I am checking some algebra from this paper
- James Anderson, John C. Doyle, Steven Low, Nikolai Matni, System Level Synthesis, 2019.
Specifically, I want to understand how transfer functions from $(\delta_x, \delta_y, \delta_u)$ to $(x, u, \hat{\delta}_x)$ in Equation (4.11) is derived in detail.
I was able to compute the first column to match the paper correctly. However, I am struggling to get anything close for transfer functions in the other columns.
For example, for the transfer function from $\delta_y$ to $x$, which would correspond to the first row and second column of the transfer function matrix, I was trying to obtain the answer using the following steps but failed.
Assume $\delta_x = 0$, and $\delta_u = 0$. From the figure, I have following equations:
$$y = x + \delta_y \tag{1}$$
$$x = \frac{1}{z}(Ax + B_{2}u) \tag{2}$$
$$u = \tilde{\Phi}_{u} \hat{\delta}_x \tag{3}$$
$$\hat{\delta}_x = y – \hat{x} \tag{4}$$
$$-\hat{x} = \tilde{\Phi}_{u} \hat{\delta}_{x} \tag{5}$$
From (2), $(zI – A)x = B_{2} u$. This gives $x = (zI – A)^{-1} B_{2} u$.
From the relation, $\tilde{\Phi}_{u} = z \Phi_u$ as seen from the figure's caption, I can substitute $u = z \Phi_u \hat{\delta}_{x}$ into the $x$ equation above.
Then, $x = (zI – A)^{-1} B_{2} z\Phi_{u}\hat{\delta}_{x}$. Subsequently, I would then substitute the equations (4), (1), and (5) to obtain the following:
$x = (zI-A)^{-1}B_{2}z\Phi_{u}(y-\hat{x})$ by substituting (4) into $\hat{\delta}_x$,
$x = (zI-A)^{-1}B_{2}z\Phi_{u}(x + \delta_{y}-\hat{x})$ by substituting (1) into $y$,
$x = (zI-A)^{-1}B_{2}z\Phi_{u}(x + \delta_{y} + \tilde{\Phi}_{u}\hat{\delta}_{x})$ by substituting (5) into $-\hat{x}$.
But then I am trapped in the loop, I don't know how to eliminate some variables here.
Edit
Calculations based on a comment from @KBS:
We have two transfer function matrices:
$$\begin{bmatrix}z_1\\w_{21}\end{bmatrix} = \begin{bmatrix}\Sigma_{11}&\Sigma_{12}\\\Sigma_{21}&\Sigma_{22}\end{bmatrix}\begin{bmatrix}w_1\\w_{12}\end{bmatrix} \tag{6}$$
$$\begin{bmatrix}z_2\\w_{12}\end{bmatrix} = \begin{bmatrix}\Gamma_{11}&\Gamma_{12}\\\Gamma_{21}&\Gamma_{22}\end{bmatrix}\begin{bmatrix}w_2\\w_{21}\end{bmatrix} \tag{7}$$
I can solve for $w_{12}$ and $w_{21}$ by computing the inverse of the matrix $\begin{bmatrix}-\Sigma_{22} & I\\
I & -\Gamma_{22}
\end{bmatrix}$,
$$
\begin{bmatrix}
w_{12}\\
w_{21}
\end{bmatrix} =
\begin{bmatrix}
-\Sigma_{22} & I\\
I & -\Gamma_{22}
\end{bmatrix}^{-1}\begin{bmatrix}
\Sigma_{21}w_1\\
\Gamma_{21}w_2\\
\end{bmatrix} \tag{8}.
$$
With the block matrix inversion, the inverse turns out to be:
$$
\begin{bmatrix}
-\Sigma_{22} & I\\
I & -\Gamma_{22}
\end{bmatrix}^{-1}
=
\begin{bmatrix}
-\Sigma_{22}^{-1}+\Sigma_{22}^{-1}(-\Gamma_{22} + \Sigma_{22}^{-1})^{-1}\Sigma_{22}^{-1} & \Sigma_{22}^{-1}(-\Gamma_{22} + \Sigma_{22}^{-1})^{-1}\\
(-\Gamma_{22} + \Sigma_{22}^{-1})^{-1}\Sigma_{22}^{-1} & (-\Gamma_{22} + \Sigma_{22}^{-1})^{-1}
\end{bmatrix}.
