Time domain response specifications• Defined based on unit step response with i.c. = 0

• Defined for closed-loop system

Prototype 2nd order system:

22

2

2 nn

n

sssH

10

:dampedUnder

:case Useful

target

Prototype 2nd order system:

22

2

2 nn

n

sssH

2 2

2 2

Given and :

d

nd

d

d

p j

2

2

1

Given an

1

d :n

n n

n

d n

p j

1cos (180 ( ))

Re( )Given p

(

ol :

)

e

n

o

d

p

angle p

p

imag p

p

2d n

= = 1

pt

21σ max 1 1tpy e e

21Overshoot : pM e

21percentag % 100e e

Settling time: 2ln 1 ln( ) 3, 4, 5s

tol tol or ort

Delay time :

0.8 0.9 1.4d

n n

t

4.5( 0.2)

Rise time : 1.8 ~ 2.2 2

rnn n

t

5%, 2%, 1%

5% or 10% or 16% or 25%

0.7 0.6 0.5 0.4

pM

Remember:

Example: Given 0.6, 5, estimate specs.n

sol: 0.6 5 3n 21 4d n

peak time: 0.8sec.pd

t

21 0.096 10%pM e

settling time for 1% :5 5

1.67sec.3st

42% : 1.33sec.

3stol t

1.8rise time: 0.36sec.r

n

t

1

k

s s +

-

Example:

When given unit step input, the output looks like:

i.e. 25%pM 3sec.pt

Q: estimate k and τ.

sol: from 25% 0.4pM

3sec.3p d

p

tt

21 n

2 21.14

1 3 1 0.4d

n

From block diagram:

22 1

k kH s

s s k s s k

2

2 2Match this against

2n

n ns s

2 21.14

12 2 0.4 1.14

n

n

k

1.09

1.42k

Solve for and , get:k

Effects of additional zerosSuppose we originally have:

0H s 1 t 0y t i.e. step response

Now introduce a zero at s = -z

01s

H s H sz

The new step response:

0

1 11

sY s H s H s

s z s

01s

Y sz

0

11

dy t y t

z dt

tytyz

ty 00

1

beginning.at rising is Typically, 0 ty

01

,0 If)1 0 tyz

z

overshoot.larger has and , before

overshoots ,n faster tha rising is

0

0

ty

tyty

01

,0 If)2 0 tyz

z

initially 0 i.e.

direction in wrong offstart may

? overshoot smaller

n slower tha rises 0

ty

tyty

:0 e.g. z

Effects: • Increased speed,• Larger overshoot,

• Might increase ts

pdr ttt ,, i.e.

pM

When z < 0, the zero s = -z is > 0,

is in the right half plane.

Such a zero is called a nonminimum phase zero.

A system with nonminimum phase zeros is called a nonminimum phase system.

Nonminimum phase zero should be

avoided in design.

i.e. Do not introduce such a zero in your controller.

Effects of additional poleSuppose, instead of a zero, we introduce

a pole at s = -p, i.e.

01

10

psHsH

p

s

s

sHs

sHsYp

s

1

1

110

sYps

p0

filter pass loworder first a is ps

p

.at frequency corner with p

L.P.F. has smoothing effect, or

averaging effect

Effects: • Slower,• Reduced overshoot,

• May increase or decrease ts

pdr ttt ,, i.e.pM

Stability• BIBO-stable:

Def: A system is BIBO-stable if any bounded input produces bounded output.

Otherwise it’s not BIBO-stable.

))((

resp. impulse)( where

finite )(stable-BIBO :Thm

1

0

sH

th

dtth

L

cancelled pole/zero allafter plane half

leftopen in the )( of poles"" all stable-BIBO sH

Asymptotically Stable

A system is asymptotically stable if for any arbitrary initial conditions, all variables in the system converge to 0 as t→∞ when input=0.

A system is marginally stable if for all initial conditions, all variables in the system remain finite, but for some initial conditions, some variable does not converge to 0 as t→∞.

A system is unstable if there are initial conditions that can cause some variables in the system to diverge to infinity.

A.S., M.S. and unstable are mutually exclusive.

