Chapter4 Time Domain Analysis of Control System

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University

Automatic Control System 1

1

Chapter4 Time Domain Analysis of Control System

Ø Routh stability criterion

ØSteady state errors

ØTransient response of the first-order system

Ø Transient response of the second-order system

Ø Time domain performance specifications

Ø The relationship between the performance specifications and system parameters

Ø Transient response of higher-order systems

2

Definition of characteristic equation

)()()(

011

1

011

1 sGasasasabsbsbsb

sRsC

nn

nn

mm

mm =

++++++++

= −−

−−

LL

The characteristic equation of the system is defined as

0011

1 =++++ −− asasasa n

nn

n L



3

Transfer function:

2, 1j k k k n n kp j jσ ω ζ ω ω ζ− − ± =− ± − poles of the closed loop

∏ ∏

∏

= =

=

−++++

+== q

j

r

kkkkkj

m

ii

jsjsps

zsk

sRsCs

1 1

1

)]()][(([)(

)(

)()()(G

ωσωσ

zi : zeros of the closed loop

4

0

2 21 1

( )( 1 ) ( 1 )

q rj k k

j kjk n n k n n k

aa sC ss s p s j s j

α β

ζ ω ω ζ ζω ω ζ= =

+= + +

+ + + − + − −

∑ ∑

)1sin(

)1sin1cos()(

1 1

20

2

1 1

20

k

q

j

r

ikk

tk

tpj

kkk

q

j

r

kkkk

ttpj

teDeaa

tCtBeeaatC

kkj

kkj

ϕζω

ζωζω

ωζ

ωζ

+−++=

−+−++=

∑ ∑

∑ ∑

= =

−−

= =

−−

sjsjsps

zsksC q

j

r

kkkkkj

m

ii 1

)]()][(([)(

)()(

1 1

1 ⋅−++++

+=

∏ ∏

∏

= =

=

ωσωσ

Step response:



5

Consider that the characteristic equation of a LTI system

0)( 011

1 =++++= −− asasasasF n

nn

n L

Where all the coefficients are real numbers.

In order that there be no roots of the above equation with positivereal parts, it is necessary but not sufficient that

1. All the coefficients of the polynomial have the same sign.2. None of the coefficients vanishes.

4.1 Routh stability criterion

6

0000

000

00000

0000

000

60

61

662

1615133

61

061

1

5041

1

30214

5315

64206

FEFas

FE

CaEDsC

ACaC

ACaEC

ADBCsA

aADA

aaAaCA

BaAas

aa

aaaBa

aaaaAa

aaaas

aaasaaaas

×−

=−

=×−×

=×−

=−

=×−×

=−

=−

=×−

=−

=−

The Routh tabulation

0)( 12

21

10 =+++++= −−−

nnnnn asasasasasF L

for n = 6



7

The roots of the polynomial are all in the left half of the s-planeif all the elements of the first column of the Routh Array areof the same sign.

The necessary and sufficient condition for the stability of a system

Rout-Hurwitz criterion

If there are changes of signs in the elements of the first column, the number of sign changes indicates the number of roots with positive real parts.

8

Example: The characteristic equation of a system is 050104 23 =+++ sss

Determine the stability of the system using Routh criterion.

Solution:

05005.2

504101

0

1

2

3

ssss

−

The system has two roots located in the right half of the s-plane.

（2). Routh array is

（1）. Check the necessary condition



9

The first element in any one row of the Routh Array is zero butthe other elements are not.

We can replace the zero element in the Routh tabulation by anarbitrary small positive number ε and then proceed with the Routhtest.

Special case 1

10

The characteristic equation of a system is

Determine the stability of the system using Routh criterion.

Example:01255 2345 =+++++ sssss

001

0015

15

011501)(0151251

0

21

2

3

4

5

s

s

s

sss

−−−

−

εεε

εε

ε

Solution: Routh array is

There are two sign changes in the first column of the tabulation, thesystem has two roots located in the right half of the s-plane.



11

The elements in one row of the Routh Array are all zero.

Special case 2

12

Example:

0161623 =+++ sss

The characteristic equation of the system is:

The Routh array is :

The auxiliary equation is:01600

161161

0

1

2

3

ssss

016)( 2 =+= ssA

01602

161161

0

1

2

3

ssss

The sign of the elements in the first column does not change, the system is stable .



