Lect9_sys Char & Routh

8/13/2019 Lect9_sys Char & Routh

http://slidepdf.com/reader/full/lect9sys-char-routh 1/15

CONTROL SYSTEM I

System Characteristics

Disturbance Rejection

Open Loop Control

+

B/A

Controller

Reference

Speed, R

)1)(1( 21 ++ ss

A

τ τ

V a

Output, Y

Load, ωDC Motor

Speed Control

2



Example

Open Loop Speed Control of DC Motor

Find K, V a, yss with and without disturbance

+

5

K

Input

r=100 rad/s

)1600 / )(160 / (

10

++ ss

VaOutput, y

-0.1 Nm

3

Solution: Open Loop Control

Assume no load condition, i.e., disturbance = 0, letopen loop DC gain =1 (from r to y), then

If disturbance = –0.1 Nm, as the DC gain of theplant A is 10, we could find

The system variation is 5%.

4

( )

( ) srad Kr y

srad V y

volt Kr V

srad volt K

ss

ass

a

/ 951051.010

/ 10010

101001.0

/ / 1.010

1

=××−+=

==

=×==

==



Example5

Closed- Loop Speed Control of DC Motor

Find yss with and without disturbance if

K = 10 volt/(rad/s);

+

5

K

Reference

Speed,

r=100

rad/s

+

+ 1

6001

60

10

ssVa

Load = -0.1Nm

+_

Output y

Solution: Closed-Loop Control

As no load condition, i.e., disturbance = 0

6

( )

srad

K ss

K

s

ssK

ssK

s RKG

KGsssY y

s

sssss

/ 991001001

100

100101

6001

60

10

lim

100

1600

160

101

1600

160

10

lim1

limlim

0

000

=×

+

⇒

⋅

+

+

+

=

⋅

+

+

+

+

+

⋅=

+

==

→

→→→



Solution

Closed Loop Control

Note: 5% speed change (100rad/s95rad/s) inopen loop vs. 0.05% (99rad/s98.96rad/s) inclosed loop, improvement by a factor of (1+AK)

7

( )

srad ss

s

ssK

ss

s

ssK

ssK

s

sKG

G R

KG

KGsssY y

s

s

ssss

/ 96.98

1001600

160

51001010lim

1.0

1600

160

101

1600

160

50

100

1600

160

101

1600

160

10

lim

1.0

1

5

1limlim

0

0

00

=

+

+

+

−××=

−

+

+

+

+

+

+⋅

+

+

+

+

+

=

−⋅

+

+

+

==

→

→

→→

Basic Properties of Feedback

Why feedback control?

Stabilize systems

Decrease sensitivity to parameter variations

Reject disturbance

Reduce steady-state error

8



Closed-Loop Characteristics

- Steady-State Error Analysis

The steady-state gain of a transfer function is found

using the final value theorem to the unit step response.

Now consider the steady-state error for the closed-

loop system

)(lim1

)(lim00

sGs

sGsgss

ss→→

=

=

G(s)

H(s)

E(s) R(s) C(s)

+

–

)(1

)()(

sGH

s Rs E

+

=

)()()()( s E sGH s Rs E −=



Applying the final value theorem

Now consider the steady-state error for different inputs,

assuming H(s) = 1 (if H(s) is not one, the steady-state

error is the control error and not the output error) .

+=

→ )(1

)(lim

0 sGH

s Rse

sss





a) For a step input with amplitude, A units (position

error)

The percent error may then be expressed in terms of

the position error constant,

Position error

constant

ssG

→

)(lim0

pK P =

( ) ( ) ps

sss

K

A

sG

A

sG

s

A

se+

=

+

=

+

=

→

→ 1lim11lim

0

0

%1001

1%100% ×

+

=×=

p

ss

K Aeerror



b) For a ramp input of A units per second (velocity error)

Velocity errorconstant

ssssGGss

→→

))((limlim00

vvK ==

Time (sec.)

A m p l i t u d e

Linear Simulation Results

0 2 4 6 8 10

0

1

2

3

4

5

6

7

8

9

ess

( ) ( ) vs

sss

K

A

ssG

A

sG

s

A

se ==

+

=

→

→

0

2

0 lim1lim





c) For a parabolic input of A/2 units per second2

(acceleration)

As before, the percent error is

acceleration

error constant

ssGs2

→

)(lim0

aK =

( ) ( )

%100%

lim1lim

2

0

3

0

×=

==

+

=

→

→

Aee

K

A

sGs

A

sG

s

A

se

ss

ss

as

sss

Classification of Systems by Type

The general form of a factored transfer function

is

The term 1/s is often referred to as a free pole

or a pole at the origin. The type of the system depends upon the order

of the free poles.

