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ME 475 Mechatronics Semester: February 2015 Dr. Sumon Saha Assistant Professor Department of Mechanical Engineering Bangladesh University of Engineering and Technology Friday, September 18, 2015
Transcript
Page 1: Pralkha

ME 475Mechatronics

Semester: February 2015

Dr. Sumon SahaAssistant Professor

Department of Mechanical Engineering

Bangladesh University of Engineering and Technology

Friday, September 18, 2015

Page 2: Pralkha

MechatronicsMechatronics

Transfer Function

Today’s topic

Page 3: Pralkha

MechatronicsMechatronics

•A transfer function (TF) relates one input and one output:

)(

)(system

)(

)(

sY

ty

sU

tu→→

Transfer function of a system, G(s), is defined as the ratio of the Laplace

Transform (LT) of the output variable, Y(s), to the LT of the input

Transfer function

Transform (LT) of the output variable, Y(s), to the LT of the input

variable, U(s), with all the initial conditions are assumed to be zero.

where:

)(

)()(

sU

sYsG =

[ ][ ])(L)(

)(L)(

tusU

tysY

=

=

Page 4: Pralkha

MechatronicsMechatronics Transfer function

Page 5: Pralkha

MechatronicsMechatronics

• Additive property

– Y(s) = G1(s)U1(s)+ G2(s)U2(s)

• Multiplicative property

– Y2(s) = G1(s)G2(s)U(s)

Properties of Transfer Functions

2 1 2

• ODE equivalence

ubdt

dubya

dt

dya

dt

yda

asasa

bsbsG

sU

sY

01012

2

2

01

2

2

01

ODE Equivalent

)()(

)(functionTransfer

+=++

+++

==

Page 6: Pralkha

MechatronicsMechatronics Example: Transfer Functions

Step Input:

Laplace transform:( ) ( ) ( ) ( )0sY s Y Y s KX sτ − + =

( )n

n

n

dxs X s

dt=

( )

( )

0 0( )

0

0 0

tx t

M t

y t t

==

>

= =

( )dyy Kx t

dtτ + =

ndt

( ) ( )( ) 1

Y s KG s

X s sτ= =

+steady state gain

time constant

K

τ

=

=

: ( )M

Step input X ss

=

( )/

( ) ( ) ( )

1

( ) 1 t

Y s G s X s

K M

s s

y t KM eτ

τ−

=

=+

⇒ = −

1

1

11

1 at

Ls

L es a

− −

=

= + MuPAD Notebook: ilaplace(K*M/(s*(tau*s + 1)), s, t)

Page 7: Pralkha

MechatronicsMechatronics Example: Transfer Functions

Laplace transform:( ) ( ) ( ) ( )0sY s Y Y s KU sτ − + =

( )n

n

n

dxs X s

dt=

( )

Ramp Input:( )dy

y Ku tdt

τ + =( )

0 0( )

0

0 0

tu t

at t

y t t

==

>

= =

ndt

( ) ( )( ) 1

Y s KG s

U s sτ= =

+

2: ( )

aRamp input X s

s=

( )2

/

( ) ( ) ( )

1

( ) 1t

Y s G s U s

K a

s s

y t Ka e Katτ

ττ −

=

=+

= − +

1

2

1L t

s

− =

MuPAD Notebook: ilaplace(K*a/(s^2*(tau*s + 1)), s, t)

Page 8: Pralkha

MechatronicsMechatronics Problem

( ) 50

50G s

s=

+

A system has a transfer function,

For step response, find the time constant, settling time and rise time.

( )Y s KAnswer: Time constant: τ = 0.02 s

Settling time: Ts = 0.08 sRise time: Tr = 0.044 s

>> c = tf([50],[1 50])

c =

50

------

s + 50

>> stepinfo(c)

ans =

RiseTime: 0.0439

SettlingTime: 0.0782

( ) ( )( ) 1

Y s KG s

X s sτ= =

+

Page 9: Pralkha

MechatronicsMechatronics Problem

Find the transfer function G(s) of the following differential

equation:

Using G(s), find the response c(t) to an input r(t) = u(t), a unit

step, assuming zero initial condition.

