The stability of linear time-invariant feedback systems

8/13/2019 The stability of linear time-invariant feedback systems

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The stability of linear time-invariant feedback systems

A. TheoryThe system is atrictly stable if

The degree of the numerator of H(s) (H(z)) the degree of the denominator of H(s)(H(z)) and/or

The poles of H(s) left half plane (interior of the unit circle for the discrete-timecase).

Wide sense stable systems

the poles of H(s), spimaginary axis j /the poles of H(z) zpunit circle r=1 (discrete-time).

Strictly stable system if and only if the denominator of the transfer function H(s) is a

Hurwitz polynomial.

Q(s) strictly/wide sense Hurwitzif Has real coefficients, (strictly or not) positive + The roots of Q(s) (poles) are in the left half plane (which means Re{z} < 0)A strictly Hurwitz polynomial has all the minors of n

thorder, n > 0 (its zeros are in the

closed left half plane).

A wide sense Hurwitz polynomial can have some minors zero k =0, and the restpositive (some of its zeros are in the closed left half plane and if there are zeros on the

imaginary axis).

1 3 5 7

0 2 4 6

1 3 5

2 4 6

1 3

... 0

... 0

0 ... 0

0 ... 0

0 0 ... 0

.. . . . ...

0 . . . ...

n

n

a a a a

a a a a

a a a

a a a

a a

a

=

( )( )

( )( ) ( )1

Y s H s

X s H s G s=

+(continuous time)

( )

( )

( )

( ) ( )1Y z H z

X z H z G z=

+(discrete time)


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( ) ( ) ( )W s H s G s= respectively ( ) ( ) ( )W z H z G z= the transfer function in open loop.

The open loop system is strictly stable if

The roots 1+W(s)=0 closed left half plane (the real part ( ) ( )( ),

H j G j

= Nyquist hodograph of the open loop system. The number of

complete rotations around the point (-1/k, 0) in counter-clockwise sense is n times where

2i cn n n= + ,

in = number of poles from the right half plane and cn = number of poles from the

imaginary axis (continuous time)

( ) ( ) ( )1

R z H z G zk

= + doesnt have zeros outside the unit circle r=1 (z=ej)

in = number of poles outside the unit circle r=1, cn = number of poles on the unitcircle r=1 (discrete time)


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B. Problems

Problem 1. Consider the amplifier with the transfer function

( ) Ga

H ss a

=+

a) Find the DC gain of the amplifier.b) What is the time constant of the system?c) The bandwidth of the amplifier is the frequency where the magnitude of the

frequency response is 2 times smaller than the DC gain. What is the bandwidth

of this amplifier?d) The amplifier is placed into a feedback loop:

What is the DC gain, the time constant and the bandwidth of the feedbacksystem?

e) Find the value k for which the bandwidth of the closed loop system is double thebandwidth in open loop.

Solution.

a) ( ) Ga

H jj a

=+

. The DC gain is ( )0H G=

b) ( )

1

1 ate ts a

+

L

.

If ( ) ( )t

h t e t

then is the time constant of the system

( ) ( )1at

h t Ga e t a

= =

c) The bandwidth is ( )2

M

GB H j= = . But ( )

2 2

GaH j

a

=+

2 2 2 T

M

Ga Ga

a

= =+

d)

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )1

Y s X s kY s H s

Y s X s H s kY s H s

Y s kH s X s H s

=

=

+ =

K

Ga

s a+

x(t) y(t)


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( )( )

( )( )

( )1

1b

GaY s H s s a H s

GaX s kH sk

s a

+= = =+ +

+

The DC gain of the feedback system is ( )0 1closedG

H kG=

+

( )( )1

closed

Ga GaH s

s a kGa s a kG= =

+ + + +

The time constant is( )

1

1a kG=

+

( )( )1

closed

GaH

j a kG=

+ +

( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( )

2 22

2

2 2 22

2 2 22 2

1 2 1

2 11

1

1 2 1

2 1 1

1

c closed

c closed

c closed

c closed

c closed

Ga G

j a kG kG

Ga G

kGa kG

a

a kG kG

a kG a kG

a kG

=+ + +

=++ +

=+ + +

+ = + +

= +

e) The cutoff frequency (bandwidth) of the closed loop system:

( )1

2 1 2c closed T a kG a k G

= + = = .

