Some software tool

Use mathematica™. a, b, cb, ca are all matrix.

$x'= b + a x$

newsystem. A change of variable defined by $x=c_b + c_a y$ then the system becomes: $c_a y' = b + a (c_b + c_a y) = (b + a c_b) + (a c_a) y$ so $y' = c_a^{-1} ( b + a c_b) + (c_a^{-1} a c_a) y$ . Return the list of {new b, new a}

swapsystem. Swap the variables $x_i$ and $x_j$ in the presentation is also a new system with correspondent x= cb + ca y. Return the list of {new b, new a, the correspondent cb, the correspondent ca}

pivotsystem. Replace $x_j$ with derivative of $x_i$ to become a new system. Return the list of {new b, new a, correspondent cb, correspondent ca}

newsystem[b_, a_, cb_, ca_]:=Block[{t}, t=Simplify[Inverse[ca], {L\>0,K\>0}]; Simplify[{t . (b + a . cb), t . a . ca }, {L\>0, K\>0}] ]; swapsystem[b_,a_, i_, j_]:=Block[{t, nb, na, cb, di, ca}, nb=b; na=a; t=nb[[i]]; nb[[i]]=nb[[j]]; nb[[j]]=t; t=na[[i]]; di=Length[t]; na[[i]]=na[[j]]; na[[j]]=t; na=Transpose[na]; t=na[[i]]; na[[i]]=na[[j]]; na[[j]]=t; na=Transpose[na]; cb=Array[0 &, {di, 1}]; ca=IdentityMatrix[di]; ca[[i]][[i]]=0; ca[[j]][[j]]=0; ca[[i]][[j]]=1; ca[[j]][[i]]=1; {nb,na,cb,ca} ]; pivotsystem[b_,a_,i_,j_]:=Block[{t, q, q1, di, cb, ca}, t=a[[i]]; q=t[[j]]; t=t/q; q1=b[[i]]/q; t[[j]]=-1/q; t=-t; q1=-q1; di=Length[t]; cb=Array[0 &, {di, 1}]; cb[[j]]=q1; ca=IdentityMatrix[di]; ca[[j]]=t; t=newsystem[b,a,cb,ca]; {t[[1]],t[[2]],cb,ca} ];

Mutual inductance

The mutual inductance matrix must be semi-positive definite and cannot be arbitrary matrix in order to satisfy physics laws. If the determinant is greater than zero, the magnetic field will absorb some energy until it is saturated and then transfers the input energy from the primary coil to the secondary coil. If the determinant is zero, the input energy will immediately be delivered from the primary coil to the secondary coil. The matrix of the symmetric architecture of the three inductors in a high frequency transformer, $a$ and $b$ being scalars, is:

$\begin{bmatrix}L& - a L& - b L\\ - a L& L& b L\\ - b L& b L& K\end{bmatrix}$

With row column operations:

$\begin{aligned} &\approx \begin{bmatrix} L&& - b L\\& L - a^2 L& b L - a b L\\ - b L& b L - a b L& K\end{bmatrix}\\ &\approx \begin{bmatrix}L&&\\& L - a^2 L& b L - a b L\\& b L - a b L& K - b^2 L \end{bmatrix}= \begin{bmatrix}L&&\\& (1 - a^2) L& b(1 - a) L\\& b(1 - a) L& K - b^2 L \end{bmatrix}\\&\approx \begin{bmatrix}L&&\\&(1 - a^2) L&\\&& K - b^2 L - \frac{b(1 - a)}{1 - a^2} b(1 - a) L \end{bmatrix}= \begin{bmatrix}L&&\\&(1 - a^2) L&\\&& K - b^2 L - \frac{b^2 (1 - a)}{1 + a} L\end{bmatrix} \\&= \begin{bmatrix}L&&\\&(1 - a^2) L&\\&& K - \left( \frac{2 b^2}{1 + a} \right) L\end{bmatrix}\end{aligned}$

Therefore $|a| \leq 1$ and $\frac{2 b^2}{1 + a} \leq \frac{K}{L}$ to make this mutual inductance matrix positive. The ideal will be $a = 1$ and $b = \sqrt{\frac{K}{L}}$ to be the mutual inductance matrix

$\begin{bmatrix}L& - L& - \sqrt{L K}\\ - L& L& \sqrt{L K}\\ - \sqrt{L K}& \sqrt{L K}& K\end{bmatrix}$

Question

I felt the electronic model of a high frequency transformer troublesome. When the MOSFET1 and MOSFET2 respectively change from ON to OFF and OFF to ON, it seems saying the current from the battery becomes the current through the ON MOSFET instantly, then the infinite inductive voltage at both the inductors in the transformer's front side (the side connected to the battery). The infinite voltage is not considered in the voltage drop equations about these two circuits. Beside, when the circuit is open, there is still some current to form the strong electric field between the G and S gates due to the inductive current. To accommodate this, I here have a thorough model to see the whole picture.