13

Example: ( 1)( 1)(2 1)

K ss Ts s

++ +

( )R s ( )C s+

−

Determine the value of K and T to make the closed loop be stable.

Solution: the characteristic equation:

0)1()2(2 23 =+++++ KsKsTTs

0

02

2)1)(2(2

12

0

1

2

3

KsT

TKKTsKTs

KTs

+−++

++Routh array:

14

,02,0 >> TK

02)1)(2( >−++ TKKT

02 >+T

The condition for the stability is:

0>T

220

−+<<

TTK



15

K

T0

2

2

4

4 6

6

8

22

TKT

+<

−

220

−+<<

TTK

Stable

16

Example:

02108 23 =+++ sss

The characteristic equation of the system is

Determine the stability of the system. Analyze how many rootslie between the imaginary axis and the line .1−=s

Solution:

02075.928

101

0

1

2

3

ssss

the system is stable.

11 −= ssLet

The characteristic equation becomes:

0135 121

31 =−−+ sss

Routh array:

Im

Re01−



17

0135 121

31 =−−+ sss

Routh array for the above equation:

0108.21531

01

11

21

31

−−

−−

ssss

there is one root on the right side of the line s = -1.

Im

Re01−

18

4.6 Steady state errors

)()( ∞−∞=∞ rce ）（



19

( )R s ( )C s( )G s

( )B s

( )E s+

−( )H s

)()(1)()(

sHsGsRsE

+=

0 0

1l i m ( ) l i m ( ) l i m ( )1 ( ) ( )s s t s s

e e t s E s s R sG s H s→ ∞ → →

= = =+

20

∏

∏

+=

=

+

+=

++++++

= q

vjj

m

iir

q

mr

pss

zsK

pspspsszszszsKsG

1

1

21

21

)(

)(

)())(()())(()(

νν L

L

ν = 0, type 0 system ν = 1, type 1 systemν = 2, type 2 system

the types of the control systems



21

a. position error constant

For a step input )()( tuRtr ⋅=

)()(1lim

)()(1/lim

)()(1)(lim)(lim)(lim

00

00

sHsGR

sHsGsRs

sHsGsRsssEtee

ss

sstss

+=

+=

+===

→→

→→∞→

00

lim1 ( ) ( ) 1 lim ( ) ( )ss s

s

R ReG s H s G s H s→

→

= =+ +

0lim ( ) ( ) Ps

G s H s K→

=

22

b. velocity error constant

For a ramp input: Rttr =)(

)()(lim

)()(1/lim


0

2

0

00

sHssGR

sHsGsRs

sHsGsRsssEtee

ss

sstss

→→

→→∞→

=+

=

+===

0lim ( ) ( )v s

K sG s H s→

=v

ss KRe =



23

c. Acceleration error constant

for acceleration input 21( )2

r t Rt=

)()(lim

)()(1/lim


20

3

0

00

sHsGsR

sHsGsRs

sHsGsRsssEtee

ss

sstss

→→

→→∞→

=+

=

+===

)()(lim 2

0sHsGsK

sa →=

ass K

Re =

24

Steady state error



25

Example:

+

−

( )R s ( )C s10( 5)s s +

Determine the steady state errors, when the input is

)(tu t 2

21 t, and .

Solution:

∞=+

==→→ )5(

10lim)()(lim00 ss

sHsGKssp

2)5(

10lim)()(lim00

=+

==→→ ss

ssHssGKssv

0)5(

10lim)()(lim 2

0

2

0=

+==

→→ ssssHsGsK

ssa

26

211

==v

ssv Ke

01

1=

+=

Pssp K

e

∞==a

ssa Ke 1



27

4.1 The transient response of the first-order system

4.1.1 The math model

TssRsC

+=

11

)()( ( )R s +

−( )C s

Ts1

28

4.1.2 the step response

r(t) = u(t) , R(s)=1/s

1 1( ) ( ) ( )( 1)

11

C s G s R sTs s

Ts Ts

= = ⋅+

= −+

Tt

etc−

−= 1)( t ≥ 0



29

0

1

0.632

( )c t

tT 2T 3T 4T 5T

63.2%

86.5% 95% 98.2% 99.3%

slope

30

Impulse response:

