Question

A system is characterized by the following state equation and output equation (\(u\): input, \(\mathbf{x}\): state vector, \(y\): output)
\[ \dot{\mathbf{x}}=\begin{bmatrix}a&b\\-a&0\end{bmatrix}\mathbf{x}+\begin{bmatrix}1\\0\end{bmatrix}u \]
\[ y=\begin{bmatrix}1&2\end{bmatrix}\mathbf{x} \]
What are the values of \(a\) and \(b\) for which the poles of the transfer function are at \(-2+j3\) and \(-2-j3\)?

• A. \(a=4,\ b=3.25\)
• B. \(a=-4,\ b=3.25\)
• C. \(a=4,\ b=-3.25\)
• D. \(a=-4,\ b=-3.25\)

Hint:

Match the characteristic polynomial of the state matrix to the quadratic whose roots are the two given poles.

Answer 1

The Correct Option is D

Step 1: Write the characteristic equation of the system.
For the state matrix
\[ A=\begin{bmatrix}a&b\\-a&0\end{bmatrix} \]
the poles of the transfer function are the eigenvalues of \(A\), found from \(\det(sI-A)=0\).

Step 2: Compute \(sI-A\).
\[ sI-A=\begin{bmatrix}s-a&-b\\a&s\end{bmatrix} \]

Step 3: Take the determinant.
\[ \det(sI-A)=(s-a)(s)-(-b)(a)=s^2-as+ab \]
So the characteristic equation is
\[ s^2-as+ab=0 \]

Step 4: Write the desired characteristic equation from the given poles.
The poles are \(-2+j3\) and \(-2-j3\). A quadratic with these roots is
\[ (s-(-2+j3))(s-(-2-j3))=(s+2)^2+9=s^2+4s+4+9=s^2+4s+13 \]

Step 5: Match coefficients.
Comparing
\[ s^2-as+ab=s^2+4s+13 \]
gives
\[ -a=4\quad\Rightarrow\quad a=-4 \]
and
\[ ab=13 \]

Step 6: Solve for \(b\).
\[ b=\frac{13}{a}=\frac{13}{-4}=-3.25 \]

Step 7: Final conclusion.
\[ \boxed{a=-4,\ b=-3.25} \]
Hence, the correct option is (D).