1. 2-DOF Pole Placement Controller - Moudgalyamoudgalya.org/digital-slides/pp-basic-1.pdf · 1....

1. 2-DOF Pole Placement Controller


Consider a plant with transfer function

G(z) = z−kB(z)

A(z),



G(z) = z−kB(z)

A(z),

• B(z) and A(z) are coprime, i.e. no common factor



G(z) = z−kB(z)

A(z),


• Want output y to be related to setpoint r as,



G(z) = z−kB(z)

A(z),



Ym(z) = γz−kBr

φclR(z)



G(z) = z−kB(z)

A(z),



Ym(z) = γz−kBr

φclR(z)

• φcl is ch. polynomial from desired region analysis



G(z) = z−kB(z)

A(z),



Ym(z) = γz−kBr

φclR(z)


• γ - chosen so that at steady state Ym = R:



G(z) = z−kB(z)

A(z),



Ym(z) = γz−kBr

φclR(z)


• γ - chosen so that at steady state Ym = R:

γ =φcl(1)

Br(1)Digital Control 1 Kannan M. Moudgalya, Autumn 2007


Look for a controller of the form,

Rc(z)U(z) = γTc(z)R(z)− Sc(z)Y (z)




Rc(z), Sc(z) and Tc(z) are polynomials in z−1,




Rc(z), Sc(z) and Tc(z) are polynomials in z−1,to be determined.




Rc(z), Sc(z) and Tc(z) are polynomials in z−1,to be determined.

yγ

Tc(z)

Rc(z)G = z−k

B(z)

A(z)

Sc(z)

Rc(z)

r u

−

Digital Control 2 Kannan M. Moudgalya, Autumn 2007


yγ

Tc(z)

Rc(z)G = z−k

B(z)

A(z)

Sc(z)

Rc(z)

r u

−


yγ

Tc(z)

Rc(z)G = z−k

B(z)

A(z)

Sc(z)

Rc(z)

r u

−

The controller has two components:


yγ

Tc(z)

Rc(z)G = z−k

B(z)

A(z)

Sc(z)

Rc(z)

r u

−


1. Feedback for internal stability.


yγ

Tc(z)

Rc(z)G = z−k

B(z)

A(z)

Sc(z)

Rc(z)

r u

−



2. Feedforward for Y to track R.


yγ

Tc(z)

Rc(z)G = z−k

B(z)

A(z)

Sc(z)

Rc(z)

r u

−



2. Feedforward for Y to track R.Known as the two degrees of freedom controller.



yγ

Tc(z)

Rc(z)G = z−k

B(z)

A(z)

Sc(z)

Rc(z)

r u

−


yγ

Tc(z)

Rc(z)G = z−k

B(z)

A(z)

Sc(z)

Rc(z)

r u

−

Y =


yγ

Tc(z)

Rc(z)G = z−k

B(z)

A(z)

Sc(z)

Rc(z)

r u

−

Y =z−kBA

1 + z−kBAScRc

γTc

RcR


yγ

Tc(z)

Rc(z)G = z−k

B(z)

A(z)

Sc(z)

Rc(z)

r u

−

Y =z−kBA

1 + z−kBAScRc

γTc

RcR

= γz−kBTc

ARc + z−kBScR



Y = γz−kBTc

ARc + z−kBScR


Y = γz−kBTc

ARc + z−kBScR

Want Y to behave like Ym:


Y = γz−kBTc

ARc + z−kBScR


Ym(z) = γz−kBr

φclR(z)


Y = γz−kBTc

ARc + z−kBScR


Ym(z) = γz−kBr

φclR(z)

Equate the two and cancel the common factors:


Y = γz−kBTc

ARc + z−kBScR


Ym(z) = γz−kBr

φclR(z)


BTc

ARc + z−kBSc=Br

φcl


Y = γz−kBTc

ARc + z−kBScR


Ym(z) = γz−kBr

φclR(z)


BTc

ARc + z−kBSc=Br

φcldeg Br < deg B



BTc

ARc + z−kBSc=Br

φcl


BTc

ARc + z−kBSc=Br

φclLet B = BgBb,


BTc

ARc + z−kBSc=Br

φclLet B = BgBb, A = AgAb


BTc

ARc + z−kBSc=Br


g: good, b: bad.


