see 1023 circuit theory chapter 4 second order circuit (11 th & 12 th week) prepared by : jaafar...

SEE 1023Circuit Theory

Chapter 4 Second Order Circuit

(11th & 12th week)

Prepared by : jaafar shafie

• In first order circuit, the RC and RL circuits are represented in first order differential equation.

• This is due to the existence of only one storage element at any particular circuit.

• In this chapter, two storage elements will exist in a particular circuit.

• Thus, this circuit are characterized by second order differential equation.

• A circuit with second order differential equation is called SECOND ORDER CIRCUIT.

Second Order Circuit

• Types of second order circuit that may exist:-

1. Series RLC circuit,

2. Parallel RLC circuit,

3. RLL circuit,

4. RCC circuit.


V

R L

C I R LC

VR2

L2L1

R1L1

C2

R2C1I

• As usual, the circuit will be analyzed in two parts:-

1. Source–free Circuit (natural response) Energy is initially stored in the element – thus no effect of current or voltage sources.

2. Circuit with source (forced response) Current or voltage sources is directly connected to the first order circuits.

• Before the circuits are being analyzed, one should find the initial and steady state value of the capacitor voltage and inductor current and it’s derivative; i.e.:-

v(0), i(0), dv(0)/dt, di(0)/dt, v(), i().


• Consider the circuit shown below where the switch has been closed for a long time. Find:-

1. v(0+), i(0+),

2. dv(0+)/dt, di(0+)/dt,

3. v(), i().


24V

4Ω

0.4F

t=0s

4Ω

1H

+v(t)

—

i(t)

Second Order Circuit1. When t=0+s=0–s, the inductor is shorted and the capacitor

is opened. The equivalent circuit is shown below,

2. When the switch is opened, the equivalent circuit is

thus,

24V4Ω

4Ω+

v(0+)-

i(0+)

v(0+)=(4/8)24=12V

i(0+)= iC (0+) =3Ai(0+)

24V4Ω 1H

0.4F

i(0+)= 24/8 =3A

and it is know that

dt

dvCi

)0()0(

V/s5.74.0

3)0()0(

C

i

dt

dv

Second Order CircuitApplying KVL to the circuit when switch is opened,

thus,

i(0+)

24V4Ω 1H

0.4F

0L )0()0()0(424 vvi

dt

div

)0()0(

LL

0L 12)0(3424 v

0L )0(v

0L

L )0()0( v

dt

di

Second Order Circuit3. The steady state value,

0)(i

i()

24V4Ω +

v()-

24V)(v

Natural Response - Series RLC Circuit• Natural response is obtained with Series RLC circuit

without source.

• Energy is initially stored in the L and C, where the initial voltage at capacitor is VO and initial current at inductor is IO.

• Applying KVL to the loop,

IO

R L

C+VO

-

01

tidt

dt

dii

CLR

Natural Response - Series RLC Circuit• To eliminate the integral, differentiate with respect to

t and rearrange the terms such that

• Now, we have second order differential equation.• It is known that in first-order differential equation,

the current is

where A and s are constants.• Substitute Aest to the second order equation.

02

2

LCL

R i

dt

di

dt

id

steAi(t)

Natural Response - Series RLC Circuit• Thus, we may write as

or

• And we should find the value of A, thus Aest must not equal to zero. The only part that should equal to zero is

0LC

As

L

ARAs2

ststst e

ee

01

LCs

L

RsA 2ste

01

LC

sL

Rs2

known as CHARACTERISTIC

EQUATION

Natural Response - Series RLC Circuit• Solve the characteristic equation, one might find the

two roots, which are

and

LC2L

R

2L

Rs1

12

LC2L

R

2L

Rs2

12

Natural Response - Series RLC Circuit• On the other hand, we may represent the two roots

as

and

where

• The two solutions for s (i.e. s1 and s2) shows that there are two values of the current, which are

and• The total response of the current would be

22O 1s

LC2L

R 1, O

22O 2s

tsei 11 1A tsei 2

2 2A

tsts eeiii(t) 2121 21 AA

• Thus, we may found three type of response which are

1. 1st Type : > ω0 ,

the response is called OVERDAMPED

2. 2nd Type : = ω0 ,

the response is called CRITICALLY DAMPED

3. 3rd Type : < ω0 ,

the response is called UNDERDAMPED

Natural Response - Series RLC Circuit

• 1st Type : > ω0 , OVERDAMPED

• In this type, > ω0, or C > 4L/R2. It is found that both roots, (i.e. s1 and s2) are negative and real. Thus, the current response is

which decays to zero when t increased.

