MESB374 Chapter 10System Modeling and Analysis

TIME DOMAIN DESIGNSystem Stability and Steady State

Response

• Stability ConceptDescribes the ability of a system to stay at its equilibrium position in the absence of any inputs.

Stability

Ex: Pendulum

where the derivatives of all states are zeros

inverted pendulum

simplependulum

hill plateau valley

– A linear time invariant (LTI) system is stable if and only if (iff) its free response converges to zero for all ICs.

Ball on curved surface

Examples (stable and unstable 1st order systems)Q: free response of a 1st order system.

05 (0)y y u t y y

15( )

0

ty t y e

Q: free response of a 1st order system.

1

5 1G

s

TF:

Pole: 0.2p

t

y

t

y

05 (0)y y u t y y

15( )

0

ty t y e

1

5 1G

s

TF:

Pole: 0.2p

• Stability Criterion for LTI Systems

Stability of LTI Systems

( ) ( 1) ( ) ( 1)1 1 0 1 1 0

11 1 0

Characteristic Polynomial

Stable All poles lie in the left-half complex plane (LHP)

All roots of ( ) 0 l

n n m mn m m

n nn

y a y a y a y b u b u b u b u

D s s a s a s a

ie in the LHP

Complex (s-plane)

Re

Im

Marginallystable/``unstable’’

RelativeStability(gain/phase margin)

AbsolutelyStable

Unstable

• Comments on LTI Stability– Stability of an LTI system does not depend on the input (why?)

– For 1st and 2nd order systems, stability is guaranteed if all the coefficients of the characteristic polynomial are positive (of same sign).

– Effect of Poles and Zeros on Stability• Stability of a system depends on its poles only.• Zeros do not affect system stability.

• Zeros affect the specific dynamic response of the system.

Stability of LTI Systems

0 02

1 0 1 2

( ) : Stable 0( ) : Stable 0 and 0

D s s a aD s s a s a a a

• Passive systems are usually stable– Any initial energy in the system is usually dissipated in real-world

systems (poles in LHP);

– If there is no dissipation mechanisms, then there will be poles on the imaginary axis

– If any coefficients of the denominator polynomial of the TF are zero, there will be poles with zero RP

System Stability (some empirical guidelines)

• Active systems can be unstable– Any initial energy in the system can be amplified by internal source

of energy (feedback)

– If all the coefficients of the denominator polynomial are NOT the same sign, system is unstable

– Even if all the coefficients of the denominator polynomial are the same sign, instability can occur (Routh’s stability criterion for continuous-time system)

In Class Exercises(1) Obtain TF of the following system:

(2) Plot the poles and zeros of the system on the complex plane.

(3) Determine the system’s stability.

L

2 5y y y u u y y y u u u 6 3 4

2 2 5s Y s sY s Y s sU s U s

2

1

2 5

Y s sG s

U s s s

Poles:

Zero:

2 2 5 0s s 1,2

2 4 201 2

2p j

1 0s 1z

Real

Img.

1 1 2p j

2 1 2p j

1z

Stable

(1) Obtain TF of the following system:

(2) Plot the poles and zeros of the system on the complex plane.

(3) Determine the system’s stability.

TF:

2

3 2

3 4

6

Y s s sG s

U s s s s

Poles:

Zeros:

3 2 6 0s s s 1 2,3

1 230,

2

jp p

2 3 4 0s s 1,2

3 7

2

jz

Real

Img.

3

1 23

2 2p j Marginally

Stable

2

1 23

2 2p j

1 0p

1

3 7

2 2z j

2

3 7

2 2z j

Example

Inverted Pendulum

(1) Derive a mathematical model for a pendulum.

(2) Find the equilibrium positions.

(3) Discuss the stability of the equilibrium positions.

B

mg

l

EOM: sinoI mgl B

is very small

Equilibrium position:

00

0

Assumption: Linearized EOM:

0o

o

K

I mgl BI B mgl

Characteristic

equation:2 0os I sB K

Poles:2

1,2

4

2o

o

B B KIp

I

Real

Img.