I saw a Youtube about the manufacturing of high frequency transformers for inverters. There are two inductors at the front side and one inductor at the rear end. The video seems saying there are mutual inductance among all the three inductors.

The electronic parts are:

1. V : battery with voltage; V.P is the positive pole, V.N is the negative pole
2. L0, L1, L2, K : inductor with inductance $L_0$ , L , L , K. L1 and K have mutual inductance. L2 and K have mutual inductance. L1 and L2 have mutual inductance. The current direction through K is defined to be such that it is increasing when current through L1 is increasing
3. C : capacitor with capacitance
4. M1, M2: MOSFET. When it is ON, there is a tiny resistance $r$ to model the tiny resistance of MOSFET. When it is off, there is a big resistance $B$ to model the open circuit and the temporarily inductive current. The two MOSFET operate four stages of duty ratio $d$. Stage 1, M1 is ON and M2 is OFF, duty ratio $\frac{d}{2}$ . Stage 2, M1 is OFF and M2 is OFF, duty ratio $\frac{1 - d}{2}$ . Stage 3, M1 is OFF and M2 is ON, duty ratio $\frac{d}{2}$ . Stage 4, M1 is OFF and M2 is OFF, duty ratio $\frac{1 - d}{2}$ . Let $T$ be the period of the MOSFET.
5. R : a resistor with resistance represents a load at the rear end of the high frequency transformer.

All circuits are described in the seven contact points:

[V.P, L0,1] [L0.2, C.1, L1.1, L2.1] [L1,2, M1.1] [L2.2, M2.1] [V.N, C.2, M1.2, M2.2] [K.1, R.1] [K.2, R.2]

A benefit of this approach is that, as no pure "open" and "close" a universal math treatment of all variables are done in one go; for Stage 3, any math corollary is the formula replacing $B$ and $r$ with $r$ and $B$; for Stage 2 and stage 4, any math corollary is the formula replacing $B$ and $r$ with $B$ and $B$.

To find the correct inductance matrix, try simple circuit without $L_0$ and $C$

$0 = \begin{bmatrix}V\\V\\0\end{bmatrix} - \begin{bmatrix}L&-a L&-b L\\-a L&L&b L\\-b L&b L&K\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}' - \begin{bmatrix}r&&\\&B&\\&&R\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$

Name this formula F0.

The so-called naive math is that for stage 1 and stage 3, $r$ is 0 and $B$ is infinite so the system is

$\begin{aligned}0 = \begin{bmatrix}V\\\frac{V}{B}\\0\end{bmatrix} - \begin{bmatrix}L&-a L&-b L\\-\frac{a L}{B}&\frac{L}{B}&\frac{b L}{B}\\-b L&b L&K\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}' - \begin{bmatrix}r&&\\&1&\\&&R\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\\ \approx \\ 0 = \begin{bmatrix}V\\0\\0\end{bmatrix} - \begin{bmatrix}L&-a L&-b L\\0&0&0\\-b L&b L&K\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}' - \begin{bmatrix}0&&\\&1&\\&&R\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} \end{aligned}$

and the system becomes:

$\begin{aligned}x_2 = 0\\0 = \begin{bmatrix}V\\0\end{bmatrix} - \begin{bmatrix}L&-b L\\-b L&K\end{bmatrix} \begin{bmatrix}x_1\\x_3\end{bmatrix}' - \begin{bmatrix}0&0\\0&R\end{bmatrix} \begin{bmatrix}x_1\\x_3\end{bmatrix}\end{aligned}$

which is easier to solve. But the naive approach cannot address well about the inductance of sudden change of currents when ON becomes OFF and the current looks like discontinuous mathematically while it is continuous in reality.

Manipulate F0, the system becomes

$0 = \begin{bmatrix}\frac{V}{1-a}\\(1 + a) V\\0\end{bmatrix} - \begin{bmatrix}L&&-\frac{b}{1 + a} L\\&(1-a^2) L&(1-a)b L\\&&K - \frac{2 b^2}{1 + a} L\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}' - \begin{bmatrix}\frac{1}{1-a^2}r&\frac{a}{1-a^2}B&\\a r&B&\\\frac{b}{1+a} r&- \frac{b}{1 + a} B& R\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$

Name this formula F4. By electricity academic, $1 - a^2 \geq 0, K - \frac{2 b^2}{1 + a} L \geq 0$. Being ideal means $K - \frac{2 b^2}{1 + a} L = 0$ , say, $a = 0 . 99 , b = \sqrt{\frac{1 .99K}{2 L}}$, so the 3rd row reads:

$0 = \frac{b}{1 + a} r x_1 - \frac{b}{1 + a} B x_2 + R x_3$

Name this formula F1. Therefore, the system is actually only $x_1$ and $x_2$ system introduced by the change of variables:

$\begin{aligned}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}1&\\&1\\-\frac{b r}{(1 + a) R}&\frac{b B}{(1 + a) R}\end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix}\\0 = \begin{bmatrix}\frac{V}{1 - a}\\(1 + a)V\end{bmatrix} - \begin{bmatrix}\left(1 + \frac{b^2 r}{(1 + a)^2 R} \right) L&- \frac{b^2 B}{(1+a^2)R} L\\-\frac{(1 - a) b^2 r}{(1 + a) R} L&\left(1 - a^2 + \frac{(1 - a) b^2 B}{(1 + a) R} \right) L\end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix}'- \begin{bmatrix}\frac{1}{1 - a^2} r&\frac{a}{1 - a^2} B\\a r&B\end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix}\end{aligned}$

Aka

$\begin{bmatrix}x_1\\x_2\end{bmatrix}'= \begin{bmatrix}\frac{\left(2 b^2 B + \left(1 + a\right)^2 R \right) V}{\left(1 - a \right) L \left(b^2 \left(B + r\right) + \left(1 + a\right)^2 R\right)}\\\frac{\left(2 b^2 r + \left(1 + a\right)^2 R\right) V}{\left(1 - a\right) L\left(b^2 \left(B + r\right) + \left(1 + a\right)^2 R\right)}\end{bmatrix}- \begin{bmatrix}\frac{r\left(b^2 B + \left(1 + a\right)R \right)}{\left(1 - a\right) L \left(b^2 \left(B + r\right) + \left(1 + a\right)^2 R\right)}&\frac{B\left(b^2 B + a\left(1 + a\right)R \right)}{\left(1 - a\right) L \left(b^2 \left(B + r\right) + \left(1 + a\right)^2 R\right)}\\\frac{r\left(b^2 r + a\left(1 + a\right)R \right)}{\left(1 - a\right) L \left(b^2 \left(B + r\right) + \left(1 + a\right)^2 R\right)}&\frac{B\left(b^2 r + \left(1 + a\right)R \right)}{\left(1 - a\right) L \left(b^2 \left(B + r\right) + \left(1 + a\right)^2 R\right)}\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}$

Name this formula F2. The $1-a$ in every entry's denominator demonstrates $a \approx 1$ will have stronger driving force. Prettier:

$\begin{bmatrix}x_1\\x_2\end{bmatrix}' \left(1 - a \right) L \left(b^2 \left(B + r\right) + \left(1 + a\right)^2 R\right)= \begin{bmatrix}\left(2 b^2 B + \left(1 + a\right)^2 R \right) V\\\left(2 b^2 r + \left(1 + a\right)^2 R\right) V\end{bmatrix}- \begin{bmatrix}r\left(b^2 B + \left(1 + a\right)R \right)&B\left(b^2 B + a\left(1 + a\right)R \right)\\r\left(b^2 r + a\left(1 + a\right)R \right)&B\left(b^2 r + \left(1 + a\right)R \right)\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}$

So the system will drive $\begin{bmatrix}x_1\\x_2\end{bmatrix}$ to the steady state $S$ such that $\begin{bmatrix}\frac{1}{1 - a^2} r&\frac{a}{1 - a^2} B\\a r&B\end{bmatrix} S = \begin{bmatrix}\frac{V}{1 - a}\\(1 + a) V\end{bmatrix}$ which is, as expected, $\begin{bmatrix}\frac{V}{r}\\\frac{V}{B}\end{bmatrix}$

Interchange of $B$ and $r$ leads to the other steady state. The alternating On-OFF causes $\begin{bmatrix}x_1\\x_2\end{bmatrix}$ move in-between and does not settle at any static numbers. Unless, the frequency is so high that the resistance leads to zero homogeneous part by any "left over" inductance, also agreeing with the linear approximate which means super short $T$ aka super high frequency.

