control systems Chapter5.pdf

8/14/2019 control systems Chapter5.pdf

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Chapter 5

Time Response

Analysis


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Test signals

Impulse

Step

Ramp


3/151

Test signals

Sinusoidal.

(frequency response)


4/151

Laplace Transforms

step u-1 (t) 1/s

Ramp t u-1 (t) 1/s2

Impulse (t) 1


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Order of a

system is ???


6/151

Order of a

system is order

of the differential

equation


7/151

zeroth order system :

C(s) / R(s) = K

a constant(algebraic equation)


8/151

dc tachogenerator

Examples of zeroth

order systems :

Potentiometer

dc amplifier


9/151

First order systems:

C(s) / R(s) = K / (s + a)


10/151

Open loop1000 / ( 100 + 200 s )

= 10 / ( 1 + 2 s )

Closed loop

1000 / ( 200 + 200 s )

= 5 / ( 1 + s )


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open loop

vt (s) = (vr (s) )(10 / ( 1+2 s ))

= (25 / s )( 10 / (1 + 2 s)Vt () = 250 V

closed loop

vt(s) = ( 25 / s)( 5 / ( 1+s ))

Vt () = 125 V.


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0

50

100

150

200

250

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

x1

x2


13/151

K : steady state value of the

function

: time constantsmaller, faster response.

In general K ( 1- e - t / )


14/151

Ramp response

of first order

systems


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Ramp Response

r (t) = t u-1 (t)

R (s) = 1 / s2


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1=KLet

s1R(s)

C(s)

t

K

+=


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)s1(s

1)2

+

=c(s

+=

s1s

1

s

1

2


19/151

)e1(-t=c(t) -t/


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0

1

2

3

4

5

6

7

8

9

0 0.5

1 1.5

2 2.5

3 3.5

4 4.5

5 5.5

6 6.5

7 7.5

8 9 9


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c

t


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23/151

Second ordersystems


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2nn2

2n

+s2s)s(R

)s(C

+

=


25/151

For unit step input

R (s ) =

1

s


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Applying final value

theorem

C(t) t =1.0


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28/151

r LKp KA+

-

+

-

KTRa

1

s(Js+B)

sKb


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Taking

KP = 0.1 V / radKA = 10 V / V

KT = Kb = 0.5 Nm/A

orV/rad/sec.


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B = 0.5 Nm / rad / sec

J = 1 Kg m2


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1+ss1

)s()s(

2r

L

+=


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n = 1 rad / sec.

= 0. 5


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s2 + s + 1 =

(((( )))) (((( ))))s + 0.52

3 2

2

++++


34/151


35/151

In general

( ) ( )222 1+s += nn

s2

+ 2 n s + n2


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2

n

n1,2

1j

s

=


37/151

= - n j d

d = n1 -2

d damped frequency damping factorn undamped

natural frequency


38/151

For 0 < < 1

roots of s are

complex conjugate


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roots are

repeating at

s = - n

For = 1


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For > 1

roots are real


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= 0 : undamped0


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dampedover:1>

dampedcritically:1

=


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Unit step response

r (t) = u-1 (t)

1R (s) = ----

s


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)()(

0

22

2

n

n

sssc

+=

=


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22

1

ns

s

s +=


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:10


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1C(s) = -

s

s + 2 n

s2

+ 2 ns +n2


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1

= -s

s + ns2 + 2 ns +n2

ns2 + 2 ns +n2


50/151


51/151

atsin22

+ asa


52/151

atsine

a+)b+s(

a

atcosea+)b+s(

b+s

bt-

22

-bt

22


53/151

22

2

)(

)(

22

dn

n

nn

n

s

s

ss

s

++

+

=++

+


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22

22

)(

.

2

dn

d

d

n

n

n

s

ss n

++

=++


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tsine.d

t-

d

n n


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tsine.

1d

t-

2

n

=


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Combine sine cos terms :

1

1

1 2-

( )

1 2-


59/151

1

12

-

sin ( t +d ))))

where cos = .


