Topic 5

First Order And Second Order Response Of RL

And RC Circuit

• Natural response of RL and RC Circuit• Step Response of RL and RC Circuit• General solutions for natural and step response• Sequential switching• Introduction to the natural and step response of

RLC circuit• Natural response of series and parallel RLC

circuit• Step response of series and parallel RLC circuit

First-Order and Second-Order Response of RL

and RC Circuit

Natural response of RL and RC Circuit

• RL- resistor-inductor• RC-resistor-capacitor• First-order circuit: RL or RC

circuit because their voltages and currents are described by first-order differential equation.

• Natural response: refers to the behavior (in terms of voltages and currents) of the circuit, with no external sources of excitation.

Natural response of RC circuit

Consider the conditions below:1. At t < 0, switch is in a closed

position for along time.2. At t=0, the instant when the

switch is opened3. At t > 0, switch is not close for

along time

•For t ≤ 0, v(t) = V0.

dtRCtv

tdvRC

tv

dt

tdvRC

tv

dt

tdvR

tv

dt

tdvC

ii Rc

1

)(

)(

)()(

0)()(

0)()(

0

RCt

eVtv

RC

t

V

tv

tRC

Vtv

dvRC

duu

dvRCu

du

tv

V

t

0

0

0

)(

0

)(

)(ln

)0(1

ln)(ln

11

1

0

voltage

For t ≥ 0:

• Thus for t > 0,

RCt

eVtv 0)(RCt

eR

V

R

tvtic

0)()(

RCt

eVCtvCtW2

20

2

2

1)(

2

1)(

The graph of the natural response of RC circuit

0

0)(

0

0

teV

tVtv

RCt

• The time constant, τ = RC and thus,

t

eVtv 0)(

• The time constant, τ determine how fast the voltage reach the steady state:

Natural response of RL circuit

Consider the conditions below:1. At t < 0, switch is in a closed

position for along time.2. At t=0, the instant when the

switch is opened3. At t > 0, switch is not close for

along time

• For t ≤ 0, i(t) = I0

For t > 0,

dvL

R

u

du

dtL

R

ti

tdi

tiRdt

tdiL

tiRdt

tdiL

tiRtv

)(

)(

)()(

0)()(

0)()(

LRt

tti

i

eiti

L

Rt

i

ti

tL

Riti

dvL

Rduu

)0()(

)0(

)(ln

)0()0(ln)(ln

10

)(

)0(

current

• Thus for t > 0,

LRteIti 0)(

LRteRI

Rtitv

0

)()(

LRteLI

tiLtw

220

2

2

1

)(2

1)(

Example…The switch in the circuit has been

closedfor along time before is opened at t=0.Find

a) IL (t) for t ≥ 0

b) I0 (t) for t ≥ 0+

c) V0 (t) for t ≥ 0+d)The percentage of the total energy

stored in the 2H inductor that is dissipated in the 10Ω resistor.

Solutiona) The switch has been closed for

along time prior to t=0, so voltage across the inductor must be zero at t = 0-. Therefore the initial current in the inductor is 20A at t = 0-. Hence iL (0+) also is 20A, because an instantaneous change in the current cannot occur in an inductor.

• The equivalent resistance and time constant:

1010402eqR

sec2.010

2

eqR

L

• The expression of inductor current, iL(t) as,

020

)0()(5

tAe

eitit

L

t

b) The current in the 40Ω resistor can be determine using current division,

4010

100 Lii

• Note that this expression is valid for

t ≥ 0+ because i0 = 0 at t = 0-. • The inductor behaves as a short

circuit prior to the switch being opened, producing an instantaneous change in the current i0. Then, 04)( 5

0 tAeti t

c) The voltage V0 directly obtain using Ohm’s law

0160

40)(5

00

tVe

itVt

d) The power dissipated in the 10Ω resistor is

02560

10)(

10

20

10

tWe

Vtp

t

•The total energy dissipated in the 10Ω resistor is

J

dtetW t

256

2560)(0

1010

•The initial energy stored in the 2H inductor is

J

iLW

40040022

1

)0(2

1)0( 2

• Therefore the percentage of energy dissipated in the 10Ω resistor is,

%64100400

256

• Natural response of RL and RC Circuit• Step Response of RL and RC Circuit• General solutions for natural and step response• Sequential switching• Introduction to the natural and step response of

RLC circuit• Natural response of series and parallel RLC

circuit• Step response of series and parallel RLC circuit

First-Order and Second-Order Response of RL

and RC Circuit

Step response of RC circuit

• The step response of a circuit is its behavior when the excitation is the step function, which maybe a voltage or a current source.

Consider the conditions below:1. At t < 0, switch is in a closed

and opened position for along time.