)())((

)())(()(

21

21

n

i

m

ps ps pss

zs zs zsK sG

+++

+++=

L

L



Classification of Systems by Type

For i = 0, 1 ,2 ⇒ type = 0, 1, 2

With i = 0, the steady state gain of G(s) is

With i = 1 or greater, gss → ∞ and ess → 0 .Therefore, the steady-state error directlydepends upon the system type.

n

mss

P p p

z zKzg

L

L

21

21=

System Type and Steady-State Error

System

Type

Step Input Ramp

Input

Parabolic

Input

0 ess=∞ ess=∞

1 ess= 0 ess =

A

K v

ess=∞

2 ess= 0 ess= 0

A

+ K p

=

1

ess = A

K a

ess



System Type and Steady-State Error

17

CONTROL SYSTEM I

System Stability



Stability Definitions

Bounded Input Bounded Output Stability:

A system is BIBO stable if, for every bounded input,

the output remains bounded with increasing time (all

system poles must lie in the left half of the s-plane).

Stability Determination

Must we first find all roots of the characteristic

equation in order to determine the stability of the

system?

19

Routh-Hurwitz Stability Criterion

A quick method for checking BIBO stability.

Assume the characteristic polynomial is

where a0 ≠ 0 .

A necessary (but not sufficient) condition for all

roots to have non-positive real parts is that all

coefficients have the same sign. For the necessary and sufficient conditions, first

form the Routh array.

011

1)( asasasasQ n

nn

n ++++= −

− L

20



Routh Array

sn

an an-2 an-4 an-6 …

sn-1

an-1 an-3 an-5 an-7 …

sn-2

b1 b2 b3 b4 …

sn-3

c1 c2 c3 c4 …

: : : :

s2

k 1 k 2

s

1

l1s

0m1

1

5412

−

−−− −=

n

nnnn

aaaaab

1

3211

−

−−− −=

n

nnnn

aaaaa

b

In Row 3

etc.

In Row 4

01

1

1)( asasasasQ n

n

n

n ++++=

−

− L

21

1

31512

1

21311

b

baabc

b baabc

nn

nn

−−

−−

−=

−

=

The elements in all subsequent rows are

calculated in the same manner.


Necessary and sufficient conditions:

If all elements in the first column of the Routh arrayhave the same sign, then all roots of the characteristicequation have negative real parts.

If there are sign changes in these elements, then thenumber of roots with non-negative real parts is equalto the number of sign changes.

Elements in the first column which are zero define aspecial case.

Example 6.2

22



Example

10532)(234

++++= sssssQ

s4 2 3 10 0

s3 1 5 0 0

s2 b1 b2 0

s1 c1 0

s0 d 1

1043.6

0)43.6(10

43.67

1035

101

0107

1

103

1

1

21

=

−

=

=

−

−−

=

=

−

=−=

−

=

d

c

bb

The characteristic equation has two roots with positive real parts

since the elements of the first column have two sign changes.

(2,1,-7,6.43,10)

23


Special Case 1:

A zero in the first column but some other elements

nonzero:

To proceed:

substitute a small positive number ε for the zero element

finish the Routh array,

let ε → 0+ and check 1st column

No sign change some roots on Imaginary axis

Sign change some roots in RHP

24



Example

If , build the Routh array as follow

s3 1 -3 0 0

s2

0(ε ) 2 0

s1 b1 0

s0 c1

23)(3

+−= sssQ

22

223

1

11

1

=⋅

=

−→

−−=

b

bc

bε ε

ε (negative)

There are two roots with positive

real parts (1, ε, -2/ ε , 2)

25

Exercise

Use Routh-Hurwitz Stability Criterion to answer:

Are all roots of the following polynomial in LHP?

1011422)( 2345

+++++= ssssssQ




Special Case 2:

An all zero row in the Routh array which correspondsto pairs of roots with opposite signs.

To proceed:

form an auxiliary polynomial from the coefficients in therow above.

Replace the zero coefficients from the coefficients of thedifferentiated auxiliary polynomial.

Complete the array and check 1st column

No a sign change the roots of the auxiliary equation define

the roots of the system on the imaginary axis. Sign change a(s) has factors of the form

27

( )( )σ σ +− ss

Example

1)(34

−−+= ssssQ

2 s)1(

2

=−

ds

sd

12

−s

Auxiliary polynomial

then

d 1 = – 1

The system has one root with a positive real part ( 1, 1, 1, 2, -1).

The root is found from the auxiliary eq. s2 – 1 = 0 , s = ± 1

s4 1 0 -1 0

s3 1 -1 0 0

s2 1 -1 0

s

1 0 0

s0 d 1

22

28

Example 6.4

Example 6.5



Assignment29

Textbook (5th edition)

5.22

6.1 a – iii, iv, and vii

Due on 11/18

Documents

Lect9_sys Char & Routh