Transfer Function Approach to Modeling Dynamic Systems

Transfer Function Approach to

Modeling Dynamic Systems

The Concept of Transfer Function

Consider the linear time-invariant system defined by the following differential

p

Consider the linear time invariant system defined by the following differential

equation :

( ) ( 1)

0 1 1

( ) ( 1)

0 1 1

( )

n n

n n

m m

m m

a y a y a y a y

b u b u b u b u n m

−

−

−

−

+ + + +

= + + + + ≥

&

L

&

L

Where y is the output of the system, and x is the input. And the Laplace transform of

the equation is,

1

0 1 1

( a S

0 ⁿ

+ a S

1 ⁿ⁻

+ + L a

_n−1

S + a Y s

_n

) ( )

(

_{n n}

) ( )

= +

−

+ + +

The Concept of Transfer Function

Th i f h L l f f h ( f i ) h L l

p

The ratio of the Lapalce transform of the output (response function) to the Laplace

Transform of the input (driving function) under the assumption that all initial

conditions are Zero.

1

0 1 1

1

0 1 1

( ) ( ) ( )

m m

m m

n n

b S b S b S b

Transfer Function Y s G s

U s a S a S a S a

−

−

−

+ + ⋅⋅⋅⋅ + +

= = =

+ + ⋅⋅⋅⋅ +

0 1 1

( )

( ) ( )

n n

U s a S a S a S a

P s n m

+ +

−

+

= ( ≥ )

( ) n m

Q s ≥

Comments on Transfer Function

1 A h i l d l

1. A mathematical model.

2. Property of system itself.

(Independent of the magnitude and nature of the input)

3. Includes the units

(But, no information about the physical structure of the system)

] [

] ) [

) ( (

) (

volt s volt

s G U

s

Y = = Y ( s ) = U ( s ) G ( s )

Comments on Transfer Function

4. Analytic method and Experimental method

system

u(t) y(t)

U(s) Y(s)

( )

) ) (

( U s

s s Y

G =

U(s) Y(s)

L.T.I

5 Different systems may have identical T F

u = T

5. Different systems may have identical T.F

ex)

J

y = ω

T b

J ω

_&

+ ω =

b Js T

U Y

= +

=ω 1

Electrical System (if b=1, J=L/R)

1 1

+ + =

=

R s R L Ls

R U

Y

R

Block Diagram

I t Si l O t t Si l

g

Input Signal Output Signal

Transfer Function

Pi k Off P i

Summing Point Pick Off Point

Closed Loop Transfer Function p

( ) ( ) ( ) ( )[ ( ) ( ) ( )]

Y s = G s E s = G s R s − H s Y s [1 + G s H s Y s ( ) ( )] ( ) = G s R s ( ) ( )

1 1

( ) ( )

( ) 1 ( ) ( ) G s

Transfer Function Y s

R s G s H s

∴ = =

+

Closed Loop Transfer Function

1 1 1

( )

F s + s X s² ( ) sX s( ) X s( )

p

ex)

m

b

s s

++

−

m k +

m

1

2

( ) k ( ) b ( ) ( )

F s X s sX s s X s

m − m − m =

[ ]

²

2 2

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ), ( ) ( ) ( )

m m m

F s kX s bsX s ms X s

ms bs X s F s kX s ms bs k X s F s

− + =

+ = − + + =

Partial Fraction Expansion with MATLAB

1 1

( ) (1) (2) ( 1)

( ) (1) (2) ( 1)

h h

n n

B s num b S b S b h

A s den a S a S a n

−

−

+ + ⋅⋅⋅⋅ + +

= =

+ + ⋅⋅⋅⋅ + +

( ) (1) (2) ( 1)

[ (1) (2) ( )], [ (1) (2) ( )]

A s den a S a S a n

num b b b h den a a a h

+ + ⋅⋅⋅⋅ + +

=

L

=

L

[r, p, k] = residue (num, den)

( ) (1) (2) ( )

B s ( ) r r r n

k

>>num=[1 8 16 9 6];

>>den=[1 6 11 6];

( ) ( ) ( ) ( )

( ) k s ( ) (1) (2) ( )

A s = + s p + s p + + s p n

− −

L

−

ex)

B s ( ) s

⁴

+ 8 s

³

+ 16 s

²

+ 9 s + 6

>>den=[1 6 11 6];

>>[r,p,k]=residue(num,den) r=

-6.0000

ex)

3 2

( )

( ) 6 11 6

A s = s s s

+ + +

^-4.0000_3.0000

p=

-3.0000 -2.0000 -1.0000 k=

1 2

6 4 3

2 3 2 1

s s s s

− −

= + + + +

+ + +

1 2

Transient Response Analysis with MATLAB

2

2

( )

d y dy du

m b k y u

dt dt dt

⎛ ⎞

= − ⎜ ⎝ − ⎟ ⎠ − −

ex)

2 2

dt dt dt

d y dy du

or m b ky b ku

dt dt dt

⎝ ⎠

+ + = +

Assume the initial condition is 0,

( ms

2

+ bs + k Y s ) ( ) = ( bs + k U s ) ( )