• A1 and A2 are determined from the initial inductor current and the rate of change of current.


tsts eeiii(t) 2121 21 AA

21 AA(0) i

2211L sAsA

(0)

L

dt

div )0(Solve forA1 and A2

Natural Response - Series RLC Circuit• For example

• A series RLC circuit has R=20Ω, L=1mH and C=100F. If i(0+)=1A and vC(0+)=18V, find the current response.

*It is clear that C > 4L/R2, and the response is the 1st type which is ‘overdamped’.

Step1: Find the value of s1 and s2

srad /167.5131

2

LC2L

R

2L

Rs1

sradk /487.191

2

LC2L

R

2L

Rs2

Natural Response - Series RLC CircuitStep2: Find the value of A1 and A2 from the initial

values.

apply KVL to the loop

thus,

21 AA(0) 1i

0L )0()0()0(20 vvi

0L 18)0(20 v

Vv 2)0( L

2211/21

)0()0(sAsAsAk

m

v

dt

di

2

LL

Natural Response - Series RLC CircuitStep3:

It is found that,

A1 = 0.9216, A2 = 78.36m

Aeei(t) tt 8.1948617.513 78.36m0.9216

-2ms 0ms 2ms 4ms 6ms 8ms 10ms 12ms 14ms 16ms 18msI(L1)

0A

0.5A

1.0A

• 2nd Type : = ω0 , CRITICALLY DAMPED

• In this type, = ω0, or C = 4L/R2. It is found that both roots, (i.e. s1 and s2) are equal to – or –R/2L. Thus, the current response is

• It can be seen that solution for A3 could not be obtained with two initial condition [i(0) & di(0)/dt)].

• Thus, there might be another method to find the response of critically damped.


tttt eeeei(t) 32121 AAAAA )(

• In this type, = ω0 = R/2L = 1/LC. Thus, the second order differential equation become

• Let , thus

• In first order differential equation, it is found that

, thus

, or ,or


02 22

2

idt

di

dt

id 0

i

dt

dii

dt

di

dt

d

idt

dif 0 ff

dt

d

tef 1A

teidt

di 1A 1A iedt

die tt 1Aie

dt

d t

• By integrating the equation, one will found the response of critically damped response as

• Thus, the derivative of the current is,

when t=0s,


tetti 21 AA)(

tt etetdt

tdi 21 AA 1)(

21 AA dt

di )0(



*It is clear that C = 4L/R2, and the response is the 2nd type which is ‘critically damped’.


sradk /101

2

LC2L

R

2L

Rs1

sradk /101

2

LC2L

R

2L

Rs2


values.


thus,

2A(0) 1i

0L )0()0()0(20 vvi

0L 18)0(20 v

Vv 2)0( L

2121

)0()0(AAk

m

v

dt

di

2

LL


It is found that,

A1 = 8k, A2 = 1

tetti 10000)( 18000

Time

-0.5ms 0ms 0.5ms 1.0ms 1.5ms 2.0msI(L1)

0A

0.5A

1.0A

• 3rd Type : < ω0 , UNDERDAMPED

• In this type, < ω0, or C < 4L/R2.

• The roots can be written as

where

and

• The response can be further written as


dO j )( 221s

dO j )( 222s

1j 22 Od

tjtj dd eei(t) )()( 21 AA

Natural Response - Series RLC Circuit• Rearranging the response such that,

• Applying Euler’s identities to the above equation, where

• Thus,

• Let B1=A1+A2 and B2=j(A1–A2), thus

tjtjt dd eeei(t) 21 AA

sincos,sincos jeje jj and

tjtei(t) ddt sincos 2121 AAAA

ttei(t) ddt sincos 21 BB

Natural Response - Series RLC Circuit• Differentiate the i(t), we have

)(sincos

)(cossin)(

tddd

t

tddd

t

ette

ettedt

tdi

2

1

B

B



*It is clear that C < 4L/R2, and the response is the 3rd type which is ‘underdamped’.


sradkjk /20101

2

LC2L

R

2L

Rs1

kandk d 2010


values.