2. At t=0, the instant when the switch is opened and closed

3. At t > 0, switch is not close and opened for along time

• For t ≤ 0, v(t)=V0

For t > 0,

s

s

s

s

Vtv

tdvdt

RC

tvV

tdvdt

RC

dt

tdvRCtvV

tRitvV

)(

)(1

)(

)(1

)()(

)()(

s

s

ss

s

VV

Vtv

RC

t

VVVtvRC

t

Vu

dudv

RC

0

0

)(ln

ln)(ln

1

t

RCt

eVVV

eVVVtv

ss

ss

0

0)(

voltage

• Thus for t >0

t

t

eVVV

VV

VV

eVVVV

sn

sf

nf

ss

0

0

Where

•Vf = force voltage or also known as steady state response

•Vn = transient voltage is the circuit’s temporary response that will die out with time

Graf Sambutan Langkah Litar RC

force

Natural

total

• The current for step response of RC circuit

t

t

t

eR

V

R

V

eVVR

eVVC

dt

dvCti

s

s

s

0

0

0

1

)(1

)(

t

eiti )0()(

Step response of RL Step response of RL circuitcircuit

Consider the conditions below:1. At t < 0, switch is in a opened

position for along time.2. At t=0, the instant when the

switch is closed3. At t > 0, switch is not open for

along time

• i(t)=I0 for t ≤ 0. For t > 0,

RV

RV

s

s

s

s

s

i

didt

L

R

ti

tdidt

L

Rdt

tdi

R

Lti

R

Vdt

tdiLtiRV

tvtiRV

)(

)(

)()(

)()(

)()(

RV

RV

RV

RV

ti

IR

V

t

RV

s

s

ss

s

s

I

ti

L

Rt

ItiL

Rt

u

dudv

L

R

u

dudv

L

R

0

0

)(

0

)(ln

ln)(ln

0

LR

ss tR

VR

V eIti 0)(

Current

•Thus,

0

0)(

0

0

teI

tIti

LR

ss tR

VR

V

0

0)(

)(

0

teIRV

tdt

tdiLtv

LRt

s

QuestionThe switch in the circuit has been open for along time. The initial

chargeon the capacitor is zero. At t = 0,

the switch is closed. Find the

expression fora) i(t) for t ≥ 0b) v(t) when t ≥ 0+

Answer (a)

• Initial voltage on the capacitor is zero. The current in the 30kΩ resistor is

mA350

)20)(5.7()0(i

• The final value of the capacitor current will be zero because the capacitor eventually will appear as an open circuit in terms of dc current. Thus if = 0.

• The time constant, τ is

ms5

10)1.0(10)3020( 63

• Thus, the expression of the current i(t) for t ≥ 0 is

0tmAe3

e3

e)0(i)t(i

t200

105t

3

t

Answer (b)

•The initial value of voltage is zero and the final value is

V150)20)(5.7(Vf

• The capacitor vC(t) is

0tVe150150

e)1500(150

eVVV)t(v

t200

t200

f0fC

t

• Thus, the expression of v(t) is

0tV)e60150(

e)3)(30(e150150)t(vt200

t200t200

• Natural response of RL and RC Circuit• Step Response of RL and RC Circuit• General solutions for natural and step response• Sequential switching• Introduction to the natural and step response of

RLC circuit• Natural response of series and parallel RLC

circuit• Step response of series and parallel RLC circuit

First-Order and Second-Order Response of RL

and RC Circuit

General solutions for natural and step response

• There is common pattern for voltages, currents and energies:

t

eVVVtv ff 0)(

t

eIIIti ff 0)(

t

eWWWtW ff

2

0)(

The general solution can be compute as:

t

exxxtx ff 0)(

Write out in words:

ttanconstime

t

e

iablevar

theofvalue

finalthe

iablevar

theofvalue

initialthe

iablevar

theofvalue

finalthe

timeoffunction

aasiablevar

unknownthe

When computing the step and natural responses of

circuits, it may help to follow these steps:1. Identify the variable of interest for the

circuit. For RC circuits, it is most convenient to choose the capacitive voltage, for RL circuits, it is best to choose the inductive current.

2. Determine the initial value of the variable, which is its value at t0.

3. Calculate the final value of the variable, which is its value as t→∞.

4. Calculate the time constant of the circuit, τ.

• Natural response of RL and RC Circuit• Step Response of RL and RC Circuit• General solutions for natural and step response• Sequential switching• Introduction to the natural and step response of

RLC circuit• Natural response of series and parallel RLC

circuit• Step response of series and parallel RLC circuit

First-Order and Second-Order Response of RL

and RC Circuit

Sequential switching

• Sequential switching is whenever switching occurs more than once in a circuit.

• The time reference for all switchings cannot be t = 0.

Example…

First switch move form a to b at t=0 andsecond switch closed at t=1ms. Find thecurrent, i for t ≥ 0.

• Step 1: current value at t=0- is determine as assume that the first switch at point a and second switch opened for along time. Therefore, the current, i(0-)=10A.

