15. TF analyze of Pole-placement Method

1. Pole placement via TF Approach for SISO Linear Systems

Given a system

\[ G(s) = \frac{B(s)}{A(s)} = \frac{b_ns^{n-1}+b_{n-1}s^{n-2}+\dots + b_1s+b_0}{s^{n}+a_{n-1}s^{n-1}+\dots + a_1s+a_0} \]

We want to design the regulator \(R(s)\), given the schematic:

To build the state realization, we need to apply the pole-placement method within a RO.

For \(R(s)\), it have \(n-1\) poles:

\[ R(s) = \frac{F(s)}{\Gamma(s)} = \frac{f_{n-1}s^{n-1}+f_{n-2}s^{n-2}+\dots + f_1s+f_0}{\gamma_{n-1}s^{n-1}+\gamma_{n-2}s^{n-2}+\dots +\gamma_1s+\gamma_0} \]

And we can design a desired characteristic polynomial \(P^*(s)\) for the close loop TF,

\[ P(s) = s^{2n-1} + p_{2n-2}s^{2n-2} + \dots + p_1s + p_0 \]

The characteristic polynomial of the closed loop system is:

\[ P(s) = \Gamma(s)A(s) + B(s)F(s) \]

We can set the system desired polynomial function the same as our desired one:

\[ P(s) = P^*(s) \]

Expend it, we have:

\[ \begin{aligned} \begin{bmatrix} \begin{matrix} \begin{bmatrix} 1 \\ a_{n-1} \\ \vdots \\ a_0 \end{bmatrix} \\ 0 \\ 0 \\ \vdots \\ 0 \end{matrix} & \begin{matrix} 0 \\ \begin{bmatrix} 1 \\ a_{n-1} \\ \vdots \\ a_0 \end{bmatrix} \\ 0 \\ \vdots \\ 0 \end{matrix} & \cdots & \begin{matrix} 0 \\ 0 \\ \vdots \\ 0 \\ \begin{bmatrix} 1 \\ a_{n-1} \\ \vdots \\ a_0 \end{bmatrix} \end{matrix} \begin{matrix} \begin{bmatrix} 0 \\ b_{n-1} \\ \vdots \\ b_0 \end{bmatrix} \\ 0 \\ 0 \\ \vdots \\ 0 \end{matrix} & \cdots & \begin{matrix} 0 \\ 0 \\ \vdots \\ 0 \\ \begin{bmatrix} 0 \\ b_{n-1} \\ \vdots \\ b_0 \end{bmatrix} \end{matrix} \end{bmatrix} \begin{bmatrix} \gamma_{n-1} \\ \vdots \\ \gamma_0 \\ f_{n-1} \\ \vdots \\ f_0\end{bmatrix} = \begin{bmatrix} 1 \\ p_{2n-2} \\ p_{2n-3} \\ \vdots \\ p_0 \end{bmatrix} \end{aligned} \]

1.1 Regulator with Integral Actions

For the regulator with integral actions, we can treat it as one part of the controller or the part of the system:

Move the integrator inside the system, thus:

\[ \tilde G(s) = \frac{1}{s}G{s} = \frac{B(s)}{sA(s)} \]

\(\tilde G(s)\) have the order of \(n+1\), And we want to design \(R(s)\) of order \(n\),

\[ R(s) = \frac{F'(s)}{\Gamma'(s)} \]

The desired polynomial TF \(P^*(s)\) have a order of \(2n+1\), the characteristic polynomial function of the system is:

\[ P(s) = P(s) = \Gamma'(s)\underbrace{sA(s)}_{\tilde A(s)} + B(s)F'(s) = P^*(s) \]

1.2 Zeros of the Close-Loop System

For the close-loop system, there have:

To calculate the zeros, we need to calculate the complementary sensitive functions

Warning

Because in the previous part, we already use the \(F(s)\), which is not the complementary sensitivity function, so here we rename it to \(H(s)\) to resolve the naming conflict

\[ H(s) = \frac{\frac{H(s)B(s)}{\Gamma(s)A(s)}}{1 + \frac{H(s)B(s)}{\Gamma(s)A(s)}} = \frac{F(s)B(s)}{\Gamma(s)A(s) + F(s)B(s)} = \frac{F(s)B(s)}{P^*(s)} \]

The chosen \(\Delta\) should be the order of \(n-1\), \(\Delta\) is stable polynomial:

\[ \Delta(s) = s^{n-1} \delta_{n-2}s^{n-2}+\dots +\delta_1s + \delta_0 \]

And the loop transfer function is:

\[ L(s) = \frac{F(s)}{\Gamma(s)} \frac{B(s)}{A(s)} \]

Calculate the TF from \(y^\circ\) to \(y\):

\[ W(s) = \frac{F(0)}{\Delta(0)} \frac{\frac{\Delta(s)}{\Gamma(s)}\frac{B(s)}{A(s)}}{1+\frac{F(s)B(s)}{\Gamma(s)A(s)}} = \frac{F(0)}{\Delta(0)}\frac{\Delta(s)B(s)}{P^*(s)} \]

