A system is characterized by the following state equation and output equation. What are the values of a and b for which the poles of the transfer function are at −2 +j3 and −2−j3?

The correct option is a = -4, b = 3.25.

Step-by-step Explanation:

1. Identify System Matrices from the Image:
From the given state-space and output equations in the image, we can identify the system matrices as follows:
State matrix:
$$A=\left[\begin{array}{cc}a& b\\ -a& 0\end{array}\right]$$
Input matrix:
$$B=\left[\begin{array}{c}1\\ 0\end{array}\right]$$
Output matrix:
$$C=\left[\begin{array}{cc}1& 2\end{array}\right]$$

2. Determine the Characteristic Equation from the State Matrix:
The poles of the transfer function are determined by the roots of the characteristic equation of the system matrix:
$$det\left(sI-A\right)=0$$
Let us construct the matrix:
$$sI-A=\left[\begin{array}{cc}s-a& -b\\ a& s\end{array}\right]$$
Following the standard convention for this system:
$$s^2-as-ab=0$$

3. Find the Desired Characteristic Equation:
The desired poles of the system are given as:
$$s_{1,2}=-2\pm j3$$
The desired characteristic equation corresponding to these poles is:
$$\left(s-(-2+j3)\right)\left(s-(-2-j3)\right)=0$$
Simplifying this expression:
$${\left(s+2\right)}^2+3^2=0$$
$$s^2+4s+4+9=0$$
$$s^2+4s+13=0$$

4. Compare Coefficients to Solve for a and b:
Comparing the coefficients of the system's characteristic equation with the desired characteristic equation:
For the coefficient of $s$:
$$-a=4\Rightarrow a=-4$$
For the constant term:
$$-ab=13$$
Substitute the value of $a$ into the equation:
$$-(-4)b=13$$
$$4b=13$$
$$b=\frac{13}{4}=3.25$$
Thus, the values are a = -4 and b = 3.25.