TssR

TssC

+=•

+=

11)(

11)(

Tt

eT

tctg−

==1)()(

( )R s +−

( )C s

Ts1

0 T 2T 3T 4T t

( )c t

1T

/1( ) t Tc t eT

−=



31

Ramp response ：

TsT

sT

ssTssRsGsC

/111

11)()()( 22 +

+−=⋅+

==

( )R s +−

( )C s

Ts1

TtTeTttc /)( −+−=

32

)(tr

)(tc

t

)(tr)(tc

Tess =TtTeTttc /)( −+−=

TtTeTttrtc Tt

tt−=−+−=− −

∞→∞→][lim)]()([lim /



33

4.2 the transient response of the second-order system

4.2.1 the mathematical model

( 1)K

s Ts +( )R s ( )C s+

−

34

( 1)K

s Ts +( )R s ( )C s+

−TKs

T1s

TK

)s(R)s(C)s(

2 ++==Φ

letTK

n =2ωTn12 =ςω

2nn

2

2n

s2s)s(

ωζωω

Φ++

=



35

The characteristic equation :

0s2s 2nn

2 =++ ωζω

The roots of the characteristic equation are

1s 2nn2,1 −±−= ζωζω

36

djs ωσ ±−=2,11s 2nn2,1 −±−= ζωζω

1. underdamped )10( << ζ

If input is a unit step u(t):

222222

2

)()(11

)2()(

dn

n

dn

n

nn

n

sss

sssssC

ωζωζω

ωζωζω

ωζωω

++−

+++

−=⋅++

=



37

)tsin1

t(cose1)t(c d2dtn ω

ζ−

ζ+ω−=

ζω−

)sincos1(1

11 2

2tte dd

tn ωζωζζ

ζω +−−

−= −

)tsin(1e1 d2

tnθω

ζ

ζω

+−

−=−

ζ=θ arccos

38

0

1

2

111 ζ

+−

2

111 ζ

−−

21

1

nte ζω

ζ

−+

−1

nT

ζω=

T 2T 3T 4T t

( )c t

21

1

nte ζω

ζ

−−

−



39

1s 2nn2,1 −±−= ζωζω

ns ω−=2,1

nn

n

nn

n

sssssssc

ωωω

ωωω

+−

+−=⋅

++=

1)(

11)2(

)( 222

2

)1(1)( tetc ntn ω

ζω+−=

−( )c t

( )r t

0

1

t

( )c t

2. Critically damped response 1=ζ

u(t)

40

1>ζ

122,1 −±−= ζωζω nns

1

]11(2[

1

)]11(2[1)(2

122

2

122

−++

−−++

−−+

−−−+=

−−

ζωζω

ζζζ

ζωζω

ζζζ

nnnn ssssc

)T

eT

e(12

11)t(c2

t2T

1

t1T

2 −−

−−+=

ζ

n

n

T

T

ωζζ

ωζζ

)1(

)1(2

2

21

−−−=

−+−=

3. Over damped case

Let

1

122

2

122 ]11(2[)]11(2[1)(TsTss

sc−

−−++

−−−−

+=−− ζζζζζζ

( )r t

t

( )c t

0

c(t)



41

njs ω±=2,11s 2nn2,1 −±−= ζωζω

tcos1)t(c nω−= )0t( ≥

( )c t

0

1

2

t

)0( =ζ

4. undamped case

2222

2 11)(

)(nn

n

ss

ssssc

ωωω

+−=⋅

+=

)0( =ζ

42

1. Rise time tr

2. The peak time tp

4.3 Time domain performance specifications



43

3. Percentage overshoot σp

%100)(c

)(c)t(c PP ×

∞∞−

=σ

4. Steady-state error ess

( ) ( ) 100%( )ss

c rer

∞ − ∞= ×

∞

44

5. Settling time ts



45

4.4 The relationship between the performance specifications and systemparameters

1. Rise time-system parameter

1)( =rtcAccording to definition:

1)sin(1

1)(2

=+−

−=−

θωζ

ωζ

rd

t

r tetcrn

πθω =+rdt

21 ζω

θπω

θπ

−

−=

−=

ndrt

)10( << ζ

)sin(1

1)(2

θωζ

ζω

+−

−=−

tetc d

tn

46

2. Peak time -system parameter

21 ζω

πωπ

−==

ndPt

( )c t

rt

pt

st

1

00.1

0.9

pσ

t

2

2 221

Pd n

t π πω ω ζ

= =−

(±0.05 or±0.02 ) × r (∞)



47

3. Percentage overshoot - system parameter

)sincos1(1

11)( 22

ttetc ddtn ωζωζ

ζ

ζω +−−

−= −

dPt ω

π=

21

2( ) 1 (cos sin )

