EML4314c Modeling 2nd Order Sys

8/3/2019 EML4314c Modeling 2nd Order Sys

1/50

This Weeks Objectives

Establish Dynamic Models of System to beControlled

Second Order SystemsObtain Solutions using LaPlace TransformsCreate Simulink Model and Generate

Simulated Results


2/50

Modeling of a Spring-Mass-Damper System

(Modeling of a Second Order System)


3/50

lets start with a one car system suppose we know:

m1 mass of the car k1, L01 spring constant and free lengthc damping coefficient

m 1

k1,L01

c

f(t)

x1


4/50

one car system suppose at start (initial condition):

x1 = L01 spring is applying zero force= 0 car is at rest

f(0) = 0 there is no applied force

m 1

k1,L01

c

f(t)

x1

1x


5/50

one car system now we want (final condition):

x1 = L01 + 3 cmwe want to move there as fast as possiblewe dont want to overshoot muchwe want the system to settle down quick

m 1

k1,L01

c

f(t)

x1

Were going to need to

properly define theseterms.


6/50

one car system we have a desired final position we can measure the position of the car we can set the value of f, the applied force, to

any value we want and constantly adjust it

m 1

k1,L01

c

f(t)

x1

Well pick a value for f that isrelated to the difference betweenthe current measured position andthe desired final position.


7/50

m 1

k1,L01

c

f(t)

x1

In Controls Terminology:BLOCK DIAGRAM


8/50

problem statement given:

m1 mass of the car k1, L01 spring constant and free length

c damping coefficientx1(0) initial position at t=0

(0) initial velocity at t=0xd desired position

generate values for the input force such thatthe car reaches the desired position within 3 secthe car will not overshoot by more than 0.25 cmthe car will settle within s 0.1 cm within 5 sec

m 1k1,L01

c

f(t)

x1

1x


9/50

how to start?lets first write the equationof motionassume the current position x 1>L01 andthat 1 is positive

m 1k1,L01

c

f(t)

x1

x

m 1ck1 (x 1- L01 )

f(t)

1x

1101111 xm)L(xkxcf(t) !


10/50

how do you solve for x 1(t) if you know f(t)as well as x 1(0) and 1(0)?

m 1k1,L01

c

f(t)

x1

f(t))L(xkxcxm 0111111 !

x


11/50

we must solve a second

order differential equation

we will write the equation in the Laplacedomain

Laplace transformations substitute easilysolved algebraic equations for differentialequations

m 1k1,L01

c

f(t)

x1

f(t))L(xkxcxm0111111

!


12/50

Review o f Laplac e transf o rms

if f(t) is some function of time, then theLaplace transformation of f(t) can be

written as

change from the time domain to the 's'domain

g

!!0

dtef(t)f(t)F(s) stL


13/50

Example:

Let f(t) = 1 for t > 0. Find F(s).

s

1

es1

es1

es1

dteF(s)

s0s

t

0t

st

st

!

!

!

!

g

g!

!

g

0


14/50

Example:

Let f(t) = eat

for t > 0 where a is aconstant. Find F(s).

as1

ea-s

1e

a-s1

ea-s

1

dtedteeF(s)

0

t

0t

a)-t(s

a)-t(s-stat

!

!

!

!!

g

g!

!

gg

00


15/50

The LaPlace transform is a linear operation.

For any functions f(t) and g(t) and for anyconstants, a and b,

L{a f(t) + b g(t)} = a L {f(t)} + b L {g(t)}


16/50

Example:

Let f(t) = cosh(at) = (eat

+ e-at

)/2for t > 0. Find F(s).

22

at-at

ass

as

1

2

1

as

1

2

1

)(e21

)(e21

at)(coshF(s)

!

!

!! LLL


17/50

LaPlace transform of the derivative of f(t).

L(f '(t)) = sL (f) - f(0)

L (f "(t)) = s 2L (f) - sf(0) - f'(0)

L (f (n)) = s nL (f) - s n-1 f(0) - s n-2 f '(0) - ...- f (n-1) (0)


18/50

LaPlace transform of the integral of f(t).

f(t)s1

d)f(t

0

!

L


19/50

Laplace transforms have been calculated

for a large variety of functions. Samplingof these are listed in Table 2.3 of the text Modern Control Systems, 9 th ed., R.C. Dorf

and R.H. Bishop, Prentice Hall, 2001 .


20/50


21/50

Inverse transformations are often of the form

where G(s) and H(s) are polynomials in s

This ratio must be re-written in terms of its partialfraction expansion.

Review the techniques for partial fractionexpansion when the polynomial H(s) hasrepeated roots or complex roots.

H(s)G(s)

F(s) !


22/50

Example:

Let a b . Find f(t)

first perform partial fraction reduction

Multiplying the left and right side of the aboveequation by (s-a)(s-b) gives

b)a)(s(s1

F(s) !

b)(s A

a)(s A

b)a)(s(s1

F(s) 21!!


23/50

1= A 1 (s-b) + A 2 (s-a)

Can pick two values for s and then solvefor A 1 and A 2 from the two equations intwo unknowns.

When s=a, 1= A 1(a-b).When s=b, 1=A 2(b-a). Thus

b)(a1- A,

b)(a1 A 21 !!


24/50

substituting A 1 and A 2 into F(s) gives

Using the linearity of the Laplacetransform,

!!

b)(s1-

a)(s1

ba1

b)a)(s(s1F(s)

-

! b)(s1-

a)(s1

ba1

f(t)11

LL


25/50

from previous example

btat eeba

1

b)(s1-

a)(s1

ba1

f(t)

!

-

! 11 LL

END OF REVIEW


26/50

car 1car 1car 2car 2 car 3car 3

Back to our problem.