2

( ) ( )

Y s bs k

Transfer Function

U s ms bs k

= = +

+ +

If, m=10kg, b=20N-s/m, k=100N/m

2 2

( ) 20 100 2 10 1

, ( )

( ) 10 20 100 2 10

Y s s s

U U s

+ +

=

₂

=

₂

( ) =

( ) 10 20 100 2 10

U s s + s + s + s + s

Transient Response Analysis with MATLAB Transient Response Analysis with MATLAB

t=0:0.01:8;

num=[2 10];

1.5

Unit Step Response

num [2 10];

den=[1 2 10];

sys=tf(num,den);

step(sys,t)

1

ut

grid

title('Unit Step Response','Fontsize',15') xlabel('t(sec)','Fontsize',15')

l b l('O ' ' i ' 1 ')

0.5

Outpu

ylabel('Output','Fontsize',15')

0

0 1 2 3 4 5 6 7 8

0

t (sec)

The law of conservation of momentum The law of conservation of momentum

By Newton's 2nd law,

( ) ( )

dv d

F ma m mv F dt d mv

dt dt

= = = → ⋅ =

2 2

1 1

2 1

( )

t v

t v

Fdt = d mv = mv − mv

∫ ∫

And if There is no input force, then we get

( ) 0, .

d mv = mv = const

The law of conservation of momentum And also,

Impulse Input Impulse Input

A bullet stuck in mass M A bullet stuck in mass M

(0 ) v

v v&

(0 ) v +

(0 ) v −

Δ

t t

Δt 0 (0 )

v + Δt

0

−A

Sudden change in the velocity and acceleration of bullet

Impulse Input

( M + m x bx ) && + & + kx = F t ( )

System equation :

Impulse Input

( ) ( ),

F t = − mv & = Δ A t δ t A t Δ

Impulse force :

0 0

: Magnitude of impulse force

0 0

0 0

( ) , (0 ) (0 )

A t δ t dt m v dt A t mv mv

+ +

− −

Δ = − Δ = − − +

∫ ∫ &

Noting that, Magnitude of impulse force is equal to change of momentum!!

(0 ) (0 )

v + = x & + =

Initial velocity of M+m

Thus the system equation becomes

( ) ( )

_y

Impulse Input p p

By taking the Laplace Transform,

[ ]

( M + m ) ⎡ ⎣ s X s

2

( ) − sx (0 ) − − x & (0 ) − + ⎤ ⎦ b sX s ( ) − x (0 ) − + kX s ( ) = mv (0 ) − − mx & (0 ) +

(0 ) (0 ) 0

x & − = x − =

Noting that we obtain

2

(0 ) (0 )

( ) ( )

mv mx

X s M m s bs k

− − +

= + + +

&

[ ]

²

[ ]

0 2

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

(0 ) lim ( ) lim ( ) lim

( )

t s s

s mv mx mv mx

x x t s sX s

M m s bs k M m

m

→ + →∞ →∞

− − + − − +

+ = = = =

+ + + +

& &

& &

(0 ) (0 )

2

x m v

M m

→ + = −

& +

2 2

( ) (0 ) 1 ( ) (0 )

( ) ( ) ( ) 2

M m x M m mv

X s M m s bs k M m s bs k M m

+ + + −

∴ = =

+ + + + + + +

&

Impulse Input p p

50 0 01 100 / 2500 / (0 ) 800 /

M k k b N k N

ex)

1 50.01 0.01 800 7.9984

( )

X s × ×

= =

50 , 0.01 , 100 / , 2500 / , (0 ) 800 /

M = kg m = kg b = Ns m k = N m v − = m s

0.02

Impulse Response of System

2 2

( ) 50.01 100 2500 50.02 50.01 100 2500

X s = s s = s s

+ + + +

num=[7.9984];

den=[50.01 100 2500];

sys=tf(num,den);

0.01 0.015

x(t)

impulse(sys) grid

title('Impulse Response of System','Fontsize',15')

-0.005 0 0.005

Response x

xlabel('t(sec)','Fontsize',15')

Ramp Response

M=10kg, b=20Ns/m, k=100N/m

2

( ) 2 10

( ) 2 10

Y s s

U

= +

Ramp Response

4 5

Unit-Ramp Response

num=[2 10];

( )

2

2 10

U s s + s +

( ) : , , 1

u t Unit ramp input u = α t α =

3.5 4 4.5

u(t)=t

num=[2 10];

den=[1 2 10];

sys=tf(num,den);

t=0:0.01:4;

2 2.5 3

t) and Input

u=t;

lsim(sys,u,t) grid

title('Unit-Ramp Response','Fontsize',15)

0.5 1 1.5

Output y(t u

y

( p p , , )

xlabel('t')

ylabel('Output y(t) and Input u(t)=t','Fontsize',15) text(0.8,0.4,'y','Fontsize',12)

text(0 4 0 8 'u' 'Fontsize' 12)

0 0.5 1 1.5 2 2.5 3 3.5 4

0

y

t (sec)

text(0.4,0.8,'u','Fontsize',12) legend('y')