thus,

1B(0) 1i

0L )0()0()0(20 vvi

0L 18)0(20 v

Vv 2)0( L

dsAkm

v

dt

di 21L BB

2

L

/21

)0()0(


It is found that,

B1 = 1, B2 = 0.4,

)20000sin()4.0)20000cos(10000 ttei(t) t (

kandk d 2010

Time

-0.4ms -0.2ms 0ms 0.2ms 0.4ms 0.6ms 0.8ms 1.0msI(L1)

-0.5A

0A

0.5A

1.0A

• In summary, the response for series RCL are

1. 1st Type : > ω0 , (OVERDAMPED)

2. 2nd Type : = ω0 , (CRITICALLY DAMPED)

3. 3rd Type : < ω0 , (UNDERDAMPED)


tsts eei(t) 2121 AA

tetti 21 AA)(

ttei(t) ddt sincos 21 BB

Natural Response - Series RLC Circuit• The comparison of the three responses are

shown below

Time

0s 0.5ms 1.0ms 1.5ms 2.0ms 2.5ms 3.0ms 3.5ms 4.0ms 4.5ms 5.0msI(L1)

-0.5A

0A

0.5A

1.0A

C=2F

C=10FC=100F

• Natural response for Series RLC circuit is obtained without any connection to source.

• Applying KCL, thus we have

• Differentiate with t,

Natural Response - Parallel RLC Circuit

01

dt

dvvdt

v tC

LR

IOR L C+VO

-

+v-

02

2

vdt

dv

dt

vd

LC

1

RC

1

• Replace the first derivative as s and the second derivative as s2. Thus, the characteristic equation is obtained as follows,

• The roots are characterized by these equations


02 LC

1

RC

1ss

LC2RC

1

2RC

1s1

12

LC2RC

1

2RC

1s2

12

Natural Response - Parallel RLC Circuit• On the other hand, we may represent the two roots

as

and

where

22O 1s

LC2RC

1 1, O

22O 2s

• 1st Type : > ω0 , OVERDAMPED

• In this type, > ω0, or L > 4R2C. The roots, (i.e. s1 and s2) are negative and real. Thus, the voltage response is

• A1 and A2 are determined from the initial inductor current and the rate of change of current.


tsts eev(t) 2121 AA

21 AA(0) v

2211 sAsA(0)

RC

R

dt

dviv L )0()0( Solve forA1 and A2

Natural Response - Parallel RLC Circuit• For example

• A parallel RLC circuit has R=20Ω, L=10mH and C=1F. If iL(0+)=0.5A and v(0+)=10V, find the current response.

Step1: Find the value of and ω0

*It is clear that > ωO , and the response is the 1st type which is ‘overdamped’.

1000011

25000)1)(20(2

1

)(10m)(1LC

2RC

1

O

Natural Response - Parallel RLC Circuit• Find s1 and s2.

• The initial condition are

kkkkO 087.2)10()25(25 2222 1s

kkkkO 9.47)10()25(25 2222 2s

10 21 AA(0)v

)(A)(AsAsA(0)

)(

)(

RC

R(0)

212211 kkdt

dv

kiv

dt

dv L

9.47087.2

1000)1(20

)5.0(2010)0()0(

Natural Response - Parallel RLC CircuitStep3:

It is found that,

A1 = -18.09, A2 = 28.09

Veev(t) tktk 9.47087.2 21.36711.367-

Time

0s 0.2ms 0.4ms 0.6ms 0.8ms 1.0ms 1.2ms 1.4ms 1.6ms 1.8ms 2.0msV(R5:2)

-10V

-5V

0V

5V

10V

• 2nd Type : = ω0 , CRITICALLY DAMPED

• In this type, = ω0, or L = 4R2C. From the series RLC circuit, the response found in parallel RLC circuit is

• The derivative is represented as

and at t=0s,


tettv 21 AA)(

tt etetdt

tdv 21 AA 1)(

21 AA dt

dv )0(



Step1: Find the value of and ω0

*It is clear that = ωO , and the response is the 2nd type which is ‘critically damped’.

1000011

10000)1)(50(2

1

)(10m)(1LC

2RC

1

O

Natural Response - Parallel RLC CircuitStep2: Find the value of A1 and A2 from the initial

values.

apply KCL to the loop

2A(0) 10v

)0(AAAA(0)

)(

)(

RC

R(0)

2121 kdt

dv

kiv

dt

dv L

1

700)1(50

)5.0(5010)0()0(

Natural Response - Parallel RLC CircuitStep3:

It is found that,

A1 = -600k, A2 = 10

tettv 10000)( 10600k-

Time

0s 0.1ms 0.2ms 0.3ms 0.4ms 0.5ms 0.6ms 0.7ms 0.8ms 0.9ms 1.0msV(R5:2)

-20V

-10V

0V

10V

• 3rd Type : < ω0 , UNDERDAMPED

• In this type, < ω0, or L < 4R2C.