27/50

equation of motion

take Laplace transform of left and right sideof equation

m 1k1,L01

c

f(t)

x1

f(t))L(xkxcxm 0111111 !

F(s)(1)Lkxkxcxm 01111111 !LLLL


28/50

m 1k1,L01

c

f(t)

x1

F(s)(1)Lk

xkxcxm

011

11111

!LLLL

(0)x(0)xs(s)Xsx

(0)x(s)sXx

1112

1

111

!

!

L

L


29/50

m 1k1,L01

c

f(t)

x1

!

21 1 1 1

1 1 1 1 1 01

m s X (s) s x (0) x (0)

1c sX (s) x (0) k X (s) k L F(s)

s

F(s)(1)Lk

xkxcxm

011

11111

!L

LLL

? A F(s)

s1

Lk(0)xc

(0)x(0)xsmkscsm(s)X

0111

11112

11

!


30/50

m 1k1,L01

c

f(t)

x1

1x

supposem 1 = 2 kg = 2 N sec 2/mk1 = 3 N/cm = 300 N/mL01 = 6 cm = 0.06 m

c = 2.5 N sec/cm = 250 N sec/mx1(0) = 4 cm = 0.04 m

(0) = -2 cm/sec = -0.02 m/sec

f(t) is a step input of 8 N starting at t=0

s

8F(s) !


31/50

m 1k1,L01

c

f(t)

x1

sN8

s1

m0.06mN

300m)(0.04msecN

250

)secm

(-0.02m)(0.04smsecN

2

mN

300smsecN

250smsecN

2(s)X2

22

1

!

-

are units consistent?

seems like s has units of 1/secand X(s) has units of m sec


32/50

? A

s1

0.06300(0.04)250

0.02s0.042s

8300s250s2(s)X 21 !

? A 9.96s0.08s

26300s250s2(s)X 21 !

s300s250s226s9.96s0.08

300s250s2

9.96s0.08s

26(s)X 23

2

21!!


33/50

x1(t) = 0.08667+ 0.00667 e -62.5 t

[-7cosh(61.29 t) 7.187sinh(61.29 t)]

how do we check the results ?

10lim ( ) 0.08667 s

sX sp

!

check boundary conditions

check steady state final position

- final value theorem:


34/50


35/50

How to model the system in Simulink?

m 1k1,L01

c

f(t)

x1


36/50

m 1k1,L01

c

f(t)

x1

1x

supposem 1 = 2 kg = 2 N sec 2/mk1 = 3 N/cm = 300 N/mL01 = 6 cm = 0.06 m

c = 2.5 N sec/cm = 250 N sec/mx1(0) = 4 cm = 0.04 m

(0) = -2 cm/sec = -0.02 m/sec

f(t) is a step input of 8 N starting at t=0


37/50

lets model this system

using Simulink

governing differential equation

for our case, f(t) = 0,(0) = -0.02 m/sec, x 1(0) = 0.04 m

thus

f(t))L(xkxcxm 0111111 !

1x

1

011111 m

)L(xkxcf(t)x !

m 1k1,L01

c

f(t)

x1


38/50

1

011111 m

)L(xkxcf(t)x !


39/50


40/50

blocks can be grouped into subsystems


41/50

x


42/50

m 1

k1,L01

c

f(t)

x1


43/50


44/50

d1 = d2 = 15 cmd3 = 80 cmm1 = 2.5 kgm2 = 5.5 kgc1 = 8 N sec/mc2 = 10 N sec/mk1 = 3 N/cm ; L01 = 6 cmk2 = 5 N/cm ; L02 = 8 cmk3 = 4 N/cm ; L03 = 10 cm

Cars 1 and 2 start at rest at their staticequilibrium position.

A constant force f 1 = 20N and f 2 = -5 N .

Obtain the motion response of the two cars, i.e.obtain x 1(t) and x 2(t).

x1

x2

d1

f 1 f 2

d2

d3

c2c1

m2m1k 2, L02k 1, L01 k 3, L03


45/50

Determine the position of the two cars at static equilibrium when noforces are applied.

x1

x2

d1

m1

k 2, L02k 1, L01

Car 1 will be in equilibrium when k 1 (x1 L01) = k 2 [x2 (x 1 + d 1 + L02 )]

x1

x2

d1 d2

d3

m2k 2, L02 k 3, L03

Car 2 will be in equilibrium when k 2 [x2 (x 1 + d 1 + L02)] = k 3 [d3 (x 2 + d 2 + L03)]

Substituting the given parameters and solving for x 1 and x 2 givesx1_equil = 17.06 cmx

2 _

equil= 46.70 cm


46/50

Now write equations of motion of the system.

x1

m1 k 2 [x2 (x 1 + d 1 + L 02)]k 1 (x 1 L01)

11xc f 1

)Lx(k xc)]Ldx(x[k f xm 011111021122111 !

x1

x2

d1 d2

d3

m2k 3 [d3 (x 2 + d 2 + L 03)]

k 2 [x2 (x 1 + d 1 + L 02)]

22 xcf 2

)]Ldx(d[k xc)]Ldx(x[k f xm 03223322021122222 !


47/50


48/50

)Lx(k xc)]Ldx(x[k f xm 011111021122111 !

)]Ldx(d[k xc)]Ldx(x[k f xm 03223322021122222 !

0)0(x,m4670.0)0(x

0)0(x,m1706.0)0(x

22

11

!!!!


49/50

time, sec

p o s

i t i o n , m

e t e r s

car 2

car 1


50/50

This Weeks Objectives

Establish Dynamic Models of System to beControlled Second Order Systems

Obtain Solutions using LaPlace TransformsCreate Simulink Model and GenerateSimulated Results

Documents

EML4314c Modeling 2nd Order Sys