transient and steady state response (cont.). example dc motor page 111 ex.1-4-3

Transient and steady state response (cont.)

Example DC Motor

• Page 111 Ex.1-4-3

Effects of a third pole and a zero on the Second-Order System Response

• For a third-order system with a closed-loop transfer function

• The s-plane is

112

12

sss

sT

Complex Axis

Effects of a third pole and a zero on the Second-Order System Response (cont.)

• The third-order system is normalized with ωn=1.

• The response of a third-order system can be approximated by the dominant roots of the second-order system – As long as the real part of the dominant roots is less than 1/10 of the

real part of the 3rd root.

• Dominant roots: The roots of the characteristic equation that represent or dominate the closed-loop transient response– Example 3-5-3 estimate the damping ratio

n

101


• If the transfer function of a system possesses finite zeros and they are located relatively near the dominant complex poles, then the zeros will significantly affect the transient response of the system.

• The transient response of a system with one zero and two poles may be affected by the location of the zero.

Relationship between steady state error and system type

n

j

jy

m

i

i

pss

zsK

sHsKG

1

1

)(

)(

)()(

The general form of the open loop transfer function of a system is given by

Here y is known as the system type, and it corresponds to the number of integrators in the system.

y

System type

0 Type zero

1 Type 1

2 Type 2

We can calculate SSE for different types of standard signals

Example (unitary feedback )

• Page 119, 120, 121


A) Step Input

s

RsR 0)(

p

n

j

j

m

i

i

n

j

j

m

i

i

s K

R

G

R

p

zK

R

ps

zsK

s

Rs

SSE

1)0(1

1

)(

)(

1

lim 00

1

1

0

1

1

0

0

Here is known as the position error constant. pK

For y=0

0

)()(

)(

lim

)(

)(

1

lim

11

1

0

0

1

1

0

0

m

i

i

n

j

jy

n

j

jy

s

n

j

jy

m

i

i

s

zsKpss

psRs

pss

zsK

s

Rs

SSE

For y>0

s

RsR 0)( !


21)(

s

RsR

B) Ramp Input

m

ii

n

jj

y

n

jj

y

s

n

jj

y

m

ii

szsKpss

psRs

pss

zsK

sR

sSSE

11

11

1

0

1

1

21

0)()(

)(

lim

)(

)(1

lim

For type 0 system the SSE =

For a type 0 system…

v

n

j

j

m

i

i

K

R

p

zK

RSSE 1

1

1

1

For a type 1 system…

velocity constant

For the steady state error is zero.1y


32)(

s

RsR

m

i

i

n

j

jy

n

j

jy

s

n

j

jy

m

i

i

s

zsKpss

psRs

pss

zsK

s

Rs

SSE

11

1

22

0

1

1

32

0

)()(

)(

lim

)(

)(

1

lim

C) Parabolic Input

For type 0 and type 1 systems, the steady state error is infinite. For type 2 systems the steady state error is given by

a

n

j

j

m

i

i

K

R

p

zK

RSSE 2

1

1

2

acceleration error constant


pK

RSSE

10

0SSEvK

R1

0SSE 0aK

R2

System Type (number of pure integrators)

Input

step ramp parabola

0

1

2

Conclusions:

1. Adding integrators (increasing system type) eliminates steady state error.

2. If steady state error is finite and not zero, then increasing the system dc gain (increasing controller proportional gain, adding poles near the origin and/or zeros far away from the origin on the LHS of s-plane), reduces steady state error.

Example Problems

Calculate the open and closed loop steady errors, ramp errors, position constants or velocity constants to step and ramp inputs, respectively, for the following cases.

KH(s) G(s)R(s) E(s)

)20)(4(

)2(10)()(

ss

ssGsKH

)20)(4(

)2(10)()(

sss

ssGsKH

Problem 1

Problem 2

Problem 3 (two problems)

Problem 4

)7040)(13(

18)()(

22

sssss

ssGsKH

)5030)(4(

210)()(

22

sss

ssGsKH

The steady-state and ramp error computations are only valid if the closed loop is stable. Check which of these examples provides a stable closed loop.

