1. Pole Placement

For a input-output model

Plant: frac{B(z)}{A(z)}

Feedback controller: frac{S(z)}{R(z)}

Feedforward controller: frac{T(z)}{R(z)}

For the close loop:,

We have close loop t.f.

H_{cl}=frac{RB}{AR+SB}frac{T}{R}=frac{TB}{AR+SB}

We have the c.p.

A_{cl}(z)=A_m(z)A_o(z)=A(z)R(z)+S(z)B(z)

If the order of the plant is n,

we will have the following in order to tune all parameters

Deg space A_{cl}=2n-1\ Deg space S=n-1\ Deg space R=n-1

2.Reduce Order by Pole-Zero Cancellation

If we want to cancel out stable zeros, we can choose

A_{cl}=ar{A}_{cl}B^+(z)\

Then

H_{cl}=frac{TB^{+}B^{-}}{ar A_{cl}B^{+}}=frac{TB^{-}}{ar A_{cl}}

But we cannot cancel unstable zeros.

and we can also cancel stable poles by choosing

T=ar TA^+

and

A_{cl}=ar{A}_{cl}A^+B^+(z)\

and we can also cancel them at the same time.

and the equation becomes

A^+A^-R+SB^+B^-=A_{cl}=ar{A}_{cl}A^+B^+(z)

in order to cancel A^+B^+

we need to choose

R=B^+ar R\ S=A^+ar S\ T=ar TA^+\

Then we have

A^-ar R+ar SB^-=ar{A}_{cl}

The order of the equation is becoming less.

3. Disturbance Rejection

we find the transfer function between disturbance and output is

H_{v}=frac{Y(z)}{V(z)}=frac{frac{B}{A}}{1+frac{SB}{RA}}=frac{RB}{AR+SB}

if the disturbance is constant, we just need $H_v(1)=0 quad R(1)=0$.

therefore, we just need

R(z)=(z-1)R

Tracking Problem

B(z) is stable, i.e. the system has stable inverse.

4. Design Steps

First build feedback controller and then build feedforward controller.

how to build feedback controller

1. Design R, if zero cancelled, R needs to contain zero. if disturbance is cancelled, R needs to contain z-1.

2. Design R and S by solving AR+BS=A_{cl}

3. Choose T or $H_{ff}$ at the final stage.

5. Example

The Plant

G(z)=frac{0.00058}{z-0.942}

The Desired Model

H_d(z)=frac{0.0952z+0.0755}{z^2-1.331z+0.5016}

We need disturbance rejection

R=z-1\ S=s_0z+s_1\ A_{cl}=z^2-1.331z+0.5016\ AR+SB=(z-0.942)(z-1)+(s_0z+s_1)(0.00058)=z^2-1.331z+0.5016

Then we have

egin{cases} s_0=1053.448\ s_1=-759.31 end{cases}

Then we have

R=z-1\ S=1053.448z-759.31

Then the feedback system is

H_{fb}=frac{(z-1)0.00058}{z^2-1.331z+0.5016}

Now we will design the feedforward controller,

H_{fb}H_{ff}=H_d\ H_{ff}=frac{0.0952z+0.0755}{(z-1)0.00058}=frac{16.41z+130.17}{z-1}

Therefore the controller is

U(z)=-H_{fb}(z)Y(z)+H_{ff}(z)U_c(z)\ H_{fb}=frac{1053.448z-759.31}{z-1}\ H_{ff}=frac{16.41z+130.17}{z-1}

Example

Plant

H(z)=frac{z+0.9}{z^2-2.5z+1}

The desired

A_m(z)=z^2-1. 8z+0.9

a) The process zero is canceled

The order of the plant is 2.

R=(z+0.9)\ S=s_0z+s_1\ A_{cl}=(z+0.9)(z^2-1.8z+0.9)

Then we have Diophantine Equation

AR+SB=(z^2-2.5z+1)(z+0.9)+(s_0z+s_1)(z+0.9)\ =A_{cl}=(z+0.9)(z^2-1.8z+0.9)\

Then we have

egin{cases} s_0=0.7\ s_1=-0.1\ end{cases}

Then the close loop system is

H_{ff}=frac{T}{R} \H_{cl}=frac{RB}{AR+SB}H_{ff}=frac{TB}{AR+SB}=frac{T}{z^2-1.8z+0.9}

We know

H_{cl}(1)=1

Then we have

T=0.1

Therefore

U(z)=-frac{S}{R}Y(z)+frac{T}{R}U(z)=-frac{0.7z-0.1}{z+0.9}Y(z)+frac{0.1}{z+0.9}U_c(z)

b) The process zero is not canceled

A_{cl}=zA_m=z^3-1.8z^2+0.9z\ S=s_0z+s_1\ R=z+r_1

Then we have Diophantine Equation

AR+SB=A_{cl}\ (z^2-2.5z+1)(z+r_1)+(s_0z+s_1)(z+0.9)=z^3-1.8z^2+0.9z

Then we have

egin{cases} r_1=0.1618\ s_0=0.5382\ s_1=-0.1798 end{cases}

Then we design the feedforward controller.

H_{cl}=frac{TB}{AR+SB}\ H_{cl}(1)=1\ T=0.05263\

Therefore, the controller is

U(z)=-frac{0.5382z-0.1798}{z+0.1618}Y(z)+frac{0.05263}{z+0.1618}U_c(z)


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