Problem 2.Determine the transfer function for each system represented below.

a)

x(t) H0(s)

G(s)

H1(s) y(t)u(t)


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b)

c)

Solution.

a)

( ) ( ) ( )0U s X s H s=

( ) ( ) ( ) ( ) ( )1Y s U s Y s G s H s=

( ) ( ) ( ) ( ) ( )1 11Y s G s H s U s H s+ =

( )( )

( )( ) ( )

1

11Y s H sU s G s H s

=+

( )( ) ( )

( )( ) ( )

1

0 11

Y s H s

X s H s G s H s =

+

( )( )

( ) ( )( ) ( )

1 0

11

Y s H s H s

X s G s H s

=

+

b)

[ ] [ ] [ ]u n x n w n= i ( ) ( ) ( )1V z U z H z=

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )1U z X z W z X z G z V z X z G z H z U z = = =

( ) ( ) ( ) ( )11U z G z H z X z+ =

( ) ( )

( ) ( )11X z

U zG z H z

=+

, but ( ) ( )

( )1

V zU z

H z=

G1(s)

H1(s)y(t)x(t)

G2(s)

H2(s)u(t) v(t) w(t)

x[n]

H0(z)

G(z)

H1(z) y[n]u[n] v[n]

w[n]


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( ) ( ) ( )

( ) ( )1

11

H z X zV z

G z H z =

+

( ) ( ) ( ) ( )

( ) ( )

( ) ( )

( ) ( )

0

1

011

Y z X z H z V z

H z X z

X z H z G z H z

= +

= ++

( )( )

( ) ( )

( ) ( )1

0

11

Y z H zH z

X z G z H z= +

+

c) ( ) ( ) ( ) ( )2U s X s G s Y s=

( ) ( ) ( )2V s U s H s=

( ) ( ) ( ) ( )1W s V s Y s G s=

( ) ( ) ( )1Y s W s H s=

( ) ( ) ( ) ( ) ( )1 1Y s H s V s Y s G s =

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ){ }1 2 1

1 2 2 1

Y s H s H s U s Y s G s

H s H s X s G s Y s Y s G s

=

=

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )1 2 2 1 1 1 21Y s H s H s G s H s G s H s H s X s+ + =

( )( )

( ) ( )( ) ( ) ( ) ( ) ( )

1 2

1 2 2 1 11

Y s H s H s

X s H s H s G s H s G s=

+ +.

Problem 3. Consider the system described by the linear constant-coefficients differentialequation

6 9 6 8y y y y x x x + + + = + +

with null initial conditions.Study the stability of the system based on the Hurwitz stability criterion.

Solution.

( ) ( ) ( ) ( ) ( ) ( ) ( )3 2 3 26 9 6 8s Y s s Y s sY s Y s s X s s X s sX s+ + + = + +

( )( )

3 2

3 2

6 8

6 9 1

Y s s s s

X s s s s

+ +=

+ + +

The numerator has positive coefficients, which means we can apply Hurwitz criterion.

0 1 2 31, 6, 9, 1a a a a= = = =

We compute the determinant:

1 3 5

3 0 2 4

1 3

6 1 0

1 9 0

0 0 6 1

a a a

a a a

a a

= =


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and the minors:

1 6 0 = >

2 54 1 53 0 = = >

3 26 9 1 1 6 0 0 1 0 0 9 0 0 6 6 1 1 1 53 = + + = =

All the minors are positive, meaning the system is strictly stable.

Problem 4. Consider the following closed loop system

Find the values for kfor which the system is stable, if

a. ( ) ( )1

1H s G s

s=

+ f. ( ) ( )

( )3

1

1H s G s

s=

+

b. ( ) ( )1

1H s G s

s=

g. ( ) ( )

( )4

1

1H s G s

s=

+(homework).

c. ( ) ( )( )

1

1 110

H s G ss

s

=

+ +

h. ( ) ( ) 21

4

sH s G s

s

+=

d. ( ) ( ) 21

1H s G s

s=

i. ( ) ( ) 2

1

2 2H s G s

s s=

+ +

e. ( ) ( )( )

2

1

1H s G s

s=

+

Solution.

a. ( ) ( )1

1H s G s

s=

+ ( ) ( )

1

1H j G j

j

=+

, sp = -1

In order to sketch the hodograph we can use the Bode plots.