Asymptotically Stable

Theorem:

A system is A.S. all eigenvalues

have real part 0

A system is M.S. at least one pole is on j -axis

all j -axis pole(s) non-repea

ted

all other pole in OLHP

A system is unstable if there is pole(s) in ORHP

or repeated j -axis poles

Asymptotically Stable vs BIBO-stable

Thm: If a system is A.S.,

then it is BIBO-stable

If a system is not BIBO-stable, then it cannot be A.S., it has to be either M.S. or unstable.

But BIBO-stable does not guarantee A.S. in general.

If there is no pole/zero cancellation, then

BIBO-stable Asymp Stable

1

1 1 0 1 01

n n m

n mn n m

d d d d dy a y a y a y b u b u b u

dt dt dt dt dt

DuCxy

BuAxx

1 01

1 1 0

( )( )

( )

mm

n nn

b s b s bY sH s

U s s a s a s a

Characteristic polynomials

Three types of models:

Assume no p/z cancellation

System characteristic polynomial is:1

1 1 0( ) det( ) n nnd s sI A s a s a s a

A polynomial

is said to be Hurwitz or stable if all of its roots are in O.L.H.P

A system is stable if its char. polynomial is Hurwitz

A nxn matrix is called Hurwitz or stableif its char. poly det(sI-A) is Hurwitz, orif all eigenvalues have real parts<0

11 1 0( ) n n

n nd s a s a s a s a

Routh-Hurwitz MethodFrom now on, when we say stability we mean

A.S. / M.S. or unstable.

We assume no pole/zero cancellation,

A.S. BIBO stable

M.S./unstable not BIBO stable

Since stability is determined by denominator, so just work with d(s)

3

1

541

1

3212

5311

642

011

1

:

:

:

)(

n

n

nnnn

n

nnnnn

nnnn

nnnnn

nn

nn

s

a

aaaa

a

aaaas

aaas

aaaas

asasasasd

:table Routh

polynomialstic characteri the

called is d(s) T.F., c.l. of den. the be

Let

0

Routh Table

Repeat the process until s0 row

Stability criterion:

1) d(s) is A.S. iff 1st col have same sign

2) the # of sign changes in 1st col

= # of roots in right half plane

Note: if highest coeff in d(s) is 1,

A.S. 1st col >0

If all roots of d(s) are <0, d(s) is Hurwitz

Example:

unstable

RHP in roots 2 changes, sign 2

- :sign col first

65.2

065.2:

05.24

64:

64:

11:

64)(

0

1

2

3

23

s

s

s

s

ssssd ←has roots:3,2,-1

unstable

roots unstb 2

changes sign 2

10:

43.6:

107:

51:

1032:

10532)(

0

1

2

3

4

234

s

s

s

s

s

sssssd(1*3-2*5)/

1=-7

(1*10-2*0)/1=10

(-7*5-1*10)/-7

A.S.

change sign no 0,col 1st

1:

2:

11:

42:

131:

1432)(

0

1

2

3

4

234

s

s

s

s

s

sssssd

Remember this

sign same have coeff all iff

A.S.is system order 2nd

:system order 2nd

cs

bs

cas

cbsassd

:

:

:

)(

0

1

2

2

3 2

3

2

1

0

3rd order system:

( )

:

:

:

:

3rd order system is A.S.

, , , all same sign iff

d s as bs cs d

s a c

s b d

bc ads

bs d

a b c d

bc ad

Remember this

A.S.

A.S.

A.S.Not 0coeff all

A.S.Not

A.S.Not

A.S.Not

A.S.Not

A.S.Not

A.S.

943)943(

943532

632632

235

1

13

2

3

15

23

23

23

23

23

23

2

2

2

2

sss

ssssss

sss

sss

ss

ss

s

ss

sse.g.