When swap $B$ and $r$, the system has a strong pull and push for $x_1$ and $x_2$ . Difficult to describe the system when $B$ is infinite, because the 2-1 entry explodes:

$\frac{B\left(b^2 B + a \left(1 + a\right) R\right)}{\left(1 - a\right) L \left(b^2 \left(B + r\right) + \left(1 + a\right)^2 R\right)}$

A change of variable to focus on total current and voltage in the rear end:

$\begin{bmatrix}y_1\\y_2\end{bmatrix} = \begin{bmatrix}1&1\\\frac{b r}{1 + a}&-\frac{b B}{1 + a}\end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix}$

Aka

$\begin{bmatrix}x_1\\x_2\end{bmatrix} = \frac{1}{B+r}\begin{bmatrix}B&\frac{1+a}{b}\\r&-\frac{1+a}{b}\end{bmatrix} \begin{bmatrix}y_1\\y_2\end{bmatrix}$

F2 becomes:

$\begin{bmatrix}y_1\\y_2\end{bmatrix}'= \begin{bmatrix}\frac{2 V}{\left(1 - a\right)L}\\-\frac{\left(1 + a\right) b \left(B - r\right) R V}{\left(1 - a\right) L \left(b^2 \left(B + r\right) + \left(1 + a\right)^2 R\right)}\end{bmatrix} + \begin{bmatrix}-\frac{2 B r}{\left(1 - a\right) L \left(B + r\right)}&\frac{\left(1 + a\right)\left(B - r\right)}{\left(1 - a\right) L \left(B + r\right)}\\\frac{\left(1 + a\right) b B r \left(B - r\right) R}{\left(1 - a\right) L \left(B + r\right) \left(b^2 \left(B + r\right) + \left(1 + a\right)^2 R\right)}&-\frac{\left(1 + a\right) \left(B^2 + r^2 - 2 a B r\right) R}{\left(1 - a\right) L \left(B + r\right) \left(b^2 \left(B + r\right) + \left(1 + a\right)^2 R\right)}\end{bmatrix} \begin{bmatrix}y_1\\y_2\end{bmatrix}$

Name this formula F3. Pretty:

$\begin{bmatrix}y_1\\y_2\end{bmatrix}'=\frac{1}{\left(1-a\right)L} \begin{bmatrix}2 V\\-\frac{\left(1 + a\right) b \left(B - r\right) R V}{b^2 \left(B + r\right) + \left(1 + a\right)^2 R}\end{bmatrix} + \frac{1}{\left(1 - a\right) L \left(B + r\right)}\begin{bmatrix}-2 B r&\left(1 + a\right)\left(B - r\right)\\\frac{\left(1 + a\right) b B r \left(B - r\right) R}{b^2 \left(B + r\right) + \left(1 + a\right)^2 R}&-\frac{\left(1 + a\right) \left(B^2 + r^2 - 2 a B r\right) R}{b^2 \left(B + r\right) + \left(1 + a\right)^2 R}\end{bmatrix} \begin{bmatrix}y_1\\y_2\end{bmatrix}$

The 4-stages average matrix are:

$\frac{1}{\left(1 - a\right) L}\begin{bmatrix}2 V\\0\end{bmatrix}$

and

$\frac{1}{\left(1 - a\right) L} \begin{bmatrix}-d \frac{2 B r}{B + r} - \left(1 - d\right) B&\\&-\frac{d \left(1 + a\right) \left(B^2 + r^2 - 2 a B r\right) R}{\left(B + r\right) \left(b^2 \left(B + r\right) + \left(1 + a\right)^2 R\right)} - \frac{\left(1 - d\right) \left(1 + a\right) \left(1 - 2 a\right) B^2}{2 B \left(2 b^2 B + \left(1 + a\right)^2 R\right)}\end{bmatrix}$

So the steady state in high frequency of $\begin{bmatrix}y_1\\y_2\end{bmatrix}$ is $\begin{bmatrix}\frac{2 V}{d \frac{2 B r}{B + r} + \left(1 - d\right) B}\\0\end{bmatrix}$ and $x_3 = 0$ according to F1; $x_3$ being alternating, $x_3 = 0$ as the approximate steady state is reasonable. $y_1$ can be as low as $\frac{2 V}{B}$ aka circuit 1 and circuit 2 are open, or as high as $\left( \frac{1}{r} + \frac{1}{B}\right) V$ aka one of curcuit 1 and circuit 2 is closed.

Specialize when B goes to infinite

By F2, the 2-2 entry will explode so it shall be described by $\begin{bmatrix}x_1\\B x_2\end{bmatrix}'$ temporarily to see the picture:

$\begin{bmatrix}x_1\\B x_2\end{bmatrix}' = \begin{bmatrix}\frac{\left(2 b^2 B + \left(1 + a\right)^2 R\right) V}{\left(1 - a\right) L \left(b^2 \left(B + r\right) + \left(1 + a\right)^2 R\right)}\\\frac{B \left(2 b^2 r + \left(1 + a\right)^2 R\right) V}{\left(1 - a\right) L \left(b^2 \left(B + r\right) + \left(1 + a\right)^2 R\right)}\end{bmatrix}- \begin{bmatrix}\frac{r \left(b^2 B + \left(1 + a\right) R\right)}{\left(1 - a\right) L \left(b^2 \left(B + r\right) + \left(1 + a\right)^2 R\right)}&\frac{b^2 B + a \left(1 + a\right) R}{\left(1 - a\right) L \left(b^2 \left(B + r\right) + \left(1 + a\right)^2 R\right)}\\\frac{r B \left(b^2 r + a \left(1 + a\right) R\right)}{\left(1 - a\right) L \left(b^2 \left(B + r\right) + \left(1 + a\right)^2 R\right)}&\frac{B \left(b^2 r + \left(1 + a\right) R\right)}{\left(1 - a\right) L \left(b^2 \left(B + r\right) + \left(1 + a\right)^2 R\right)}\end{bmatrix} \begin{bmatrix}x_1\\B x_2\end{bmatrix}$

Then it becomes:

$\begin{bmatrix}x_1\\B x_2\end{bmatrix}'= \begin{bmatrix}\frac{2 V}{\left(1 - a\right) L}\\\frac{\left(2 b^2 r + \left(1 + a\right)^2 R\right) V}{\left(1 - a\right) b^2 L}\end{bmatrix}- \begin{bmatrix}\frac{r}{1 - a L}&\frac{1}{\left(1 - a\right) L}\\\frac{r \left(b^2 r + a \left(1 + a\right) R\right)}{\left(1 - a\right) b^2 L}&\frac{b^2 r + \left(1 + a\right) R}{\left(1 - a\right) b^2 L}\end{bmatrix} \begin{bmatrix}x_1\\B x_2\end{bmatrix}$

As $B x_2$ is bounded (actualy, it goes to zero), $x_2$ is 0 as $B$ goes infinite

Summary

By F4, if not ideal inductance matrix, it becomes, $k \equiv \frac{K}{L}$ :

$\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}'= \frac{V}{L} \begin{bmatrix}\frac{1}{1 - a}\\\frac{1}{1 - a}\\0\end{bmatrix} + \frac{1}{\left(k + a k - 2 b^2\right) L} \begin{bmatrix}\frac{\left(b^2 - k\right) r}{1 - a}&\frac{\left(b^2 - a k\right) B}{1 - a}&-b R\\\frac{\left(b^2 - a k\right) r}{1 - a}&\frac{\left(b^2 - k\right) B}{1 - a}&b R\\- b r&b B&-\left(1 + a\right) R\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$

Name this formula F5. Note the zero denominator possibility suggests the condition of ideal inductance matrix $0 = k + a k - 2 b^2$ or $0 = 1 - a$. In the general situations of not ideal, the steady state for each stage 1, 2, 3, 4 is $\begin{bmatrix}\frac{V}{r}\\\frac{V}{B}\\0\end{bmatrix},\begin{bmatrix}\frac{V}{B}\\\frac{V}{B}\\0\end{bmatrix},\begin{bmatrix}\frac{V}{B}\\\frac{V}{r}\\0\end{bmatrix}, \begin{bmatrix}\frac{V}{B}\\\frac{V}{B}\\0\end{bmatrix}$ . When the linear system of each stage converges, the overall system convergs too. See 直流電路的最終電壓電流. Suppose the over all 4 stages, denote the cycled steady state ${}_0x,{}_1x,{}_2x ,{}_3x$, the linear system is ${}_4x = \mathcal{B} + \mathcal{A} \cdot {}_0x$ where 𝒜 and ℬ have two additional parameters $d$ and $T$ and the seady state is $(I - \mathcal{A})^{- 1} \mathcal{B}$ . Different $d$ leads to different $\mathcal{B}$ and $\mathcal{A}$ . When $d = 0$ , it corresponds to a system with $R = \infty$ , aka no load, the steady state is $x_1 = x_2 = \frac{V}{B}$ and $x_3 = 0$ . The supper-short period $T$ leads the steady state answer to their linear average approximate; even with tiny resistance and tine inductance, a super short period aka super high frequency leads to super high resistance and this is why linear approximation reports $x_3 = 0$ in this case. The caveat is that when some entry in the matrix explode, like here $B$, the $T$ can not be thought as infinite small and the linear approximate solution comes as the result. Experiments with Mathematica™ about this confirm the observation. The experiments show that the $R x_3$ number can be far away from if specific $± b VT$ and $d$ are arranged.

