How to create the Smith Mcmillan form of a polynomial matrix?

Question

Here is the matrix:

$$ G(s) = \begin{bmatrix} \frac{4}{(s+1)(s+2)} & \frac{-0.5}{s+1} \\ \frac{1}{s+2} & \frac{2}{(s+1)(s+2)}\\ \end{bmatrix} $$

And this should be the final answer: $$ M(s) = \bar L(s)G(s) \bar R(s) = \begin{bmatrix} \frac{1}{(s+1)(s+2)} & 0\\ 0 & \frac{s^2+3s+18}{(s+1)(s+2)}\\ \end{bmatrix} $$

Where: $$ \bar L(s) = \begin{bmatrix} \frac{1}{4} & 0\\ -2(s+1) & 8\\ \end{bmatrix} \bar R(s) = \begin{bmatrix} 1 & \frac {s+2}{8}\\ 0 & 1\\ \end{bmatrix} $$

I also can't grasp the idea how the uni-modular matrices $\bar L(s)$ and $\bar R(s)$ where derived to setup the smith form M(s).

Would appreciated any answer to this question :)

Answer 1

Smith McMillan is a variation of the Smith Normal Form (SNF) for rational matrices. Basically all we have to do is factor out the least common multiple of the denominators to get a polynomial matrix and apply the SNF

$$ P = \begin{bmatrix} \frac{4}{(s+1)(s+2)} & \frac{-0.5}{s+1} \\ \frac{1}{s+2} & \frac{2}{(s+1)(s+2)}\\ \end{bmatrix} = \frac{1}{(s+1)(s+2)} \begin{bmatrix} 4 & -\frac{1}{2}(s+2) \\ s+1 & 2\\ \end{bmatrix} = \frac{1}{(s+1)(s+2)}L^{-1}(s) \begin{bmatrix} 1 & 0 \\ 0 & s^2 + 3s +18\\ \end{bmatrix} R^{-1}(s) $$

And to get the SNF of a matrix what you do is you multiply elementary matrices from the left and right. I.e. row/col swaps or adding a multiple of a row/col onto another. Obviously any constant entry will do.

You start by swapping an non-zero element to the 1-1 position that divides all other elements of the matrix.$^1$ $$\begin{bmatrix} 4 & -\frac{1}{2}(s+2) \\ s+1 & 2\\ \end{bmatrix}$$

So now you we can eliminate the other entries of the first row and column by multiplying the matrix $L_1$ that adds $ -\frac 14 (s+1)$ times the first row to the second and the matrix $R_1$ that adds $\frac 18(s+2) $ times the first column to the second from the left/right:

$$ \underbrace{\begin{bmatrix} 1 & 0 \\-\frac 14 (s+1) & 1 \end{bmatrix}}_{L_1} \cdot \begin{bmatrix} 4 & -\frac{1}{2}(s+2) \\ s+1 & 2\\ \end{bmatrix} \cdot \underbrace{\begin{bmatrix} 1 & \frac 18(s+2) \\0 & 1 \end{bmatrix}}_{R_1} = \begin{bmatrix} 4 & 0 \\ 0 & \frac 18 (s+1)(s+2) + 2 \end{bmatrix} $$

And now we only need to normalize by multplying the first row/col by $\frac 14$ and $8$ respectively, which we do by setting

$$ L = \begin{bmatrix} \frac{1}{4} & 0 \\ 0 & 8 \end{bmatrix} \cdot L_1 = \begin{bmatrix} \frac{1}{4} & 0\\ -2(s+1) & 8 \end{bmatrix} $$

$^1$If such an element does not exist, we take a non-zero element of minimal degree and perform Euclids algorithm and use it to reduce all other elements, then try again.

Answer 2

Multiplying $P$ by $(s+1)(s+2)$ gives the matrix $$ Q=\begin{pmatrix} 4 & -\frac{s+2}{2} \cr s+1 & 2 \end{pmatrix} $$ The Smith normal form then obviously is $$ Q_S=\begin{pmatrix} 1 & 0 \cr 0 & s^2+3s+18 \end{pmatrix} $$ where the transforming matrix $S$ with $SQ_SS^{-1}=Q$ is given by $$ S=\begin{pmatrix} 4 & 0 \cr s+1 & \frac{1}{4} \end{pmatrix} $$ Now dividing again by $(s+1)(s+2)$ we obtain your result.