(

)(1)(

2

21

21

2

2

21

LCsRCs

ss

eAeAtv

tvLCdt

tvd

tsts

:EQUATION STICCHARACTERI the for solutions are and Where

:Solution

0dt

dv(t)

RC

1:equation Describing

is called the “characteristic equation” because it characterizes the circuit.

A. True

B. False

0)1()1(2 LCsRCs



rad/s) in frequency radian resonant (the and

rad/s) in frequency neper (the where

0LC

RC

LCRCRCs

LCRCRCsLCsRCs

1

2

1

)1()21()21(

2

)1(4)1()1(;0)1()1(

2

0

22

2,1

2

2,1

2

The two solutions to the characteristic equation can be calculated using the quadratic formula:

So far, we know that the parallel RLC natural response is given by

A. The value of

B. The value of 0

C. The value of (2 - 02)

and where 0LCRC

s

eAeAtv tsts

1

2

1

)(

2

0

2

2,1

2121

There are three different forms for s1 and s2. For a parallel RLC circuit with specific values of R, L and C, the form for s1 and s2 depends on

Natural Response – Overdamped Example

Given V0 = 12 V and I0 = 30 mA, find v(t) for t ≥ 0.

rad/s rad/s,

case! overdamped the is this so

rad/s

rad/s

2

0

000,2050007500000,12

)000,10()500,12(500,12

000,10)2.0)(05.0(

11

500,12)2.0)(200(2

1

2

1

21

222

0

2

2,1

2

ss

s

LC

RC

o



circuit the in equation the in


:circuit the in conditions initial the

satisfy to equation the in tscoefficien the use must we Now

for V,

00

00

000,20

2

5000

1

)()(

)()(

0)(

tt

tt

tt

dt

tdv

dt

tdv

tvtv

teAeAtv



21

)0(000,20

2

)0(5000

1

21

0

21

)0(000,20

2

)0(5000

1

000,205000

)000,20()5000()0(

12

12)0(

)0(

AA

eAeAdt

dv

AA

Vv

AAeAeAv

:Equation

V :Circuit

:Equation

Now we need the initial value of the first derivative of the voltage from the circuit. The describing equation of which circuit component involves dv(t)/dt?

A. The resistor

B. The inductor

C. The capacitor



000,450000,205000

000,450200

1203.0

2.0

1)0()0(

1)0(

))0()0((1

)0(1)0()(

)(

000,205000

)000,20()5000()0(

21

21

)0(000,20

2

)0(5000

1

AA

R

vi

Cdt

dv

iiC

iCdt

dv

dt

tdvCti

AA

eAeAdt

dv

L

RLCC

V/s

:Circuit

:Equation



(OK)

(OK) V :Checks

for V,

Thus,

usly,simultaneo Solving

for V,

0)(

122614)0(

02614)(

26,14

000,450000,205000;12

0)(

000,205000

21

2121

000,20

2

5000

1

v

v

teetv

AA

AAAA

teAeAtv

tt

tt



You can solve this problem using the Second-Order Circuits table:

1. Make sure you are on the Natural Response side. 2. Find the parallel RLC column. 3. Use the equations in Row 4 to calculate and 0. 4. Compare the values of and 0 to determine the

response form (given in one of the last 3 rows). 5. Use the equations to solve for the unknown coefficients. 6. Write the equation for v(t), t ≥ 0. 7. Solve for any other quantities requested in the problem.

The values of the __________ determine whether the response is overdamped, underdamped, or critically damped

A. Initial conditions

B. R, L, and C components

C. Independent sources



Recap:

teBteBtv

jAjAteAAte

tjteAtjteAtv

xjxexjxe

eeAeeAeAeAtv

js

eAeAtv

LCRCsLCsRCs

d

t

d

t

d

t

d

t

dd

t

dd

t

jxjx

tjttjttjtj

dd

tsts

dddd

sincos)(

)(sin)(cos

)sin(cos)sin(cos)(

sincos;sincos

)(

:

)(:

1

2

1;;0)1()1(

21

2121

21

21

)(

2

)(

1

22

02,1

2

0

2

21

2

0

2

2

0

2

2,1

2

21

:identity sEuler' Note

where so d,underdampe

so ,overdamped

; 0

When the response is underdamped, the voltage is given by the equation In this equation, the coefficients B1 and B2 are

A. Real numbers

B. Imaginary numbers

C. Complex conjugate numbers

teBteBtv d

t

d

t sincos)( 21

Natural Response – Underdamped Example

Given V0 = 0 V and I0 = 12.25 mA, find v(t) for t ≥ 0.

