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Chapter 5

Time Response

Analysis

Test signals• Impulse

• Step

• Ramp

Test signals

• Sinusoidal.

(frequency response)

Laplace Transforms

step u-1 (t) 1/s

Ramp t u-1 (t) 1/s2

Impulse δδδδ (t) 1

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Order of a

system is ???

Order of a

system is order

of the differentialequation

zeroth order system :

C(s) / R(s) = K

a constant

(algebraic equation)

dc tachogenerator

Examples of zeroth

order systems :

Potentiometer

dc amplifier

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First order systems:

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

Open loop

1000 / ( 100 + 200 s )

= 10 / ( 1 + 2 s )

Closed loop

1000 / ( 200 + 200 s )

= 5 / ( 1 + s )

open loop

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

= (25 / s )( 10 / (1 + 2 s)

Vt (∞∞∞∞) = 250 Vclosed loop

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

Vt (∞∞∞∞) = 125 V.0

50

100

150

200

250

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

x1

x2

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K : steady state value of the

function

: time constant

smaller, faster response.

In general K ( 1- e - t / )τ

ττ

Ramp response

of first order

systems

Ramp Response

r (t) = t u-1

(t)

R (s) = 1 / s21=KLet

s1R(s)

C(s)

t 

K 

+

=

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R G C

s

s R

ssG

2

1)(

1

1)(

=

+

=

τ   

)s1(s1)

2 τ +=c(s

 

 

 

 

+−−= τ 

τ 

τ   s1s1

s

12

)e1(-t=c(t)-t/ τ 

τ    −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

ττττc

t

ττττ

Second order

systems 2nn 22

n

 +s2s)s(R

)s(C

ω ζω 

ω 

+=

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For unit step input

 R (s ) =1

s

Applying final valuetheorem

C(t) t →→→→ ∞∞∞∞  =1.0

Second order

example

- position control.

θr   θLKp KA

+

-

+

-

KTRa

1

s(Js+B)

sKb

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Taking

KP = 0.1 V / rad

KA = 10 V / V

KT = Kb = 0.5 Nm/A

or

V/rad/sec.

B = 0.5 Nm / rad / sec

J = 1 Kg m2

 1+ss

1

)s(

)s(2

r

L

+=θ 

θ    ∴∴∴∴ ωωωωn = 1 rad / sec.

ζζζζ = 0. 5

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s2 + s + 1 =

(((( ))))   (((( ))))s + 0.5 2 3 22

++++××××

××××  j 3 2

- j 3 2

- 0.5

In general

( )   (   )222 1+s   ζ ω ζω    −+=   nn

s2 + 2 ζζζζ ωωωωn s + ωωωωn2

2

n

n1,2

1 j

s

ζ ω 

ζ 

−

±−=∴

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= - ζζζζ  ωωωωn  ±±±±  j ωωωωd ωd = ωn √1 - ζ2

ωωωωd damped frequencyζζζζ  →→→→  damping factorωωωω n →→→→   undamped

natural frequency

For 0 < ζζζζ  < 1roots of s are

complex conjugate

roots are

repeating at

s = - ωωωωn

For ζζζζ  = 1 For ζζζζ > 1

roots are real

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ζ = 0 : undamped0 

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22

1

ns

s

s   ω +

−=ζζζζ = 0

c ( t ) = 1 - cos ωωωωnt

(pure oscillation)

 :10  

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1= -

s

s + ζ ωns2 + 2ζ ωns +ωn2

ζ ωns2 + 2ζ ωns +ωn2

we know

at cosas

s

)t(us

1

22

1-

→

+

→

atsin22 →+ as

a

atsinea+)b+s(

a

atcosea+)b+s(

b+s

bt-

22

-bt

22

→

→

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22

2

)(

)(

22

d n

n

nn

n

s

s

ss

s

ω ζω ζω 

ω ζω 

ζ  

++

+

=

++

+

t e d t n ω ζω  cos−→

22

22

)(.

2

d n

d 

d 

n

n

n

s

ss  n

ω ζω 

ω 

ω 

ζω 

ω ζω 

ζ  

++

=

++

 tsine. dt-

d

n n

ω ω 

ζω    ζω 

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tsin e.

