Question

Consider a linear time-invariant system represented by the state-space equation: \[ \dot{x} = \begin{bmatrix} a & b -a & 0 \end{bmatrix} x + \begin{bmatrix} 1 0 \end{bmatrix} u \] The closed-loop poles of the system are located at \(-2 \pm j3\). The value of the parameter \(b\) is: 

• A. 3.25
• B. -3.25
• C. 13
• D. -13

Hint:

For any 2x2 matrix \(A = \begin{bmatrix} p & q
r & t \end{bmatrix}\), the characteristic equation is always \(s^2 - (p+t)s + (pt-qr) = 0\), where \((p+t)\) is the trace and \((pt-qr)\) is the determinant of \(A\). This can be a faster way to find the equation. In this case, trace = \(a+0=a\), determinant = \(a(0) - b(-a) = ab\). The equation is \(s^2 - (a)s + (ab) = 0\), which matches our result.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the state-space representation of a system with a state matrix \(A\). The poles of a linear time-invariant system are the eigenvalues of its state matrix \(A\). We are given the locations of these poles and need to find the value of an unknown parameter \(b\) in the matrix \(A\).
Step 2: Key Formula or Approach:
The poles of the system are the roots of the characteristic equation, which is given by \(\det(sI - A) = 0\), where \(s\) is the Laplace variable, \(I\) is the identity matrix, and \(A\) is the state matrix.
Step 3: Detailed Explanation:
1. Formulate the characteristic equation from the matrix A:
The state matrix is \(A = \begin{bmatrix} a & b \\ -a & 0 \end{bmatrix}\). \[ sI - A = s\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} a & b \\ -a & 0 \end{bmatrix} = \begin{bmatrix} s-a & -b \\ a & s \end{bmatrix} \] The determinant is: \[ \det(sI - A) = (s-a)(s) - (-b)(a) = s^2 - as + ab \] So, the characteristic equation is \(s^2 - as + ab = 0\).
2. Formulate the characteristic equation from the given poles:
The poles are located at \(-2 + j3\) and \(-2 - j3\). The characteristic equation can be formed from its roots: \[ (s - (-2 + j3))(s - (-2 - j3)) = 0 \] \[ ((s+2) - j3)((s+2) + j3) = 0 \] Using the identity \((x-y)(x+y) = x^2 - y^2\): \[ (s+2)^2 - (j3)^2 = 0 \] \[ (s^2 + 4s + 4) - (j^2 \cdot 9) = 0 \] \[ s^2 + 4s + 4 - (-9) = 0 \] \[ s^2 + 4s + 13 = 0 \] 3. Compare coefficients to find a and b:
Now we equate the two forms of the characteristic equation: \[ s^2 - as + ab = s^2 + 4s + 13 \] By comparing the coefficients of the powers of \(s\):

• Coefficient of \(s\): \(-a = 4 \implies a = -4\)
• Constant term: \(ab = 13\)

Substitute the value of \(a\) into the second equation: \[ (-4)b = 13 \] \[ b = -\frac{13}{4} = -3.25 \] Step 4: Final Answer:
The value of the parameter \(b\) is -3.25.