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: Relating Poles to the Characteristic Equation
The closed-loop poles are given as \(s = -2 \pm j3\). The characteristic equation can be reconstructed from these roots: \[ (s - (-2 + j3))(s - (-2 - j3)) = 0 \] \[ ((s+2) - j3)((s+2) + j3) = (s+2)^2 - (j3)^2 = s^2 + 4s + 4 + 9 = s^2 + 4s + 13 = 0 \]
Step 2: Deriving the Characteristic Equation from the State Matrix
The eigenvalues of the matrix \(A = \begin{bmatrix} a & b \\ -a & 0 \end{bmatrix}\) are the poles of the system. We find them using \(\det(sI - A) = 0\): \[ \det\left( \begin{bmatrix} s & 0 \\ 0 & s \end{bmatrix} - \begin{bmatrix} a & b \\ -a & 0 \end{bmatrix} \right) = \det \begin{bmatrix} s-a & -b \\ a & s \end{bmatrix} = 0 \] \[ (s-a)(s) - (-b)(a) = s^2 - as + ab = 0 \]
Step 3: Comparing Coefficients to find \(b\)
We now have two expressions for the characteristic equation: 1) \(s^2 + 4s + 13 = 0\) 2) \(s^2 - as + ab = 0\) By comparing the coefficient of \(s\): \(-a = 4 \implies a = -4\). By comparing the constant term: \(ab = 13\). Substitute \(a = -4\): \((-4)b = 13 \implies b = -\frac{13}{4} = -3.25\).