1. Stability

Jurys Stability Test

For the system

frac{1}{z^2+a_1z+a_2}

Using Jurys Stability Test

egin{array}{ccc} 1 & a_1 & a_2\ a_2 & a_1 & 1 & imes -a_2\ hline 1-a_2^2 & a_1-a_1a_2\ a_1-a_1a_2 & 1-a_2^2 & & imes-frac{a_1}{1+a_2}\ hline frac{(1-a_2)[(1+a_2)^2-a_1^2]}{1+a_2} end{array}

If the system is stable, we have

1-a_2^2 >0\ frac{(1-a_2)[(1+a_2)^2-a_1^2]}{1+a_2}>0

2. Controllability

Controllable: The system can be driven from any initial state to any other destination in finite time.

For the system

x(k+1)=Phi x(k)+Gamma u(k)\ y(k)=Cx(k)\

We have the initial state $x(0)$

Then

x(1)=Phi x(0)+Gamma u(0)\ x(2)=Phi x(1)+Gamma u(1)=Phi(Phi x(0)+Gamma u(0))+Gamma u(1)\ =Phi^2 x(0)+PhiGamma u(0)+Gamma u(1)\ vdots\ x(n)=Phi^nx(0)+Phi^{n-1}Gamma u(0)+Phi^{n-2}Gamma u(1)+dotsm+Gamma u(n-1)

If the system can be driven to any state in finite steps, we just need to ensure

W_c=egin{bmatrix} Gamma & PhiGamma & Phi^2Gamma & dotsm & Phi^{n-1}Gamma end{bmatrix}

is NONSINGULAR.

If we cannot get to one state with n-1 steps, we cannot to get to one state with n steps.

Proof

A(lambda)=lambda^n+a_1lambda^{n-1}+a_2lambda^{n-2}+dotsm+ a_n \

Then we have

A(Phi)=Phi^n+a_1Phi^{n-1}+a_2Phi^{n-2}+dotsm+ a_nI \

Similarly,

We know Phi^n is linear combination of Phi^0, quad Phi,quaddotsmquadPhi^{n-1} .

Therefore, we still cannot get to arbitrary desired state.

Controllable Canonical Form

Phi= egin{bmatrix} -a_{n} & -a_{n-1} & dotsm & -a_{2} & -a_1\ 1 & 0 & dotsm & 0 & 0\ 0 & 1 & dotsm & 0 & 0\ vdots & vdots & ddots & vdots& vdots\ 0 & 0 & dotsm & 1 & 0\ end{bmatrix} Gamma= egin{bmatrix} 1\ 0\ 0\ vdots\ 0\ end{bmatrix} C= egin{bmatrix} b_n & b_{n-1} & dotsm & b_1 end{bmatrix}

3. Observability

Observer

hat x(k+1)=Phihat x(k)+ Gamma u(k)+K(y(k)-hat y(k))\ hat y=Chat x(k) \ e(k+1)=Phi e(k)-KCe(k)=(Phi-KC)e(k)

Observability Matrix

M_o^{-1}=egin{bmatrix} C\CPhi\CPhi^2\vdots\CPhi^{n-1} end{bmatrix}

Observable Canonical Form

Phi= egin{bmatrix} -a_{n} & 1 & 0 & dotsm & 0\ -a_{n-1} & 0 & 1 & dotsm & 0\ vdots & vdots & vdots & ddots & vdots\ -a_{2} & 0 & dotsm & 0 & 1\ -a_{1} & 0 & dotsm & 0 & 0\ end{bmatrix} Gamma= egin{bmatrix} b_n\ b_{n-1}\ b_{n-2}\ vdots\ b_1\ end{bmatrix} C= egin{bmatrix} 1 & 0 & 0 & dotsm & 0 end{bmatrix}

