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The Performance of Feedback Control Systems
Objectives:
- Specify the measures of performance (time-domain) Æ the first steps in the design process Percent overshoot / Settling time ( Ts ) / Time to rise / Steady-state error ( ess )
- Input signals such as the step to test the response of the control system.
- Correlation between the system performance and the location of T.F. poles and zeros in the s-plane
- Relationships between the performance specifications and the natural frequency ( ω ) and damping ratio ( ζ ) for 2nd-order systems
<Performance of a 2nd-order sys >
Y(s)
=
( )= 2 R ( s ) s ps K
Y s K
If we can select p & K, we may be able to provide desired
ζ , ωn , and/or the combination ( design a controller! )
s1, s 2 = - ζ ωn ± jωn 1 −^ ζ^2 = σ ± j ω
G(s)
- Sign of the real part ( σ) Æ stability
R(s) E
( ) 2 2 2
2 R s s (^) ns n
n ζω ω
ω
ζ : damping ratio & ω: natural frequency
s ( p )
k
( ) ( ) R s G s
G s Y s
s =
Math. Derivations
If R(s) = (^) s
1 , a unit step input Æ Y(s) = (^) ( 2 2 2 )
2
n n
n s s ξω ω
ω
L.T. Table
Peak: to derive the peak, differentiate y(t)
L ⎥⎦
⎤ ⎢⎣
⎡ dt
dy ( t ) = (^22)
2 2
( ) n n
n s s
s Y s ξω ω
ω
⋅ =
L.T. Table
e t B
y t n^ nt ω n β
ω (^) ξω ( )= − ⋅sin
To get the peak = ⇒ ω^ β =^ π
y ( t ) (^0) n t
∴ (^) Peak time Tp = (^2) ω 1 ξ
π ω β
π −
= n n
The peak response
/ 1 2 1
−ξπ − ξ M (^) pt = + e Æ The percent overshoot / 1 2
.. 100
−ξπ − ξ PO = ⋅ e
sin( )
1 ( ) 1 ω β θ β
y t = − e −ζω^ g + n nt
β = 1 −ξ^2 , θ =cos−^1 ξ, 0 <ξ < 1
Settling time :
Textbook: 5% Æ ζω^ n Ts =^3 Æ n
Ts τ (^) ξω
β ζ
β
ω β θ ω β π π θ θ
⇒ − =−
−
=
sin(tan )
sin( )| sin( ) sin
1
n t n t
Therefore,
)
1 (
4 4
- 02 ( 2 %) 4
n n
s
n s
T
T
e n s T
ξω
τ ξω
τ
ζω ζω
∴ = = =
− < ⇒ =
θ
ξ^2 +β^2 = 1
ξ
p
w t
e t T
y t e n
−
−
/ 1 2 1
ξπ ζ
β = 1 − ξ^2
θ = cos −^1 ξ, 0 < ξ < 1
Observability
All the roots of the characteristic equation can be placed where desired in the s -plane if, and only if, a system is observable and controllable. Observability refers to the ability to estimate a state variable. Thus we say a system may be observable if the output has a component due to each state variable.
A system is observable if, and only if, there exist a finite time T such that the initial state x (0) can be determined from the observation history y ( t ) given the control u(t).
Consider the single-input, single-output system x = Ax + Bu and y = C x ,
⋅
where C is a row vector, and x is a column vector. This system is observable when the determinant of Q is nonzero, where
which is an [ n x n ] matrix.
Example: Observability of a two-state system Consider the system given by
In order to check system’s controllability and Observability, need to evaluate Pc and Q matrices. The matrices for Pc are
Therefore we have
And determinant Pc=0. Thus the system is not controllable. To determine Q we obtain
(3)
C = [ 1 1 ] and CA =[ 1 1 ]
Therefore we obtain
And determinant Q = 0. Therefore the system is not observable.