$$
Then, I now have explicit expressions for $w_{12}$ and $w_{21}$ as follows:
$$
w_{12} = \bigg(-\Sigma_{22}^{-1}+\Sigma_{22}^{-1}(-\Gamma_{22} + \Sigma_{22}^{-1})^{-1}\Sigma_{22}^{-1}\bigg)\Sigma_{21}w_1 + \bigg(\Sigma_{22}^{-1}(-\Gamma_{22} + \Sigma_{22}^{-1})^{-1}\bigg)\Gamma_{21}w_2.
$$
$$
w_{21} = \bigg((-\Gamma_{22} + \Sigma_{22}^{-1})^{-1}\Sigma_{22}^{-1}\bigg)\Sigma_{21}w_1 + \bigg((-\Gamma_{22} + \Sigma_{22}^{-1})^{-1}\bigg)\Gamma_{21}w_2.
$$
$w_{12}$ and $w_{21}$ are now in terms of transfer functions $\Sigma_{ij}$ and $\Gamma_{ij}$, and can be substituted into $z_1$ and $z_2$, resulting in the following expressions:
$$
z_1 = \Sigma_{11}w_1 + \Sigma_{12}\bigg(-\Sigma_{22}^{-1}+\Sigma_{22}^{-1}(-\Gamma_{22} + \Sigma_{22}^{-1})^{-1}\Sigma_{22}^{-1}\bigg)\Sigma_{21}w_1 + \bigg(\Sigma_{22}^{-1}(-\Gamma_{22} + \Sigma_{22}^{-1})^{-1}\bigg)\Gamma_{21}w_2.
$$
$$
z_2 = \Gamma_{11}w_2 + \Gamma_{12}\bigg((-\Gamma_{22} + \Sigma_{22}^{-1})^{-1}\Sigma_{22}^{-1}\bigg)\Sigma_{21}w_1 + \bigg((-\Gamma_{22} + \Sigma_{22}^{-1})^{-1}\bigg)\Gamma_{21}w_2.
$$
So, I can see that I can get the general map from $w_1$ and $w_2$ to $z_1$ and $z_2$ as KBS suggested.
However, this leaves me with defining $\Sigma_{ij}$ and $\Gamma_{ij}$, which I find equally difficult.
I tried to obtain $\Sigma_{21}$ and $\Sigma_{22}$ below, and wondering if I am in the right direction:
From the figure, I have
$$x = \frac{1}{z}(\delta_x + Ax + B_2 u),$$
which gives me $$x = (zI – Ax)^{-1}(\delta_{x} + B_{2}u).$$
By matching with the equation $x = \Sigma_{21}\delta_x + \Sigma_{22}u$, which is from $z_{1} = \Sigma_{21}\delta_x + \Sigma_{22}u$, I can deduce that
$\Sigma_{21} = (zI-A)^{-1},$ and $\Sigma_{22} = (zI – A)^{-1}B_2.$
Best Answer
I give the general idea. The rest will be left to you. Assume that we have the following systems
$$ \begin{bmatrix} z_1\\ w_{21} \end{bmatrix}=\begin{bmatrix} \Sigma_{11} & \Sigma_{12}\\ \Sigma_{21} & \Sigma_{22} \end{bmatrix}\begin{bmatrix} w_1\\ w_{12} \end{bmatrix} $$ and $$ \begin{bmatrix} z_2\\ w_{12} \end{bmatrix}=\begin{bmatrix} \Gamma_{11} & \Gamma_{12}\\ \Gamma_{21} & \Gamma_{22} \end{bmatrix}\begin{bmatrix} w_2\\ w_{21} \end{bmatrix} $$
and you want to find the map $(w_1,w_2)\mapsto(z_1,z_2)$. So, all you have to is eliminate the loop signals $w_{12}$ and $w_{21}$ from the interconnection. You can find the general expression for this mapping.
In your case, you have $w_1=\delta_x$, $w_2=(\delta_u,\delta_y)$, $z_1=x$, $z_2=(u,\hat\delta_x)$, $w_{12} = u$ and $w_{21} = x$. So, define the different operators $\Sigma_{ij}$ and $\Gamma_{ij}$, $i=1,2$, accordingly and apply the general formula you have previously obtained.
Edit for help.
The equations relating $w_{21}$ and $w_{12}$ can be combined into $$ \begin{bmatrix} -\Sigma_{22} & I\\ I & -\Gamma_{22} \end{bmatrix}\begin{bmatrix} w_{12}\\ w_{21} \end{bmatrix}=\begin{bmatrix} \Sigma_{21} & 0\\ 0 & \Gamma_{21}\\ \end{bmatrix}\begin{bmatrix} w_1\\ w_2 \end{bmatrix}. $$
For the interconnection to be well-posed, it is necessary that the matrix on the left be invertible, so you can assume that here but this needs to be also checked when applying this on a particular interconnection. I guess it is easy to continue from there.
