MESB374 System Modeling and Analysis

Transfer Function Analysis

Transfer Function Analysis• Dynamic Response of Linear Time-Invariant (LTI)

Systems– Free (or Natural) Responses

– Forced Responses

• Transfer Function for Forced Response Analysis– Poles

– Zeros

• General Form of Free Response – Effect of Pole Locations

– Effect of Initial Conditions (ICs)

• Obtain I/O Model based on Transfer Function Concept

( ) ( ) (0)Y s U s y

( )

y y Ku

y y u

K

£ £

£ £ K £

Dynamic Responses of LTI SystemsEx: Let’s look at a stable first order system:

Y s

y y Ku

( ) ( ) (0)1 1

KY s U s y

s s

– Solve for the output:

– Take LT of the I/O model and remember to keep tracks of the ICs:

– Rearrange terms s.t. the output Y(s) terms are on one side and the input U(s) and IC terms are on the other:

0sY s y U s

1s K

Free ResponseForced Response

Time constant

Free & Forced Responses• Free Response (u(t) = 0 & nonzero ICs)

– The response of a system to zero input and nonzero initial conditions.

– Can be obtained by

• Let u(t) = 0 and use LT and ILT to solve for the free response.

• Forced Response (zero ICs & nonzero u(t))– The response of a system to nonzero input and zero initial conditions.

– Can be obtained by

• Assume zero ICs and use LT and ILT to solve for the forced response (replace differentiation with s in the I/O ODE model).

In Class ExerciseFind the free and forced responses of the car suspension system without tire model:

2 4 2 4 , , ry y y u u y z u x

2 2 4 ( ) 2 4 ( ) 2 (0) (0)s s Y s s U s s y y

2

2 4 2 4

2 4 2 4

0 0 2 0 4 2 0 4

y y u u

y y y u u

s Y s sy y sY s y Y s sU s u U s

£ £

£ £ £ £ £

2 2

2 (0) (0)2 4( ) ( )

2 4 2 4

s y ysY s U s

s s s s

– Solve for the output:

– Rearrange terms s.t. the output Y(s) terms are on one side and the input U(s) and IC terms are on the other:

Free ResponseForced Response

– Take LT of the I/O model and remember to keep tracks of the ICs:

Given a general n-th order system model:

The forced response (zero ICs) of the system due to input u(t) is:– Taking the LT of the ODE:

Forced Response & Transfer Function

( ) ( 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

( ) ( ) WHY?n ny s Y s £1

1 1 01

1 1 0

1 11 1 0 1 1 0

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( )

n nn

m mm m

n n m mn m m

D s N s

s Y s a s Y s a sY s a Y s

b s U s b s U s b sU s b U s

s a s a s a Y s b s b s b s b

( )U s

ForcedResponse Transfer Function Inputs Transfer

FunctionInputs

11 1 0

11 1 0

( )

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

( )

m mm m

n nn

G s

b s b s b s b N sY s U s U s G s U s

s a s a s a D s

= =

Transfer FunctionGiven a general nth order system:

The transfer function of the system is:

– The transfer function can be interpreted as:

( ) ( 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

11 1 0

11 1 0

( )m m

m mn n

n

b s b s b s bG s

s a s a s a

DifferentialEquation

u(t)

Input

y(t)

Output

Time Domain

G(s)U(s)

Input

Y(s)

Output

s - Domain

zero I.C.s

( )Y s

G sU s

Static gain

Transfer Function Matrix

For Multiple-Input-Multiple-Output (MIMO) System with m inputs and p outputs:

( ) ij p mG s G s

zero I.C.s

iij

j

Y sG s

U s

Inputs

1U s

2U s

mU s

Outputs

1Y s

2Y s

pY s

1, ,k p

1

1 1 2 2

m

k kj jj

k k km m

Y s G s U s

G s U s G s U s G s U s

1 11 1 1

1

m

p p pm m

U sY s G s

Y s G s G s U s

Y s G s G s U s

Poles and Zeros

• PolesThe roots of the denominator of the TF, i.e. the roots of the characteristic equation.

Given a transfer function (TF) of a system:1

1 1 01

1 1 0

( )( )

( )

m mm m

n nn

b s b s b s b N sG s

s a s a s a D s

11 1 0

1 2

( )

( )( ) ( ) 0

n nn

n

D s s a s a s a

s p s p s p

1 2 , , , np p p

1 0 1 2

1 0 1 2

( )( ) ( )( )( )

( ) ( )( ) ( )

mm m m

nn

b s b s b b s z s z s zN sG s

s a s a D s s p s p s p

n poles of TF

• ZerosThe roots of the numerator of the TF.N s b s b s b s b

b s z s z s zm

mm

m

m m

( )

( )( ) ( )

11

1 0

1 2 0

1 2 , , , mz z z

m zeros of TF

(2) For car suspension system:

Find TF and poles/zeros of the system.