08.979sin8.979cos

sincos)(

8.979)200()1000(

1000)125.0)(8(

11

200)125.0)(000,20(2

1

2

1

200

2

200

1

21

222

0

2

tteBteB

teBteBtv

LC

RC

tt

d

t

d

t

d

o

V,

rad/s

case! dunderdampe the is this so

rad/s

rad/s

2

2

0

Now we evaluate v(0) and dv(0)/dt from the equation for v(t), and set those values equal to v(0) and dv(0)/dt from the circuit, solving for B1 and B2. The values for v(0) and dv(0)/dt from the circuit do not depend on whether the response is overdamped, underdamped, or critically damped.

A. True

B. False


0

0)0(

)0(8.979sin)0(8.979cos)0(

1

0

1

)0(200

2

)0(200

1

B

Vv

BeBeBv

V :Circuit

:Equation



V/s

V/s

:Circuit

:Equation

000,988.979200

000,98000,20

0)01225.0(

125.0

1

1)0(

1)0(

8.979200

)0(8.979cos8.979)0(8.979sin)200(

)0(8.979sin8.979)0(8.979cos)200()0(

21

00

2121

)0(200

2

)0(200

2

)0(200

1

)0(200

1

BB

R

VI

Ci

Cdt

dv

BBBB

eBeB

eBeBdt

dv

CC

d



08.979sin100)(

100000,981

8.979200)0(

;0)0(

sincos)(

200

200

2121

01

21

ttetv

BRVIC

BBBBdt

dv

VBv

teBteBtv

t

d

d

t

d

t

V,




Recap:

0:

sincos)(:

)(:

1

2

1;;0)1()1(

2,1

2

0

2

22

0

21

2

0

2

21

2

0

2

2

0

2

2,1

2

21

s

teBteBtv

eAeAtv

LCRCsLCsRCs

d

d

t

d

t

tsts

so damped, Critically

where

so d,underdampe

so ,overdamped

; 0

When the response is critically damped, a reasonable expression for the voltage is

A. True

B. False

0)( 21 teAeAtv tt V,



When the circuit’s response is critically damped, the assumed form of the solution we have been using up until now does not provide enough unknown coefficients to satisfy the two initial conditions from the circuit. Therefore, we use a different solution form:

tttsts eDteDeDteDtv

2121

2

0

2

21)(

: so damped Critically

Natural Response – Critically damped Example


0)(

500)10)(4.0(

11

500)10)(100(2

1

2

1

500

2

500

121

2

teDteDeDteDtv

LC

RC

tttt

o

V,

case! damped critically the is this so

rad/s

rad/s

2

0


5050)0(

)0()0(

20

2

)0(500

2

)0(500

1

DVv

DeDeDv

V :Circuit

:Equation


Use the initial conditions from the equation and from the circuit to solve for the unknown coefficients.


V/s

V/s

:Circuit

:Equation

000,75500

000,75100

5025.0

10

1

1)0(

1)0(

500

)500()0)(500()0(

21

00

21

)0(500

2

)0(500

1

)0(500

1

DD

R

VI

Ci

Cdt

dv

DD

eDeDeDdt

dv

CC



050000,50)(

000,50000,781

500)0(

;50)0(

)(

500500

100

2121

02

500

2

500

1

tetetv

DRVIC

DDDDdt

dv

VDv

eDteDtv

tt

tt

V,


Natural Response of Parallel RLC Circuits – Summary

Use the Second-Order Circuits table: 1. Make sure you are on the Natural Response side. 2. Find the parallel RLC column. 3. Use the equations in Row 4 to calculate and 0. 4. Compare the values of and 0 to determine the

response form (given in one of the last 3 rows). 5. Use the equations to solve for the unknown coefficients. 6. Write the equation for v(t), t ≥ 0. 7. Solve for any other quantities requested in the problem.