Unstable

Example (Simple Pendulum)

B

mg

EOM: sinoI mgl B

is very small

Equilibrium position:

00

0

Assumption: Linearized EOM:

0o

o

K

I mgl BI B mgl

Characteristic

equation:2 0os I sB K

Poles:2

1,2

4

2o

o

B B KIp

I

Real

Img.

stable2 4 0oB KI

Real

Img.

stable2 4 0oB KI

Real

Img.

stable

2 4 0oB KI

How do the positions of poles change when K increases?

(root locus)

Transient and Steady State ResponseEx:

5u t t

5 10y y u

to a ramp input:

Let’s find the total response of a stable first order system:

with I.C.: 0 2y

- total response

20

Transfer FunctionFree Response

Forced Response

10 5 12

5 5 y

U s

Y ss s s

- PFE 31 22

2

5 5

aa aY s

s s s s

22 forced

0( ) 10

sa s Y s

0

2 12

1 forced 22 1 00

1 50 50( ) 2

2 1 ! 5 5s

ss

d da s Y s

ds ds s s

3 forced 55 ( ) 2

sa s Y s

5 5

3Transient responseSteady state response Transient responsefree responsefrom Forced response from Forced response

2 10 2t ty t t a e e

Transient and Steady State Response

to a input u(t) can be decomposed into two parts

In general, the total response of a STABLE LTI system

Transient Response Steady State Response

T SSy t y t y t

( ) ( 1) ( ) ( 1)1 1 0 1 1 0

n n m mn m my a y a y a y b u b u b u b u

where

• Transient Response – contains the free response of the system plus a portion of forced response

– will decay to zero at a rate that is determined by the characteristic roots (poles) of the system

• Steady State Response– will take the same (similar) form as the forcing input

– Specifically, for a sinusoidal input, the steady response will be a sinusoidal signal with the same frequency as the input but with different magnitude and phase.

Transient and Steady State ResponseEx:

5sin 3u t t

4 3 6y y y u

to a sinusoidal input:

Let’s find the total response of a stable second order system:

with I.C.: 0 0, 0 2y y

- total response

2 2 2 2

Forced Response Free Response

6 5 3 2 4 2

4 3 3 4 3

sY s

s s s s s

- PFE

31 2 4 1 2

3 1 3 3 3 1

aa a a b bY s

s s s j s j s s

2

9

2a 1

5

2a

3

11

2a j

3 33 1 1 2 2

Steady state response Transient response

1 3

2 Re

7 155 sin 3 tan 2

2 2

jt t t

t t

y t a e a b e a b e

t e e

4

11

2a j

2 3b 1 1b

Steady State Response

f f t sF st s

( ) lim ( ) lim ( ) 0

4 12 4 3y y y u u

2

4 3 5

4 12

sY s

s s s

• Final Value Theorem (FVT)

Given a signal’s LT F(s), if all of the poles of sF(s) lie in the LHP, then f(t) converges to a constant value as given in the following form

Ex.

(1). If a constant input u=5 is applied to the sysetm at time t=0, determine whether the output y(t) will converge to a constant value?

(2). If the output converges, what will be its steady state value?

We did not consider the effects of IC since •it is a stable system•we are only interested in steady state response

A linear system is described by the following equation:

0

5( ) lim ( ) lim ( )

4t sy y t sY s

Steady State ResponseGiven a general n-th order stable system

( ) ( 1) ( ) ( 1)1 1 0 1 1 0

n n m mn m my a y a y a y b u b u b u b u

1

1 1 01

1 1 0

m mm m

n nn

b s b s b s bG s

s a s a s a

11 1 0

( )( )Free n n

n

F sY s

s a s a s a

Free Response

Transfer Function

Steady State Value of Free Response (FVT)

10 01 1 0

00

( )lim ( ) lim

0 (0)lim 0

SS Free n ns sn

s

sF sy sY s

s a s a s a

F

a

In SS value of a stable LTI system, there is NO contribution from ICs.