It is the time response of a system to an input that sets the criteria for our control systems. Many...

It is the time response of a system to an input that sets the criteria for our control systems.

Many quantitative criteria have been defined to characterise the time response of a system.

The time constant and system gain of a first order system are useful in its analysis, but other criteria describe the time response more accurately to an engineer.

A first order system may be written as:

or where K=1 and T = 1/a

→ and for a unit step

→ using Laplace tables

we obtain the standard response

©2000, John Wiley & Sons,

Inc. Nise/Control Systems

Engineering, 3/e

sT)+(1

K

sR

sC

First Order Time Response

s)+(a

a

sR

sC

s)+(a

a

sR

sC s)+(a

asRsC

ssR

1

s)+(a

a1

ssC

a)+(s

11

ssC

atnf etctctc 1

First-order system response to a unit step

Rise Time, TrRise time is defined as the time for the response to go from 0.1 to 0.9 of its final value:

Settling Time, TsSettling time is defined as the time for the response to reach, and stay within, 2% of its final value (2% is standard)



Engineering, 3/e


aaaTr

2.211.031.2

aTs

4



Engineering, 3/e




Engineering, 3/e

Second Order Time Response

Second-order step response components generated by complex poles

Consider the characteristic equation ofto give roots

The real part generates:The complex part generates:

9)2s+(s

92

sR

sC

812,1' s

te

tk 8cos

Damping Ratio

Consider a second order system:

with unity in front of the s2 term and k such that

note that the constants a`, b`, c` are not equivalent to a, b

Natural Frequency:When there is no damping in the system (a = 0 in this case)We obtain with poles

We know that as a complex number.

So the frequency of oscillation , which is termed the natural frequency

Exponential Decay Frequency:Considering an underdamped system

The real part of , is , where is termed the exponential decay frequency

Damping ratio :The damping ratio is defined to be

= Exponential decay frequency = = Natural frequency

c`)b`s+(a`s

k2

sR

sC

b)as+(s

b2

sR

sC

b)+(s

b2

sR

sCbjs 2,1

nb

js

b)as+(s

b2

sR

sC

js 2

a

n

n

a

2

Damping Ratio

Consider a second order system:

Written in terms of damping ratio and natural frequency:

Consider the roots of the characteristic equation:

c)bs+(as

c2

ksR

sC

2

n2

2n

s2+s nsR

sC

122,1 nns

j

j

j

j

Second-order underdamped responses for damping ratio values



Engineering, 3/e


0 2 4 6 8 10 12 140

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Time t (seconds)

f(t)

Second-order underdamped response specifications

Can not use equations for first order system!

Still would like to consider:Rise time, Settling time, but alsoTime to peak,Percentage overshoot (%OS),Steady-state error,



Engineering, 3/e


Time to Peak

Determine the time to the first peak (zero slope):

Differentiate the time response to a step input and set to zero:[ this is straightforward thanks to Laplace!]

→

But it is easy to differentiate this equation: the differential is

Now we complete the square to utilise the Laplace tables:

We now need to find something familiar in the table:

2

n2

2n

s2+s nsR

sC

2n

2

2n

s2+s

1

nssC

2

n2

2n

s2+s n

ssC

22n

2n

2n

2n

2n

2n

2

2n

1+s?+ss2+s

n

21

nK

21 n

Time to Peak

Determine the time to the first peak (zero slope):Differentiate the time response to a step input and set to zero:[ this is straightforward thanks to Laplace!]

We can return to the time domain:

We can set this to zero and note that sin-10 = nπ:

→

We now no the time to all n peaks, but only need n=1:

22n

2n

2n

1+s

21

nK 21 n

teteK ntnat n 2

21sin

1sin**

ntn 2121

n

nt

21

n

pT

Similarly:

Rise time, calculated by iterating c(t), see final value theorem for cfinal

Settling time,

Time to peak,

[note cmax = c(tp)]

Percentage overshoot (%OS),

[note %OS is a function of damping ratio only ]

Steady-state error, is the difference between the input and output for a prescribed test input as


21

n

pT

nsT

4

100%21

eOS

t

Normalized rise time vs. damping ratio for a second-order underdamped response



Engineering, 3/e

Second Order Rise Time

Step responses of second-order underdamped systems as poles move:a. with constant real part;b. with constant imaginary part;c. with constant damping ratio



Engineering, 3/e


Controllers

s

sT+1C

s

sP+I

s

IP

K

(1+ sT )(1+ sT )1 2

O

E

CK

s(1+ sT )1

O

I

CK

s T + s + CK21

Another form of controller is the P+I controller Its transfer function can be written as

Here C = I and T = P/I.P and I can be chosen so the 1+sT term (controller zero) cancels Plant pole.

Suppose Plant

If apply P+I to this Plant, and make T = T2,

then

So

Note, the I term means that the steady state value is 1.

Controllers

P Ds C(1 sT)

K

s(1+ sT)

O

E

CK

s

CKs

CK

I

O

s

TTs)sTT(1C

s

Ds+Ps+I=Ds+

s

I+P 21

221

2

O

E

CK

s

P+D controller

Make 1+s T cancel Plant pole.If Plant is and P+D is applied, then

so

PID controller - can cancel two lags in a plant:

then

In all these examples, by careful arrangement, systems is first/second order.Cancellation may not give best response, but analysis of systems is easier!

Controllers

Suppose the system is one with controller C and plant G.

We want

• Rise time Fast

• Settling time within two seconds

• Time to peak < 0.5

• Percentage overshoot (%OS) < 90%

• Steady-state error < 0.1

But have:

I OGC

0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4

Time t (seconds)

f(t)

Gain Controller

Try a gain controller Kp.

• Rise time Fast





But have:

I OGC

0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Time t (seconds)

f(t)

Controllers

Try a gain controller 1 +Ki/s.

We want

• Rise time Fast





But have:

I OGC

0 1 2 3 4 5 6 7 8-1

-0.5

0

0.5

1

1.5

2x 10

4

Time t (seconds)

f(t)

Controllers

Try a gain controller Kp +Ki/s.

We want

• Rise time Fast





But have:

I OGC

0 1 2 3 4 5 6 7 8-1

-0.5

0

0.5

1

1.5

2x 10

4

Time t (seconds)

f(t)

0 1 2 3 4 5 6 7 8-4

-2

0

2

4

6

8

10

12

14

16

Time t (seconds)

f(t)

0 1 2 3 4 5 6 7 80

2

4

6

8

10

12

14

16

18

20

Time t (seconds)

f(t)

0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Time t (seconds)

f(t)

D Controller

Try a gain controller Kds.

We want

• Rise time Fast





But have:

I OGC

0 1 2 3 4 5 6 7 80

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time t (seconds)

f(t)

PD Controller

Try a gain controller 1 +Kds.

We want

• Rise time Fast





But have:

I OGC

0 1 2 3 4 5 6 7 80

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time t (seconds)

f(t)

PD Controllers

Try a gain controller Kp +Kds.

We want

• Rise time Fast





But have:

I OGC

0 1 2 3 4 5 6 7 80

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time t (seconds)

f(t)

Pole Cancellations

Try a lead/lag controller (1 +K1s)/(1+K2s).

We want

• Rise time Fast





But have:

I OGC

0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4

Time t (seconds)

f(t)

It is the time response of a system to an input that sets the criteria for our control systems. Many...

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