# Jordan Canonical Form matrix

I apologize in advance if I didn't give enough information. I don't really have the mathematical background to this type of problem.

Given the following equation:

$$G(s) = \frac{2s+3}{s^4 + 3s^2 + 3s^2 +s}$$

I need to put this into Jordan matrix form.

$$\dot{{x}}=Ax+Bu$$ $$y=Cx$$

First I did the partial fraction expression:

$$G(s) = 2s+3 = s^3(D+C) + s^2(B+2C+3D) + s(A+B+C+3D) + D$$

$$G(s) = \frac{-1}{(s+1)^3}+\frac{-3}{(s+1)^2}+\frac{-3}{(s+1)}+\frac{3}{s}$$

From this point, I'm not too sure if I'm correct about the $B$ matrix. Since there are 3 repeated roots, would the $B$ matrix contain three 1's?

$$\dot{{x}} = \begin{bmatrix}-1&1&0&0\\0&-1&1&0\\0&0&-1&0\\0&0&&0\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\\end{bmatrix} + \begin{bmatrix}0\\1\\1\\1\\\end{bmatrix}u$$

$$y = \begin{bmatrix}-1&-3&-3&3\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\\end{bmatrix}$$

It would be $\left( \begin{array}{cccc} 0 & 0 & 1 & 1 \\ \end{array} \right)^T$.

The reasoning is as follows. From the output equation, we know that

$$X_1(s)=\frac{1}{(s+1)^3}U(s) \ \ \ X_2(s)=\frac{1}{(s+1)^2}U(s) \ \ \ X_3(s)=\frac{1}{s+1}U(s) \ \ \ X_4(s)=\frac{1}{s}U(s)$$

This can be simplified as: $$X_1(s)=\frac{1}{s+1}X_2(s) \ \ \ X_2(s)=\frac{1}{s+1}X_3(s) \ \ \ X_3(s)=\frac{1}{s+1}U(s) \ \ \ X_4(s)=\frac{1}{s}U(s)$$

Now we are beginning to see that the input only appears in the last two equations.

Converting these to the time domain, using the inverse Laplace transform, we get the following equations:

$$\dot{x}_1+x_1=x_2$$ $$\dot{x}_2+x_2=x_3$$ $$\dot{x}_3+x_3=u$$ $$\dot{x}_4=u$$

which is the desired Jordan form.