99-1

Chapter 9 Transient response9.1 First-order transients

zero-input response, step response, pulse response, switched DC transients

9.2 Switched AC transients9.3 Second-order natural response

second-order circuit equations, overdamped response, underdamped response, critically damped response

9.4 Second-order transientsinitial conditions, switched DC transients

99-2

9.1 First-order transients Basics1. Zero-input response

1

1

1

1

' 0,(or ' 0), . . ( ) ( )

s-domain solution:let characteristic equation 01 , : time constant

t-domain solution:zero-input

o

o o

o o o ost

o

o

ot

st sto o o

a y a y y y i c y t y t Y

y Ae a s aa asa a

Ae Y A Y e Y e

+

+ = + = = =

= + =

= = =

= = =

( )response, ( ) ,

( )

o

o

t ts t t

o o ot

o o

y t Y e Y e t t

Y e u t t

= = >

=

'( ) : initial slopeooYy t

+ =

99-3

2. Step response: transient response with step function excitation

t t

t t0)(

0 t10 t0

)(function stepunit

0

>

= = >

9-4

9

4. Switched dc transientsgiven: DC source switched at t=to, steady-state response y(t)=Yssfind: transient response of first-order circuitsapproach: for each time interval to write the corresponding

transient response based on the initial value, steady-state value,and time constant

' , . . ( ) ,

sol: ( )

( ) ( )

solution : ( ) ( ) ,

( ) ( )

o o

o

ss o o ot

sst t

o ss o o sst t

ss o ss ot

ss o ss o

y y Y i c y t Y t t

y t Y Ae

y t Y Ae Y A Y Y e

y t Y Y Y e t t

Y Y Y e u t t

+

+

+ = = >

= +

= + = =

= + >

= +

(general expression)

9-5

99-6

Discussion1. RC circuit (zero-input response)

, , ,, 0 0,(or 0), ( ) ( )

s-domain solution:let characteristic equation 1 0,1 1 , : time constant

t-domain solution: ( ) , ( )o

C C C C C C C C C o C o ost

C

t to

C o C

i Cv Ri v RCv v v v v t v t V

v Ae RCs

s RCRC

Vv t V e i tR

+

= + = + = + = = =

= + =

= = =

= = ,ot t

oe t t

>

continuous

RV

Rtvti ooCoC ==+

+ )()(

9

(express using iC)

,

,

,

1, 0 0

( )0, ( )

t-domain solution: ( ) ,

D.E. is the same form as 0, ( )

o

C C C C C C

C o oC C C o

t to

C o

C C C o o

i Cv Ri v Ri i dtC

v t VRCi i i tR R

Vi t e t tR

RCv v v t V

+

+

= + = + =

+ = = =

= >

+ = =

9-6

9

2. RL circuit (zero-input response)

, , ,, 0 0, (or 0), ( ) ( )

s-domain solution:let characteristic equation 0,1 , : time constant

t-domain solution: ( ) , ( )o

L L L L L L L L L o L o ost

L

t t

L o L o

v Li Ri v Li Ri i i i t i t I

i Ae Ls RR LsL R

i t I e v t RI e

+

= + = + = + = = =

= + =

= = =

= = ,ot t

ot t >

iL(t)Io

vL(t)