1d

t-

2

n ω 

ζ 

ζ    ζω  

−

=

Combine sine cos terms :

1

1

1 2- ζζζζ

( )ζζζζ

ζζζζ1 2-

θθθθ

1

1 2- ζζζζωωωω θθθθ  sin ( t +d   ))))

where cos θθθθ  = ζζζζ .)+t(sin

1

 e -1

= (t)c

d2

t- n

θ ω ζ 

ζω 

−

∴

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C(t)

1.0

Time(t)

1+e- ζωnt

√ 1- ζ2

1-e-ζωnt

√ 1-ζ2

Unit step response of 2nd order system for ζ

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C(t)

1.0

tp

Mp

tr

Time

Tolerance band

Time response spceification

Peak overshoot (Mp)

dc

dt

  ==== 0

⇒⇒⇒⇒ sin (ωωωωdt + θθθθ). cos θθθθ

- cos (ωωωωdt + θθθθ).Sin θθθθ = 0

⇒ sin (ωωωωdt) = 0

ωωωωdt = ππππ

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Time at peakovershoot ( tp )

tp = π / ωd

At t = tp ,

)+(sin

1

1)(

2

.

θ π 

ζ 

ζω 

−

−=

−  pn t 

 p

et c

21

21.

1

ζ ζπ 

ζ ω π ζω 

−−

=∴

+=  −−

e M 

e

 p

nn

%.100

21

×=

−−   ζ ζπ 

e

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As damping factor

increases …

peak overshoot ??

As damping factor

increases …

peak overshoot reduces

ζζζζ Mp0 1000.1 730.2 53

0.3 37

0.4 25

0.5 16

ζζζζ Mp

0.7 4.60.8 1.5

0.9 0.02

1.0 0

0.6 9.5

0

10

20

30

40

50

60

70

80

90

100

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

ζ

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Settling time (ts)Assume tolerance

band   ±±±± 5 %0.05 = e- ζωnts

n

snn   t l

ζω 

ζ 

3 =or t

)05.0(

s

−=

n

s

4 

ise toleranc2%fort

ζω 

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Rise time (tr)

)+t(sin

1

 e -1=1= )(tc

rd

2

t-

r

rn

θ ω 

ζ 

ζω 

−

or

sin (ωωωωdtr + θθθθ) = 0ωωωω

d

tr

+ θθθθ = ππππ

trd

====  −−−−ππππ θθθθ

ωωωω

or Steady stateerror for step

input = 0

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Steady state errorof a second order

system with ramp

input:

1   ωn2= 1-

s2 (s2 + 2ζ ωns + ωn2)

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

2

n

2

n

2

2 +s2s

2.

1

n

ss

s   ω ζω 

ζω 

+

+

=

nω 

ζ  2=

e(t) = Lt sE(s)t ∞ s 0

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For ζζζζ  = 0.5 &  ωωωωn  = 1

  ess  = 1 unit

Some more MATLAB

commands .

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

>> printsys (ng,dg)

num/den =

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

++++

2S

1S 

>> z = roots(ng)

z =

-1

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>> p = roots(dg)p =

-2

>>

>>pzmap(ng,dg)

>>title(‘pole-zero map’)

>>[nt,dt] =

cloop(ng,dg,-1)

>>zc = roots(nt)

zc =

-1

>>pc = roots(dt)

pc = -1.5

>>

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>>ng = [1];

dg = [1 1 1];

>>p = roots(dg)

p =

-0.5 + j 0.866

-0.5 - j 0.866

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

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

>>plot(t,y), grid

>>xlabel(‘Time[sec]’)

>>ylabel(‘output’)

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Let K = 100

  ωωωωn  = 10 rad / sec

  ζζζζ  = 0.05

∴∴∴∴ ====

××××−−−− −−−−

M

e

p

ζπζπζπζπ ζζζζ1 2

100%

= 85 %

ess for ramp input

= 2ζζζζ / ωωωωn= 0.1 / 10

= 0.01

While the steady state

error is acceptable

Mp is not. - Damping

to be increased to say,

0.5

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1

0.52

1 = 

1=2Then

n

n

=

×

∴ ω 

ζ  ess increases to 1 - notacceptable. So we go for

changing the configuration

of the amplifier - derivative

error compensation.

R(s)+

-

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

Error coefficients We know

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

and

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

( )(s)G+1(s)R = (s)E∴

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(s)EsLt=

)(eerrorstateSteady

0s

ss

→

step input

R (s) = 1 /s

(constant position)

)(1

1

.