BTc

ARc + z−kBSc=Br


g: good, b: bad. Example: good ≡ roots inside unit circle.


BTc

ARc + z−kBSc=Br



Tc = AgT1,


BTc

ARc + z−kBSc=Br



Tc = AgT1, Rc = BgR1,


BTc

ARc + z−kBSc=Br



Tc = AgT1, Rc = BgR1, Sc = AgS1,


BTc

ARc + z−kBSc=Br




The last equation becomes

BbBgAgT1

AbAgBgR1 + z−kBbBgAgS1

=Br

φcl


BTc

ARc + z−kBSc=Br




The last equation becomes

BbBgAgT1

AbAgBgR1 + z−kBbBgAgS1

=Br

φclCancelling good common factors,

BbT1

AbR1 + z−kBbS1

=Br

φclDigital Control 6 Kannan M. Moudgalya, Autumn 2007

7. Aryabhatta’s Identity


Recall from the previous slide:

BbT1

AbR1 + z−kBbS1=Br

φcl



BbT1

AbR1 + z−kBbS1=Br

φcl

Equate numerator and denominator:

Br = BbT1



BbT1

AbR1 + z−kBbS1=Br

φcl


Br = BbT1

AbR1 + z−kBbS1 = φcl



BbT1

AbR1 + z−kBbS1=Br

φcl


Br = BbT1


• Known as Aryabhatta’s Identity



BbT1

AbR1 + z−kBbS1=Br

φcl


Br = BbT1



• It can be solved for R1 and S1



BbT1

AbR1 + z−kBbS1=Br

φcl


Br = BbT1



• It can be solved for R1 and S1

• One choice for T1 is T1 = 1Digital Control 7 Kannan M. Moudgalya, Autumn 2007

8. 2-DOF: Closed Loop Transfer Function


yγ

Tc(z)

Rc(z)G = z−k

B(z)

A(z)

Sc(z)

Rc(z)

r u

−


yγ

Tc(z)

Rc(z)G = z−k

B(z)

A(z)

Sc(z)

Rc(z)

r u

−

Desired closed loop relation:


yγ

Tc(z)

Rc(z)G = z−k

B(z)

A(z)

Sc(z)

Rc(z)

r u

−


Ym(z) = γz−kBr

φclR(z)


yγ

Tc(z)

Rc(z)G = z−k

B(z)

A(z)

Sc(z)

Rc(z)

r u

−


Ym(z) = γz−kBr

φclR(z) = γz−k

BbT1

φclR(z)


yγ

Tc(z)

Rc(z)G = z−k

B(z)

A(z)

Sc(z)

Rc(z)

r u

−


Ym(z) = γz−kBr

φclR(z) = γz−k

BbT1

φclR(z)

• If Bb 6= 1,


yγ

Tc(z)

Rc(z)G = z−k

B(z)

A(z)

Sc(z)

Rc(z)

r u

−


Ym(z) = γz−kBr

φclR(z) = γz−k

BbT1

φclR(z)

• If Bb 6= 1, it will appear as a part of closed looptransfer function.


yγ

Tc(z)

Rc(z)G = z−k

B(z)

A(z)

Sc(z)

Rc(z)

r u

−


Ym(z) = γz−kBr

φclR(z) = γz−k

BbT1

φclR(z)

• If Bb 6= 1, it will appear as a part of closed looptransfer function.

• That is, bad zeros of original transfer function can-not be changed by feedback.