Routh CriteriaRegular case: (1) A.S. 1st col. all same sign

(2)#sign changes in 1st col. =#roots with Re(.)>0

Special case 1: one whole row=0Solution: 1) use prev. row to form aux. eq. A(s)=0

2) get:

3) use coeff of to replace 0-row 4) continue as usual

)(sAds

d

Example

5 4 3 2

5

4

3

2

1

2

2 2

( ) 4 8 8 7 4

: 1 8 7

: 4 1 8 2 4 1

: 6 1 6 1

: 4 1 4 1

: 0

prev. row: : 4 4

( ) 4 4 4( 1)

d s s s s s s

s

s

s

s

s

s

A s s s

←whole row=0

5 4 3 2

5

4

3

2

1

2

2 2

( ) 4 8 8 7 4

: 1 8 7

: 4 8 4

: 6 6

: 4 4

: 0

prev. row: : 4 4

( ) 4 4 4( 1)

d s s s s s s

s

s

s

s

s

s

A s s s

2

2

1

0

1( )differentiate: 8

2

: 4 4

: 8

: 4

No sign change in 1 col. no roots in R.H.P

But Not A.S. since we did have 0 in 1st col. originally.

marginally stable

sdA ss

ds s

s

s

s

2 2

5 4

Fact: The roots of are all roots of

original ( ) ( ) ( )

e.g.: in prev example:

( ) 4 4 4( 1)

has roots:

&these are roots of ( ).

Indeed ( ) 4 8

A(s)

d s A s

A s s s

s j

d s

d s s s

3 28 7 4

has roots: 1.5 1.3229

0

1

s s s

j

j

-

7

)474)(1()(

0

44

44

7

747

44

874

474478841

232

2

2

3

23

24

234

35

23

23452

sssssd

s-

s

ss-

sss

ss-

sss

ss-

sssssssss

0

11

44

44)(

)(121

00

121

121

122)(

1

2

3

3

244

3

4

5

2345

:s

:s

:s

ssds

sdA

sAsss

:s

:s

:s

ssssssd

:row From

e.g.

0)Re( withroots no

col. 1st in change sign No

:row From

1:

2:

2)(

1)(

0

1

22

s

s

sds

sdA

ssAs

unstable. is

roots. double are &

0))fence.(Re( the on roots areThey

at root double

at root double

of roots are

of roots But

)(

)()()1(12)(

).(

12)(

222224

24

sd

js

js

jsjsssssA

sd

sssA

e.g.

continue & 0by "0" replace :solution

0row wholebut

0col 1st in #an :2 case Special

3:

32:

3:

30:

21:

321:

322)(

0

1

2

2

3

4

234

s

-s

s

s

s

s

sssssd

Replace by

0)Re( have two these

:roots has :Verify

0)Re( i.e. RHP, in roots 2

col. 1st in changes sign 2

0 assume wesince

2928140570

902.009057.0

0)(

03

2

.j.

j

sd

Useful case: parameter in d(s)

How to use: 1) form table as usual

2) set 1st col. >0

3) solve for parameter range for A.S.

2’) set one in 1st col=0

3’) solve for parameter that leads to M.S. or leads to sustained oscillation

Example

s+3

s(s+2)(s+1) Kp

pp

p

p

plc

p

KsKss

sKssssd

sKsss

sKsG

K

3)2(3

)3()1)(2()(

)3()1)(2(

)3()(

23

..

char.poly

:Sol

stability for of range find:Q

+

0

03)2(3

03

0

3:3

3)2(3:

33:

21:

0

1

2

3

p

pp

p

p

pp

p

p

K

KK

K

Ks

KKs

Ks

Ks

col. 1st : A.S.For

:table Routh

03

7)1(

03

42

0463

41)2(3

202

003

4)2(3)(

2

2

2

23

k

kk

kk

kk

kk

kk

skksssd

2)

1)

:need we: A.S.For

prod. outer two mid of prod 2)

0coeff all 1)

criteria? Routh order 3rd remember

e.g.

A.S.for

need weall over

also. and but

or

528.013

7

20

13

7

3

71

13

7

3

71

3

7)1( 2

k

kk

kk

kk

k

)137

(3

4

3

4

3

4

0)(43

)(,

13

7

4)2(3

22

1

k

kjs

sAkss

s

sdk

k

kk

:freq osci

noscillatio sustained to leads

:row From

0row And

M.S. is this At

:get we

set weIf

Q: find region of stability in K- plane.

2

2

3 2

( ) ( 2)( )

( 1) ( ) ( 2)

( ) ( 1) ( ) ( 2)

( ( 2) 1) 2

:

0

1( 2) 1 0

22 0 0

1 2( ( 2) 1) 2

s K sH s

s s s K s

d s s s s K s

s Ks K s K

RouthCriteria

K

K K

K

K K K K

2

K