For linear approximation, average matrix are:

$\bar{B} = \frac{V}{L} \begin{bmatrix}\frac{1}{1 - a}\\\frac{1}{1 - a}\\0\end{bmatrix}$

and

$\bar{A}= \frac{1}{\left(k + a k - 2 b^2 \right) L} \begin{bmatrix}\frac{\left(b^2 - k\right) \left(\frac{d}{2} r + \left(1 - \frac{d}{2}\right) B\right)}{1 - a}&\frac{\left(b^2 -a k\right) \left(\frac{d}{2} r + \left(1 - \frac{d}{2}\right) B\right)}{1 - a}&-b R\\\frac{\left(b^2 -a k\right) \left(\frac{d}{2} r + \left(1 - \frac{d}{2}\right) B\right)}{1 - a}&\frac{\left(b^2 - k\right) \left(\frac{d}{2} r + \left(1 - \frac{d}{2}\right) B\right)}{1 - a}&b R\\-b \left(\frac{d}{2} r + \left(1 - \frac{d}{2}\right) B\right)&b \left(\frac{d}{2} r + \left(1 - \frac{d}{2}\right) B\right)&-\left(1 + a\right) R\end{bmatrix}$

So

$-\bar{A}^{-1} \bar{B} = \begin{bmatrix}\frac{2 V}{\left(2 - d\right) B + d r}\\\frac{2 V}{\left(2 - d\right) B + d r}\\0\end{bmatrix}$

which is very small irrelevant of $d$. All ${}_0x,{}_1x,{}_2x,{}_3x$ are somewhat around and away from, as there are exploding entries in the matrix of the four stages dismiss the thinking of infinite small $T$.

Complete picture

With addition L0 inductor and capacitor. Let $x_0$ be the current going through L0.

$\begin{array}{l}0 =\begin{bmatrix}V\\0\\0\\0\\0\end{bmatrix}-\begin{bmatrix}&L_0&&&\\C&&&&\\&&L&-a L&-b L\\&&-a L&L&b L\\&&-b L&b L&k L\end{bmatrix}\begin{bmatrix}y\\x_0\\x_1\\x_2\\x_3\end{bmatrix}'-\begin{bmatrix}1&&&&\\&-1&1&1&\\-1&&r&&\\-1&&&B&\\&&&&R\end{bmatrix}\begin{bmatrix}y\\x_0\\x_1\\x_2\\x_3\end{bmatrix}\\\\\begin{bmatrix}y\\x_0\\x_1\\x_2\\x_3\end{bmatrix}'=\begin{bmatrix}0\\\frac{V}{L_0}\\0\\0\\0\end{bmatrix} +\begin{bmatrix}&\frac{1}{C}&-\frac{1}{C}&-\frac{1}{C}&\\-\frac{1}{L_0}&&&&\\\frac{1}{(1 - a) L}&&\frac{(b^2 - k) r}{(1 - a)(k + a k - 2 b^2) L}&\frac{(b^2 - k) B}{(1 - a)(k + a k - 2 b^2) L}&- \frac{b R}{(k + a k - 2 b^2) L}\\\frac{1}{(1 - a) L}&&\frac{(b^2 - a k) r}{(1 - a)(k + a k - 2 b^2) L}&\frac{(b^2 - k) B}{(1 - a)(k + a k - 2 b^2) L}&\frac{b R}{(k + a k - 2 b^2) L}\\&&- \frac{b r}{(k+ a k - 2 b^2) L}&\frac{b B}{(k + a k - 2 b^2) L}& -\frac{(1 + a) R}{(k + a k - 2 b^2) L}\end{bmatrix}\begin{bmatrix}y\\x_0\\x_1\\x_2\\x_3\end{bmatrix}\end{array}$

Name this formula F6

$\begin{array}{l} 1 - a> 0\\(1 + a) k - 2 b^2 >0\\\bar{M}=\begin{bmatrix}0\\\frac{V}{L_0}\\0\\0\\0\end{bmatrix}\\\bar{A}=\begin{bmatrix}&\frac{1}{C}&-\frac{1}{C}&-\frac{1}{C}&\\-\frac{1}{L_0}&&&&\\\frac{1}{(1 - a) L}&&\frac{(b^2 - k) \left( \frac{d}{2} r + \left( 1 - \frac{d}{2} \right) B \right)}{(1 - a)(k + a k - 2 b^2) L}&\frac{(b^2 - a k) \left( \frac{d}{2} r + \left( 1 - \frac{d}{2} \right) B \right)}{(1 - a)(k + a k - 2 b^2) L}&-\frac{b R}{(k + a k - 2 b^2) L}\\\frac{1}{(1 - a) L}&&\frac{(b^2 - a k) \left( \frac{d}{2} r + \left( 1 - \frac{d}{2} \right) B \right)}{(1 - a)(k + a k - 2 b^2) L}&\frac{(b^2 - k) \left( \frac{d}{2} r + \left( 1 - \frac{d}{2} \right) B \right)}{(1 - a)(k + a k - 2 b^2) L}&\frac{b R}{(k + a k - 2 b^2) L}\\&&-\frac{b \left(\frac{d}{2} r + \left(1 - \frac{d}{2}\right) B \right)}{(k + a k - 2 b^2) L}&\frac{b \left(\frac{d}{2} r + \left(1 - \frac{d}{2}\right) B \right)}{(k + a k - 2 b^2) L}&-\frac{(1 + a) R}{(k + a k - 2 b^2) L}\end{bmatrix}\\-\bar{A}^{-1} \bar{M}= \begin{bmatrix}V\\\frac{4 V}{(2 - d) B + d r}\\\frac{2 V}{(2 - d) B + d r}\\\frac{2 V}{(2 - d) B + d r}\\0\end{bmatrix}\end{array}$

