lecture 12 first order transient response

Lecture 12Lecture 12First Order Transient Response

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Chapter 4Chapter 4Transients

1 S l fi t d RC RL i it1. Solve first-order RC or RL circuits.

2 U d t d th t f t i t2. Understand the concepts of transient response and steady-state response.


3. Relate the transient response of first-order circuits to the time constant.

4. Solve RLC circuits in dc steady-state conditions.

5. Solve second-order circuits.

6. Relate the step response of a second-order system to its natural frequency and damping ratio.


TransientsTransients

The time-varying currents and voltages resulting from the sudden application ofresulting from the sudden application of sources, usually due to switching, are called transients. By writing circuit equations, we obtain integro-differential q gequations.


Discharge of a Capacitance through g ga Resistance

KCL at the top node of the circuit:iC iR

dtdvC

dtdqiCvq

vqC C ==→=→=

tv )(

( ) ( ) ( )R

tvi CR

)(=

( ) ( ) 0=+R

tvdt

tdvC CC ( ) ( ) 0=+ tvdt

tdvRC CC


Rdt dt

Discharge of a Capacitance through a Resistance

( ) ( )tdv ( )tdv 1( ) ( ) 0=+ tvdt

tdvRC C

C ( ) ( )tvRCdt

tdvC

C 1−=

( )

We need a function vC(t) that has the same form as it’s derivative.

( ) stC Ketv =

S b i i hi i f ( )

0stst KRCK

Substituting this in for vc(t)


0=+ stst KeRCKse

Discharge of a Capacitance through a

1−Resistance

RCs 1=Solving for s:

( ) RCtC Ketv −=Substituting into vc(t): ( )g c( )

Initial Condition: Full Solution:

( ) RCteVtv −=( ) iC Vv =+0

Initial Condition: Full Solution:


( ) iC eVtv =( ) iC Vv +0


( ) RCtC Ketv −=

To find the unknown constant K, we need to use the boundary conditions at t=0. At t=0 the capacitor is initially charged to a voltage Vi and then discharges through thecharged to a voltage Vi and then discharges through the resistor.

( ) RCtiC eVtv −=( ) iC Vv =+0


Discharge of a Capacitance through g ga Resistance

• The time interval t= RC is called the time constant of the circuit

• RC circuits can be used for timing applications (e g garage door light)applications (e.g. garage door light)


Larry Light-Bulb Experimenty g

iR(t)

RCtiR e

RV

Rtvti /)()( −==


R RR

Charging a Capacitance from a DC Source through a Resistance



KCL at the node that joins the resistor and the capacitor

Current into the capacitor: dv

C cpdt

C c

Current through the i t )(resistor:

RVtv SC −)(

0)()(

=−

+R

Vtvdt

tdvC SCC


Rdt


0)()(

=−

+R

Vtvdt

tdvC SCC

Rearranging:

SCC Vtvdt

tdvRC =+ )(

)(dt

This is a linear first-order differential equation with constant coefficients


constant coefficients.


The boundary conditions are given by the fact that the voltage across the capacitance cannot change instantaneously:instantaneously:

0)0()0( =−=+ CC vv



stC eKKtv 21)( +=Try the solution:

Substituting into the differential equation:

SCC Vtv

tdvRC =+ )(

)(

Substituting into the differential equation:

SC Vtvdt

RC + )(Gives:

st VKeKRCs =++ )1( SVKeKRCs =++ 12)1(



Sst VKeKRCs =++ 12)1(

For equality, the coefficient of est must be zero:

RCsRCs 101 −=→=+

Which gives K1=VS



RCtst eKVeKKtv /)( −++

Substituting in for K1 and s:

SC eKVeKKtv 221)( +=+=

Evaluating at t=0 and remembering that vC(0+)=0

sSSC VKKVeKVv −=→=+=+=+ 220

2 0)0(

RCtSS

stC eVVeKKtv /

21)( −−=+=

Substituting in for K2 gives:


SSC eVVeKKtv 21)( +


( ) τtVV −


( ) τtssC eVVtv −=


If the initial slope is extended from t=0, when does it intersect the final value VS?

( )τ

τ

SCtSC

tssC

Vdt

dve

Vdt

dveVVtv

=→=

−=

−

−

/A line with this slope would intersect VS after a


ττ tdtdt =0S

time τ

Larry Light-Bulb Experimenty g

tRtiV 0)()( ++

eVeVVVtvVRti

tvRtiV

RCt

RCtS

tSSSCS

CS

/

/)()()(

0)()(−− =−−=−=

=++−

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.R

eVti

RCtS

/)(

−

=

DC Steady Statey

tdvC )(dt

tdvCti C

C)(

)( =

In steady state, the voltage is constant, so the current through the capacitor is zero, so it behaves as an open circuitas an open circuit.


DC Steady Statey

tdiL )(dt

tdiLtv LL

)()( =

In steady state, the current is constant, so the voltage across and inductor is zero, so it behaves as a short circuitas a short circuit.


DC Steady StateDC Steady State

The steps in determining the forced response for RLC circuits with dc sources are:

1. Replace capacitances with open circuits.

2 Replace ind ctances ith short circ its2. Replace inductances with short circuits.

3. Solve the remaining circuit.


DC Steady Statey


DC Steady Statey

Find the steady state current ia and voltage va


DC Steady Statey

• Open circuit the capacitor


DC Steady Statey

VARiv aa 50)25)(2( =Ω==

Aia 2=

• Short circuit the inductor


DC Steady Statey

Find the steady state currents i1, i2, i3


DC Steady Statey

• Open circuit the capacitor


DC Steady Statey

10 Ω 5 Ω10 Ω

1A

1A

• Short circuit the inductor• i1=20V/10Ω=2A


1

• Current divider gives i2=i3=1A

lecture 12 first order transient response

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