s ⇔^ ∫
1
<Control Design: 2nd^ order System > G(s) Y(s)
=
( )= 2 R ( s ) s ps K
Y s K
If we can select p & K, we may be able to design a controller to satisfy with performance requirements given
by^ ζ^ , ωn , and/or the combination ( Ts = 4 /^ ζ^ ωn) since
p = 2^ ζ^ ωn & K =
2
ω n
- Find C.E.: s^2 + p s + K
2. Compared with the general C.E.: s^2 +2^ ζ^ ωns +
2
ω n
- Select K & p to satisfy with the performance requirement - p = 2^ ζ^ ωn K =
2
ω n
- Natural frequency ωn should be 3 [Hz] Æ K = 9
- Damping ratio should be^ ζ^ = 1 Æ p = 6
R(s) E
( ) 2 2 2
2 R s s (^) ns n
n ζω ω
ω
ζ : damping ratio & ω: natural frequency
s ( p )
k
( ) ( ) R s G s
G s Y s
s =
EX (^) G(s)
- Find C.E.:1. Find C.E.: ss^2 + d s + p
- Compared with the general C.E.: s2. Compared with the general C.E.: s^2 +
(^2) + d s + p
ns +^
2
ω n
- Select p & d to satisfy with the performance
requirement Æ d = 2^ ζ^ ωn p =
2
ω n
s^2 + d s + p = s^2 +2^ ζ^ ωns +
2
ω n
Requirements: 1) Ts = 4 /^ ζ^ ωn = 2 [sec] and 2) ωn = 4
[Hz] From 2), p = 16 and
From 1), Ts = 4 /^ ζ^ ωn = 4 / 4^ ζ^ = 2 Æ^ ζ^ = 0.5 Æ d =
2 ζ^ ωn = 4
R(s) E
p Y(s)
( ) 2 2 2
2 R s s (^) ns n
n ζω ω
ω
s ( s + d ) 1 ( ) ( )
( ) ( ) R s G s
G s Y s
=
ss1,1, ss 22 = -= - ζζ ωωnn ±± jjωωnn 1 − ζ^2 = σ ± j ω
Q(s) = s^2 + 2 ζω n s + ω^2 n = s^2 + 2 s + K = 0
Design Requirements: ζ should be greater than 0.707 and Ts should be faster than 4 [sec]
C.E. of a 2nd^ order sys is given by
Q(s) = s^2 + ds + p = s^2 + 2 ζω n s + ω^2 n Æ d = 2 ζω^ n ,^ p =^ ω n^2
Æ Proper choices of p & d can provide required performance specified by^ ζ^ or/and ωn (setting time)
- Ts = 4 / ζω n < 4 [sec] Æ ζω n > 1 Æ d = 2 ζω n > 2
- ζω n > 1 Æ ω n > 1 / ζ = 2 Æ p = ω^2 n > 2
REMARK:
- ζω n ↑ ( poles go away to the left ) Æ Ts ↓ ( faster )
- Since θ = cos-1ζ (cos-10.707 = 45^0 ) θ ↓ Æ ζ ↑ s (^) 1, s 2 = - ζ ωn ± jωn^1 −^ ζ^2 = σ ± j ω
G(s) R(s) E
( ) 2 2 2
2 R s s s
n ζω ω
ω
s ( s d )
p
( ) ( ) R s G s
G s Y s
=
Y(s)
PID controlPID control
s 2 +2^ ζ^ w s+n w n^2
s 2 +2^ ζ^ w s+n w n^2
E(s) (^) G (^) c(s) G (^) p (s) U(s) U(s) R(s)
Controller G (^) c(s)= kp + kI (^) s
1
Control input
U(s) = [k (^) p + kI (^) s
1
If G (s)=p (^) s s s 2 s
1 ( 2 )
1
dt
det u t kpe t kI e t dt kd
- P control: G (s) = k
T(s)=
c p
p
p c p
Gc ⋅ G p s s k
k G G + +
= 1 + (^22)
Q(s)= s 2 +2s+k =
- PD control: G (s)=k +k s
T(s)=
p
c p D
D p
k (^) D s + k p s^2 + ( 2 + k ) s + k Q(s)= s 2 +(2+ k )s + k =D p
- PI control: (^) s
k s k s
G (^) c s kp kI p I
( )= +^1 =
T(s) = (^) p I
p I s s k s k
k s k
2 ( 2 )
( ) 3 2
= (^) ( )( )
( ) 1 2 s a s^2 k s k
k (^) ps kI
- PID control: G (^) c(s) = k (^) P +k (^) I (^) s^1 +k (^) D s = (^) s
k (^) D s^2 + kps + kI
T(s)= (^) D p I
D p I s k s k s k
k s k s k
3 2
2
( 2 )
( )(^212 )
2
s a s k s k
kDs kps kI
=
- Find C.E.: 1+ G(s)H(s) = 1+ G (^) c(s)Gc(s)H(s) = 0
2. Compared with the general C.E.: s^2 +2^ ζ^ ωn s +
2
ω n
- Select KP , KI , & KD to satisfy with the required performance
Select two poles (dominant poles) near to the origin for the comparison if there are more than two poles Æ compare the real part of poles
EX> if poles are at -1± j2, -
Æ select -1± j2 Æ s^2 +2s +5 = s^2 +2^ ζ^ ωn s + = 0
2
ω n