Symbolically with polynomial of $T$ up to 3 degree, the formula, whose constant term is $-\bar{A}^{-1} \bar{B}$ , is still quite different from the numerical results. The reason is that the $T$ is supposed to be so small that even $B$ is not significant. There needs some way to handle the calculation of $\alpha(M)$ where some row or column of $M$ has exploding rows or columns, aka, the terms of the power series need to be more; hardly get that formula.

Beyond that. If the MOSFET period is not that short, the steady states are reached at "different speed" based on their diagonal entry value in the matrix. Using stage 1 for example. By F6, the most negative diagonal is 4-4 entry, followed by 5-5 entry, followed by 3-3 entry, it means $x_2$ will be settled down first, then $x_3$ then $x_1$ . The harmonic frequency $\frac{1}{\sqrt{L_0 C}}$ is low by design. So, by the length of a MOSFET period, three cases:

==== case 1 ====

The period is such that "too long for $x_2$ but not too long for other variables", then the $x_2$ at the time can be calculated by set the 4th row's derivative zero in the linear equation:

$0 =\begin{bmatrix}\frac{1}{(1 - a) L}&0&\frac{(b^2 - a k) r}{(1 - a)(k + a k - 2 b^2) L}&\frac{b R}{(k + a k - 2 b^2) L}\end{bmatrix} \begin{bmatrix}y\\x_0\\x_1\\x_3\end{bmatrix} + \frac{(b^2 - k) B}{(1 - a)(k + a k - 2 b^2) L} \begin{bmatrix}x_2\end{bmatrix}$

which as expected, $x_2 = \begin{bmatrix}\frac{k + a k - 2 b^2}{(k - b^2) B}&0&\frac{(b^2 - a k) r}{(k - b^2) B}&\frac{(1 - a) b R}{(k - b^2) B}\end{bmatrix} \begin{bmatrix}y\\x_0\\x_1\\x_3\end{bmatrix}$ is almost zero already at the time.

==== case 2 ====

The period is such that "too long for $x_2$ and $x_3$ but not too long for other variables", then the state can be calculated by set the 4th and 5th row's derivative zero and results in the linear equation:

$0 =\begin{bmatrix}\frac{1}{(1 - a) L}&0&\frac{(b^2 - a k) r}{(1 - a)(k + a k - 2 b^2) L}\\0&0&-\frac{b r}{(k + a k - 2b^2)L}\end{bmatrix}\begin{bmatrix}y\\x_0\\x_1\end{bmatrix} + \begin{bmatrix}\frac{(b^2 - k) B}{(1 - a)(k + a k - 2 b^2) L}&\frac{b R}{(k + a k - 2 b^2) L}\\\frac{b B}{(k + a k - 2 b^2) L}&-\frac{(1 + a)R}{(k + a k - 2 b^2) L}\end{bmatrix} \begin{bmatrix}x_2\\x_3\end{bmatrix}$

which solved to

$\begin{bmatrix}x_2\\x_3\end{bmatrix} = \begin{bmatrix}\frac{1 + a}{B}&-\frac{a r}{B}\\\frac{b}{R}&-\frac{b r}{R}\end{bmatrix}\begin{bmatrix}y\\x_1\end{bmatrix}$

As one can see, $x_3 = \frac{b(y - r x_1)}{R}$ simply means the voltage transformer across the loop1 inductor primary side, aka $y - r x_1$ , raised by $b$ ratio to the secondary side at the time and consumed in loop3. This is the period that the inverter's design targets. Plug in the formula for $x_2$ and $x_3$ , the system now at the time becomes:

$\begin{bmatrix}y\\x_0\\x_1\end{bmatrix}'= \begin{bmatrix}0\\\frac{V}{L_0}\\0\end{bmatrix} + \begin{bmatrix}-\frac{1 + a}{B C}&\frac{1}{C}&-\frac{B - a r}{B C}\\-\frac{1}{L_0}&0&0\\\frac{1}{L}&0&-\frac{r}{L}\end{bmatrix} \begin{bmatrix}y\\x_0\\x_1\end{bmatrix}$

