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### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

### Modified nodal analysis

We add extra unknowns for

- branch currents for voltage sources
- all elements for which it is impossible or inconvenient to write \( i = f(v_1, v_2) \).
Imagine, for example, that we have a component illustrated in fig. 1.

\begin{equation}\label{eqn:multiphysicsL4:20}

v_1 – v_2 = 3 i^2

\end{equation} - any current which is controlling dependent sources, as in fig. 2.
- Inductors

\begin{equation}\label{eqn:multiphysicsL4:40}

v_1 – v_2 = L \frac{di}{dt}.

\end{equation}

# Solving large systems

We are interested in solving linear systems of the form

\begin{equation}\label{eqn:multiphysicsL4:60}

M \overline{{x}} = \overline{{b}},

\end{equation}

possibly with thousands of elements.

## Gaussian elimination

\begin{equation}\label{eqn:multiphysicsL4:80}

\begin{bmatrix}

M_{11} & M_{12} & M_{13} \\

M_{21} & M_{22} & M_{23} \\

M_{31} & M_{32} & M_{33} \\

\end{bmatrix}

\begin{bmatrix}

x_1 \\

x_2 \\

x_3 \\

\end{bmatrix}

=

\begin{bmatrix}

b_1 \\

b_2 \\

b_3 \\

\end{bmatrix}

\end{equation}

It’s claimed for now, to be seen later, that back substitution is the fastest way to arrive at the solution, less computationally complex than completion the diagonalization.

Steps

\begin{equation}\label{eqn:multiphysicsL4:100}

(1) \cdot \frac{M_{21}}{M_{11}} \Rightarrow

\begin{bmatrix}

M_{21} & \frac{M_{21}}{M_{11}} M_{12} & \frac{M_{21}}{M_{11}} M_{13} \\

\end{bmatrix}

\end{equation}

\begin{equation}\label{eqn:multiphysicsL4:120}

(2) \cdot \frac{M_{31}}{M_{11}} \Rightarrow

\begin{bmatrix}

M_{31} & \frac{M_{31}}{M_{11}} M_{32} & \frac{M_{31}}{M_{11}} M_{33} \\

\end{bmatrix}

\end{equation}

This gives

\begin{equation}\label{eqn:multiphysicsL4:140}

\begin{bmatrix}

M_{11} & M_{12} & M_{13} \\

0 & M_{22} – \frac{M_{21}}{ M_{11} } M_{12} & M_{23} – \frac{M_{21}}{M_{11}} M_{13} \\

0 & M_{32} – \frac{M_{31}}{M_{11}} M_{32} & M_{33} – \frac{M_{31}}{M_{11}} M_{33} \\

\end{bmatrix}

\begin{bmatrix}

x_1 \\

x_2 \\

x_3 \\

\end{bmatrix}

=

\begin{bmatrix}

b_1 \\

b_2 – \frac{M_{21}}{M_{11}} b_1 \\

b_3 – \frac{M_{31}}{M_{11}} b_1

\end{bmatrix}.

\end{equation}

Here the \( M_{11} \) element is called the {pivot}. Each of the \(

M_{j1}/M_{11} \) elements is called a {multiplier}. This operation can be written as

\begin{equation}\label{eqn:multiphysicsL4:160}

\begin{bmatrix}

M_{11} & M_{12} & M_{13} \\

0 & M_{22}^{(2)} & M_{23}^{(3)} \\

0 & M_{32}^{(2)} & M_{33}^{(3)} \\

\end{bmatrix}

\begin{bmatrix}

x_1 \\

x_2 \\

x_3 \\

\end{bmatrix}

=

\begin{bmatrix}

b_1 \\

b_2^{(2)} \\

b_3^{(2)} \\

\end{bmatrix}.

\end{equation}

Using \( M_{22}^{(2)} \) as the pivot this time, we form

\begin{equation}\label{eqn:multiphysicsL4:180}

\begin{bmatrix}

M_{11} & M_{12} & M_{13} \\

0 & M_{22}^{(2)} & M_{23}^{(3)} \\

0 & 0 & M_{33}^{(3)} – \frac{M_{32}^{(2)}}{M_{22}^{(2)}} M_{23}^{(2)} \\

\end{bmatrix}

\begin{bmatrix}

x_1 \\

x_2 \\

x_3 \\

\end{bmatrix}

=

\begin{bmatrix}

b_1 \\

b_2 – \frac{M_{21}}{M_{11}} b_1 \\

b_3 – \frac{M_{31}}{M_{11}} b_1

– \frac{M_{32}^{(2)}}{M_{22}^{(2)}} b_{2}^{(2)} \\

\end{bmatrix}.

\end{equation}

## LU decomposition

Through Gaussian elimination, we have transformed the system from

\begin{equation}\label{eqn:multiphysicsL4:200}

M x = b

\end{equation}

to

\begin{equation}\label{eqn:multiphysicsL4:220}

U x = y.

\end{equation}

Writing out our Gaussian transformation in the form \( U \overline{{x}} = b \) we have

\begin{equation}\label{eqn:multiphysicsL4:240}

U \overline{{x}} =

\begin{bmatrix}

1 & 0 & 0 \\

-\frac{M_{21}}{M_{11}} & 1 & 0 \\

\frac{M_{32}^{(2)}}{M_{22}^{(2)}}

\frac{M_{21}}{M_{11}}

–

\frac{M_{31}}{M_{11}}

&

–

\frac{M_{32}^{(2)}}{M_{22}^{(2)}}

& 1

\end{bmatrix}

\begin{bmatrix}

b_1 \\

b_2 \\

b_3

\end{bmatrix}.

\end{equation}

We can verify that the operation matrix \( K^{-1} \), where \( K^{-1} U = M \) that takes us to this form is

\begin{equation}\label{eqn:multiphysicsL4:260}

\begin{bmatrix}

1 & 0 & 0 \\

\frac{M_{21}}{M_{11}} & 1 & 0 \\

\frac{M_{31}}{M_{11}} &

\frac{M_{32}^{(2)}}{M_{22}^{(2)}} &

1

\end{bmatrix}

\begin{bmatrix}

U_{11} & U_{12} & U_{13} \\

0 & U_{22} & U_{23} \\

0 & 0 & U_{33} \\

\end{bmatrix}

\overline{{x}} = \overline{{b}}

\end{equation}

Using this {LU} decomposition is generally superior to standard Gaussian elimination, since we can use this for many different \(\overline{{b}}\) vectors using the same amount of work.

Our steps are

\begin{equation}\label{eqn:multiphysicsL4:280}

b

= M x

= L \lr{ U x}

\equiv L y.

\end{equation}

We can now solve \( L y = b \), using substitution for \( y_1 \), then \( y_2 \), and finally \( y_3 \). This is called {forward substitution}.

Finally, we can now solve

\begin{equation}\label{eqn:multiphysicsL4:300}

U x = y,

\end{equation}

using {back substitution}.

Note that we produced the vector \( y \) as a side effect of performing the Gaussian elimination process.