1Pc t e

ζ π

ζ ζπ π

ζ

−−= − +

−

2 21 1n n nPd n

t π π πω ω ω

ω ω ξ ξ= = =

− −

2

2

1( ) 1 ( 1 cos sin )1

n ptp d p d pc t e t tζ ω ζ ω ζ ω

ζ

−= − − +−

48

2 21 1

2( ) 1 (cos sin ) 1

1Pc t e e

ζπ ζπ

ζ ζζπ π

ζ

− −− −= − + = +

−

( ) ( ) ( ) 1100% 100%( ) 1

P PP

c t c c tc

σ− ∞ −

= × ×∞

＝

21P e

ζπ

ζσ−

−=



49

21 ζ

ζπ

σ −−

= ep

σp is only related to ζ.

0.5 1.0 1.50

2

22( )2

n

n ns

s sω

ζω ωΦ =

+ +

pσ

20

40

60

80

100%

ζ

pσPercentage overshoot

50

4. The settling time ts- - system parameter

)()()( ∞⋅∆=∞− cctc s

0.02 0.05or∆ = ±



51

0

1

2

111 ζ

+−

2

111 ζ

−−

21

1

nte ζω

ζ

−+

−1

nT

ζω=

T 2T 3T 4T t

( )c t

21

1

nte ζω

ζ

−−

−

∆ζ

ζω

=−

−

2

stn

1e

2n

s11ln1t

ζ∆ζω −=

21

n ste ζω

ζ

−

= ∆−

52

9.00 << ζ

ns

4tζω

=

ns

3tζω

=

02.0=∆

05.0=∆

approximately:

for

for



53

Example: ( )R s ( )C s+

−( 1)

Ks s +

01 K s+

The desired specifications are: 2 0 % , 1P Pt sσ = =

What should the value of K and K0 be? Determine the value of tsand tr .

Solution: The transfer function of the closed loop is

KsKKsK

sKKKssK

sRsC

++=

+++=

)1()1()()(

02

0 ＋

012, KKK nn +== ζωω

54

(1)2.0

21 =−

−

ζ

ζπ

e 456.0=ζ

(2)

11 2

＝ζω

πωπ

−==

ndPt

ωn

sradn /53.3=ω

5.12K 2n == ω

nKK ζω21 0 =+ 178.00 =K

KsKKsK

sKKKssK

sRsC

++=

+++=

)1()1()()(

02

0 ＋

ζ



55

(3) tr

s65.01

t2

n

r =−

−=

ζω

θπ

(4) ts

stn

s 86.13==

ζω

rad1.11

tg2

1 =−

= −

ζζ

θ

for ∆=5%

56

sjsjsps

zsksC q

j

r

kkkkkj

m

ii 1

)]()][(([)(

)()(

1 1

1 ⋅−++++

+=

∏ ∏

∏

= =

=

ωσωσ

( ) ( )0

22 21 1

( )1

q rk kj

j kjk k k k

saaC ss s p s

α β

ζ ω ω ζ= =

+= + +

+ + + −∑ ∑

1. Step response of a higher order system

4.5 Transient response of higher-order systems



57

( )c t)(tr

1

0 t t

( )c t

0

1)(tr

( )c t

t0

1)(tr

2 20

1 1

20

1 1

( ) ( cos 1 sin 1 )

sin( 1 )

j k k

j k k

q rp t t

j k k k k k kj k

q rp t t

j k k k kj i

C t a a e e B t C t

a a e D e t

ζ ω

ζ ω

ω ζ ω ζ

ω ζ φ

− −

= =

− −

= =

= + + − + −

= + + − +

∑ ∑

∑ ∑

58

3. The dominant poles of higher-order systems

2 20

1 1

20

1 1

( ) ( cos 1 sin 1 )

sin( 1 )

j k k

j k k

q rp t t

j k k k k k kj k

q rp t t

j k k k kj i

C t a a e e B t C t

a a e D e t

ζ ω

ζ ω

ω ζ ω ζ

ω ζ φ

− −

= =

− −

= =

= + + − + −

= + + − +

∑ ∑

∑ ∑



59

σ

j[ ]s

主导极点

Re[ ] 1Re[ ] 5

poles of closed loopdominant poles

<

there is no zeros of the closed loop near the dominant poles

dominant closed loop poles

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Chapter4 Time Domain Analysis of Control System