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)+t(sin1

e

-1

=(t)c

d2

t- n


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C(t)

1.0

Time(t)

1+e-nt

1-2

1-e-nt

1-2

Unit step response of 2nd order system for


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where ( n) is thereal part of roots

- n


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d is the imaginarypart of rootsd =

2n -1j


64/151

Cos =

1 -n 2

- nn


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C(t)

1.0

tp

Mp

trTime

Tolerance band

Time response spceification


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67/151

sin (dt + ). cos

- cos (dt + ).Sin = 0


68/151

sin (d

t) = 0

dt =


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70/151

At t = tp ,

)+(sin1

1)( 2

.

=

pnt

p

e

tc


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2

1

21.1

=

+=

eM

e

p

nn


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%.10021

=

e


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As damping factor

increases

peak overshoot ??


74/151

As damping factor

increases

peak overshoot reduces

M


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p

0 100

0.1 730.2 53

0.3 370.4 25

0.5 16

Mp

0.7 4.60.8 1.5

0.9 0.02

1.0 0

0.6 9.5

100


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0

10

2030

405060

70

8090

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

%

Mp


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Settling time (ts)

Assume tolerance

band 5 %


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0.05 = e-nts


79/151

n

snn

tl

3=or t

)05.0(

s

=


80/151

n

s

4

isetoleranc2%fort

Rise time (t )


81/151

r

)+t(sin

1

e-1=1=)(tc

rd

2

t-

r

rn


82/151

orsin (dtr + ) = 0

dtr + =


83/151

trd

====

or


84/151


85/151


86/151

1 n2

= 1-s2 (s2 + 2 ns + n2)

E(s) = R(s) - C(s)


87/151

2n

2

n2

2 +s2s

2

.

1

n

ss

s

++

=


88/151

n2=

e(t) = Lt sE(s)

t s 0


89/151

Some more MATLAB

d


90/151

commands .

>> ng = [1 1]; dg =[1 2];

>> printsys (ng,dg)

num/den =

++++ 1S


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>>>>>>>>++++++++

2S

1S

t ( )


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>> z = roots(ng)

z =

-1

>> t (d )


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>> p = roots(dg)

p =

-2

>>

>> ( d )


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>>pzmap(ng,dg)

>>title(pole-zero map)

>>[nt,dt] =


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cloop(ng,dg,-1)

>>zc = roots(nt)

zc =

-1

>> t (dt)


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>>pc = roots(dt)

pc = -1.5

>>


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>>t = [0:0 5:10];


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>>t = [0:0.5:10];

>>[y,x,t] = step(ng,dg,t);

>>plot(t,y), grid


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>>xlabel(Time[sec])


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>>xlabel(Time[sec])

>>ylabel(output)


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An example :


102/151

R +

-

K 1s(s+1)

C

KG


103/151

Kss

KRC

ss

KG

++=

+=

2

)1(

K=


104/151

K

Kn

2

1=

n

Let K = 100


105/151

Let K = 100

n = 10 rad / sec

= 0.05

====Mp


106/151

M

e

p

1 2

100%= 85 %

e for ramp input


107/151

ess for ramp input

= 2 /n= 0.1 / 10

= 0.01

While the steady state


108/151

While the steady stateerror is acceptable

Mp is not. - Damping

to be increased to say,

0.5

1=2Then


109/151

10.52

1

=

1=2Then

n

n

=

ess increases to 1 - not


110/151

ess increases to 1 not

acceptable. So we go forchanging the configuration

of the amplifier - derivativeerror compensation.

