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: Use trace and determinant instead of expanding the determinant symbolically.
For a $2\times2$ matrix, the characteristic equation can always be written as
\[ s^2-\text{trace}(A)\,s+\det(A)=0 \]

Step 2: Compute the trace and determinant of the given matrix.
\[ A=\begin{bmatrix}a & b\\-a & 0\end{bmatrix} \]
\[ \text{trace}(A)=a+0=a \]
\[ \det(A)=a(0)-b(-a)=ab \]
So the characteristic equation is $s^2-as+ab=0$, matching the general formula.

Step 3: Use Vieta's formulas on the target poles.
The poles are $-2+j3$ and $-2-j3$. Their sum is
\[ (-2+j3)+(-2-j3)=-4 \]
and their product is
\[ (-2+j3)(-2-j3)=(-2)^2-(j3)^2=4-(-9)=13 \]

Step 4: Match sum and product to trace and determinant.
For $s^2-as+ab=0$, the sum of roots equals $a$ and the product of roots equals $ab$. So
\[ a=-4 \]
\[ ab=13 \]

Step 5: Solve for $b$.
\[ b=\frac{13}{-4}=-3.25 \]

Step 6: Conclude.
\[ \boxed{a=-4,\ b=-3.25} \]