Answer: Taking Laplace transform assuming zero initial

2dc

c rdt

+ =

Answer: Taking Laplace transform assuming zero initial

condition,

( ) ( ) ( )

( ) ( )( )

2

1

2

sC s C s R s

C sG s

R s s

+ =

= =+

Page 10: Pralkha

MechatronicsMechatronics Problem

Find the transfer function G(s) of the following differential

equation:

Using G(s), find the response c(t) to an input r(t) = u(t), a unit

step, assuming zero initial condition.

Answer: Now,

2dc

c rdt

+ =

Answer: Now,

Taking inverse Laplace transformation,

( ) ( ) ( ) 1 1 1/ 2 1/ 2

2 2C s G s R s

s s s s= = = −

+ +

( ) ( ) 21

2

tc t u t e

−= −

( )1

1

11 or

1 at

L u ts

L es a

− −

=

= + MuPAD Notebook: ilaplace(1/(s*(s + 2)), s, t)

Page 11: Pralkha

MechatronicsMechatronics Second order system

Write down the response equation for this system.

What is the transfer function for this system?

m

k

m

c

Page 12: Pralkha

MechatronicsMechatronics Second order system

Question: Obtain the complete solution for equation:

With ζ = 0 using Laplace transformation. Assume all initial

conditions are zero.

2

2 2

22

n n n

d f dff K

dt dtςω ω ω+ + =

Taking Laplace transform: ( ) ( ) ( )2

2 22 n

n n

Ks F s sF s F s

s

ωζω ω+ + =

ζ = 0 (No damping)

( ) ( ) ( )

( ) ( )2

2 2

2

2

n n

n

n n

s F s sF s F ss

KF s

s s s

ζω ω

ωζω ω

+ + =

=+ +

( ) ( )2

2 1 2

2 2 2 2

1 22 2

1, ,

n

n

n n

n n

K C CF s K

s s s s

sHere C C

ωω

ω ω

ω ω

= = + + +

= = −

Page 13: Pralkha

MechatronicsMechatronics Second order system

Question: Obtain the complete solution for equation:

With ζ = 0 using Laplace transformation. Assume all initial

conditions are zero.

2

2 2

22

n n n

d f dff K

dt dtςω ω ω+ + =

( ) 1 sF s K

= −

Taking inverse Laplace transform,

( )2 2

1

n

sF s K

s s ω

= − +

( ) ( )1 cosn

f t K tω= −1

2 2cos

sL t

α− = +

MuPAD Notebook: ilaplace(wn*wn*K/(s*(s^2+wn^2)), s, t)

Page 14: Pralkha

MechatronicsMechatronics Second order system

Question: Obtain the complete solution for equation:

With ζ = 1 using Laplace transformation. Assume all initial

conditions are zero.

2

2 2

22

n n n

d f dff K

dt dtςω ω ω+ + =

Taking Laplace transform: ( ) ( ) ( )2

2 22 n

n n

Ks F s sF s F s

s

ωζω ω+ + =

ζ = 1 (Critical damping)

( ) ( ) ( )

( ) ( )2

2 2

2

2

n n

n

n n

s F s sF s F ss

KF s

s s s

ζω ω

ωζω ω

+ + =

=+ +

( ) ( ) ( )

( )

2 2

22 2

2 31 2

2 1 2 32 2

2

1 1 1, , ,

n n

n n n

n

n n n nn

K KF s

s s s s s

CC CK Here C C C

s s s

ω ωω ω ω

ωω ω ω ωω

= =+ + +

= + + = = − = −

+ +

Page 15: Pralkha

MechatronicsMechatronics Second order system

Question: Obtain the complete solution for equation:

With ζ = 1 using Laplace transformation. Assume all initial

conditions are zero.