Then, after some pivotsystem and swapsystem to replace $y$ and $x_0$ :

$x_1''' = \frac{V}{C L L_0} - \frac{L + a L + B C r}{B C L} x_1'' - \left(\frac{1}{C L_0} + \frac{B + r}{B C L} \right) x_1' - \frac{r}{C L L_0} x_1$

At the period, $x_1$ moves like $0 = \frac{V}{C L L_0} - \left(\frac{1}{C L_0} + \frac{B + r}{B C L} \right) x_1' - \frac{r}{C L L_0} x_1$ aka $x_1' = \frac{B V}{B (L + L_0) + L_0 r} - \frac{B r}{B (L + L_0) + L_0 r} x_1$ . Since $x_1$ at the period is far away from its steady state $\frac{V}{r}$ , there is positive $x_1'$ at $\frac{B V}{B (L + L_0) + L_0 r}$

==== case 3 ====

The period is such that "too long for all variables", then zero derivatives in the linear system and solved to $\begin{bmatrix}y\\x_0\\x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}V\\\left(\frac{1}{B} + \frac{1}{r} \right) V\\\frac{V}{r}\\\frac{V}{B}\\0\end{bmatrix}$ and $x_3$ drops to 0 from $\frac{b (y - r x_1)}{R}$ of the previous period. This period is not what interests inverter's designers and is never reached and the loop1 MOSFET is OFF to next stage.

Typically 33k Hz is the period frequency for an inverter's MOSFET with respect to its C and L.

Ideal transformer

When $a=1,k=b^2$

$0 =\begin{bmatrix}V\\0\\0\\0\\0\end{bmatrix}-\begin{bmatrix}&L_0&&&\\C&&&&\\&&L&-L&-b L\\&&-L&L&b L\\&&-b L&b L&b^2 L\end{bmatrix}\begin{bmatrix}y\\x_0\\x_1\\x_2\\x_3\end{bmatrix}'-\begin{bmatrix}1&&&&\\&-1&1&1&\\-1&&r&&\\-1&&&B&\\&&&&R\end{bmatrix}\begin{bmatrix}y\\x_0\\x_1\\x_2\\x_3\end{bmatrix}$

becomes

$0 =\begin{bmatrix}V\\0\\0\\0\\0\end{bmatrix}-\begin{bmatrix}&L_0&&&\\C&&&&\\&&0&0&0\\&&0&0&0\\&&-b L&b L&b^2 L\end{bmatrix}\begin{bmatrix}y\\x_0\\x_1\\x_2\\x_3\end{bmatrix}'-\begin{bmatrix}1&&&&\\&-1&1&1&\\-1&&r&&\frac{R}{b}\\-1&&&B&-\frac{R}{b}\\&&&&R\end{bmatrix}\begin{bmatrix}y\\x_0\\x_1\\x_2\\x_3\end{bmatrix}$

i.e. a system of $y,x_0,x_3$ by

$\begin{bmatrix}y\\x_0\\x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}1&&\\&1&\\\frac{1}{r}&&-\frac{R}{b r}\\\frac{1}{B}&&\frac{R}{b B}\\&&1\end{bmatrix} \begin{bmatrix}y\\x_0\\x_3\end{bmatrix}$

the system becomes

$0 =\begin{bmatrix}V\\0\\0\end{bmatrix}-\begin{bmatrix}&L_0&\\C&&\\b L(\frac{1}{B}-\frac{1}{r})&&L(\frac{R}{r}+\frac{R}{B}+b^2)\end{bmatrix}\begin{bmatrix}y\\x_0\\x_3\end{bmatrix}'-\begin{bmatrix}1&&\\\frac{1}{B}+\frac{1}{r}&-1&\frac{R}{b} (\frac{1}{B}-\frac{1}{r})\\&&R\end{bmatrix}\begin{bmatrix}y\\x_0\\x_3\end{bmatrix}$

aka

$\begin{bmatrix}y\\x_0\\x_3\end{bmatrix}'=\begin{bmatrix}0\\\frac{V}{L_0}\\0\end{bmatrix}-\begin{bmatrix}\frac{B+r}{B r C}&-\frac{1}{C}&\frac{(r-B)R}{b B r C}\\\frac{1}{L_0}&&\\\frac{b(B+r)(B-r)}{C(b^2 B r+(B+r)R)B r}&\frac{b(r-B)}{C(b^2 B r+(B+r)R)}&\frac{(B^2 r^2 C-L(B-r)^2)R}{L C(b^2 B r+(B+r)R)B r}\end{bmatrix} \begin{bmatrix}y\\x_0\\x_3\end{bmatrix}$