Question

Given the state-space system:
\[ \dot{x} = \begin{bmatrix} -4 & -1.5 \\ 4 & 0 \end{bmatrix} x + \begin{bmatrix} 2 \\ 0 \end{bmatrix} u \] \[ y = \begin{bmatrix} 0.15 & 0.625 \end{bmatrix} x \] Find the Transfer Function (TF).

Hint:

For state-space to transfer function conversion, always use $G(s) = C(sI-A)^{-1}B + D$. Compute determinant carefully to avoid sign errors.

Answer 1

Step 1: Use Transfer Function Formula.
For a state-space system:
\[ G(s) = C (sI - A)^{-1} B + D \] Here $D=0$.
Given:
\[ A = \begin{bmatrix} -4 & -1.5 \\ 4 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 2 \\ 0 \end{bmatrix}, \quad C = \begin{bmatrix} 0.15 & 0.625 \end{bmatrix}. \]
Step 2: Compute $sI - A$.
\[ sI - A = \begin{bmatrix} s+4 & 1.5 \\ -4 & s \end{bmatrix}. \]
Step 3: Compute Determinant.
\[ |sI-A| = (s+4)s - (1.5)(-4) \] \[ = s^2 + 4s + 6. \]
Step 4: Compute Adjoint Matrix.
\[ \text{adj}(sI-A) = \begin{bmatrix} s & -1.5 \\ 4 & s+4 \end{bmatrix}. \]
Thus, \[ (sI-A)^{-1} = \frac{1}{s^2+4s+6} \begin{bmatrix} s & -1.5 \\ 4 & s+4 \end{bmatrix}. \]
Step 5: Multiply $(sI-A)^{-1} B$.
\[ \begin{bmatrix} s & -1.5\\ 4 & s+4 \end{bmatrix} \begin{bmatrix} 2 \\ 0 \end{bmatrix} = \begin{bmatrix} 2s \\ 8 \end{bmatrix}. \]
Hence, \[ (sI-A)^{-1}B = \frac{1}{s^2+4s+6} \begin{bmatrix} 2s \\ 8 \end{bmatrix}. \]
Step 6: Multiply by $C$.
\[ G(s) = \frac{1}{s^2+4s+6} \left[ 0.15(2s) + 0.625(8) \right]. \]
\[ = \frac{1}{s^2+4s+6} \left[ 0.3s + 5 \right]. \]
Final Transfer Function:
\[ \boxed{ G(s) = \frac{0.3s + 5}{s^2 + 4s + 6} }. \]