2

2 2

22

n n n

d f dff K

dt dtςω ω ω+ + =

( ) 1 1nF s K

ω = − −

Taking inverse Laplace transform,

( )( )2

1 1n

n n

F s Ks s s

ωω ω

= − − + +

( ) ( )1t tn n

nf t K e te

ω ωω− −= − − ( )1

2

1 atL te

s a

− −

= +

MuPAD Notebook: ilaplace(wn*wn*K/(s*(s+wn)^2), s, t)

Page 16: Pralkha

MechatronicsMechatronics Second order system

Question: Obtain the complete solution for equation:

With ζ < 1 using Laplace transformation. Assume all initial

conditions are zero.

2

2 2

22

n n n

d f dff K

dt dtςω ω ω+ + =

Taking Laplace transform: ( ) ( ) ( )2

2 22 n

n n

Ks F s sF s F s

s

ωζω ω+ + =

ζ < 1 (Under damping)

( ) ( ) ( )

( ) ( )2

2 2

2

2

n n

n

n n

s F s sF s F ss

KF s

s s s

ζω ω

ωζω ω

+ + =

=+ +

( ) ( )2

2 31

2 2 2 22 2

n

n n n n

K C s CCF s K

s s s s s s

ωζω ω ζω ω

+= = + + + + +

Page 17: Pralkha

MechatronicsMechatronics Second order system

Question: Obtain the complete solution for equation:

With ζ < 1 using Laplace transformation. Assume all initial

conditions are zero.

2

2 2

22

n n n

d f dff K

dt dtςω ω ω+ + =

( ) 2

21

1n n

ζω ω ζζ

+ + − −n

a ζω= −

Taking inverse Laplace transform,

( )( )

( ) ( )

( ) ( )( )

2

2 2 2

2 2

111

1

/1

n n

n n

d d

d

s

F s Ks s

s a aK

s s a

ζω ω ζζ

ζω ω ζ

ω ω

ω

+ + − − = − + + −

− −= −

− +

( ){ }1 cos / sinat

d d df K e t a tω ω ω = − −

21d n

ω ω ζ= −

( )1

2 2cosats a

L e ts a

αα

− −

= − +

( )1

2 2sinat

L e ts a

αα

α−

= − +

Page 18: Pralkha

MechatronicsMechatronics Second order system

Question: Obtain the complete solution for equation:

With ζ > 1 using Laplace transformation. Assume all initial

conditions are zero.

2

2 2

22

n n n

d f dff K

dt dtςω ω ω+ + =

Taking Laplace transform: ( ) ( ) ( )2

2 22 n

n n

Ks F s sF s F s

s

ωζω ω+ + =

ζ > 1 (Over damping)

( ) ( ) ( )

( ) ( )2

2 2

2

2

n n

n

n n

s F s sF s F ss

KF s

s s s

ζω ω

ωζω ω

+ + =

=+ +

( ) ( )( )( )

2 2

31 2a b K CC C

F s Ks s a b s a b s s a b s a b

− = = + + − − − + − − − +

2 1

n

n

a

b

ζω

ω ζ

= −

= −

Page 19: Pralkha

MechatronicsMechatronics Second order system

Question: Obtain the complete solution for equation:

With ζ > 1 using Laplace transformation. Assume all initial

conditions are zero.

2

2 2

22

n n n

d f dff K

dt dtςω ω ω+ + =

( ) 1 1 1a b a bF s K

− + = + −

Taking inverse Laplace transform,

( )( ) ( )

1 1 1

2 2

a b a bF s K

s b s a b b s a b

− + = + − − + − −

1 1 atL e

s a

− = −

( ){ } ( ) ( ){ } ( )1 / 2 / 2

a b t a b tf K a b b e a b b e

+ − = + − − +

MuPAD: ilaplace(wn*wn*K/(s*(s^2+2*z*wn*s+wn^2)), s, t)

Page 20: Pralkha

MechatronicsMechatronics Problem

For the system shown below do the following:

33 /

15 /

3

k N m

c Ns m

m kg

=

=

=m

kx

i. Find the transfer function G(s) = X(s) / F(s).

ii. Find the system equation for transient response subjected

to a unit step input.

iii. Find ωn, ζ, %OS, Ts, Tp, and Tr.