1

0   sGss Lt e sss

+= →

Let

coeff.)error(position

)(0

  ps

K sG Lt    =→

Then

 pss K e += 1

1

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Ramp input

R (s) = 1/s2

(Constant velocity)

Steady state errorin position

vs

s

K ssG Lt 

sGss Lt 

1

)(

1

)(1

1.

1

0

20

≅

+=

→

→

K v = velocity error coefficient

parabolic input

R (s) = 1/s3

(constant acceleration)

Steady state error

in position

as

s

K sGs Lt 

sGss Lt 

1)(

1

)(1

11.

20

30

≅=

+=

→

→

Ka = acceleration error

coefficient

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systemtheof type

)(....))((

)(....))(()(

21

21

→

++

++=

n

 ps pss

 zs zssG

n

Type ‘0’ system

 valuefinite

1

1e

 valuefinite)(

ss

0

=

+

=∴

==→

 p

s p

K 

sG Lt K 

Type ‘0’ system

∞=∴

==→

ss

0

e

0)(ssG Lt K s

v

Type ‘0’ system

∞=∴

==→

ss

2

0

e

0)(sGs Lt K s

a

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Type ‘1’ system

0e

)(

ss

0

=∴

∞==→

sGt  LK s

 p

Type ‘1’ system

 finiteK 

 finite

sGst  LK 

v

sv

==∴

=

=→

1e

 

)(

ss

0

Type ‘1’ system

∞=∴

==→

ss

2

0

e

0)(sGs Lt K 

s

a

Type ‘2’ system

0e

)( 

ss

0

=∴

∞==→

sGt  LK s

 p

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Type ‘2’ system

0e

 (s)GstLK

ss

0sv

=∴

∞==→

Type ‘2’ system

 finiteK 

 finite

a   ==∴

==→

1e

 (s)GsLtK

ss

2

0sa

input

Typet

u t

2

12   −−−−  ( )

1

1 ++++ K p

1

K v

aK 

1

0

1

2

∞   ∞

0

0 0

∞

u-1(t) tu-1(t)When feedback is non unity,

)()(Lt

)()(Lt

)()(Lt

2

0s

0s

0s

s H sGsK 

s H sGsK 

s H sGK 

a

v

 p

→

→

→

=

=

=

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R +

-K

1s(s+1)

C

Derivative error

Compensation   θrθL

θL

Kp KA+-

+

-KT /Ra

1s(Js+B)

sKb

with

K = 100

ωωωωn = 10 rad/secζ= 0.05Mp = 85% (not acceptable)

.sec84%)2(   == nst    ζω 

ess for ramp input(type ‘1’ system)

= 0.01

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Objective : increase the

damping factor

from 0.05 to 0.5.

Change the amplifier

transfer functionto K + s KD

)1()(

++=∴sssK K sG   D

 D

 D

K sK ss

K sK 

sG

sG

 1)(

)(

2 +++

+=

+

ωωωωn  = K  ( no change)  = 10 rad/sec

  2ζζζζ ωωωωn  = 1 + KD∴∴∴∴ for ζζζζ = 0.5,  KD  = 9

ess for ramp inputremains the same

at 0.01

ts

n

  sec.( .2%)4

0 8==== ====ζωζωζωζω

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R +

-K

1s(s+1)

C

sKt

+

-

Derivative output Compensation

Introduce another feed back loop

K K ss

K sGsG

K ss

K sG

t 

t 

+++

=+∴

++=

)1(

))(1()(

)1()(

2

Specifications

ζζζζ = 0.5

ess  for ramp input

  = 0.01

K Lt sG s

K

K

vs

t

====

==== ++++   ====

→→→→0

1 100

( )

  - - - - - ( 1 )

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)2(1K 

12

−−−+=

+=

t 

t n

K or 

K ω ζ 

From (1) & (2),

.99 

10000 

100 

=

=

=

t K 

K or 

K 

Comparing both methods…

derivative output

•transducer required•gain reduced due to

feedback 

+

-K

  K

sAI

++++

   

   

    

   

1

1s s( )++++

Integral Control

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)1()(

2 +

+=

ss

K sK sG   I  A

type 2 →→→→

∴∴∴∴ ess  for step input = 0

  ess for ramp input = 0

  ess for parabolic

  input is finite

However system

becomes of 3 rd order

- stability problem -

will be discussed inlater chapters