• For a third-order system with a closed-loop transfer function
• The s-plane is
T
  s
2

2
 s
1

1
   s

1

Complex Axis

• The third-order system is normalized with ω n
=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 3 rd root.

1

10
 n
• 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

• 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.

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.
KG ( s ) H ( s )

K m  i

1
( s s y j n 

1
( s

 z i
) p j
)
System type y
We can calculate
SSE for different types of standard signals
0
1
Type zero
Type 1
2 Type 2

Relationship between steady state error and system type
A) Step Input
R ( s )

R
0 s
R ( s )

R
0 s
For y=0
SSE
 lim s

0
1
 s


R
0 s
K i m 

1
( s


 j n 

1
( s
 z i
) p j
)

1

R
0
K i m 

1 z i j n 

1 p j

1

R
0
G ( 0 )

1

R
0
K p
K
For y>0
SSE
 s lim

0
1
 s



R
0 s
K m  i

1
( s


 z i
) s y j n 

1
( s
 p j
) s y
R
0 j n 

1
( s
 p j
)
 lim s

0 s y j n 

1
( s
 p j
)

K m  i

1
( s
 z i
)

0
!

Relationship between steady state error and system type
B) Ramp Input
R ( s )

R
1 s
2
For a type 0 system…
SSE
 s lim

0
1
 s
R
1 s
2 s
K m  i

1
( s y j n 

1
( s

 z i
) p j
) s y

1 R
1 n  j

1
( s

 lim s

0 s y j n 

1
( s
 p j
)
 p j
)
K m  i

1
( s
 z i
)
 
For type 0 system the SSE =

For a type 1 system…
SSE

R
1
K i m 

1 z i j n 

1 p j

R
K v
1 velocity constant

Relationship between steady state error and system type
C) Parabolic Input
R ( s )

R
2 s
3
SSE
 s lim

0
1

R
2 s
 s
3
K m  i

1
( s s y j n 

1
( s

 z i
) p j
) s y

2
R
2 j n 

1
( s
 p j
)
 lim s

0 s y j n 

1
( s
 p j
)

K m  i

1
( s
 z i
)
For type 0 and type 1 systems, the steady state error is infinite. For type 2 systems the steady state error is given by
SSE

R
2
K i m 

1 z i j n 

1 p j

R
K a
2 acceleration error constant

Relationship between steady state error and system type
Input System Type
(number of pure integrators) step ramp parabola
0
1
2
SSE

1

R
0
K p
SSE

0
SSE

0 0

R
1
K v

R
2
K a

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 splane), 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.
Problem 1
Problem 2
Problem 3
(two problems)
Problem 4
R ( s )
E ( s )
KH ( s )
KH ( s ) G ( s )

( s
10

( s
4 )( s

2

)
20 )
KH ( s ) G ( s )
 s ( s
10 ( s


4 )( s
2 )

20 )
KH ( s ) G ( s )
 s ( s 2 
3 s

8 s
1 )( s

1
2 
40 s

70 )
KH ( s ) G ( s )

( s 2 

10 s
4 )( s 2

2

30 s

50 )
G ( s )
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.

q

• 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

• 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:
Y

1

G
 
GH
    if GH

1

Y

H
1
   
– 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.

Sensitivity of Control Systems to Parameter
Variations (cont.)
• 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
Open

Y
 
-

Loop System

G
   
Closed
Y
 

Y

When

Y

GH

-

Y

Loop System

1
1
 


1

 
GH



G
G
 
GH

GH

G
GH
 
  
2 s
R
 


G
 
G

G
 
 
 
 
 
1

GH
    

• 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.

Sensitivity of Control Systems to Parameter
Variations (cont.)
• The system sensitivity
System Sensitivit y
 percentage change in the system transfer function percentage change of the process transfer function
– System Transfer Function is
System Sensitivit y

S

T
 

T

G

Y
R
 
 
   
   
• For small incremental changes:
S



T
G
/
/
T
G


 ln ln
T
G

Sensitivity of Control Systems to Parameter
Variations (cont.)
• The system transfer function of the closed-loop system is:
T

1

G
 
GH
 
• The sensitivity of the feedback system is:
S
T
G


T

G
G
T


1

1
GH

2
G /

1
G

GH

S
T
G

1

G
1
– 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.

Sensitivity of Control Systems to Parameter
Variations (cont.)
• The sensitivity of the feedback system to changes in the feedback element H(s) is:
S
T
H


T

H
H
T

G
1

GH
2
G /


1

H
GH

S
T
H

1


GH
GH
– When GH is large, the sensitivity approaches unity (1)
– The changes in H(s) directly affect the output response
S T
H

1
• It is important to use feedback components that will not vary with environmental changes or that can be maintained constant.