9. Recall: Magnetically Suspended Ball

−

+

g

R L

V

M

hi

• Current through coil induces magnetic force

•Magnetic force balances gravity

• Ball is suspended in midair - 1 cm from core

•Want to move to another equilibrium


10. Magnetically Suspended Ball - Ctd.

−

+

g

R L

V

M

hi

Force balance:

Md2h

dt2= Mg − Ki

2

h

Voltage balance

V = Ldi

dt+ iR


11. Magnetically Suspended Ball - Ctd.

In matrix form:

d

dt

x1x2x3

=

0 1 0

K

M

i2sh2s

0 −2K

M

is

hs

0 0 −RL

x1x2x3

+

001L

uwhere,

x14= ∆h x2

4= ∆h

x34= ∆i u

4= ∆V


12. Assign Values to Magnetically Suspended Ball

d

dt

x1

x2

x3

=

0 1 0

K

M

i2sh2s

0 −2K

M

is

hs

0 0 −RL

x1

x2

x3

+

001L

uM Mass of ball 0.05 KgL Inductance 0.01 HR Resistance 1 ΩK Coefficient 0.0001g Acceleration due to gravity 9.81 m/s2

hs Equilibrium Distance 0.01 mis Current at equilibrium 7A


13. Model of Magnetically Suspended Ball


d

dt

x1x2x3

=

0 1 0981 0 −2.8010 0 −100

x1x2x3

+

00

100

u


d

dt

x1x2x3

=

0 1 0981 0 −2.8010 0 −100

x1x2x3

+

00

100

uy = Cx,


d

dt

x1x2x3

=

0 1 0981 0 −2.8010 0 −100

x1x2x3

+

00

100

uy = Cx,

C =[1 0 0

]


d

dt

x1x2x3

=

0 1 0981 0 −2.8010 0 −100

x1x2x3

+

00

100

uy = Cx,

C =[1 0 0

]x14= ∆h, x2

4= ∆h, x3

4= ∆i

Only displacement is measured.


d

dt

x1x2x3

=

0 1 0981 0 −2.8010 0 −100

x1x2x3

+

00

100

uy = Cx,

C =[1 0 0

]x14= ∆h, x2

4= ∆h, x3

4= ∆i

Only displacement is measured. G = C(sI−A)−1B:


d

dt

x1x2x3

=

0 1 0981 0 −2.8010 0 −100

x1x2x3

+

00

100

uy = Cx,

C =[1 0 0

]x14= ∆h, x2

4= ∆h, x3

4= ∆i

Only displacement is measured. G = C(sI−A)−1B:

G(s) =−280.14

s3 + 100s2 − 981s− 98100Digital Control 13 Kannan M. Moudgalya, Autumn 2007

14. Magnetically Suspended Ball - Discrete Model


• Sample continuous time model using Ts = 0.01s



• Using myc2d.sci,




G(z) =

z−1(−3.7209× 10−5 − 1.1873× 10−4z−1 − 2.2597× 10−5z−2)

1− 2.4668z−1 + 1.7721z−2 − 0.3679z−3




G(z) =

z−1(−3.7209× 10−5 − 1.1873× 10−4z−1 − 2.2597× 10−5z−2)

1− 2.4668z−1 + 1.7721z−2 − 0.3679z−3

= z−1−3.7209× 10−5(1 + 2.9877z−1)(1 + 0.2033z−1)

(1− 1.3678z−1)(1− 0.7311z−1)(1− 0.3679z−1)




G(z) =

z−1(−3.7209× 10−5 − 1.1873× 10−4z−1 − 2.2597× 10−5z−2)

1− 2.4668z−1 + 1.7721z−2 − 0.3679z−3

= z−1−3.7209× 10−5(1 + 2.9877z−1)(1 + 0.2033z−1)

(1− 1.3678z−1)(1− 0.7311z−1)(1− 0.3679z−1)

= z−kB

A

15. Magnetically Suspended Ball - Splitting



z−kB

A= z−1−3.7209× 10−5(1 + 2.9877z−1)(1 + 0.2033z−1)

(1− 1.3678z−1)(1− 0.7311z−1)(1− 0.3679z−1)



z−kB

A= z−1−3.7209× 10−5(1 + 2.9877z−1)(1 + 0.2033z−1)

(1− 1.3678z−1)(1− 0.7311z−1)(1− 0.3679z−1)

