0

Code:

$$x(t) = 
\left[ \begin{array}{c}
i_{L1}(t)\\
i_{L2}(t)\\
v_{C1}(t)\\
v_{C2}(t)\\
i_l(t)
\end{array}\right]
$$
Recall that the input voltage $V_g$ is an independent voltage source and the load impedence is $Z_l = R_l + sL_l$. Apparently, the load is short-circuited by $S_2$ in Mode 1.\\
The circuit equations in Mode 1 can be written in the state space form $K\cdot x = A_1\cdot x + B_1\cdot u$, that is,

$$\left[ \begin{array}{ccccc}
L_1 & 0 & 0 & 0 & 0\\
0 & L_2 & 0 & 0 & 0\\
0 & 0 & C_1 & 0 & 0\\
0 & 0 & 0 & C_2 & 0\\
0 & 0 & 0 & 0 & L_l
end{array}\right] \frac{d}{dt} \left[ \begin{array}{c}
i_{L1}(t)\\
i_{L2}(t)\\
v_{C1}(t)\\
v_{C2}(t)\\
i_l(t)
\end{array}\right] = \left[ \begin{array}{ccccc}
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
-1 & 0 & 0 & 0 & 0\\
0 & -1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & -R_l
\end{array}\right] \left[ \begin{array}{c}
i_{L1}(t)\\
i_{L2}(t)\\
v_{C1}(t)\\
v_{C2}(t)\\
i_l(t)
\end{array}\right]$$

Error:

! Extra \right.
l.25 end{array}\right]
                       \frac{d}{dt} \left[ \begin{array}{c}
Reddy
  • 31

1 Answers1

3

It's easier to set the matrices with package amsmath. This also improves the horizontal spacing. Things like the missing \ or not using $$ are already covered by the question comments.

\documentclass{article}
\usepackage{amsmath}

\begin{document}
\[
  x(t) = 
  \begin{bmatrix}
    i_{L1}(t)\\
    i_{L2}(t)\\
    v_{C1}(t)\\
    v_{C2}(t)\\  
    i_l(t)
  \end{bmatrix}
\]
Recall that the input voltage $V_g$ is an independent voltage source and the
load impedence is $Z_l = R_l + sL_l$. Apparently, the load is
short-circuited by $S_2$ in Mode~1.\\
The circuit equations in Mode~1 can be written in the state space form
$K\cdot x = A_1\cdot x + B_1\cdot u$, that is,
%
\[
  \begin{bmatrix}
    L_1 & 0 & 0 & 0 & 0\\
    0 & L_2 & 0 & 0 & 0\\
    0 & 0 & C_1 & 0 & 0\\
    0 & 0 & 0 & C_2 & 0\\
    0 & 0 & 0 & 0 & L_l
  \end{bmatrix}
  \frac{d}{dt}
  \begin{bmatrix}
    i_{L1}(t)\\
    i_{L2}(t)\\ 
    v_{C1}(t)\\ 
    v_{C2}(t)\\ 
    i_l(t) 
  \end{bmatrix}
  =
  \begin{bmatrix}
    0 & 0 & 1 & 0 & 0\\
    0 & 0 & 0 & 1 & 0\\
    -1 & 0 & 0 & 0 & 0\\
    0 & -1 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & -R_l
  \end{bmatrix}
  \begin{bmatrix}
    i_{L1}(t)\\  
    i_{L2}(t)\\  
    v_{C1}(t)\\  
    v_{C2}(t)\\  
    i_l(t)
  \end{bmatrix}
\]
\end{document}

Result

Heiko Oberdiek
  • 271,626