Sensitivity of Control Systems to Parameter
Variations (cont.)
• The sensitivity to α:
S
T



S
T
G
S
G

: a parameter within the transfer function of the block G
• The transfer function of the system T(s) is a fraction of the form
T
 s ,
 

N
D
 s ,

 s ,
 


: a parameter that may be subject to variation due to the environmen t
• The sensitivity to α:
S
T



 ln ln
T



 ln ln
T


0
– α is a nominal value of the parameter

 ln
 ln
D


0

S

N 
S

D

Sensitivity of Control Systems to Parameter
– -K a
: gain of amplifier
Variations (cont.)
– Output voltage: v
 
0
K a v in
– We add feedback using a potentiometer R p
.
– The transfer function of the amplifier without feed back is:
– The sensitivity to changes is the amplifier gain is:
T
S
K a
T
 
K a

• Example: 1-7-3

STABILITY
Examining the closed loop poles
The zeros of the denominator
1

KG c
G p
H

0 will determine the impulse and step response stability.
The inverse Laplace transform applied to b s
 p gives time signal be pt  be
( a
 bj ) t  be at
(cos bt
 j sin bt )
For stability all closed loop poles must have negative real parts.

• The characteristic equation in the Laplace variable is:
   a s n n  a n

1 s n

1
...
a s
1
 a
0

0
In factored form: a n
 s
 r
1
 s
 r
2
  s
 r n
 
0
Multipliplying the factors together:
   a s n n  a n
 r
1 r
2

 a n
 r r
1 2
 r r
2 3
 r r
1 3
...
 s n

2 a n
 r r r
1 2 3
 r r r
1 2 4

...
 r n
 s n

3 s n

1

...
 a n
  n r r r
1 2 3
...
r n

0

• For the nth-degree equation: q
 a


 n a a s n n  a
 sum n
 sum n a n
 sum of of of
the
the all roots products products
   product of
 s of of n

1
the
the all n roots
 roots taken

0
2 at a time roots taken 3 at a time
 s
 s n
 n

3
2

...
– 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.

Ruth-Hurwitz Criterion Procedure
Does this equation have only stable roots?
(I.e do all solutions have negative real parts? ) a n s n  a n

1 s n

1 
......
 a
1 s
 a o

0
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
..
s
1 s
0 s n s n

1 s n

2 s n

3 a n a n

1 c
1 d
1
...
j
1 k
1 a n

2 a n

3 c
2 d
2
...
a n

4 a n

5 c
3 d
3
...
...
...
...
..
...
c
1
 a n

1 a n

2
 a n a n

3 c c
2
3 d
1
 a n

1 a n a n

1

4
 a n a n

5

 a n

1 a n a n

1

6
 a n a n

7 c
1 a n

3 a n

1
 a n

1 c
2 d
2
 c
1 a n

5 c
1
 a n

1 c
3 c
1
.....

Ruth-Hurwitz Criterion Procedure
Case 2 : there is a zero in the first column but that row is not zero everywhere
Replace the 0 by a small
 positive and carry on in the
 usual way
0 s s n n

1 s n

2 a n a n

1
 a a n

3 c n

2
2 a n

4 a n

5 c
3 zero replacement s n

3
..
d
1
...
d
2
...
d
3
...
...
...
...
..
...
s
1 j
1 s
0 k
1
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

Ruth-Hurwitz Criterion Procedure
Case 2 : there is a zero in the first column and that row is zero everywhere auxiliary polynomial is c
1 s n

2  c
2 s n

4 
...
s n s n

1 s n

2 s n

3 a n a n

1 c
1
0 a n

2 a n

3 c
2
0 a n

4 a n

5 c
3
0
...
...
...
zero row
..
..
s
1
...
j
1
...
...
...
The auxiliary polynomial (order is always even) gives the number of symmetrical root pairs (to the origin).
s
0 k
1
It divides the polynomial and long division can be used to obtain the other factor.
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 .)

Examples: Ruth-Hurwitz criterion procedure
Example 2
G ( s )
 s 4 
3 s 3
K

3 s 2 
2 s

K s
1 s
0 s
4 s
3 s
2
1
3
3
2
7
3
2

9
7
K
K
K
K
0
Condition of stability:
0

K

14
9

Examples: Ruth-Hurwitz criterion procedure
Example 3 : zero in first column (but not all zero in row ) s
4
 s
3

2 s
3

2 s

5

0 small positive number s
4 s
3 s
2 s
1 s
0
1 2 5
1

2

5

5
2
5
0
0
0 large negative number

Examples: Ruth-Hurwitz criterion procedure
Example 4 : a zero row s
4

2 s
3

11 s
3

18 s

18

0
Coefficients for auxiliary equation s
4 s
3 s
2 s
1 s
0
2
2
1 11
18
18
18
0
0
0 0 zero row
2 s
2 
18

0
2 ( s

3 j )( s

3 j )

0
Long division of original polynomial by auxiliary polynomial: s
4 
2 s
3 
11 s
2 
18 s

18 s 4 
2 s 3 
9 s 2 
18 s
2 s
2 
18
: s
2 
9 s 2
2

2 s s
4 
2 s
3 
11 s
3 
18 s

18


( s
2

9 )( s
2

2 s

2 )
Conclusion: unstable