• The roots can be written as

where

and

• The response can be further written as


dO j )( 221s

dO j )( 222s

1j 22 Od

tjtj dd eev(t) )()( 21 AA

Natural Response - Parallel RLC Circuit• Applying the same method as in series, the

response of the parallel is

• Let B1=A1+A2 and B2=j(A1–A2), thus

tjtev(t) ddt sincos 2121 AAAA

ttev(t) ddt sincos 21 BB

Natural Response - Parallel RLC Circuit• Differentiate the v(t), we have

)(sincos

)(cossin)(

tddd

t

tddd

t

ette

ettedt

tdv

2

1

B

B

• In summary, the responses for parallel RCL are





tsts eev(t) 2121 AA

tettv 21 AA)(

ttev(t) ddt sincos 21 BB

• Step response is obtained in Series RLC circuit with source.

• Applying KVL to the loop,

,where

• Substitute i to the derivative terms,

Step Response - Series RLC Circuit

SVLR vdt

dii

i

R L

C+v-

CVS

dt

dvi C

LCLCL

R SVv

dt

dv

dt

vd

2

2 2nd order diff. equ for capacitor

voltage

Step Response - Series RLC Circuit• To obtain the total response, consider the KVL of

the loop

• Differentiate the equation, we get

• It can be seen that the characteristic equation is the same with source-free series RLC.

• Thus, by looking at the roots of the characteristic equation, one may find the type of the response (either overdamped, critically damped or underdamped).

0C

LR i

dt

id

dt

di2

2

SVC

LR

tidt

dt

dii

1

Step Response - Series RLC Circuit• Furthermore, it can be concluded that the total

response can be represented in terms of ‘transient’ and ‘steady-state’ value,

where vt(t) is the ‘transient response’ and vss(t) is the ‘steady-state’ response.

• The vt(t) is the part which will determine the type of the response, and it reflects to the response for source-free series RLC circuit.

)()()( tvtvtv SSt

Step Response - Series RLC Circuit• The transient responses are:

• While vss(t) is the final voltage value (i.e. when t=) across the capacitor. Normally, it will equal to Vs.

tstst ee(t)v 21

21 AA

tt ettv 21 AA)(

tte(t)v ddt

t sincos 21 BB

overdamped

critically damped

underdamped

SV )(v(t)vSS

Step Response - Series RLC Circuit• Thus, the complete responses for series RLC with

source are:




tsts eev(t) 2121S AAV

tettv 21S AAV)(

tdd ettv(t) sincos 21S BBV

Step Response - Series RLC CircuitExample• A series RLC circuit is shown below. If i(0+)=1A and

vC(0+)=18V, find i(t) and v(t).

Step 1: Find and ω0. Determine type of response.

It is found that > ω0, type of response overdamped.

i(t)

20Ω 1mH

100F+

v(t)-

20V

sradk /102L

R sradk /162.30 LC

1

sradk /487.19,167.51320

2 1,2s

Step Response - Series RLC CircuitStep 3: The steady state voltage across capacitor is

Step2: Find the value of A1 and A2 from the initial values.

the derivative of the response at initial is

thus,

2AA

AA2018(0)

21

21

v

2211/10100

1)0()0(sAsAsVk

i

dt

dv

C

0.473A1.527,A 21

Vv(t)vSS 20)( SV

Step Response - Series RLC CircuitStep3:

It is found that the complete response is

• Next, find the voltage response if R is changed to 6.324Ω and then to 2Ω

Veev(t) tt 8.1948617.513 0.4731.52720

Aee

dt

tdvCi(t)

tt 8.1948617.513

)(

921.68m78.361m

Step Response - Series RLC Circuit• Comparison of the three responses with different R

value.

Time

0s 1ms 2ms 3ms 4ms 5ms 6ms 7ms 8ms 9ms 10msV(L2:2)

18V

19V

20V

21V

22V

20Ω

2Ω

6.324Ω

• Step response is obtained in Parallel RLC circuit with source.