Re-edit. So, using the signals I have defined we have that
$$z_1=\underbrace{(zI-A)^{-1}}_{\mbox{$\Sigma_{11}$}}w_1+\underbrace{(zI-A)^{-1}B_2}_{\mbox{$\Sigma_{12}$}}w_{12}, w_{21}=\underbrace{(zI-A)^{-1}}_{\mbox{$\Sigma_{21}$}}w_1+\underbrace{(zI-A)^{-1}B_2}_{\mbox{$\Sigma_{22}$}}w_{12},$$
$$ z_2=\underbrace{\begin{bmatrix} 1 & K\\ 0 & G \end{bmatrix}}_{\mbox{$\Gamma_{11}$}} \begin{bmatrix} \delta_u\\ \delta_y \end{bmatrix}+\underbrace{\begin{bmatrix} K\\ G \end{bmatrix}}_{\mbox{$\Gamma_{12}$}}w_{21}, $$ and $$w_{12}=\underbrace{\begin{bmatrix} 1 & K \end{bmatrix}}_{\mbox{$\Gamma_{21}$}} \begin{bmatrix} \delta_u\\ \delta_y \end{bmatrix}+\underbrace{K}_{\mbox{$\Gamma_{22}$}}w_{21}$$
where $G=(I-\tilde\Phi_x)^{-1}$ and $K=\tilde\Phi_uG$. I have also assumed positive feedback in the loops.
So, the loop is well-posed if and only if
$$ \begin{bmatrix} -\Sigma_{22} & I\\ I & -\Gamma_{22} \end{bmatrix}=\begin{bmatrix} -(zI-A)^{-1} & I\\ I & -K \end{bmatrix} $$ is invertible. Assuming that this is the case, we obtain that
$$ \begin{array}{rcl} \begin{bmatrix} w_{12}\\ w_{21} \end{bmatrix}&=&\begin{bmatrix} -(zI-A)^{-1}B_2 & I\\ I & -K \end{bmatrix}^{-1}\begin{bmatrix} (zI-A)^{-1} & | & 0 & 0\\ 0 & | & 1 & K\\ \end{bmatrix}\begin{bmatrix} w_1\\ \hline w_2 \end{bmatrix}\\ &=&\begin{bmatrix} K(I-(zI-A)^{-1}B_2K)^{-1} & (I-K(zI-A)^{-1}B_2)^{-1}\\ (I-(zI-A)^{-1}B_2K)^{-1} & (zI-A)^{-1}B_2(I-K(zI-A)^{-1}B_2)^{-1} \end{bmatrix}\begin{bmatrix} (zI-A)^{-1} & | & 0 & 0\\ 0 & | & 1 & K\\ \end{bmatrix}\begin{bmatrix} w_1\\ \hline w_2 \end{bmatrix}. % \end{array} $$ We have that $$ (I-(zI-A)^{-1}B_2K)^{-1}(zI-A)^{-1}=(zI-A-B_2K)^{-1} $$
which yields
$$ \begin{bmatrix} w_{12}\\ w_{21} \end{bmatrix}=\begin{bmatrix} K(zI-A-B_2K)^{-1} & (I-K(zI-A)^{-1}B_2)^{-1}\begin{bmatrix}1 & K\end{bmatrix}\\ % (zI-A-B_2K)^{-1} & (zI-A)^{-1}B_2(I-K(zI-A)^{-1}B_2)^{-1}\begin{bmatrix}1 & K\end{bmatrix} \end{bmatrix}\begin{bmatrix} w_1\\ \hline w_2 \end{bmatrix} $$
which yields
$$z_1=((zI-A)^{-1}+(zI-A)^{-1}B_2K(zI-A-B_2K)^{-1})w_1+(zI-A)^{-1}B_2(I-K(zI-A)^{-1}B_2)^{-1}w_2 $$
and
$$z_2=\left(\begin{bmatrix} 1 & K\\ 0 & G \end{bmatrix}+\begin{bmatrix} K\\ G \end{bmatrix}(zI-A)^{-1}B_2(I-K(zI-A)^{-1}B_2)^{-1}\begin{bmatrix}1 & K\end{bmatrix}\right)w_2+\begin{bmatrix} K\\ G \end{bmatrix}(zI-A-B_2K)^{-1}w_1. $$
Which is the final result. It could be further simplified, I suppose, in certain cases. But this is the most general one.