( ) ( )2

1

1H j G j

=+

( ) ( ) ( )2

220log 10log 1 10log 11

H j G j

= + = +

( ) ( ){ } ( )arg arg 1 arctg arctg1

H j G j j

= + = =

x(t)k

G(s)

H(s)y(t)


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=0

=

=1

= -1Counter-clockwise


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The open loop system is stable the hodograph doesnt have to make rotations around

the point of coordinates1

,0k

( )1 1

0 0 or 1 1,0k kk k

< > >

The closed loop system is stable for k>-1!

b. ( ) ( )1

1H s G s

s=

The open loop system is unstable with one pole in the left half plane sp=1

( ) ( )2

1

1H j G j

=+

( ) ( ){ } ( )arg arg 1 arctg arctg1

H j G j j

= + = + = +

( ) ( ) 2 2

2 2

Re Im

1 1 11 1 1

1

1 1

j jH j G jj

j

+ += = = +

= +

+ +

For 0> ( ) ( ){ }Re 0H j G j < , ( ) ( ){ }Im 0H j G j <

( ) ( ){ }Re H j G j ( ) ( ){ }Im H j G j 0 -1 0

1 -1/2 -1/2

0 0

The Nyquist hodograph makes rotations in counter-clockwise sense. The system is stable

if the point of coordinates1

,0k

is encircled one time in counterclockwise sense:

11 1

11

0 0

kk

k

k

k

> >

> < >

. The system is stable for k>1.


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=0=

=1

counterclockwise


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c. ( ) ( )( )

1

1 110

H s G ss

s

=

+ +

. The open loop system is stable.

( ) ( )

( )

1

1 110

H j G jj

j

=

+ +

( ) ( )2

2

1

1 110

H j G j

=

+ +

( ) ( )2 2

20log 10log 1 10log 11 10

H j G j

= + +

( ) ( ){ } ( )arg arctg arctg10

H j G j

=

10

1

1 1

kk

k

>

.

The sense of rotation of the hodograph isnt counterclockwise the point1

,0k

shouldnt be encircled.


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d. ( ) ( )2

1

1H s G s

s=

1,2 1ps = . The open loop system is unstable.

( ) ( ) ( ) ( ){ }2 21 1

Re1 1

H j G j H j G j

= = = +

; ( ) ( ){ }Im 0H j G j =

0 1

( ) ( ){ }Re H j G j -1 -1/2 0

=0

=1

=10

=0 =

counterclockwise

=

Sensul e antiorar

=1


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The open loop system has a pole in the right half plane; the closed loop system is stable if

the critical point is encircled one time in counter-clockwise:

11 0 1k

k < < >

e. ( ) ( )( )

( ) ( )( )

2 2

1 1

1 1H s G s H j G j

s j

= =+ +

( ) ( ) 21

1H j G j

=+

( ) ( )2

20log 20log 11

H j G j

= +

( ) ( ){ } ( )arg 2arctgH j G j =

The open loop system is stable, therefore the critical point shouldnt be encircled:

1 10 sau 1 0 sau 1 0 1k k k

k k

< > > < < >


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f. ( ) ( )( )

( ) ( )( )

3 3

1 1

1 1H s G s H j G j

s j

= =+ +

( ) ( )

( )23

1

1

H j G j

=

+( ) ( )

23

20log 10log 11

H j G j

= +

( ) ( ){ } ( )arg 3arctgH j G j = .Counterclockwise sense

( ) ( )1 1 1

1 or 1 0 or 8 ,8 1,08

k k kk k > < < < <

=0

=1

=10

Counterclockwise


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g. homework.

h. ( ) ( ) { }1,2 121

2 Re 04

p p

sH s G s s s

s

+= = >

The open loop system is unstable.

( ) ( ) ( ) ( ){ }

( ) ( ){ }

2 2

2

1 1Re ,

4 4

Im4

jH j G j H j G j

H j G j

+= =

+

= +

( ) ( ){ }Re H j G j ( ) ( ){ }Im H j G j 0 -1-4 0

1 -1/5 -1/5

0 0

The system is stable for

1 10 4 and 0 4

4k k k

k < < > > > .

=0

=1

=

= 3

Re=-1/8

Counterclockwise


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i. ( ) ( ) { }1,2 1,221

1 Re 02 2

p pH s G s s j ss s

= =

> >

=0

=1

Counter clockwise

=


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=0=

= 2

Counter-clockwise

The stability of linear time-invariant feedback systems

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