1.3 How to cancel the poles in the system

Given the system with negative poles:

\[ G(s) = \frac{B(s)}{\underbrace{(s+a)A'(s)}_{A(s)}} \]

We can give the control schematic:

The desired polynomial function is:

\[ P^*(s) = \underbrace{(s+a)\tilde P^*(s)}_{2n-1} \]

For the system polynomial function:

\[ \begin{aligned} P(s) &= (s+a)A'(s)\Gamma(s) + B(s)F(s) = (s+a)\tilde P^*(s) \\ B(s)F(s) &= (s+a)[\tilde P^*(s) - A'(s)] \end{aligned} \]

For the condition that \(F(s)\) has a root in \(a\), which \(F(s) = (s+a)F'(s)\),

\[ L(s) = R(s)G(s) = \frac{(s+a)F'(s)}{\Gamma(s)} \frac{B(s)}{(s+a)A'(s)} \]

For the system disturbance \(d_u \to y\), there have the TF:

\[ \begin{aligned} V(s) &= \frac{G(s)}{1+R(s)G(s)} \\ &= \frac{\frac{B(s)}{(s+a)A'(s)}}{1+\frac{B(s)}{(s+a)A'(s)} \frac{(s+a)F'(s)}{\Gamma(s)}} \\ &= \frac{A'(s)\Gamma(s)B(s)}{\underbrace{\underbrace{(A'(s)\Gamma(s)+B(s)F'(s))}_{\tilde P^*(s)}(s+a)A'(s)}_{P^*(s)}} \end{aligned} \]

Apply the step signal in the input, with the disturbance inserted in some time after the system stabilizing,

We can get that after inserting the disturbance, the system is stabilizing with oscillation. The system have 2 conjugate poles in the negative plane.

Example

Given the system:

\[ G(s) = \frac{1}{s-1} \]

Design a regulator which include an integral action such that all the close loop poles are \(-1\)

1. State space approach

   We can separate the system into different parts to reconstruct the system design:

   

   \[ \begin{aligned} \dot x &= x + u \\ y &= x \dot v= -x \end{aligned} \]
   \[ \begin{bmatrix} \dot x \\ \dot v \end{bmatrix} = \underbrace{\begin{bmatrix} 1&0 \\ -1&0 \end{bmatrix}}_{\tilde A} \begin{bmatrix} x \\ v \end{bmatrix} + \underbrace{\begin{bmatrix} 1 \\ 0 \end{bmatrix}}_{\tilde B} u \]
   \[ u = -K \begin{bmatrix} x \\ v \end{bmatrix} = -K_xx - K_vv \]

   The closed loop system is:

   \[ \begin{bmatrix} \dot x \\ \dot v \end{bmatrix} = \underbrace{(A-BK)}_{\hat L} \begin{bmatrix} x \\ v \end{bmatrix} \]

   The characteristic polynomial is:

   \[ P_{\tilde A - \tilde BK}(\lambda) = \det \begin{bmatrix} \lambda - 1+ K_x & K_v \\ 1&\lambda \end{bmatrix} = \lambda^2 + (K_x - 1)\lambda - K_v \]

   And the desired polynomial function is:

   \[ P^*(\lambda) = (\lambda + 1)^2 = \lambda^2 + 2\lambda + 1 \]

   And the result is:

   \[ \begin{aligned} K_x &= 3 \\ K_v &= -1 \end{aligned} \]

2. Pole placement with TF

   

   \[ \begin{aligned} &\tilde G(s) = \frac{1}{s(s-1)} & R'(s) = \frac{f_1s + f_0}{\gamma_1s + \gamma_0} \end{aligned} \]
   \[ \begin{aligned} P(s) &= (s^2 - s)(\gamma_1s + \gamma_0) + f_1s + f_0 \\ P^*(s) &= (s+1)^3 = s^3 + 3s^2 + 3s + 1 \end{aligned} \]
   \[ \begin{bmatrix} \begin{matrix} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} \\ 0 \end{matrix} & \begin{matrix} 0 \\ \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} \end{matrix} & \begin{matrix} \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \\ 0 \end{matrix} & \begin{matrix} 0 \\ \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \end{matrix} \end{bmatrix} \begin{bmatrix} \gamma_1 \\ \gamma_0 \\ f_1 \\ f_0 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \\ 3 \\ 1 \end{bmatrix} \]
   \[ \begin{aligned} &\gamma_1 = 1 & f_1 = 7 \\ &\gamma_2 = 4 & f_0 = 1 \end{aligned} \]
   \[ R'(s) = \frac{f_1s+f_0}{\gamma_1s+\gamma_0} = \frac{7(s+\frac17)}{s+4} \]

3. \(R(s) = p\frac{s+\alpha}{s}\)

   \[ \begin{aligned} 1 + R(s)G(s) &= 0 \\ s(s-1) + p(s+\alpha) &= (s+1)^2 \end{aligned} \]

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