4. How to Create State Space From Difference Equation

Example

y(k+1)=-a_1y(k)-a_2y(k-1)-a_3y(k-2)+b_1u(k)+b_2u(k-1)+b_3u(k-2)\

First, we choose

x_1(k)=y(k)\

Then we have

x_1(k+1)=y(k+1)=-a_1x_1(k)-a_2y(k-1)-a_3y(k-2)+b_1u(k)+b_2u(k-1)+b_3u(k-2)\

Then we choose

x_2(k)=-a_2y(k-1)-a_3y(k-2)+b_2u(k-1)+b_3u(k-2)\ x_2(k+1)=-a_2y(k)-a_3y(k-1)+b_2u(k)+b_3u(k-1)\ x_3(k)=-a_3y(k-1)+b_3u(k-1)\ x_3(k+1)=-a_3y(k)+b_3u(k)\

Therefore, we have

x_1(k+1)=-a_1x_1+x_2(k)+b_1u(k)\ x_2(k+1)=-a_2x_1(k)+x_3(k)+b_2u(k)\ x_3(k+1)=-a_3x_1(k)+b_3u(k)\

Therefore,

Phi=egin{bmatrix} -a_1 & 1 & 0\ -a_2 & 0 & 1\ -a_3 & 0 & 0 end{bmatrix} Gamma=egin{bmatrix} b_1\b_2\b_3 end{bmatrix} C=egin{bmatrix} 1 & 0 & 0 end{bmatrix}\

5. Some Useful Conclusions

1. The minimal realization has the same order as that of the transfer function. If higher, it is called non-minimal realization.

2. The minimal realization can be always cast into the form of observable canonical form. Therefore, the minimal realization is always observable.

3. If there are common zeros and poles, uncontrollable, otherwise, controllable.

6. Example

Example 1

y(k+1)=3y(k)-2y(k-1)+u(k-1)+2u(k-2)\

a) What is the Transfer Function?

zY(z)=3Y(z)-2z^{-1}Y(z)+z^{-1}U(z)+2z^{-2}U(z)\ frac{Y(z)}{U(z)}=frac{z^{-1}+2z^{-2}}{z-3+2z^{-1}}=frac{z+2}{z^3-3z^2+2z}=frac{z+2}{z(z-1)(z-2)}

b) is it possible to realize the system such that it is observable but not controllable.

It is not controllable, therefore, it should have common zeros and poles.

G(z)=frac{(z+2)(z+1)}{z(z-1)(z-2)(z+1)}=frac{z^2+3z+2}{z^4-2z^3-z^2+2z}\

The system is observable, therefore, we realize the system in observable canonical form.

Phi= egin{bmatrix} 2 & 1 & 0 & 0\ 1 & 0 & 1 & 0\ -2 & 0 & 0 & 1\ 0 & 0 & 0 & 0\ end{bmatrix} Gamma= egin{bmatrix} 0\ 1\ 3\ 2\ end{bmatrix} C= egin{bmatrix} 1 & 0 & 0 & 0 end{bmatrix}\

M_C= egin{bmatrix} 0 & 1 & 5 & 13\ 1 & 3 & 3 & 3\ 3 & 2 & -2 & -10\ 2 & 0 & 0 & 0\ end{bmatrix} M_o= egin{bmatrix} 1 & 0 & 0 & 0 \ 2 & 1 & 0 & 0\ 5 & 2 & 1 & 0\ 10 & 5 & 2 & 1 end{bmatrix}\

Therefore, the system is observable but not controllable.

However, we need to check if this space state is that transfer function. It is easy to find that it is exactly the G(z) instead of the transfer function in the question. Therefore, it is **impossible** to realize this system in observable but not controllable form.

c) is it possible to realize the system such that it is controllable but not observable.

Same as the above, it is possible to realize.

d) is it possible to realize the system such that it is both controllable and observable.

Obviously, it is possible.

Example 2.

y(k+1)=3y(k-1)-2y(k-2)+u(k)-2u(k-1)+u(k-2)\

The questions are same as the above.

a)

zY(z)=3z^{-1}Y(z)-2z^{-2}Y(z)+U(z)-2z^{-1}U(z)+z^{-2}U(z)\ frac{Y(z)}{U(z)}=frac{z^2-2z+1}{z^3-3z-2}=frac{(z-1)^2}{(z-1)^2(z+2)}\

b) it is possible observable but not controllable

c) it is not possible controllable but not observable

d) it is not possible controllable and observable


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