• Applying KCL to the loop,

,where

• Substitute i to the derivative terms,

Step Response - Parallel RLC Circuit

SCR

I idt

dvvdt

div L

LCLCRC

1 SIi

dt

di

dt

id

2

2 2nd order diff. equ for capacitor

voltage

i

R L+v-

CIS

Step Response - Parallel RLC Circuit• It has the same characteristic equation to that of

natural response of parallel/series RLC.• Furthermore, it can also be concluded that the total

response can be represented in terms of ‘transient’ and ‘steady-state’ value,

where it(t) is the ‘transient response’ and iss(t) is the ‘steady-state’ response.

• The it(t) is the part which will determine the type of the response, and it reflects to the response for source-free parallel RLC circuit.

)()()( tititi SSt

Step Response - Parallel RLC Circuit• The transient responses are:

• While iss(t) is the final voltage value (i.e. when t=) across the capacitor. Normally, it will equal to Is.

tstst ee(t)i 21

21 AA

tt etti 21 AA)(

tte(t)i ddt

t sincos 21 BB

overdamped

critically damped

underdamped

SI)( i(t)iSS

Step Response - Parallel RLC Circuit• Thus, the complete responses for parallel RLC with

source are:




tsts eei(t) 21I 21S AA

tetti 21S AAI)(

tdd etti(t) sincosI 21S BB

General Second-Order Circuit• Previously, only the second-order series and

parallel circuits are considered.• If the circuits were neither in series nor in parallel,

what method should be used?• In this topic, we would concentrate to find the

response for a second-order circuit which is neither a series nor parallel circuit.

• The response might be in terms of voltage or current. Thus, generally the response are characterized as x(t).

• In general 2nd-order circuit, the most important part is to find the characteristic equation and find the two roots (s1 and s2). From the two roots, one should know the type of the response.

General Second-Order Circuit• The type of the response are the same with series

and parallel RLC circuit.• Finally, find the initial and steady state value (x(0),

dx(0)/dt and x()) to determine the constant value; i.e. A1 and A2.

• Generally, the response is the summation of the ‘transient’ and ‘steady-state’ value and can be expressed as

where x is either in voltage or current.

)()()( txtxtx SSt

General 2nd-Order Circuit – step by step

Turn off the independent source to find the secondorder differential equation using KCL and/or KVL.

Find the roots and determine the type of the response (i.e. O-D, C-D or U-D)

The 2nd order differential equation would determine whether the response is voltage or current!

Find the steady-state and initial valueto determine A1 and A2.

General Second-Order Circuit - eg• Consider a circuit shown below. Find the complete

response of the voltage v(t)?

Step 1: Turn off the independent source. Find the second order differential equation.

i

4Ω 1H

+v(t)-

0.5F12V2Ω

t >0s

using KCL and KVL to find the 2nd order

differential equations,i

4Ω 1H

+v(t)-

0.5F2Ω

v

General Second-Order Circuit - egKCL at node ‘v’ KVL at mesh ‘i’

We just concern on ‘v’. Thus, substitute ‘i’ to the right hand side equation

dt

dvvdt

dvvi

iii CR

2

1

4

CR

01

0

vdt

dii

vdt

dii

)(4

LR

02

2

vdt

dv

dt

vd65

Thus, the characteristic equation is,

02 65ss

General Second-Order Circuit - egStep 2: The two roots are s1 = -2, and s2 = -3

*It is obvious that the two roots are negative and real. Thus, the transient response is the 1st type (OVERDAMPED)

Step 3: The standard response is

The steady state voltage is

The transient voltage is

ttt ee(t)v 32 21 AA

)()()( tvtvtv SSt

VvtvSS 4)0()(

General Second-Order Circuit - egStep 4: The initial voltage and current is,

At just after the switch is closed, the circuit is shown as

Vvv 12)0()0(

sVdt

dv

dt

dvvi

iii CR

/12)0(

)0()0()0(

)0()0()0(

2

1

2

iC

i

iR

4Ω 1H

+v(t)-

0.5F12V 2Ω

Aii 0)0()0(

General Second-Order Circuit - egThe overall response is

From the overall response, at initial

It is found that A1 = 12 and A2 = -4.

Thus,

123A2A 21

dt

dv )0(

12AA4 21 )0(v

tt eev(t) 32 21 AA4

tt eev(t) 32 4124

see 1023 circuit theory chapter 4 second order circuit (11 th & 12 th week) prepared by : jaafar...

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