Examples(1) Recall the first order system:

Find TF and poles/zeros of the system.

y y Ku y By K y Bu Ku

1 ( ) ( )s Y s K U s

1

Y s KG s

U s s

Pole:

Zero:

1 0s 1

s

No Zero

2 ( ) ( )s Bs K Y s Bs K U s

2

Y s Bs KG s

U s s Bs K

Pole:

Zero:

2 0s Bs K 2

1,2

4

2

B B Kp

0Bs K 1

Kz

B

System ConnectionsCascaded System

2 1G s G s G sInput Output

1G s 2G s 1U s U s 1 2Y s U s 2Y s Y s

Parallel System

1 2G s G s G s

InputOutput

1G s

2G s

1

2

U sU s

U s

1Y s

Y s++

2Y s

Feedback System

1

1 21

G sG s

G s G s

Input Output 1G s

2G s

U s 1Y s Y s+

-

2U s 2Y s

1U s

Given a general nth order system model:

The free response (zero input) of the system due to ICs is:

– Taking the LT of the model with zero input

(i.e., )

General Form of Free Response( ) ( 1) ( ) ( 1)

1 1 0 1 1 0n n m m

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

( ) ( 1)1 1 0 0n n

ny a y a y a y

1 1 2 ( 1)1 1 0 1 1

( )

( ) (0) (0)n n n n nn Free n

F s

s a s a s a Y s s a s a y y

( ) ( 1)

1 ( 1) 1 2 ( 2)1

1

( ) (0) (0) ( ) (0) (0)

( ) (0)

n n

n n n n n nn

y y

s Y s s y y a s Y s s y y

a sY s y

£ £

£0 ( ) 0

yy

a Y s

£

11 1 0

( ) ( )Free n n

n

F sY s

s a s a s a

A Polynominal of s that depends on ICsFree Response(Natural Response)

=Same Denominator as TF G(s)

Free Response (Examples)Ex: Find the free response of the car suspension system without tire model (slinker toy):

Ex: Perform partial fraction expansion (PFE) of the above free response when:

(what does this set of ICs means physically)?

2 2

2 (0) (0)2 4( ) ( )

2 4 2 4

s y ysY s U s

s s s s

2 4 2 4 , , ry y y u u y z u x

(0) 0 and (0) 1y y

Q: Is the solution consistent with your physical intuition?

2 2

2 2( ) (0)

2 4 2 4

s sY s y

s s s s

2 2 22 2 2

12 1 3( )

31 3 1 3 1 3

ssY s

s s s

11 2sin 3 cos 3 sin 3 tan 3

3 3t ty t e t t e t

Decaying rate: damping, mass

Frequency: damping, spring, mass

phase: initial conditions

The free response of a system can be represented by:

Free Response and Pole Locations

ip tiAe

11 1 0 1 2

1 2

1 2

( ) ( )( )

( )( ) ( )Free n nn n

n

n

F s F sY s

s a s a s a s p s p s p

A A A

s p s p s p

1 2

For simplicity, assume that

np p p

1 211 2( ) ( ) np tp t p t

Free Free ny t Y s A e A e A e

£

Real

Img.

,

0

is real 0

0

0

0

0

i

i i

i

i j

p

p p

p

p j

0ip

exponential decrease

iA constant

exponential increaseip tiAe

0ip 0ip

2 Re2 cos

i i i

i

p t p t p ti i i

ti i

Ae Ae AeA e t

ip j

jp j

decaying oscillation

Oscillation with constant magnitude

increasing oscillation

0

t

• Complete Response

Q: Which part of the system affects both the free and forced response ?

Q: When will free response converges to zero for all non-zero I.C.s ?

Complete Response

G sN s

D s( )

( )

( )U(s)

Input

Y(s)

Output( 1)ICs: (0) , (0) , , (0)ny y y

A Polynomial of depending on I.C.s

( ) ( ) ( )

( )

Forced Free

sD s

Y s Y s Y s

N sU s

D s

Denominator D(s)

All the poles have negative real parts.

Obtaining I/O Model Using TF Concept (Laplace Transformation Method)

• Noting the one-one correspondence between the transfer function and the I/O model of a system, one idea to obtain I/O model is to:

– Use LT to transform all time-domain differential equations into s-domain algebraic equations assuming zero ICs (why?)

– Solve for output in terms of inputs in s-domain to obtain TFs (algebraic manipulations)

– Write down the I/O model based on the TFs obtained

• Step 1: LT of differential equations assuming zero ICs

1 1 1 1 1 2 1 1 1 2

2 2 1 1 1 2 2 1 1 1 2 2 2 2

0

p p

M x B x B x K x K x

M x B x B B x K x K K x B x K x

Example – Car Suspension System

• Step 2: Solve for output using algebraic elimination method

1. # of unknown variables = # equations ?

2. Eliminate intermediate variables one by one. To eliminate one intermediate variable, solve for the variable from one of the equations and substitute it into ALL the rest of equations; make sure that the variable is completely eliminated from the remaining equations

L

21 1 1 1 1 2 1 1 1 2

22 2 1 1 1 2 2 1 1 1 2 2 2 2

0

p p

M s X s B sX s B sX s K X s K X s

M s X s B sX s B B sX s K X s K K X s B sX s K X s

21 1 1 1 1 1 2

22 1 2 1 2 2 1 1 1 2 2

0

p

M s B s K X s B s K X s

M s B B s K K X s B s K X s B s K X s

xp

1K

2K

g

2x

2B

1B

1x1M

2M

Example (Cont.)

• Step 3: write down I/O model from TFs

2

1 1 1

2 11 1

M s B s KX s X s

B s K

from first equation

Substitute it into the second equation

2

1 1 122 1 2 1 2 1 1 1 1 2 2

1 1p

M s B s KM s B B s K K X s B s K X s B s K X s

B s K

1 1 1 2 2

22 22 1 2 1 2 1 1 1 1 1p

X s B s K B s K

X s M s B B s K K M s B s K B s K

1

1 1 1 2 2

22 22 1 2 1 2 1 1 1 1 1

x pp

X s B s K B s KG

X s M s B B s K K M s B s K B s K

1 1 2 1 3 1 4 1 5 1 1 2 3p p pa x a x a x a x a x b x b x b x