Step Response of Second-order RLC Circuits

The problem – find the response for t ≥ 0. Note that there may or may not be initial energy stored in the inductor and capacitor!

The circuit for the parallel RLC step response is repeated here. Consider how this circuit behaves as t ∞. Which component’s final value is non-zero?

A. The resistor

B. The inductor

C. The capacitor

D. All of the above

Step Response of a Parallel RLC Circuits

As t ∞:

response) natural the of form (the FL Iti )(

The only component whose final value is NOT zero is the inductor, whose final current is the current supplied by the source. We now have to construct a response form that can satisfy two initial conditions and one non-zero final value. We can satisfy the final value directly if we specify the inductor current as the response we will solve for:

Step Response of a Parallel RLC Circuit

The problem – there is no initial energy stored in this circuit; find i(t) for t ≥ 0.

To begin, find the initial conditions and the final value. The initial conditions for this problem are both zero; the final value is found by analyzing the circuit as t ∞.



t ∞:

mA 24FI



Next, calculate the values of and 0 and determine the form of the response:

rad/s nm

rad/s n

000,40)25)(25(11

000,50)25)(400(2

1

2

1

0

LC

RC

We just calculated = 50,000 rad/s and 0 = 40,000 rad/s, so the form of the response is

A. Overdamped

B. Underdamped

C. Critically damped

Once we know the response form is overdamped, we know we have to calculate

A. d

B. s1 and s2

C. Nothing additional



Since the response form is overdamped, calculate the values of s1 and s2:

0024.0)(

000,80000,20

000,30000,50

000,40000,50000,50

000,80

2

000,20

1

21

222

0

2

2,1

teAeAti

ss

s

tt

L A,

rad/s and rad/s

rad/s



Next, set the values of i(0) and di(0)/dt from the equation equal to the values of i(0) and di(0)/dt from the circuit.

0)0(

024.0)0(

0

21

Ii

AAi

L

L

:circuit the From

:equation the From

0024.0)( 000,80

2

000,20

1 teAeAti tt

L A,



0)0()0(

000,80000,20)0(

0

21

L

V

L

v

dt

di

AAdt

di

LL

L

:circuit the From

:equation the From

0024.0)( 000,80

2

000,20

1 teAeAti tt

L A,



083224)(

832

0000,80000,20

0024.0

000,80000,20

21

21

21

teeti

AA

AA

AA

tt

L mA,

mA mA;

and

:Solve

0024.0)( 000,80

2

000,20

1 teAeAti tt

L A,

We can check this result at t = 0 and as t ∞; from the equation we get

A. iL(0)=32 mA, iL(∞)=0

B. iL(0)=0, iL(∞)=0

C. iL(0)=0, iL(∞)=24 mA

083224)( 000,80000,20 teeti tt

L mA,

If we now want to find v(t) for t ≥ 0, we need to

A. Find the derivative of the current and multiply by L

B. Find the integral of the current and divide by C

C. Multiply the current by R

083224)( 000,80000,20 teeti tt

L mA,

Step Response of RLC Circuits – Summary

Use the Second-Order Circuits table: 1. Make sure you are on the Step Response side. 2. Find the appropriate column for the RLC circuit topology. 3. Find the values of the two initial conditions and the one

non-zero final value.Make sure the initial conditions and final value are defined exactly as shown in the figure!

4. Use the equations in Row 4 to calculate and 0. 5. Compare the values of and 0 to determine the

response form (given in one of the last 3 rows). 6. Write the equation for vC(t), t ≥ 0 (series) or iL(t), t ≥ 0

(parallel), leaving only the 2 coefficients unspecified. 7. Use the equations provided to solve for the unknown

coefficients. 8. Solve for any other quantities requested in the problem.

Step Response of a Series RLC Circuit

The problem –find vC(t) for t ≥ 0.