Split A and B into good and bad parts:



z−kB

A= z−1−3.7209× 10−5(1 + 2.9877z−1)(1 + 0.2033z−1)

(1− 1.3678z−1)(1− 0.7311z−1)(1− 0.3679z−1)


A = AgAb



z−kB

A= z−1−3.7209× 10−5(1 + 2.9877z−1)(1 + 0.2033z−1)

(1− 1.3678z−1)(1− 0.7311z−1)(1− 0.3679z−1)


A = AgAb

Ag = (1− 0.7311z−1)(1− 0.3679z−1)

Ab = (1− 1.3678z−1)



z−kB

A= z−1−3.7209× 10−5(1 + 2.9877z−1)(1 + 0.2033z−1)

(1− 1.3678z−1)(1− 0.7311z−1)(1− 0.3679z−1)


A = AgAb

Ag = (1− 0.7311z−1)(1− 0.3679z−1)

Ab = (1− 1.3678z−1)

Bg = −3.7209× 10−5(1 + 0.2033z−1)

Bb = (1 + 2.9877z−1)



z−kB

A= z−1−3.7209× 10−5(1 + 2.9877z−1)(1 + 0.2033z−1)

(1− 1.3678z−1)(1− 0.7311z−1)(1− 0.3679z−1)


A = AgAb

Ag = (1− 0.7311z−1)(1− 0.3679z−1)

Ab = (1− 1.3678z−1)

Bg = −3.7209× 10−5(1 + 0.2033z−1)

Bb = (1 + 2.9877z−1)

• Good parts have their roots inside the unit circle



z−kB

A= z−1−3.7209× 10−5(1 + 2.9877z−1)(1 + 0.2033z−1)

(1− 1.3678z−1)(1− 0.7311z−1)(1− 0.3679z−1)


A = AgAb

Ag = (1− 0.7311z−1)(1− 0.3679z−1)

Ab = (1− 1.3678z−1)

Bg = −3.7209× 10−5(1 + 0.2033z−1)

Bb = (1 + 2.9877z−1)

• Good parts have their roots inside the unit circle

• We also have k = 1


16. Recall: Small Fall Time in Error

e(n) = ρn cosωn, 0 < ρ < 1

Error becomes zero, i.e.,

e(n) = 0.

for the first time when

ωn =π

2

⇒ n =π

2ω

As want n < Nr, some given value, we get

π

2ω< Nr ⇒ ω >

π

2Nr

n

e(n)

n

y(n)

Desired region:Im(z)

Re(z)


17. Recall: Small Undershoot

e(n) = ρn cosωn

When does it reach first min.? de/dn = 0

• Not applicable, because n is an integer

• Look for a simpler expression

• Reaches min. approx. when ωn = π

e(n)|ωn=π = ρn cosωn|ωn=π

= −ρn|ωn=π = −ρπ/ω

User specified maximum deviation = ε:

ρπ/ω < ε, ρ < εω/π

n

e(n)

n

y(n)

Desired region:Im(z)

Re(z)


18. Determination of φcl


Ts = 0.01.


Ts = 0.01. Transient specifications for a step input:



Steady state error ≤ 2%,




Overshoot = ε ≤ 5%





Settling time ≤ 0.5s






Guess






Guess

Rise time ≤ 0.15s






Guess

Rise time ≤ 0.15s

N ≤ rise time/Ts






Guess

Rise time ≤ 0.15s

N ≤ rise time/Ts = 15






Guess

Rise time ≤ 0.15s


Choose






Guess

Rise time ≤ 0.15s


Choose

N = 15






Guess

Rise time ≤ 0.15s


Choose

N = 15

ω =π

2N






Guess

Rise time ≤ 0.15s


Choose

N = 15

ω =π

2N=π

30






Guess

Rise time ≤ 0.15s


Choose

N = 15

ω =π

2N=π

30= 0.1047



Overshoot constraint:



