ch-5 time res webpage

7/31/2019 Ch-5 Time Res Webpage

1/48

Test signals Impulse

Step

Ramp

Sinusoidal (frequency response)

Impulse (t) 1

step u(t) 1/s

Ramp t u(t) 1/s2

Laplace Transforms

Chapter 5:Time Response Analysis:


2/48

Examples of zeroth order systems :

Potentiometerdc amplifier

dc tachogenerator

zeroth order system :

C(s) / R(s) = K ; a constant

(algebraic equation)


3/48

First order systems:C(s) / R(s) = K / (1+s )

K : steady state value of the Function : time constant,

smaller, faster response.

C(t) = K ( 1- e-t / ) ;for unit step

Step Response


4/48

0

50

100

150

200

250

300

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

t1

t2

Time constant : t2 < t1

dc/dt = 1/ t=0K=1

Error e(t) = e-t /

ess = 0

c

t

If K 1 and input is not unitythen ess ?


5/48

Ramp response of first order systems

r (t) = t u(t); R (s) = 1 / s2

1=KLet;ts1

K

R(s)

C(s)

+=

)s1(s

1)

2 +=c(s


6/48

+= s1s1s1)s(C 2

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

Error e(t) = (1-e-t /)

ess

=

If K 1 and input is not unity

then ess ?


7/48


8/48

2

nn

2

2

n

+s2s)s(R)s(C

+

=

Second order systems

Second order example

position control


9/48

For unit step input

R ( s ) =1

s

C(t) t =1.0

Applying final value theorem

Steady state error for step input = 0


10/48

r LKp KA+

-

+

-

KTRa

1

s(Js+B)

sKb

position control Example


11/48

Taking KP = 0.1 V / rad

KA = 10 V / VKT (Nm/A) = Kb (V/rad/sec. ) = 0.5

J = 1 Kg m2

B = 0.5 Nm / rad / sec

1+ss

1

)s(

)s(2

r

L

+=

n = 1 rad / sec and = 0. 5

s2 + s + 1 = (((( )))) (((( ))))s + 0.52 3

22++++

j 3 2

- j3

2

- 0.5


12/48

In general

( ) ( )22n2n 1+s +=s2 + 2 n s + n 2

2

nn1,2 1js =

= - n j d ; d = n 1 - 2

d damped freq of oscillations damping factor

n undamped natural frequency


13/48

roots of s are complex conjugate

= 0 : undamped

dampedunder:10

roots of s are real and distinct


14/48

Unit step response

r (t) = u(t) ; R(s) =1/s

)s(s)s(C 2n2

2

n

+

=

c ( t ) = 1- cos nt (pure oscillation)

Case 1: =0

n2

C(s) =s (s2 + 2 ns + n2)

Case 2: 0 < < 1


15/48

1 -n 2

- n

n

Where - n isreal part and d is the imaginary part of the roots

t)(coset)(sine

1

-1=(t)c dt-

d

t-

2

nn +

)+t(sin1

e-1=(t)c d

2

t- n

where cos

=


16/48

C(t)

1.0

Time(t)

1+e- nt

1- 2

1-e-nt

1-2

Unit step response of 2nd order system for


17/48

C(t)

1.0

tp

Mp

trTime

Tolerance band

Time response specification

ts


18/48

sin (dt) = 0, dt =

Time at peak overshoot ( tp ) = /d

Peak overshoot (Mp):d c

d t==== 0

At t = tp ,

)+(sin

1

1)(2

.

= pn t

p

etc

21

eMp

=


19/48

Settling time (ts):

Assume tolerance band 5%

0.05 = e

- nts

n

ssnn

3=or tt)05.0(l

=

n

s4isbandetoleranc2%fort


20/48

Delay time (tr):

)+t(sin1

e

-1=0.5=)(tc dd2

t-

d

dn

So delay time (td) (1.1+.125+.469 2)/n

Rise time (tr):

)+t(sin1

e-1=1=)(tc rd

2

t-

r

rn

d

r -t =


21/48

Second order system response

with ramp input

n2

s2 (s2 + 2 ns + n2)C(s) =

)+t(sin1

e2-t)t(c d

2

n

t-

n

n

+

=


22/48

1 n2

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

Steady state error of a second order

system with ramp input:

2

nn

2

n

2

2 +s2s

s2s.

s

1

++

=

n

2

=

For the example taken,

= 0.5 & n = 1

ess = 1 unitess


23/48

R(s) +

-

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

Error coefficients

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

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

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

H (s)

(s)EsLt=)(eerrorstateSteady 0sss


24/48

Step input

R (s) = 1 /s (constant position)

)s(H)s(G1

1.

s

1sLte

0sss +

=

coeff.)error(positionK)S(H)s(GLt p0s

=

p

ss

K

e

+

=

1

1So


25/48

Ramp input

R (s) = 1 /s2 (constant Velocity)

)s(H)s(G11.