-RIo

continuous

RIRtitv ooLoL ==++ )()(

9-7

9

3. Ex.9.1 find 0for t (t)(t) >, vi, ,

33

, (40 10) 0 (40 10) 0

60 10 251.2 10 , 0 2.550 10

( ) 2.5 , ( ) 40 ( ) 100

L L L L L L

L

t t

L L

v Li i v Li i

L i ( )R

i t e v t i t e

= + + = + + =

= = = = =

= = =

4. RC circuit (step response)

,

( ), ( ) ( ) 0

, ,

(1 ),o o

oc ss o C o C o

o eq C C ss eq C C ss eq

t t tt

C ss ss C ss o

v V u t t v t v t

t t R i v V R Cv v V R C

v V Ae A V e v V e t t

+

= = =

> + = + = =

= + = = >

9-8

RL circuit (step response): dual circuit

,

, , ,

(1 ),o

C L eq eq eqeq

t t

L L ss L ss oeq

Lv i C L R G R CR

L i i I i I e t tR

=

+ = = >

9

5. Ex. 9.2 vc(0-)=0, C=50uF, find step response 0for t (t)(t) >, vi

10

10

60 3 // 6 2 , 12 8 , 0.13 6

( ) (1 ) (1 ) ( ) 8(1 ) ( )8 ( )( ) 4 ( )2000

o

eq oc ss

t t tt

ss ss o

t

t R k v V RC

v t V e V e u t t e u tv ti t e u t mA

> = = = = = = =+

= = =

= =

9-9

9

6. Ex. 9.3 iL(0-)=0, to find the relay last time with iL reaching 150mA till falling to 40mA

30ms)]-u(t- )([5(t) tuvs =

1

2

16 161 1

3016

2 2 2 1

400 510 15 25, 16 , 20025 25

0 30 : ( ) 200(1 ) ( ) 200(1 ) 150 22.230 : (30) 169

30 : ( ) 169 40 53.1 , 30.9

eq sseq

tt

L L

Lt

L

L mR ms I mAR

t ms i t e mA i t e t mst ms i mA

t ms i t e t ms t t ms

= + = = = = = =

< < = = = == =

> = = = =

9-10

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7. Ex.9.4 Sequential switched transients, to find v(t), i(t)

Step 1: for the circuitfind

0=t(0), (0), (0)C Ly v i

Step 2: for the circuit find from the continuity

of capacitor voltage and inductor current

0>toYy =

+ )0(

0)0(,0)0( == ivC

(0 ) (0 ) 0

16 (0 )(0 ) 28

o C C

o

V v v

vI i mA

+

++

= = =

= = =

Step 3: find the steady state value of Yss

24( 16) 128 24

16 0.58 24

ss

ss

V

I mA

= = +

= =

+

C is OC

C is SC

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Step 4: find the time constant Step 5: find the response for

each time interval

0.6

0.6

(1) 0 1(8 // 24 )100 0.6

( ) ( )

12 12(1) 9.73

( ) ( )

0.5 1.5(1) 0.783

t

C ss o sst

Ct

ss o sst

C

tk k s

v t V V V e

ev

i t I I I e

e mAi mA

< = =

= +

= +=

= +

= =

mAi 2816

=

=

99-13

10.4

(2) 1(12 // 8 ) // 24 4.8 // 24 4

60 1660 12 14.412 8

4 100 0.4(1 ) 9.73

14.4 (1 ) 5.034.8

2414.4 124.8 24

14.4 0.54.8 24

( ) ( )

12 21.73

( )

o

th

th

o C

Co

ss

ss

t t

C ss o sst

tR k k k k k k

V

kV v

vI mA

V

I mA

v t V V V e

e

i t

+

+

>= = =

+= =

+ = =

= =

= =

= =+

= =+

= +

=

10.4

( )

0.5 4.53

ot t

ss o sst

I I I e

e mA

= +

= +

mAi 03.58.4

)73.9(4.14=

=

9

9.2 Switched AC transientsBasics1. First-order circuit with a sinusoidal excitation

given: sinusoidal source switched at t=to, initial value y(to)=Yofind: transient response of first-order circuitapproach: for each time interval, write the corresponding transient

response based on the initial value, time constant, and resultingphasor value

( ) ( ) , , ( ) cos( ), . . ( )

( ) ( ) [ ( )]

solution ( ) ( ) [ ( )]

o o

o

t

F o F m o ot t

o F o o o F ot t

F o F o

y t y t Ae t t y t Y wt i c y t Y

y t y t Ae Y A Y y t e

y t y t Y y t e

+

+ + +

+

= + > = + =

= + = =

= + (general expression)

9-14

99-15

Discussion1. Ex. 9.5 find i(t), v(t),

> =

= = = =

= + ringing

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4. Ex. 9.12 Find the step response for t>0, L=0.1H, C=1/640F

>