ρ ≤ εω/π



ρ ≤ εω/π = 0.050.1047/π



ρ ≤ εω/π = 0.050.1047/π = 0.905



ρ ≤ εω/π = 0.050.1047/π = 0.905

Choose



ρ ≤ εω/π = 0.050.1047/π = 0.905

Choose

ρ = 0.905



ρ ≤ εω/π = 0.050.1047/π = 0.905

Choose

ρ = 0.905

• Desired closed loop poles are at ρe±jω



ρ ≤ εω/π = 0.050.1047/π = 0.905

Choose

ρ = 0.905


• From the previous slide, ω = 0.1047



ρ ≤ εω/π = 0.050.1047/π = 0.905

Choose

ρ = 0.905



φcl(z) = (z − ρejω)(z − ρe−jω)



ρ ≤ εω/π = 0.050.1047/π = 0.905

Choose

ρ = 0.905




= 1− 2z−1ρ cosω + ρ2z−2



ρ ≤ εω/π = 0.050.1047/π = 0.905

Choose

ρ = 0.905




= 1− 2z−1ρ cosω + ρ2z−2

= 1− 1.8z−1 + 0.819z−2


20. Controller Design


Ag = (1− 0.7311z−1)(1− 0.3679z−1)

Ab = (1− 1.3678z−1)

Bg = −3.7209× 10−5(1 + 0.2033z−1)

Bb = (1 + 2.9877z−1)

k = 1


Ag = (1− 0.7311z−1)(1− 0.3679z−1)

Ab = (1− 1.3678z−1)

Bg = −3.7209× 10−5(1 + 0.2033z−1)

Bb = (1 + 2.9877z−1)

k = 1

φcl = 1− 1.8z−1 + 0.819z−2


Ag = (1− 0.7311z−1)(1− 0.3679z−1)

Ab = (1− 1.3678z−1)

Bg = −3.7209× 10−5(1 + 0.2033z−1)

Bb = (1 + 2.9877z−1)

k = 1

φcl = 1− 1.8z−1 + 0.819z−2

Want to solve Aryabhatta’s identity: AbR1+z−kBbS1 = φcl.


Ag = (1− 0.7311z−1)(1− 0.3679z−1)

Ab = (1− 1.3678z−1)

Bg = −3.7209× 10−5(1 + 0.2033z−1)

Bb = (1 + 2.9877z−1)

k = 1

φcl = 1− 1.8z−1 + 0.819z−2


(1− 1.3678z−1)R1 + z−1(1 + 2.9877z−1)S1

= 1− 1.8z−1 + 0.819z−2


Ag = (1− 0.7311z−1)(1− 0.3679z−1)

Ab = (1− 1.3678z−1)

Bg = −3.7209× 10−5(1 + 0.2033z−1)

Bb = (1 + 2.9877z−1)

k = 1

φcl = 1− 1.8z−1 + 0.819z−2


(1− 1.3678z−1)R1 + z−1(1 + 2.9877z−1)S1

= 1− 1.8z−1 + 0.819z−2

Using xdync.sci,


Ag = (1− 0.7311z−1)(1− 0.3679z−1)

Ab = (1− 1.3678z−1)

Bg = −3.7209× 10−5(1 + 0.2033z−1)

Bb = (1 + 2.9877z−1)

k = 1

φcl = 1− 1.8z−1 + 0.819z−2


(1− 1.3678z−1)R1 + z−1(1 + 2.9877z−1)S1

= 1− 1.8z−1 + 0.819z−2

Using xdync.sci, we get S1 = 0.0523, R1 = 1− 0.4845z−1


21. ball basic.sce

21. ball basic.sce

1 / / U p d a t e d ( 5 − 8 − 0 7 )

2 / / 9 . 1

3

4 / / M a g n e t i c a l l y s u s p e n d e d b a l l p r o b l e m

5 / / O p e r a t i n g c o n d i t i o n s

6

7 M = 0 . 0 5 ; L = 0 . 0 1 ; R = 1 ; K = 0 . 0001 ; g = 9 . 8 1 ;8

9 / / E q u i l i b r i u m c o n d i t i o n s

10 hs = 0 . 0 1 ; i s = sqrt (M∗g∗hs /K) ;11

12 / / S t a t e s p a c e m a t r i c e s

13 a21 = K∗ i s ˆ2/M/hs ˆ2 ; a23 = − 2∗K∗ i s /M/hs ; a33 = − R/L ;14 b3 = 1/L ;15 a1 = [0 1 0 ; a21 0 a23 ; 0 0 a33 ] ;16 b1 = [ 0 ; 0 ; b3 ] ; c1 = [1 0 0 ] ; d1 = 0 ;17

