SNUMSE

K.H. Oh

Chapter 4Transients

The time-varying currents and voltages resulting from the sudden application of sources, usually due to switching, are called transients.

SNUMSE

K.H. Oh

1. First-order RC or RL Circuits.

2. Concepts of Transient Response and Steady-State Response.

3. Transient Response of First-Order Circuits to Time Constant.

4. RLC Circuits in DC Steady-State Conditions.

5. Second-Order Circuits.

6. Step Response of a Second-Order System to its Natural Frequency and Damping Ratio.

Goal

SNUMSE

K.H. OhFirst-Order RC Circuits

( ) ( ) 0=+R

tvdt

tdvC CC

( ) stC Ketv =( ) ( ) 0=+ tv

dttdvRC C

C

0=+ stst KeRCKse RCs 1−=

( ) RCtC Ketv −=

Discharge of a Capacitance through a Resistance

( ) RCtiC eVtv −=

( ) iC Vv =+0

KCL at Node

SNUMSE

K.H. Oh

( ) RCtiC eVtv −=

The time interval τ = RC is called the time constant of the circuitIn one time Constant, Voltage decays by factor 1/e

SNUMSE

K.H. Oh

( ) τtssC evvtv −−=

Charging Capacitance from DC Source through a Resistance

( ) ( ) 0=−

+R

vtvdt

tdvC sCC

( ) ( ) sCC vtvdt

tdvRC =+

( ) ( ) 000 =−=+ CC vv

KCL at Node

I.C.

( ) stC eKKtv 21 +=

sst vKeKRCs =++ 12)1(

RCs 1−= svK =1

( ) RCtsC eKvtv /

2−+= svK −=2

:sv Steady State Response (Forces Response)τt

sev −

RC=τ

: Transient Response

SNUMSE

K.H. Oh

( ) τtssC evvtv −−=

SNUMSE

K.H. OhDC STEADY STATE

Steady State (Forced) Response for RLC circuits with DC sources are:

1. Replace Capacitances with open circuits.

2. Replace Inductances with short circuits.

3. Solve the remaining circuit.

( ) 0)( ==dt

tdvCti Cc

( ) 0)( ==dt

tdiLtv LL

At Steady State of DC Circuit,

- Voltage is Constant

- Current is Constant

SNUMSE

K.H. OhExample of DC STEADY STATE

If Time is Sufficiently Large, Steady State becomes.

ARR

vi sx 1

5510

21

=+

=+

=

ViRv xx 52 ==

SNUMSE

K.H. OhRL Circuits

To solve circuits with DC sources, Resistances, & One energy-storage element (Inductance or Capacitance) are:

1. Apply KCL & KVL to write the Circuit Equation.

2. If Integrals, Differentiate to Differential Equation.

3. Assume a Solution form K1 + K2est.

4. Substitute Solution into Differential equation to determine K1 and s

5. Use the initial conditions to determine the value of K2.

6. Write the final solution.

(Alternatively, we can determine K1 by solving the circuit in steady state )

SNUMSE

K.H. OhRL Transient Analysis (Series & Switch On)

Step 1. Apply KCL & KVL to write the circuit equation (No Step 2)( )

svdt

tdiLtRi =+)(

Step 3. Assume a solution of the form K1 + K2est.steKKti 21)( +=

Initial Condition00)( <= tti

SNUMSE

K.H. Oh

( ) LtRs eKRvti −+= 2/

Time constant isRL

=τ

Step 4. Substitute the solution into the differential equation

sst vesLKRKRK =++ )( 221

= Zero= vs

RvK s=1 L

Rs −=

Step 5. Use the initial conditions to determine the value of K2.

0/)0( 2 =+=+ KRvi s ( ) )1(/ LtRs eRvti −−=

Step 6. Write the final solution.

( ) )1(2 τteti −−=250

100==

Rvs ms

RL 2

501.0===τ

SNUMSE

K.H. Oh

( ) )1(2 τteti −−=

( )dt

tdiLtv LL =)(

( ) LtRsL evtv −=( ) )1(/ LtR

s eRvti −−=

( ) τtL etv −=100

Using

In given Circuit Condition, R=100Ω, L=0.1H

SNUMSE

K.H. OhRL Transient Analysis (Parallel & Switch Off)

Initial Condition

00)(&)(1

<== ttvRvti s

Before Switch Off, Current Circulates through R1 , vs & InductorAfter Switch Off, Current from vs disappear

Current flows through inductor return to R2Voltage appears across R2 & Inductor, causing to Current to decaySteady-State Solution becomes Zero State

SNUMSE

K.H. OhStep 1. Apply KCL & KVL Circuit equation (No Step 2)

( )sv

dttdiLtRi =+)(

Step 3&4. Assume Solution form K1 + K2est steKKti 21)( +=

Steady State is Zero State : K1 = 0 steKti 2)( =

τ/2)( teKti −=

2RL

=τ

Step 5. Use the initial conditions to determine the value of K2.

