Time Respons

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System Response

This module is concern with the response ofLTI system.

L.T. is used to investigate the response of first

and second order systems. Higher ordersystems can be considered to be the sum ofthe response of first and second order system.

Unit step, ramp, and sinusoidal signal playimportant role in control system analysis. It istherefore we will investigate this signals.


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(1)

2. Convolution

Output of LTI system is the convolution

of its input and its impulse response:

h(t)r(t) c(t)

t

dthrthtrtc )()()(*)()(

t

dhtr )()(

Taking the L.T. of (1) yields

C(s) = R(s)H(s) (2)Where C(s)is the L.T. of c(t)

R(s) is the L.T. of r(t)

H(s) is the L.T. of h(t)

H(s) is called the transfer function (T.F)

3. Derivative

If the input is r(t) then the output is c(t)

wherer(t) denotes the derivative ofr(t)

Review of some LTI properties

We will express system as in figure below,system input is r(t), output is c(t), and

impulse response is h(t)

4. Integral

If the input is r(t)dtthen the output isc(t)dt

1. Impulse response

Impulse response, denoted by h(t), is the

output of the system when its input is

impulse (t). h(t) is called the impulse

response of the system or the weighting

function5. Poles and ZeroT.F.isusually rational and therefore can be

expressed as N(s)/D(s). Poles is the values

ofs resulting T.F to be infinite. Zeroes is

the values ofs resulting T.F to be zero

System Response


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Poles, Zeros, and System Response The output response of a system is the sum of two

responses: theforced response andthe natural response The poles of a transfer function are (1) the values of

the Laplace transform variable, s, that cause the

transfer function to become infinite or (2) any roots of

the denominator of the transfer function that are

common to roots of the numerator

The zeros of a transfer function are (1) the values of the

Laplace transform variable, s, that cause the transferfunction to become zero, or (2) any roots of the

numerator of the transfer function that are common to

roots of the denominator.


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Pole dan Zero Pada System Orde Satu

Fungsi alih sistem orde 1,

Untuk mencari A dan B,

Persamaan respon sistemnya,

Secara grafis,

INTRO


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Time Response of the First Order Systems.

We can found the differential equation

first we write (1) as

The diff. Eq. is the inverse L.T. of (2)

Now we take the L.T of (3) and include

the initial condition term to get

1)(

)()(

s

K

sR

sCsG (1)

)(

)(

1sR

KsCs

(2)

)()(1

)( tr

Ktc

tc (3)

)(

)(

1)0()( sR

KsCcssC (4)

Solving for C(s) yields

)/1(

)R(s)/(

)/1(

)0()(

s

K

s

csC (5)

Note that the initial condition as an

input has a Laplace transform ofc(0),which is constant.

The inverse L.T of a constant is impulse

(t). Hence the initial condition appearsas the impulse function

Here we can see that the impulse

function has a practical meaning, eventhough it is not a realizable signal

1

s

K

1

1

s

R(s)

c(0)

+

+

C(s)

Here we will investigate the time response

of the first order systems.

The transfer function of a general first

order system can be written as:

The eq. can be shown in the block diagram as

shown in the figure bellow.


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Time Response of the First Order Systems.

Unit step response

For unit step input R(s)=1/s, then

Since we usually ignore the initial

condition in block diagram, we use the

system block diagram as shown bellow.

)()/1(

)/()( sR

s

KsC

(1)

Suppose that the initial condition is

zero then

1

s

KR(s) C(s)

/1s

1

)/1(

)/()(

s

K

s

K

s

KsC (2)

Taking the inverse L.T of (2) yields

)1()( t

eKtc (3)

The first term originates in the pole of input

R(s) and is called the forced response or steady

state response The second term originates inthe pole of the transfer function G(s) and is

called the natural response Figure below plot

c(t)

t

c(t)

K1/

)1( t

eK

The final value or the steady state value ofc(t)

is K that is

lim c(t)= Kt

c(t) is considered to reach final value after

reaching 98% of its final value. The parameter is called the time constant. The smaller the time

constant the faster the system reaches

the final value.


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Time Response of Second Order System

22

2

2)(

)()(

nn

n

sssR

sCsG

(1)

The standard form second order system is

Where = damping ratio, n = naturalfrequency, or undamped frequency.

Consider the unit step response of this system

sssRsGsC

nn

n

)2()()()(

22

2

(2)

Case 1: 0


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General Second-Order System

Natural Frequency,n

The natural frequency of a second-order system is the frequency of

oscillation of the system without damping

Damping Ratio,

So,

= =


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PROBLEM:

Given the transfer function of Eq. , find and n


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Time Response Specification in Design

The typical of unit step response of a system

is as in figure below

nt

c(t)

Mpt

1.0

0.9

0.1

Tr Tp

1+ d

1d

css

Ts

Rise time Tr

is the time required for the

response to rise from 10% to 90% of the final

value css. Mpt is the peak value, and Tpis the

time required to reach Mpt.

Mptcsspercent overshoot =100%

css

Settling time, Ts,is the time required for the

output to settle within a certain percent of its

final value. Common values are 5% and 2%.

Settling time is proportional to the ,

kTs = k =

n

If 2% is used to specify the settling value thenk=4. T

r

, Ts

,and css

, are equally meaningful for,over, critically, or under-damped cases, whileTp,and Mp,are meaningful only for under-damped cases.

Under-damped cases.

To find Tp,we differentiate c(t) and equalizing

the result to zero and solve the equation for t.

We will find that2

1/1

1

eMT p

n

p

%100shootover%2

1/ e


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Time Response Specification in DesignRecall that

1

or

1

pn

n

p TT

and

%100shootover%2

1/ eBoth %over-shootand nTp are functionsonly of and can be plotted as in thefollowing figure

%over-shoot

n

Tp

100

80

60

40

20

0

5.0

4.6

4.2

3.8

3.4

3.00.2 0.4 0.6 0.8 1

%over-s

hoot

nTp

Example:

Servomotor is used to control the position of plotter

pen as in the following figure.

a

a

a

a

Kss

K

ssK

ssK

sT 5.02

5.0

)2(/5.01

)2(/5.0

)( 2

ann K5.0and222

Here we have

Suppose that we want to have =1, the fastest

response with no overshoot thenn= 1 and Ka=2.

Note that the settling time is

Tp= 4/ n= 4sIt is not fast enough. If we want the faster response

First we have to choose a different motor, then

Redesign the compensator.

0.5/s(s+2)

motorR(s)

Kp

Amplifier

+


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TRANSIENT RESPONS ANALYSIS WITH MATLAB(2)

Example 2 : find the unit impuls- respons for following equation,


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Sumber Pustaka

Norman S. Nise, Control Systems Engineering, Fourth Edition,

2004, Wiley & Sons..

Time Respons

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