18 / / T r a n s f e r f u n c t i o n s

19 G = s y s l i n ( ’ c ’ , a1 , b1 , c1 , d1 ) ; Ts = 0 . 0 1 ;Digital Control 21 Kannan M. Moudgalya, Autumn 2007

20 [B ,A , k ] = myc2d (G, Ts ) ;21

22 / / p o l y n o m i a l s a r e r e t u r n e d

23 [ Ds , num , den ] = s s2 t f (G ) ;24 num = clean (num ) ; den = clean ( den ) ;25

26 / / T r a n s i e n t s p e c i f i c a t i o n s

27 r i s e = 0 . 1 5 ; e p s i l o n = 0 . 0 5 ;28 ph i = d e s i r e d (Ts , r i s e , e p s i l o n ) ;29

30 / / C o n t r o l l e r d e s i g n

31 [ Rc , Sc , Tc ,gamm] = pp b a s i c (B,A , k , ph i ) ;32

33 / / S e t t i n g u p s i m u l a t i o n p a r a m e t e r s f o r b a s i c . c o s

34 s t = 0 . 0 001 ; / / d e s i r e d c h a n g e i n h , i n m .

35 t i n i t = 0 ; / / s i m u l a t i o n s t a r t t i m e

36 t f i n a l = 0 . 5 ; / / s i m u l a t i o n e n d t i m e

37

38 / / S e t t i n g u p s i m u l a t i o n p a r a m e t e r s f o r c s s c l . c o s

39 N var = 0 ; x I n i t i a l = [0 0 0 ] ; N = 1 ; C = 0 ; D = 1 ;40

41 [ Tc1 , Rc1 ] = c o s f i l i p (Tc , Rc ) ; / / T c / R c


42 [ Sc2 , Rc2 ] = c o s f i l i p ( Sc , Rc ) ; / / S c / R c

43

44 [ Tcp1 , Tcp2 ] = c o s f i l i p (Tc , 1 ) ; / / T c / 1

45 [ Np , Rcp ] = c o s f i l i p (N, Rc ) ; / / 1 / R c

46 [ Scp1 , Scp2 ] = c o s f i l i p ( Sc , 1 ) ; / / S c / 1

47 [ Cp ,Dp ] = c o s f i l i p (C ,D) ; / / C / D


22. desired.sci

22. desired.sci

1 / / U p d a t e d ( 2 6 − 7 − 0 7 )

2 / / 9 . 4

3 funct ion [ ph i , dph i ] = d e s i r e d (Ts , r i s e , e p s i l o n )4

5 Nr = r i s e /Ts ; omega = %pi/2/Nr ; rho = e p s i l o n ˆ( omega/%pi ) ;6 ph i = [1 −2∗rho ∗cos ( omega ) rho ˆ 2 ] ; dph i = length ( ph i )−1;7 endfunction ;


23. pp basic.sci

23. pp basic.sci

1 / / U p d a t e d ( 2 6 − 7 − 0 7 )

2 / / 9 . 5

3 / / f u n c t i o n [ R c , S c , T c , g a m m a ] = p p b a s i c ( B , A , k , p h i )

4 / / c a l c u l a t e s p o l e p l a c e m e n t c o n t r o l l e r

5 / / −−−−−−

6

7 funct ion [ Rc , Sc , Tc ,gamm] = pp b a s i c (B,A, k , ph i )8

9 / / S e t t i n g u p a n d s o l v i n g A r y a b h a t t a i d e n t i t y

10 pause11 [ Ag , Ab ] = p o l s p l i t 2 (A ) ; dAb = length (Ab) − 1 ;12 \pause13 [ Bg , Bb ] = p o l s p l i t 2 (B ) ; dBb = length (Bb) − 1 ;14 pause15