Example (unitary feedback )

• Look at Table 1-3 on page 126 for a summary of Steady-State Errors

• The power of s on the denominator (sq) denotes the type of system

Effect of feedback!

• Time constant (rate of sytem response) – How make it smaller or bigger by feedback using!

• Figures: 25-3, 26-3

• External disturbance– Always living in our systems!– Two type

• Load dist. Or offset• Random noise

– Figures 27-3, 28-3– SNR: signal-to-noise-ratio

• Sensitivity

Sensitivity of Control Systems to Parameter Variations

• With an open-loop system, all changes and errors at the output are ignored, resulting in a changing and inaccurate output.

• A closed-loop system senses the changes in the output and attempts to correct the output.

• The sensitivity of a control system to parameter variations is very important.– A closed-loop system can reduce the system’s sensitivity.

Sensitivity of Control Systems to Parameter Variations (cont.)

• For the closed-loop case:

– The output is only affected by H(s).– If H(s)=1, we have the desired result

• Caution: The requirement that GH(s)>>1 may cause the system response to be highly oscillatory and even unstable.

• As we increase the magnitude of the loop transfer function G(s)H(s), we reduce the effect of G(s) on the output.

• The first advantage of a feedback system is that the effect of the variations of the process, G(s), is reduced.

sRsH

sYsGH

sRsGH

sGsY

11 if

1


• Illustration of the parameter variations– Let’s consider a change in the process so that the new process is G(s)

+ΔG(s).

– The change in the transform of the output is

sRsGsY System Loop-Open

sRsGH

sGsY

sGHsGH

sRsGHsGHsGH

sGsY

sRsHsGsG

sGsGsYsY

21

When

11

1

System Loop-Closed


• For the closed-loop system– The change in the output of the closed-loop system

us reduced by the factor [1+GH(s)]– This is usually much greater than 1 one the range of

complex frequencies on interest.


• The system sensitivity

– System Transfer Function is

• For small incremental changes:

function transfer process theof change percentage

function transfer system in the change percentageySensitivit System

sGsG

sTsTS

/

/ySensitivit System

sR

sYsT

G

T

GG

TTS

ln

ln

/

/


• The system transfer function of the closed-loop system is:

• The sensitivity of the feedback system is:

– The sensitivity of a system may be reduced below that of the open-loop system by increasing GH(s) over the frequency range of interest.

sGH

sGsT

1

sHsG

GHG

G

GHT

G

G

T

1

1S

1/1

1S

TG

2TG


• The sensitivity of the feedback system to changes in the feedback element H(s) is:

– When GH is large, the sensitivity approaches unity (1)

– The changes in H(s) directly affect the output response

• It is important to use feedback components that will not vary with environmental changes or that can be maintained constant.

GH

GH

GHG

H

GH

G

T

H

H

T

1S

1/1S

TH

2TH

1STH


• The sensitivity to α:

• The transfer function of the system T(s) is a fraction of the form

• The sensitivity to α:

– α is a nominal value of the parameter

Gblock theoffunction transfer within theparameter a :

SSS GTG

T

tenvironmen the todue variationsubject to bemay that parameter a :

,

,,

sD

sNsT

DN SSDTT

00

ln

ln

ln

ln

ln

lnST


– -Ka: gain of amplifier

– Output voltage:

– We add feedback using a potentiometer Rp.

– The transfer function of the amplifier without feed back is:

– The sensitivity to changes is the amplifier gain is:

• Example: 1-7-3

inavKv 0

aKT

1TKa

S

STABILITYExamining the closed loop poles

01 HGKG pc

The zeros of the denominator

will determine the impulse and step response stability.

The inverse Laplace transform applied to

gives time signal

ps

b

)sin(cos)( btjbtbebebe attbjapt

For stability all closed loop poles must have negative real parts.