3m kg=

c

Page 21: Pralkha

MechatronicsMechatronics Problem

For the system shown below do the following:

33 /

15 /

3

k N m

c Ns m

m kg

=

=

=m

kx

i. Find the transfer function G(s) = X(s) / F(s).

=

c

( ) ( )2

2 2 2

/ 1

2 3 5 11

n

n n

kG s

s s s s

ωζω ω

= =+ + + +

11n

k

mω = =

5

2 2 11c

c c

c mkζ = = =

Page 22: Pralkha

MechatronicsMechatronics Problem

For the system shown below do the following:

33 /

15 /

3

k N m

c Ns m

m kg

=

=

=m

kx

ii. Find the system specifications for transient response

subjected to a unit step input.

=

c

( ) ( )2

2 2

1

2

n

n n

X sk s s s

ωζω ω

=+ +

2

5 / 2

1 19 / 2

n

d n

a ω ζ

ω ω ζ

= − = −

= − =( ) ( ){ }1

1 cos / sinat

d d dx t e t a t

kω ω ω = − −

Page 23: Pralkha

MechatronicsMechatronics Problem

For the system shown below do the following:

33 /

15 /

3

k N m

c Ns m

m kg

=

=

=m

kx

ii. Find the system specifications for transient response

subjected to a unit step input.

=

c

2

5 / 2

1 19 / 2

n

d n

a ω ζ

ω ω ζ

= − = −

= − =

( ) ( ) ( )5 /21 51 cos 19 / 2 sin 19 / 2

33 19

tx t e t t− = − +

Page 24: Pralkha

MechatronicsMechatronics Problem

For the system shown below do the following:

33 /

15 /

3

k N m

c Ns m

m kg

=

=

=m

kx

iii. Find ωn, ζ, %OS, Ts, Tp, and Tr.

=

c2/ 1

% 100 2.72

41.6 sec

s

n

OS e

T

ζπ ζ

ζω

− −= × =

= =2

1.44 sec1

1.11 sec

p

n

r

d

T

T

π

ω ζ

π βω

= =−

−= = 1tan 0 .7 2 raddω

βσ

− = =

Page 25: Pralkha

MechatronicsMechatronics Problem

For the system shown below do the following:

33 /

15 /

3

k N m

c Ns m

m kg

=

=

=m

kx

=

c

( ) ( )2

2 2 2

/ 1

2 3 5 11

n

n n

kG s

s s s s

ωζω ω

= =+ + + +

>> g = tf([1],[3 15 33])

>> damp(g)

>> stepinfo(g,'Risetimelimits',[0 1])

>> step(g)

RiseTime: 1.1133

SettlingTime: 1.7204

SettlingMin: 0.0303

SettlingMax: 0.0311

Overshoot: 2.7224

Undershoot: 0

Peak: 0.0311

PeakTime: 1.4411

Page 26: Pralkha

MechatronicsMechatronics

Question: Consider the transfer function for a feedback system

given below

Calculate the transient response specifications for the system,

when the system subjected to a unit step input. Here, ωd = ωn √(1-

ζ2) and σ = ζωn.

2

25( )

6 25G s

s s=

+ +

Example: Transfer Function

ζ2) and σ = ζωn.

2

5, 0 .6 1

1 4, 3

n

d n n

ω ζ

ω ω ζ σ ζ ω

= = <

= − = = =

Underdamped system

2/ 1

2

% 100 9.5

41.33 sec

0.785 sec1

0.55 sec

s

n

p

n

r

d

OS e

T

T

T

ζπ ζ

ξω

π

ω ζ

π βω

− −= × =

= =

= =−

−= =

1tan 0 .9 3 raddωβ

σ− = =

Page 27: Pralkha

MechatronicsMechatronics Home Work

For the system shown below do the following:

a) Find the transfer function G(s) = X(s) / F(s).

b) Find ωn, ζ, %OS, Ts, Tp, and Tr.