• When t=0, an RL circuit is obtain as .ms1

R

L

• Thus the current for 0 ≤ t ≤ 1ms is,

Ae10i t1000

• At t=t1=1ms,

A

eti

68.3

10)( 11

When switch is closed at t=1ms, the equivalent resistance is 1Ω. Then,

ms21

2

R

L1

• Thus i for t ≥ 1ms is

Ae

etiitt

tt

1

1

1

1

)(

)(

1

68.3

)(

The graph of current for t ≥ 0

• Natural response of RL and RC Circuit• Step Response of RL and RC Circuit• General solutions for natural and step response• Sequential switching• Introduction to the natural and step response of

RLC circuit• Natural response of series and parallel RLC

circuit• Step response of series and parallel RLC circuit

First-Order and Second-Order Response of RL

and RC Circuit

Second order response for RLC c

• RLC circuit: consist of resistor, inductor and capacitor

• Second order response : response from RLC circuit

• Type of RLC circuit:1. Series RLC2. Parallel RLC

Natural response of parallel RLC

•Summing all the currents away from node,

01

0 0 t

dt

dvCIvd

LR

V

•Differentiating once with respect to t,

01

2

2

dt

vdC

L

v

dt

dv

R

01

2

2

LC

v

dt

dv

RCdt

vd

•Assume thatsteAv

02 ststst eLC

Ae

RC

AseAs

0LC

1

RC

sseA

equationsticcharacteri

2st

•Characteristic equation is zero:

012

LCRC

ss

•The two roots:

LCRCRCs

1

2

1

2

12

1

LCRCRCs

1

2

1

2

12

2

•The natural response of series RLC:

tsts eAeAv 2121

• The two roots:

RC2

1

20

21 s

20

22 s

•where:

LC

10

•Summary

20

22

20

21

s

s

RC2

1

0LC

10

Parameter Terminology Value in natural response

s1, s2Charateristic

equation

α Neper frequency

Resonant radian frequency

• The two roots s1 and s2 are depend on α and ωo value.

• 3 possible condition is: 1. If ωo < α2 , the voltage response

is overdamped2. If ωo > α2 , the voltage response

is underdamped3. If ωo = α2 , the voltage response

is critically damped

Overdamped voltage Overdamped voltage responseresponse

• overdamped voltage solution

tsts eAeAv 2121

• The constant of A1 dan A2 can be obtain from,

21)0( AAv

2211

)0(AsAs

dt

dv

• The value of v(0+) = V0 and initial value of dv/dt is

C

i

dt

dv C )0()0(

The process for finding the overdamped

response, v(t) :1. Find the roots of the

characteristic equation, s1 dan s2, using the value of R, L and C.

2. Find v(0+) and dv(0+)/dt using

circuit analysis.

3. Find the values of A1 and A2 by solving equation below simultaneously:

4. Substitute the value for s1, s2, A1 dan A2 to determine the expression for v(t) for t ≥ 0.

21)0( AAv

2211

)0()0(AsAs

C

i

dt

dv C

•Example of overdamped voltage response for v(0) = 1V and i(0) = 0

Underdamped voltage response

•When ωo2 > α2, the roots of

the characteristic equation are complex and the response is underdamped.

•The roots s1 and s2 as,

• ωd : damped radian frequency

dj

j

s

220

2201 )(

djs 2

•The underdamped voltage response of a parallel RLC circuit is

tsineB

tcoseB)t(v

dt

2

dt

1

•The constants B1 dan B2 are real not complex number.

10)0( BVv

211

)0()0(BB

C

i

dt

dvd

C

The two simultaneous equation that determine B1 and B2 are:

Example of underdamped voltage response for v(0) = 1V

and i(0) = 0

Critically Damped voltage response

•A circuit is critically damped when ωo

2 = α2 ( ωo = α). The two roots of the characteristic equation are real and equal that is,

RCss

2

121

• The solution for the voltage is tt eDetDtv 21)(

21

20

)0()0(

)0(

DDC

i

dt

dv

DVv

C

•The two simultaneous equation needed to determine D1 and D2 are,

Example of the critically damped voltage response for

v(0) = 1V and i(0) = 0

The step response of a The step response of a parallel RLC circuitparallel RLC circuit

The step response of a The step response of a parallel RLC circuitparallel RLC circuit

•From the KCL,

Idt

dvC

R

vi

Iiii

L

CRL

•Because dt

diLv

2

2

dt

idL

dt

dv L•We get

•Thus,

Idt

idLC

dt

di

R

Li LLL

2

2

LC

I

LC

i

dt

di

RCdt

id LLL 1

2

2

•There is two approach to solve the equation that is direct approach and indirect approach.