s1sLte2

0sss +

=

coeff.)error(VelocityK)S(H)s(sGLt v0s

=

vss K

1e =So


26/48

Parabolic input

R (s) = 1 /s3 (constant Acceleration)

)s(H)s(G11.

s1sLte3

0sss +

=

coeff.)erroron(AccleratiK)S(H)s(GsLt a2

0s=

ass K

1e =So


27/48

systemtheoftypen

)(....)ps)(ps(s)(....)zs)(zs()s(H)s(G21

n21

++++=


28/48

Type 0 system

valuefiniteK1

1e

valuefinite)s(H)s(GLtK

p

ss

0sp

=+

=

==

===

ss0s

v e0)s(H)s(sGLtK

===

ss

2

0s

a e0)s(H)s(GsLtK


29/48

Type 1 system

0e)s(H)s(GLtK ss0s

p ===

finiteK1e

finiteH(s))s(sGLtK

vss

0sv

==

==

===

ss

2

0s

a e0)s(H)s(GsLtK


30/48

Type 2 system

0e)s(H)s(GLtK ss0s

p ===

0e)s(H)s(sGLtK ss0sv ===

finiteK1e

finite)s(H)s(GsLtK

ass

2

0sa

==

==


31/48

)t(u2

t2

11++++Kp

1

Kv

aK

1

0

1

2

0

0 0

u(t) t u(t)InputType


32/48

Compensation Techniques:

R +

-

K 1s(s+1)

C

Example

Kss

K

)s(R

)s(Cand

)1s(s

KG

2

++

=

+

=


33/48

K2

1

,Kn ==

If K = 100n = 10 rad / sec, = 0.05

Mp= 85 %

ess for ramp input = 2/n = 0.01 unit

.sec84%)2( == nst


34/48

While the steady state error is acceptable

Mp is not.Damping () to be increased to say, 0.5

10.521=1=2Then nn =

ess

for ramp input = 2/n

= 1unit

ess increases to 1 - not acceptable.

change theamplifierconfiguration.


35/48

R +

-K+sKD

1

s(s+1)

C

Derivative error Compensation

Required Specifications:

= 0.5 and ess for ramp input = 0.01


36/48

)1s(s

sKK

)s(H)s(G

D

+

+

=

KKsss

KsK

)s(R

)s(C

D

2 D ++++

=

n =K

= 10 rad/sec2n = 1 + KD

for = 0.5, KD = 9

sec.8.04

%)2(t

n

s =

= ess for ramp input

remains the same


37/48

R +-

K 1s(s+1) C

sKt

+

-

Derivative output Compensation

Introduce another feed back loop


38/48

K)K1(ss

K

)s(R)s(C

)K1s(s

K

)s(H)s(G

t

2

t

+++=

++=

Required Specifications:

= 0.5 and ess for ramp input = 0.01


39/48

(1)-----100

K1

K

)s(H)s(sGLtK

t

0sv

=

+

=

=

)2(K1KorK12 ttn +=+=

99K

10000K100K

t =

==


40/48

Comparing both methods

derivative output compensation:additional transducer required

gain reduced due to feedback sohigh gain amplifier required and will

have a higher n and hence lesser ts

I t l C t l


41/48

+

-K

K

sA

I++++

1

1s s( )++++

Integral Control

)1s(s

KsK)s(H)s(G

2

IA

++=


42/48

type 2 ess for step input = 0

ess for ramp input = 0ess for parabolic input is finite

However system becomes of 3 rd order -

stability problem arises and will bediscussed in later chapters


43/48

Addition of Zeros and Poles to the system

2

nn

2

z

2

n

2

nn

2

2

n

2

nn

2

z

2

n

+s2s

sT

+s2s)s(R

)s(C

+s2s

)sT1(

)s(R

)s(C

+

+

+

=

+

+=

y(t)= y1(t)+ Tz dy1/dt

Zero added to closed loop Transfer function


44/48

Unit step response when zero is added to closed loop Transfer

function at various mentioned positions


45/48

Unit step response showing the effect of zero addition to closed

loop Transfer function.

Zero added to forward path


46/48

Zero added to forward path

Unit step response when zero is added to Forward path at various positions

6s)T62(s3s

)sT1(6

)s(R

)s(C

2p,1p,6K,)ps)(ps(s

)sT1(K)s(G

z

23

z

21

21

z

+++++

=

===++ +=

Pole added to forward path


47/48

Pole added to forward path

Unit step response when pole is added to Forward path at various mentioned positions

2nn

2pn

3p

2

n

pn

2

n

s2+)sT21(sT)s(R

)s(C

s)T)(12s(s)s(R)s(C

+++

=

++=

Pole added to closed loop Transfer Function


48/48

Pole added to closed loop Transfer Function

Unit step response when pole is added to closed loop transfer function at

various mentioned positions.

s)T)(1s2(s)s(R

)s(C

p

2

nn

2n

2 +++=

ch-5 time res webpage

Documents