Find the initial conditions by analyzing the circuit for t < 0:

A

V

V k k

k

0

50

)80(159

15

0

0

I

V



Find the final value of the capacitor voltage by analyzing the circuit as t ∞:

V 100FV



Use the circuit for t ≥ 0 to find the values of and 0:

rad/s

dunderdampe

rad/s

rad/s

0

6000

8000000,10

000,10

)2)(005.0(11

8000)005.0(2802

2222

0

2

0

2

d

LC

LR



Write the equation for the response and solve for the unknown coefficients:

06000sin67.666000cos50100)(

67.6650

060008000)0(

50100)0(

06000sin6000cos100)(

80008000

21

210

21

101

8000

2

8000

1

ttetetv

BB

BBC

IBB

dt

dv

BVBVv

tteBteBtv

tt

C

dC

FC

tt

C

V,

V V,

V,

Natural Response of Series RLC Circuits

The problem – given initial energy stored in the inductor and/or capacitor, find i(t) for t ≥ 0.

The difference(s) between the analysis of series RLC circuit and the parallel RLC circuit is/are:

A. The variable we calculate.

B. The describing differential equation.

C. The equations for satisfying the initial conditions

D. All of the above



0)(1)()(

0)(

)(1)(

0)()(1)(

2

2

2

2

0

0

tiLCdt

tdi

L

R

dt

tidL

dt

tdiRti

Cdt

tidL

tRiVdxxiCdt

tdiL

t

:form standard in place to by sides both Divide

:integral the remove to sides both ateDifferenti

:KVL

The describing differential equation for the series RLC circuit is Therefore, the characteristic equation is

A. s2 + (1/RC)s + 1/LC = 0

B. s2 + (R/L)s + 1/LC = 0

C. s2 + (1/LC)s + 1/RC = 0

0)(1)()(

2

2

tiLCdt

tdi

L

R

dt

tid



rad/s) in frequency radian resonant (the and

rad/s) in frequency neper (the where

0LC

L

R

sLCsLRs

1

2

;0)1()( 2

0

2

2,1

2

The two solutions to the characteristic equation can be calculated using the quadratic formula:

Natural Response Series RLC Problems

You can solve these problem using the Second-Order Circuits table:

1. Make sure you are on the Natural Response side. 2. Find the series RLC column. 3. Use the equations in Row 4 to calculate and 0. 4. Compare the values of and 0 to determine the

response form (given in one of the last 3 rows). 5. Use the equations to solve for the unknown coefficients. 6. Write the equation for i(t), t ≥ 0. 7. Solve for any other quantities requested in the problem.


Natural Response Series RLC Example

The capacitor is charged to 100 V and at t = 0, the switch closes. Find i(t) for t ≥ 0.

09600sin9600cos)(

9600

000,10)1.0)(1.0(

11

2800)1.0(2

560

2

2800

2

2800

1

22

0

2

tteBteBti

LC

L

R

tt

d

o

A,

rad/s

case! dunderdampe the is this so

rad/s

rad/s

2

0





:circuit the in conditions initial the

satisfy to equation the in tscoefficien the use must we Now

A,

00

00

2800

2

2800

1

)()(

)()(

09600sin9600cos)(

tt

tt

tt

dt

tdi

dt

tdi

titi

tteBteBti

The following quantities used to calculate the unknown coefficients are defined by different equations in both the series and parallel RLC natural response problems:

A. The initial values of voltage or current from the equation.

B. The initial values of voltage or current from the circuit.

C. The initial values of the derivative of voltage or current from the equation.

D. The initial values of the derivative of voltage or current from the circuit.


0

0)0(

)0(

09600sin9600cos)(

1

0

1

2800

2

2800

1

B

Ii

Bi

tteBteBti tt

:Circuit

)case! parallel the as (same :Equation

A,



09600sin104.0)(

104.010009600)0(2800

1000)0(560)100(1.0

1

1)0()0(

1)0(

1)0(

)0(

2800

22

00

21

tteti

BB

RIVL

vvL

vLdt

di

BBdt

di

t

RCL

d

A,

A/s

:Circuit

)case! parallel the as (same :Equation


Natural Response of RLC Circuits – Summary

Use the Second-Order Circuits table: 1. Make sure you are on the Natural Response side. 2. Find the appropriate column for the RLC circuit topology. 3. Make sure the initial conditions are defined exactly as

shown in the figure! 4. Use the equations in Row 4 to calculate and 0. 5. Compare the values of and 0 to determine the

response form (given in one of the last 3 rows). 6. Use the equations to solve for the unknown coefficients. 7. Write the equation for v(t), t ≥ 0 (parallel) or i(t), t ≥ 0

(series). 8. Solve for any other quantities requested in the problem.

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