principles & applications of el t i l e i ielectrical...

Konkuk UniversityKonkuk University

Principles & Applications ofEl t i l E i iElectrical Engineering

Rizzoni, 5th Ed.

Ch 5 Transient Analysis

,

Ch. 5 Transient Analysis: first/second order circuits excited by switched DC source: provides general nature of D.E. solution method

ContentsContents Learning objectives

Understanding of transientsgSections 5.1.

Write differential equations for circuits containing q ginductors and capacitorsSections 5.2.

Determine the DC steady state solution of circuits containing inductors and capacitors Section 5.3.

Write differential equations of the first order circuits in standard form, and determine the complete solution of the first order circuits excited by switched DC sources

S ti 5 4

Dept. of Aerospace Information Engineering2

Section 5.4

Write differential equations of the second order circuits in standard form, and determine the complete solution pof second order circuits excited by switched DC sourcesSection 5.5

Understand analogies between electric circuits and hydraulic, thermal, and mechanical systems.y , , y


5.1 Transient analysis5.1 Transient analysis Transient analysis

Is to describe the behavior of a voltage or current during Is to describe the behavior of a voltage or current during the transition between two distinct steady state conditionsconditions

For instance, in the first order circuit,A decaying exponentialA decaying exponentialA rising exponential

1 0.368e


An exemplary representation of general R, L, C circuit with switched DC excitationThe required analysis is regarding ‘Transient

response’ using initial and final conditionsp g


5.2 Writing D.E. for circuit containing L and C5.2 Writing D.E. for circuit containing L and C Differential equations for circuits with dynamic

elementselements

vvdv

preferable1R

RCv

RCv

dtdv SCC


Exam 5.1Derive diff equation of the circuitDerive diff. equation of the circuit


p. 220 standard formulationp. 220 standard formulation First order system equation

Circuit containing only one dynamic elementCircuit containing only one dynamic elementComposed of time constant and DC gain

( ) ( )Sdx x t K f tdt

1 0 0( ) ( )dxa a x t b f tdt

Second order system equationCircuit containing two dynamic elementsg yComposed of natural frequency, damping ratio and DC

gaing2

2 2

1 2 ( ) ( )Sd x dx x t K f tdt dt

2

2 1 0 02 ( ) ( )d x dxa a a x t b f tdt dt


n ndt dt 0 02dt dt

Variables for first and second order D.E.Time constantTime constantDC gain

N t l fNatural frequencyDamping ratio


Exam 5.2Derive differential equation of RLC circuitDerive differential equation of RLC circuit

Use node voltage method and mesh current method, in orderin order

2

( ) ( )L Ld i diR CL R R C L R R i


1 1 2 1 22 ( ) ( )L LL SR CL R R C L R R i v

dt dt

5.3 DC steady state solution of circuits with Inductors and 5.3 DC steady state solution of circuits with Inductors and Capacitors Capacitors –– Initial and final conditionsInitial and final conditions

DC steady state solution of circuits with Inductors and Capacitorsand CapacitorsDC steady state solution provides the initial and final

conditions for the differential equationq

( ) ( )Sdx x t K f t Sx K f as t ( ) ( )S fdt S f

2 21 2 ( ) ( )Sd x d x x t K f t

x K f as t

The DC S S solution assumes steady state i e there

2 2 2 ( ) ( )Sn n

x t K f tdt dt

Sx K f as t

The DC S.S. solution assumes steady state, i.e., there is no change with respect to timeTime derivative of S.S solution zero


For capacitor case,steady state capacitor current by switched DC y p y

source( )( ) C

Cdv ti t C

dt

For inductor case

( ) 0C

dti t as t

For inductor case,steady state inductor voltage by switched DC source

( )di t( )( )

( ) 0

LL

L

di tv t Ldt

v t as t

At DC steady state, all capacitors behave as opencircuit and all inductors as short circuit

( )L


circuit and all inductors as short circuit.