Ruth-Hurwitz Criterion Procedure

• The characteristic equation in the Laplace variable is:

11 1 0

1 2

11 2

21 2 2 3 1 3

31 2 3 1 2 4

... 0

In factored form:

... 0

Multipliplying the factors together:

...

...

... ...

n nn n

n n

n nn n n

nn

nn

q s a s a s a s a

a s r s r s r

q s a s a r r r s

a r r r r r r s

a r r r r r r s

1 2 3 1 ... 0n

n na r r r r

Ruth-Hurwitz Criterion Procedure• For the nth-degree equation:

– Note that all the coefficients of the polynomial must have the same sign if all the roots are in the left-hand plane.

– For a stable system all the coefficients must be nonzero.– Both of these requirements must be sufficient for the system to be

stable.– If the are satisfied we can proceed to check for other conditions to prove

that the system is stable.

0rootsn all ofproduct 1

... timeaat 3 taken roots theof products theof sum

timeaat 2 taken roots theof products theof sum

roots all of sum

3

2

1

nn

nn

nn

nn

nn

a

sa

sa

sasasq


0...... 11

1 o

nn

nn asasasa

Does this equation have only stable roots?

(I.e do all solutions have negative real parts? )

For first and second order systems (n=1,2) the necessary and sufficient condition for stability is that the coefficients of the polynomial are non-zero and all have the same sign.

For higher order

systems produce

this table

10

11

3213

3212

5311

42

..............

..

...

...

...

ks

js

ddds

cccs

aaas

aaas

n

nnnn

nnnn

n

.....1

31512

1

21311

1

7613

1

5412

1

3211

c

caacd

c

caacd

a

aaaac

a

aaaac

a

aaaac

nn

nn

n

nnnn

n

nnnn

n

nnnn


Case 2 : there is a zero in the first column but that row is not zero everywhere

10

11

3213

322

5311

42

..............

..

...

...

...

ks

js

ddds

ccs

aaas

aaas

n

n

nnnn

nnnn

Replace the 0 by a small positive and carry on in the usual way

0

zero replacement

Again the number of sign changes in the first column will determine the number of unstable zeros.

1. Product polynomial at (s+1) start roth table

2. Replace s with 1/x

For the rest of this class practice the application of the Ruth-Hurwitz procedure to decide on stability for as many of the examples as you can.

(The solutions involve the computation of the closed loop transfer functions first .)

Case 2 : there is a zero in the first column and that row is zero everywhere

10

11

3

3212

5311

42

..............

..000

...

...

...

ks

js

s

cccs

aaas

aaas

n

n

nnnn

nnnn

auxiliary polynomial

is

zero row


...42

21 nn scsc

The auxiliary polynomial (order is always even) gives the number of symmetrical root pairs (to the origin).

It divides the polynomial and long division can be used to obtain the other factor.

Examples: Ruth-Hurwitz criterion procedure

Example 2

Ks

s

Ks

s

Ks

K

0

1

2

3

4

7

92

3

7

023

31

Kssss

KsG

233)(

234

Condition of stability:

9

140 K


0522 334 ssss

Example 3 : zero in first column (but not all zero in row)

5

0

05

021

521

0

1

2

3

4

52

s

s

s

s

s

large negative number

small positive

number

01818112 334 ssssExample 4 : a zero row

0

1

2

3

4

00

0182

0182

18111

s

s

s

s

s

zero row

Coefficients for auxiliary equation


0)3)(3(2

0182 2

jsjs

s

Long division of original polynomial by auxiliary polynomial:

2182

21892

9:1818112

2

2234

2234

s

ssssss

sssss

)22)(9(

181811222

334

sss

ssss

Conclusion: unstable

transient and steady state response (cont.). example dc motor page 111 ex.1-4-3

Documents

system types

order system response

steady state error

steady state response

system type b ramp input

system velocity constant

system dc gain

system type c parabolic