21/)0( KRvi s ==+ ( ) τts eRvti −= 1/

( )dt

tdiLtv LL =)( ( ) 0

1

>−= − teRLvtv ts τ

τ

( ) 00 <= ttv

SNUMSE

K.H. Oh

( ) τts eRvti −= 1/

( )dt

tdiLtv LL =)(

( ) 01

>−= − teRLvtv ts τ

τ

( ) 00 <= ttv

SNUMSE

K.H. OhRC & RL Circuits with General Sources

When Inductor is Linked with General SourceUsing Thévenin Theorem Voltage Source + Resistor

Note : Voltage Source & Current are varying with time

KVL ( ) )()( tvtRi

dttdiL t

L =+( )

Rtvti

dttdi

RL tL )()( =+

( ) )()( tftxdt

tdx=+τGenerally τ : Time Constant

f(t) : Forcing function

SNUMSE

K.H. OhSolution of 1st Order Differential Equation

( ) )()( tftxdt

tdx=+τ

)()(' trytpy =+ ∫+∫∫=−−

∫dttpdttpdttp

eCdttreey)()()(

)(Appendix

The particular solution (called the forced response) is any expression that satisfies the equation.

The complementary solution (called the natural response) is obtained by solving the homogeneous equation.

)()()( txtxtx cp +=

( ) )(1)(1 tftxdt

tdxττ

=+ )(1)(&1)( tftrtpττ

==

ττττττ

ττ

tttdtdtdteCdttfeeeCdttfeey−−−−

∫∫ +=∫+∫∫= )(1)(1 111

( ) 0)(1=+ tx

dttdx

cc

τ( )

τ1

)(/

−=tx

dttdx

c

c ( ) cttxc +−=τ

ln ( ) τt

c Ketx−

=

SNUMSE

K.H. OhStep-by-Step Solution

Circuits with a resistance, a source, and an inductance (or a capacitance)

1. Write Circuit Equation as 1st Order Differential Equation.

2. Find a Particular Solution. The details of this step depend on the form of the forcing function.

3. Obtain the Complete Solution by adding the particular solution to the Complementary Solution

4. Use Initial Conditions to find the value of K.

SNUMSE

K.H. Oh

0)200sin(2)0()(1)(0

=−++ ∫ tvdttiC

tRi c

t

Use KVL

Step 1. Write Circuit Equation as 1st-Order Differential Equation.

)200cos(400)(1)( ttiCdt

tdiR =+ )200cos(400)()( tCtidt

tdiRC =+

Step 2. Find a particular solution.

)200sin()200cos()( tBtAtip +=Using General Solution orGuessing from Math.

)200cos(10400)()(105 63 ttidt

tdi −− ×=+× Atttip µ)200sin(200)200cos(200)( +=

A=B=200µA

SNUMSE

K.H. OhStep 3. Obtain the Complete Solution

0)()(=+ ti

dttdiRC ( ) τ

tRCt

c KeKeti−−

==

RCt

KeAttti−

++= µ)200sin(200)200cos(200)(

Step 4. Use initial conditions to find the value of K.

AKAR

vi R µµ )200(2005000

1)0()0( +=−=−

=+

=+

AK µ400−=

AeAttti RCt

µµ−

−+= 400)200sin(200)200cos(200)(

msRC 25==τ

SNUMSE

K.H. Oh

AeAttti RCt

µµ−

−+= 400)200sin(200)200cos(200)(

SNUMSE

K.H. OhSummary of 1st Order Diff. Equation

τττ

τ

ttt

eCdttfeeti−−

∫ += )(1)(

General Voltage SourceConstant Voltage Source (DC)

st

ttt

eKK

eCKdteeti

21

1)(

+=

+=−−

∫ τττ

τ

τττ

τ

ttt

eCdttfeeti−−

∫ += )(1)(

SNUMSE

K.H. OhSecond-Order Circuits

( ) ( ) ( ) ( ) ( )tvvdttiC

tRidt

tdiL sC

t

=+++ ∫ 01

0

Using KVL

Taking Derivative & Divide by L

( ) ( ) ( ) ( )dt

tdvL

tiLCdt

tdiLR

dttid s11

2

2

=++

SNUMSE

K.H. Oh

( ) ( ) ( ) ( )tftidt

tdidt

tid=++ 2

02

2

2 ωα

LR

2=α

LC1

0 =ω

( )dt

tdvL

tf s1)( =

: Damping Coefficient

: Undamped Resonant Frequency

: Forcing function

Linear 2nd Order Differential Equation

Here We know that )()()( txtxtx cp +=

)(txp

)(txc

( ) ( ) ( ) ( )tftxdt

tdxdt

txdp

pp =++ 202

2

2 ωα

( ) ( ) ( ) 02 202

2

=++ txdt

tdxdt

txdc

cc ωα

:

:

: Particular Solution

: Complementary Solution

SNUMSE

K.H. Oh

20

21 ωαα −+−=s

20

22 ωαα −−−=s

0ωαζ =

( ) ( ) ( ) ( )tftxdt

tdxdt

txdp

pp =++ 202

2

2 ωα)(txp : Particular Solution :

In Electric Circuit, forcing function is Primarily DC & Sinusoidal SourcesDC Source : Steady State Solution

Can be obtained by Replacing Inductor to Short Circuit & Capacitor to Open Circuit

Sinusoidal Source : Complex Number Analysis (Chapter 5)

( ) ( ) ( ) 02 202

2

=++ txdt

tdxdt

txdc

cc ωα)(txc : Complementary Solution :

Let xc(t)=Kest 0)2( 22 =++ sto Kess ωα

02 22 =++ oss ωαGenerally : Characteristic Equation

Two Solutions :

: Damping Ratio

SNUMSE

K.H. Oh2nd Order Circuit Analogy

( ) ( ) ( ) ( )dt

tdvL

dttiLCdt

tdiLR

dttid s11

2

2

=++( ) ( ) ( ) )(12

2

tfm

txmk

dttdx

mC

dttxd

=++

SNUMSE

K.H. Oh

1. Overdamped case : ζ > 1 or α > ω0 : Real & Distinct Two Roots

( ) tstsc eKeKtx 21

21 +=

2. Critically damped case : ζ = 1 α = ω0 : Real & Equal Roots.

( ) tstsc teKeKtx 11

21 +=

3. Underdamped case : ζ < 1 or α < ω0 : Complex Roots

nn jsjs ωαωα −−=+−= 21 &

Natural frequency is given by 220 αωω −=n

( ) ( ) ( )teKteKtx nt

nt

c ωω αα sincos 21−− +=

In electrical engineering, we use j rather than i to stand for square root of -1, because of i for current.

Solution Case of 2nd Order Equation

SNUMSE

K.H. OhAnalysis of 2nd Order Circuit with DC Source

( )dt

tdvCti c=)(KVL : ( ) ( ) ( ) sC VtvtRi

dttdiL =++

( ) ( ) ( ) sCcc Vtvdt

tdvRCdt

tvdLC =++2

2 ( ) ( ) ( )LCVtv

LCdttdv

LR

dttvd s

Ccc =++

12

2

LR

2=α

LC1

0 =ω

( ) ( ) ( )LCVtv

dttdv

dttvd s

Cocc =++ 2

2

2

2 ωα

SNUMSE

K.H. OhParticular Solution

Set Inductor as Short CircuitCapacitor as Open Circuit

scp Vtv =)(

5.1&105.1 4 =×= ζα

40 101

==LC

ωComplementary Solution

R=300Ωtsts

c eetv 21 708.11708.110)( −+=

41 10618.2 ×−=s

42 10382.0 ×−=s

1&104 == ζαR=200Ωtsts

c teetv 21 5101010)( −−=

421 10−== ss

5.0&105.0 4 =×= ζαR=100Ω)sin(774.5)cos(1010)( tetetv n

tn

tc ωω αα −− −−=

8660=nω

SNUMSE

K.H. Oh

R=300Ω R=200Ω

R=100Ω

SNUMSE

K.H. OhGeneral RLC Circuit with DC Source

3,2,1,5.0,1.0=ζ

To avoid Overshoot,Design Damping Ratio to 1

SNUMSE

K.H. OhRLC Circuit with Parallel L & C

( ) ( ) ( ) ( ) ( )tiidttvL

tvRdt

tdvC nL

t

=+++ ∫ 011

0

Using KCL

Taking Derivative & Divide by C

( ) ( ) ( ) ( )dt

tdiC

tvLCdt

tdvRCdt

tvd n1112

2

=++

SNUMSE

K.H. Oh

( ) ( ) ( ) ( )tftvdt

tdvdt

tvd=++ 2

02

2

2 ωα

RC1

=α

LC1

0 =ω

( )dt

tdiC

tf n1)( =

: Damping Coefficient

: Undamped Resonant Frequency

: Forcing function

Linear 2nd Order Differential Equation

)()()( txtxtx cp +=

All treatments are same as Serial !