Discrete Riccati equation in discrete $h-\infty$ controller design

I want to design an discrete H infinty controller， I get this reference

consider a discrete system with the state space representation like this $$x_k+1 = Ax_K + B1w_k + B2u_k,X_0 = 0$$ $$z_k = C1x_k + D12u_k$$ $$y_k = C2x_k + D21w_k$$ and for simplicity ,the following assumptions are made: $$D12^TD12>0$$, $$C1^TD12 =0$$$$B1D21^T =0$$,$$D21D21^T>0$$

there is a Riccati equation$$M =A^TMP^{-1}A +C1^TC1$$ , $$P = I+(B2(D12^T D12)^{-1}*B2^T-\gamma^{-2}B1B1^T)M$$

where $$A,B1,B2,C1,C2$$ are appropriate matrix , there is a problem, it does't like the normal discrete equation we can use dare to solve it,

$$A^TXA−X−A^TXB(B^TXB+R)^{−1}B^TXA+Q=0$$

I don't know how to solve this kind of riccati equation , besides how to choose the parameter $$\gamma$$, I konw how to solve the reccati equation mentioned above.l afraid that my presentation is not clearly, so I attached the image of the Riccati equation, l hope this will help you understand more clearly, enter image description hereenter image description hereenter image description here I alredy have a discrete system:

 clc;clear;

A = [0 0 -0.06;
1  0  0.25;
0  1    1];

B1 = [0; 0; 1]
B2 = [0; 0; 1];

C1 = [sqrt(2) 0 0;
0    1   0;
0    0   1;
0    0  0];

D12 = [0 0 0 1];

C2 = [0 0 1],

D21 =0;

OB = rank(obsv(A,C2))

CO = rank(ctrb(A,B2))


so the riccati equation can be simplified with $$D12^TD12 =I$$: $$M =A^TM(I+(B2B2^T-\gamma^{-2}B1B1^T)M)^{-1}A +C1^TC1$$ I am learning discrete H infinty control theory, and stuck with this question for a while,l already konw there are other approachs to deal with output feedback discrete H infinity controller design such that bilinear transformation ,while acording to my homework requirement l have to use this method,

I appreciate any guidance or suggestion , Thanks in advance!

You can convert the stated matrix equation into the standard DARE form by using the Woodbury matrix identity, which states

$$\left(A + U\,C\,V \right)^{-1} = A^{-1} - A^{-1} U \left(C^{-1} + V\,A^{-1} U \right)^{-1} V\,A^{-1}. \tag{1}$$

Using $$(1)$$ together with $$A = I$$, $$U = F(\gamma) = B_2 B_2^\top - \gamma^{-2} B_1 B_1^\top$$, $$C = I$$ and $$V = M$$ yields the following

$$\left(I + F(\gamma)\,M\right)^{-1} = I - F(\gamma)\,(I + M\,F(\gamma))^{-1} M. \tag{2}$$

When assuming that $$F(\gamma) \geq 0$$ it should also be possible find a $$L(\gamma)$$ such that $$F(\gamma) = L(\gamma)\,L(\gamma)^\top$$. One way to find such $$L(\gamma)$$ would be for example by using the Cholesky decomposition. Substituting this in $$F(\gamma)\,(I + M\,F(\gamma))^{-1}$$ and partially applying the push-through identity yields

$$F(\gamma)\,(I + M\,F(\gamma))^{-1} = L(\gamma)\,(I + L(\gamma)^\top M\,L(\gamma))^{-1} L(\gamma)^\top. \tag{3}$$

Substituting this in $$(2)$$ and its resulting expression in the original matrix equation yields

$$M = A^\top M\,A - A^\top M\,L(\gamma)\,(I + L(\gamma)^\top M\,L(\gamma))^{-1} L(\gamma)^\top M\,A + C_1^\top C_1, \tag{4}$$

which can be solved for $$M$$ using a DARE solver using $$R = I$$ and $$B = L(\gamma)$$.

• Thanks a lot ! @Kwin van der Veen. I have read a lot of your answers! Your profound knowledge is admirable！ – David peter Jan 4 at 4:42