Error coefficients


111/151

R(s)+

-

E(s) C(s)G(s)

We know

C (s) = R (s) G (s) / ( 1+G (s) )


112/151

C (s) = R (s) . G (s) / ( 1+G (s) )

andE (s) G (s) = C (s)

( )(s)G+1(s)R=(s)E


113/151

(s)EsLt=

)(eerrorstateSteady

0s

ss

step input

R (s) = 1 /s


114/151

R (s) = 1 /s

(constant position)

)(1

1.1

0 sGs

sLtes

ss

+

=

Let


115/151

coeff.)error(position

)(0 ps KsGLt =

Then

1


116/151

p

ssK

e+

=1

1

Ramp input


117/151

R (s) = 1/s

2

(Constant velocity)

Steady state error

in position

GsLt

)(1

112=


118/151

vs

s

KssGLt

sGssLt

1)(

1

)(1.

0

20

+=

K v = velocity error coefficient

parabolic input

R (s) = 1/s3


119/151

R (s) = 1/s3

(constant acceleration)

Steady state error

in position

sGsLt

)(1

11. 30 +=


120/151

as

s

KsGsLt

sGs

1

)(

1

)(1

20

30

=

+=

Ka = acceleration errorcoefficient

)( ))(( ++ zszs


121/151

systemtheoftype

)(....))((

)(....))((

)( 21

21

++

++

=

n

pspss

zszs

sG n

Type 0 system


122/151

valuefinite1

1e

valuefinite)(

ss

0

=+=

==

p

sp

K

sGLtK

Type 0 system


123/151

=

==

ss

0

e

0)(ssGLtKs

v

Type 0 system


124/151

=

==

ss

20

e

0)(sGsLtKs

a

Type 1 system


125/151

0e

)(

ss

0

=== sGtLK sp

Type 1 system

sGstLK )(


126/151

finiteK

finite

sGstLK

v

sv

===

=

1e

)(

ss

0

Type 1 system


127/151

=

==

ss

2

0

e

0)(sGsLtK sa


128/151

Type 2 system


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0e

(s)GstLK

ss

0sv

===

Type 2 system


130/151

finiteK

finite

a==

== 1e

(s)GsLtK

ss

2

0sa

input

Type

tu t

2

12 ( )

1

u-1(t) tu-1

(t)


131/151

1

1++++Kp

1Kv

aK1

0

1

2

0

0 0

When feedback is non unity,

)()(Lt sHsGK =


132/151

)()(Lt

)()(Lt

)()(Lt

2

0s

0s

0s

sHsGsK

sHsGsK

sHsGK

a

v

p

=

=

Derivative error

Compensation


133/151

R +

-

K1

s(s+1)

C

r

Kp KA+ -


134/151

L

L+-

KT

/Ra

1

s(Js+B)

sKb


135/151

ess for ramp input


136/151

e p p(type 1 system)

= 0.01

Objective : increase the

damping factor


137/151

from 0.05 to 0.5.Change the amplifier

transfer function

to K + s KD

)1()(

++= sssKKsG D


138/151

)1( +ss

D

DKsKss

KsK

sG

sG

1)(

)(2 +++

+=+

n = K ( no change)

= 10 rad/sec


139/151

2n = 1 + KD

for = 0.5,

KD = 9

ess for ramp inputremains the same


140/151

eremains the sameat 0.01

ts

n

sec.( .2%)4

08==== ====

R + 1 C+

Derivative output Compensation


141/151

R +

- K1

s(s+1)

sKt

-

Introduce another feed back loop

KsG =)(


142/151

KKss

K

sGsG

Kss

sG

t

t

+++=+

++

)1())(1()(

)1(

)(

2

Specifications

= 0.5


143/151

ess for ramp input

= 0.01

K Lt sG sv s==== 0 ( )


144/151

K

Kt====

++++ ====

1100 - - - - - (1 )


145/151

)2(1K

12

+=

+=t

tn

Kor

K

From (1) & (2),


146/151

.99

10000

100

=

=

=

tK

Kor

K


147/151

Integral Control


148/151

+

-

KK

sAI++++

1

1s s( )++++


149/151

)1(

)(2

++=

ss

KsKsG IA

type 2

ess for step input = 0


150/151

essfor ramp input = 0

essfor parabolic input is finite

However system

becomes of 3 rd order


151/151

- stability problem -

will be discussed inlater chapters

control systems Chapter5.pdf

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