Indirect approachIndirect approach

•From the KCL:

Idt

dvC

R

vvd

L

t0

1

•Differentiate once with respect to t:

01

2

2

dt

vdC

dt

dv

RL

v

01

2

2

LC

v

dt

dv

RCdt

vd

• The solution for v depends on the roots of the characteristic equation:

tsts eAeAv 2121

tsineB

tcoseBv

dt

2

dt

1

tt eDetDv 21

•Substitute into KCL equation :

tstsL eAeAIi 21

21

teB

teBIi

dt

dt

L

sin

cos

2

1

ttL eDetDIi 21

Direct approach

•It is much easier to find the primed constants directly in terms of the initial values of the response function.

212121 ,,,B,, DDBAA

•The primed constants could be find from

and dt

diL )0()0(Li

•The solution for a second-order differential equation equals the forced response plus a response function identical in form to natural response.

• If If and Vf is the final value of the response function, the solution for the step function can be write in the form,

responsenaturaltheas

formsametheofFunctionIi f

responsenaturaltheas

formsametheoffunctionVv f

Natural response of a series RLC

• The procedures for finding the natural response of a series RLC circuit is the same as those to find the natural response of a parallel RLC circuit because both circuits are described by differential equations that have same form.

Series RLC circuit

•Summing the voltage around the loop,

01

00 Vdi

Cdt

diLRi

t

•Differentiate once with respect to t ,

02

2

C

i

dt

idL

dt

diR

02

2

LC

i

dt

di

L

R

dt

id

•The characteristic equation for the series RLC circuit is,

012 LC

sL

Rs

•The roots of the characteristic equation are,

LCL

R

L

Rs

1

22

2

2,1

@

20

22,1 s

•Neper frequency (α) for series RLC,

sradL

R/

2

And the resonant radian frequency,

sradLC

/1

0

The current response will be overdamped, underdamped or critically damped according to,

220 22

0

220

• Thus the three possible solutions fo the currents are,

tsts eAeAti 2121)(

teB

teBti

dt

dt

sin

cos)(

2

1

tt eDetDti 21)(

Step response of series RLC

•The procedures is the same as the parallel circuit.

Series RLC circuit

Cvdt

diLiRv

•Use KVL,

•The current, i is related to the capacitor voltage (vC ) by expression,

dt

dvCi C

•Differentiate once i with respect to t

2

2

dt

vdC

dt

di C

•Substitute into KVL equation,

LC

V

LC

v

dt

dv

L

R

dt

vd CCC 2

2

•Three possible solution for vC are,

tstsfC eAeAVv 21

21

teB

teBVv

dt

dt

fC

sin

cos

2

1

ttfC eDetDVv 21

Example 1 (Step response of parallel RLC)

The initial energy stored in the circuit is zero.

At t = 0, a DC current source of 24mA isapplied to the circuit. The value of the

resistoris 400Ω.

1. What is the initial value of iL?

2. What is the initial value of diL/dt? 3. What is the roots of the characteristic

equation?4. What is the numerical expression for iL(t)

when t ≥ 0?

Solution1. No energy is stored in the

circuit prior to the application of the DC source, so the initial current in the inductor is zero. The inductor prohibits an instantaneous change in inductor current, therefore iL(0)=0 immediately after the switch has been opened.

2. The initial voltage on the capacitor is zero before the switch has been opened, therefore it will be zero immediately after. Because

dt

diLv L thus 0

)0(

dt

diL

3. From the circuit elements,

812

20 1016

)25)(25(

101

LC

srad

RC

/105

)25)(400)(2(

10

2

1

4

9

82 1025

•Thus the roots of the characteristic equation are real,

srad

s

srad

s

/00080

103105

/00020

103105

442

441

4. The inductor current response will be overdamped.

tstsfL eAeAIi 21

21

•Two simultaneous equation:

0)0(

0)0(

2211

21

AsAsdt

di

AAIi

L

fL

mAAmAA 832 21

• Numerical solution:

0

8

3224)(

80000

20000

tfor

mAe

eti

t

t

L

Example 2 (step response of series RLC)

• No energy is stored in the 100mH inductor or 0.4µF capacitor when switch in the circuit is closed. Find vC(t) for t ≥ 0.

Solution• The roots of the characteristic

equation:

sradjs

sradj

s

/48001400

/48001400

4.01.0

10

2.0

280

2.0

280

2

62

1

•The roots are complex, so the voltage response is underdamped. Thus:

04800sin

4800cos48

14002

14001

tteB

teBv

t

tC

•No energy is stored in the circuit initially, so both vC(0) and dvC(0+)/dt are zero. Then:

12

1

140048000)0(

480)0(

BBdt

dv

Bv

C

C

•Solving for B1’and B2’yields,

VB

VB

14

48

2

1

•Thus, the solution for vC(t),

0

4800sin14

4800cos4848)(

1400

1400

tfor

Vte

tetv

t

t

C