16 [ zk , dzk ] = zpowk ( k ) ;17

18 [N, dN ] = po lmul (Bb , dBb , zk , dzk ) ;19 dph i = length ( ph i ) − 1 ;20


21 [ S1 , dS1 , R1 , dR1 ] = xdync (N, dN , Ab , dAb , phi , dph i ) ;22 pause23

24 / / D e t e r m i n a t i o n o f c o n t r o l l a w

25 Rc = convol (Bg , R1 ) ; Sc = convol (Ag , S1 ) ;26 Tc = Ag ; gamm = sum( ph i )/sum(Bb ) ;27

28 endfunction ;


24. Controller Performance


0

2e-05

4e-05

6e-05

8e-05

0.0001

0.00012

0 0.1 0.2 0.3 0.4 0.5

y(n)

t(s)

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0 0.1 0.2 0.3 0.4 0.5

u(n)

t(s)


0

2e-05

4e-05

6e-05

8e-05

0.0001

0.00012

0 0.1 0.2 0.3 0.4 0.5

y(n)

t(s)

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0 0.1 0.2 0.3 0.4 0.5

u(n)

t(s)

• Overshoot constraint, settling time constraints are met


0

2e-05

4e-05

6e-05

8e-05

0.0001

0.00012

0 0.1 0.2 0.3 0.4 0.5

y(n)

t(s)

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0 0.1 0.2 0.3 0.4 0.5

u(n)

t(s)


• Settling time constraint also is met


0

2e-05

4e-05

6e-05

8e-05

0.0001

0.00012

0 0.1 0.2 0.3 0.4 0.5

y(n)

t(s)

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0 0.1 0.2 0.3 0.4 0.5

u(n)

t(s)



• The auxiliary condition on rise time is not met, however


0

2e-05

4e-05

6e-05

8e-05

0.0001

0.00012

0 0.1 0.2 0.3 0.4 0.5

y(n)

t(s)

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0 0.1 0.2 0.3 0.4 0.5

u(n)

t(s)



• The auxiliary condition on rise time is not met, however

• Try reducing the required rise timeDigital Control 27 Kannan M. Moudgalya, Autumn 2007

25. Controller Design with Rise Time = 0.1s


Ag = (1− 0.7311z−1)(1− 0.3679z−1)

Ab = (1− 1.3678z−1)

Bg = −3.7209× 10−5(1 + 0.2033z−1)

Bb = (1 + 2.9877z−1)

k = 1


Ag = (1− 0.7311z−1)(1− 0.3679z−1)

Ab = (1− 1.3678z−1)

Bg = −3.7209× 10−5(1 + 0.2033z−1)

Bb = (1 + 2.9877z−1)

k = 1

ε = 0.05

r = 0.8609

ω = 0.1571


Ag = (1− 0.7311z−1)(1− 0.3679z−1)

Ab = (1− 1.3678z−1)

Bg = −3.7209× 10−5(1 + 0.2033z−1)

Bb = (1 + 2.9877z−1)

k = 1

ε = 0.05

r = 0.8609

ω = 0.1571

φcl = 1− 1.7006z−1 + 0.7411z−2


Ag = (1− 0.7311z−1)(1− 0.3679z−1)

Ab = (1− 1.3678z−1)

Bg = −3.7209× 10−5(1 + 0.2033z−1)

Bb = (1 + 2.9877z−1)

k = 1

ε = 0.05

r = 0.8609

ω = 0.1571

φcl = 1− 1.7006z−1 + 0.7411z−2

R1 = 1− 0.3984z−1

S1 = 0.0657Digital Control 28 Kannan M. Moudgalya, Autumn 2007


0

2e-05

4e-05

6e-05

8e-05

0.0001

0.00012

0 0.1 0.2 0.3 0.4 0.5

y(n)

t(s)

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0 0.1 0.2 0.3 0.4 0.5

u(n)

t(s)

All requirements have been met


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