Exam 5.4Determine the inductor current just before the switch isDetermine the inductor current just before the switch is

opened


‘Continuity’ of inductor current and capacitor voltagevoltage inductor current and capacitor voltage cannot change

instantaneously as this abrupt change will cause infiniteinstantaneously as this abrupt change will cause infinite amount of power.

This leads to the consequence thatThis leads to the consequence that Value of an inductor current or capacitor voltage just

prior to the closing (or opening) of a switch is equalprior to the closing (or opening) of a switch is equal to the value just after the switch has been closed (or opened)opened)

(0 ) (0 )C Cv v


(0 ) (0 )L Li i

Exam. 5.5 Initial and final value of inductor current? Initial and final value of inductor current?


5.4 Transient response of 15.4 Transient response of 1stst order D.E.order D.E. First order systems

ElectricalElectricalSingle dynamic element + resistor

H d liHydraulicFlow resistance + fluid capacitance (fluid mass

t )storage)Mechanical

Mass + dampingThermal

Storage + heat dissipation


Analogy in other engineering fieldHydraulic tankHydraulic tank

dq q q in out stored

out

q q qghqR

storeddhq Adt

h dh

( ) 0

gh dhAR dtRA dh h

( ) 0hg dt


5.4 Transient response of 15.4 Transient response of 1stst order D.E.order D.E. First order response

General form of D E ( ) ( )dx x t K f t General form of D.E.Assume the forced input is DC function (i.e.,f(t)=F)

N t l

( ) ( )Sx t K f tdt

Natural responseGiven when there exists no forced input

/

( ) ( ) 1( ) 0 ( )N NN N

dx t dx tx t x tdt dt

Forced response

/( ) tNx t e

pGiven when there exists a forced input of DC

( )dx t


( ) ( ) ( ) ( )FF S F S

dx t x t K F x t K F xdt

Complete responseComplete solution is sum of natural responseComplete solution is sum of natural response

(homogeneous solution) and forced response (particular solution)(particular solution)

( ) ( ) ( )N Fx t x t x t

f

In other form,

/( ) [ (0) ( )] ( )tx t x x e x ( ) [ (0) ( )] ( )x t x x e x


Example 5.7Find out capacitor voltage?Find out capacitor voltage?

Compute S.S. solution and initial condition Write D E and find time constant Write D.E. and find time constant Complete solution

/( ) [ (0) ( )] ( )t RC /( ) [ (0) ( )] ( )t RCC C C Cv t v v e v


Energy storage in capacitor and inductorCapacitorCapacitor

21( ) ( )W t Cv t

I d t

( ) ( )2 CCW t Cv t

Inductor

21( ) ( )W t Li t( ) ( )2L LW t Li t


/( ) tR LL Bi t I e

Equivalent circuit is considered to make a simple first order circuits, only when the action on switch is ybehaved.

Solve the capacitor voltage when switch is turned-on in the following figure.g g


After switched-on, we can use the equivalent circuit transformation as given belowg


Example 5.10Derive capacitor voltage equation?Derive capacitor voltage equation?

Switch closes at t=0, and open at t=50ms again. R1=R2=1k R3=500 R1 R2 1k, R3 500

1( ) ( ) [ (0) ( )] 0 50 sect

v t v v v e t m

2

'

( ) ( ) [ (0) ( )] , 0 50 sec

( ') ( ) [ ( ' 0) ( )] , 0 ' 50 sec( ) ?

C C C Ct

C C C C

v t v v v e t m

v t v v t v e t t mv t

( ) ?Cv t


5.5 Transient response of 25.5 Transient response of 2ndnd order systemorder system Circuit containing RLC

elementselementsContaining both types of

energy storage elementsenergy storage elements 2 simple configuration

L d C i ll lL and C in parallel connection

Diff t t / h• Different current amp/phase

L and C in series connectionconnection

• Different voltage amp/phase


Derivation of 2nd order D.E of RLC circuitsApply three basic equationsApply three basic equations

( ) ( ) ( ) 0( ) ( ) ( ) 0

T T S Cv t R i t v ti t i t i t

( ) ( ) ( ) 0( ) ( ) 0

S C L

L C

i t i t i tv t v t


Further equation development( ) ( ) ( )v t v t dv t( ) ( ) ( ) ( ) 0

( )

T C CL

T

v t v t dv tC i tR dt

di t

( )( ) LC

di tv t Ldt

( ) ( ) ( )t di t di tL d

2

( ) ( ) ( )( ) ( )

( ) ( ) ( )

T L LL

T T

v t di t di tL dC L i tR R dt dt dt

v t d i t di tL

2

( ) ( ) ( ) ( )T L LL

T T

v t d i t di tLLC i tR dt R dt

2

2

( ) ( )( )) ( )C CTC

d v t dv tdv tL Lcf LC v tR dt dt R dt


T TR dt dt R dt

Solution of second order circuits (normal differential equation form)q )

2

2 1 0 02

( ) ( ) ( ) ( )d x t dx ta a a x t b f tdt dt

From an engineering point of view (standard form)dt dt

22 1

2

( ) ( )2 ( ) ( )nn sd x t dx t x t K f t

dt dt

Natural frequency : 0 2/n a a

Damping ratio : DC gain :

1 0 2

0 0

( / 2) 1//

a a aK b a


g0 0/sK b a

The coefficients in the second order differential equation characterizes the system dynamics in terms of q y yoscillation frequency, overshoot characteristics, time response to steady state, and steady state DC gain


Zeta characterizes the overshoot and transient characteristicsDamping coefficient


Natural response of second order system2 ( ) ( )d d

S l ti d t b ti l f

22 1

2

( ) ( )2 ( ) ( )nn sd x t dx t x t K f t

dt dt

Solution assumed to be an exponential form as( ) st

Nx t eBy substituting the solution into governing equation,

2 2 12 0nst st st

n s e s e e Then the characteristic polynomials is given by

nn

2 2 1 2 22 1 0 2 0n nn ns s s s

Then the natural response is obtained using the solution of the characteristic polynomial

n nn n


solution of the characteristic polynomial 1 2

1 2( ) s t s tNx t e e

where2 2 21 (2 ) 4 1

Roots of second order system obtained as three cases

2 2 21,2 (2 ) 4 1

2n n n n ns

Roots of second order system obtained as three casesReal and distinct roots:

O d d1

• Overdamped case

Real and repeated roots:C iti ll d d

1 • Critically damped case

Complex conjugate roots:U d d d

1

• Underdamped case


Overdamped solution2 2

1 2

( 1) ( 1)1 2

/ /

( ) n n n nt tN

t t

x t e e

e e

Critically damped solution

1 2e e

Critically damped solution/ /

1 2( ) t tNx t e te

Underdamped solution Underdamped solution2 2( 1 ) ( 1 )

1 2( ) n n n nj t j tNx t e e


In the underdamped case, for instance If the same coefficient If the same coefficient

2 21 1( ) ( )n nn j t j ttNx t e e e

22 cos( 1 )ntne t

Exponentially decaying envelope

Sinusoidal signal

p y y g p


Natural response of over-damped second order system 2 2( 1) ( 1)( ) t t system

1 2

( 1) ( 1)1 2

/ /1 2

( ) n n n nt tN

t t

x t e e

e e


Natural response of critically damped second order system / /order system / /

1 2( ) t tNx t e te


Natural response of under-damped second order systemsystem

2 2( 1 ) ( 1 )1 2( ) n n n nj t j t

Nx t e e

2 2( ) 2 cos( 1 )ntN nx t e t


Forced response2 ( ) ( )d x t dx t2 1

2

( ) ( )2 ( ) ( )nn sd x t dx t x t K f t

dt dt

In this chapter assume only DC input force case. ( )f t F( )( ) ( )F s

fx t x K F


Complete responseSum of natural response (homogeneous solution) andSum of natural response (homogeneous solution) and

forced response (particular solution)Overdamped case:Overdamped case:

2 2

( ) ( ) ( )N Fx t x t x t 2 2( 1) ( 1)

1 2 ( )n n n nt te e x

Critically damped case( ) ( ) ( )N N Fx t x t x t

/ /1 2 ( )

N N F

t te te x 1

n


Underdamped case

( ) ( ) ( )x t x t x t 2 2( 1 ) ( 1 )

1 2

( ) ( ) ( )

( )n n n n

N F

j t j t

x t x t x t

e e x

To obtain unknown variables of alpha1 and alpha2,I iti l diti f (t 0) d d i ti f (t 0) Initial condition of x(t=0) and derivative of x(t=0)need to be substituted into the solution form


Automotive suspensionBecause of road surfaceBecause of road surface

profile, there exists displacement in thedisplacement in the automotive body.

What would be theWhat would be the dynamic response in accordance with damper acco da ce t da pestate??

Derive differentialDerive differential equation


Time response of displacementThe governing equation and transient responseThe governing equation and transient response

2

2

( )( ) ( ) ( ) ( ) RoadRoad

dx td x t dx tm b kx t x t bdt dt dt

dt dt dt


Reminders Identify the objective of the circuit equation Identify the objective of the circuit equationObtain the steady state values

I iti l/fi l diti Initial/final conditionsDerive differential equationDetermine the solution formFix the constants and complete solution


Example 5.14Determine complete response by solving differentialDetermine complete response by solving differential

equation for the current initial assumption (0 ) 5[ ]v V initial assumption (0 ) 5[ ]Cv V


Steady state response Initial conditions Initial conditions

DC final states1 2

(0 )(0 ) , LL

dii c cdt

DC final statesDifferential equation

21

2

( ) ( )2 ( ) 0nSL L

LdVd i t di tLC i t C

dt dt dt

Write the complete solution

Solve for the constants using initial conditions


solution3 208.7 3 4791.3( ) 4.36 10 4.36 10 0t t

Li t e e t


Example 5.16Determine complete response by solving differentialDetermine complete response by solving differential

equation for the current Initially capacitor is charged such that (0 ) 5[ ]v V Initially, capacitor is charged such that (0 ) 5[ ]Cv V


Steady state response

Initial conditions

Differential equation2

2

( ) ( ) ( ) 0SL LL

dVd i t di tLC RC i t Cdt dt dt

Write the complete solution

Solve for the constants using initial conditions


Final solution200( ) 0.05 sin 400 0t

Li t e t t ( ) 0.05 sin 400 0Li t e t t


SummarySummary Summary

Write differential equations for circuits containing inductor and icapacitor

Applying Kirchhoff’s law and constitutive i-v relationship of inductor and capacitorinductor and capacitor

Determine the DC steady state solution of circuits containing inductors and capacitors DC S.S. condition says that inductor behaves as short circuit

and capacitor as open circuit Write differential equations of the first order circuits in standard

form, and determine complete solution of first order circuits excited by switched DC sourcesby switched DC sources Commonly characterized by 2 constants, DC gain and time

constant


SummarySummaryWrite differential equations of the second order circuits

in standard form, and determine complete solution of psecond order circuits excited by switched DC sourcesCommonly characterized by 3 constants, natural y y ,

frequency, damping ratio, and DC gain

Understand analogies between electric circuits, hydraulic, thermal, and mechanical system.hydraulic, thermal, and mechanical system.Use the same standard form of differential equation


ExerciseExercise Example 5.11

First order circuit for motorFirst order circuit for motorFind motor voltage, vm


ExerciseExercise Prob. 5.81

Solve v(t)Solve v(t)Apply KCL and capacitor voltage to find out solution


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