Let's solve each part step by step: --- ## **Given:** \[ A = \begin{bmatrix} & 1 \\ -6 & -5 \end{bmatrix}, \quad B = \begin{bmatrix} \\ 1 \end{bmatrix}, \quad C = \begin{bmatrix} 1 & \end{bmatrix}, \quad D = \] --- ## **i) State Transition Matrix** \(\Phi(t) = e^{At}\): **Laplace inverse method:** \[ \Phi(t) = \mathcal{L}^{-1}\{ (sI - A)^{-1} \} \] ### **Step 1: Find \( sI - A \)** \[ sI - A = \begin{bmatrix} s & \\ & s \end{bmatrix} - \begin{bmatrix} & 1 \\ -6 & -5 \end{bmatrix} = \begin{bmatrix} s & -1 \\ 6 & s + 5 \end{bmatrix} \] ### **Step 2: Find \( (sI - A)^{-1} \)** For a 2x2 matrix \( M = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \): \[ M^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] - **Determinant:** \[ \det(sI - A) = (s)(s+5) - (-1)(6) = s^2 + 5s + 6 \] - **Inverse:** \[ (sI - A)^{-1} = \frac{1}{s^2 + 5s + 6} \begin{bmatrix} s+5 & 1 \\ -6 & s \end{bmatrix} \] ### **Step 3: Find Laplace inverse** Factor denominator: \[ s^2 + 5s + 6 = (s+2)(s+3) \] - **First Element (\(s+5\)):** \[ \frac{s+5}{(s+2)(s+3)} = \frac{A}{s+2} + \frac{B}{s+3} \] \[ s+5 = A(s+3) + B(s+2) \] Let \(s = -3\): \(2 = B(-1) \implies B = -2\) Let \(s = -2\): \(3 = A(1) \implies A = 3\) So, \[ \frac{s+5}{(s+2)(s+3)} = \frac{3}{s+2} - \frac{2}{s+3} \] - **Second Element (\(1\)):** \[ \frac{1}{(s+2)(s+3)} = \frac{C}{s+2} + \frac{D}{s+3} \] \[ 1 = C(s+3) + D(s+2) \] Let \(s=-3\): \(1 = D(-1) \implies D = -1\) Let \(s = -2\): \(1 = C(1) \implies C = 1\) So, \[ \frac{1}{(s+2)(s+3)} = \frac{1}{s+2} - \frac{1}{s+3} \] - **Third Element (\(-6\)):** \[ \frac{-6}{(s+2)(s+3)} = -6 \left(\frac{1}{s+2} - \frac{1}{s+3}\right) \] - **Fourth Element (\(s\)):** \[ \frac{s}{(s+2)(s+3)} = \frac{E}{s+2} + \frac{F}{s+3} \] \[ s = E(s+3) + F(s+2) \] Let \(s=-3\): \(-3 = F(-1) \implies F = 3\) Let \(s=-2\): \(-2 = E(1) \implies E = -2\) So, \[ \frac{s}{(s+2)(s+3)} = \frac{-2}{s+2} + \frac{3}{s+3} \] ### **Step 4: Inverse Laplace** \[ \mathcal{L}^{-1}\left\{\frac{1}{s+a}\right\} = e^{-at} \] So, \[ \Phi(t) = e^{At} = \begin{bmatrix} 3e^{-2t} - 2e^{-3t} & e^{-2t} - e^{-3t} \\ -6e^{-2t} + 6e^{-3t} & -2e^{-2t} + 3e^{-3t} \end{bmatrix} \] --- ## **ii) Time Response \(y(t)\) for \(u(t) = e^{-t}\), \(x() = \begin{bmatrix} 1 \\ \end{bmatrix}\):** General solution: \[ x(t) = \Phi(t)x() + \int_^t \Phi(t-\tau) B u(\tau) d\tau \] \[ y(t) = Cx(t) \] ### **Step 1: Compute \(\Phi(t)x()\):** \[ \Phi(t)x() = \begin{bmatrix} 3e^{-2t} - 2e^{-3t} \\ -6e^{-2t} + 6e^{-3t} \end{bmatrix} \] ### **Step 2: Compute Forced Response** \[ \int_^t \Phi(t-\tau) B u(\tau) d\tau \] Here, \(B = \begin{bmatrix} \\ 1 \end{bmatrix}\), so only the second column of \(\Phi(t-\tau)\) matters. \[ \int_^t \begin{bmatrix} e^{-2(t-\tau)} - e^{-3(t-\tau)} \\ -2e^{-2(t-\tau)} + 3e^{-3(t-\tau)} \end{bmatrix} e^{-\tau} d\tau \] Let’s write the first entry (which is what \(y(t)\) will be, since \(C = [1 \quad ]\)): \[ y(t) = [1 \quad ] \Bigg( \begin{bmatrix} 3e^{-2t} - 2e^{-3t} \\ -6e^{-2t} + 6e^{-3t} \end{bmatrix} + \int_^t \begin{bmatrix} e^{-2(t-\tau)} - e^{-3(t-\tau)} \\ -2e^{-2(t-\tau)} + 3e^{-3(t-\tau)} \end{bmatrix} e^{-\tau} d\tau \Bigg) \] So, \[ y(t) = 3e^{-2t} - 2e^{-3t} + \int_^t [e^{-2(t-\tau)} - e^{-3(t-\tau)}] e^{-\tau} d\tau \] \[ = 3e^{-2t} - 2e^{-3t} + \int_^t \left(e^{-2(t-\tau)} e^{-\tau} - e^{-3(t-\tau)} e^{-\tau}\right) d\tau \] \[ = 3e^{-2t} - 2e^{-3t} + \int_^t \left(e^{-2t+\tau} e^{-\tau} - e^{-3t+2\tau} e^{-\tau}\right) d\tau \] \[ = 3e^{-2t} - 2e^{-3t} + \int_^t \left(e^{-2t+\tau} e^{-\tau} - e^{-3t+\tau}\right) d\tau \] But actually, let's keep it simple: Note that \(e^{-2(t-\tau)} e^{-\tau} = e^{-2t+2\tau} e^{-\tau} = e^{-2t+\tau}\) Similarly, \(e^{-3(t-\tau)} e^{-\tau} = e^{-3t+3\tau} e^{-\tau} = e^{-3t+2\tau}\) So, \[ y(t) = 3e^{-2t} - 2e^{-3t} + \int_^t \left( e^{-2t+\tau} - e^{-3t+2\tau} \right) d\tau \] Evaluate the integrals: - For \(e^{-2t+\tau}\): \[ \int_^t e^{-2t+\tau} d\tau = e^{-2t} \int_^t e^{\tau} d\tau = e^{-2t} [e^{\tau}]_^t = e^{-2t} (e^{t} - 1) \] - For \(e^{-3t+2\tau}\): \[ \int_^t e^{-3t+2\tau} d\tau = e^{-3t} \int_^t e^{2\tau} d\tau = e^{-3t} \left[ \frac{1}{2} e^{2\tau} \right]_^t = e^{-3t} \left( \frac{1}{2} e^{2t} - \frac{1}{2} \right) \] Plug back in: \[ y(t) = 3e^{-2t} - 2e^{-3t} + \left[ e^{-2t} (e^{t} - 1) - e^{-3t} \frac{1}{2} (e^{2t} - 1) \right] \] \[ = 3e^{-2t} - 2e^{-3t} + e^{-2t} e^{t} - e^{-2t} - \frac{1}{2} e^{-3t} e^{2t} + \frac{1}{2} e^{-3t} \] \[ = 3e^{-2t} - 2e^{-3t} + e^{-t} - e^{-2t} - \frac{1}{2} e^{-t} + \frac{1}{2} e^{-3t} \] Now, combine like terms: - \(e^{-t}\): \(e^{-t} - \frac{1}{2} e^{-t} = \frac{1}{2} e^{-t}\) - \(e^{-2t}\): \(3e^{-2t} - e^{-2t} = 2e^{-2t}\) - \(e^{-3t}\): \(-2e^{-3t} + \frac{1}{2} e^{-3t} = -\frac{3}{2} e^{-3t}\) So, **final answer:** \[ \boxed{ y(t) = \frac{1}{2} e^{-t} + 2 e^{-2t} - \frac{3}{2} e^{-3t} } \] --- ## **iii) Transfer Function** \[ H(s) = C (sI - A)^{-1} B + D \] Recall: \[ (sI - A)^{-1} = \frac{1}{(s+2)(s+3)} \begin{bmatrix} s+5 & 1 \\ -6 & s \end{bmatrix} \] and \(B = \begin{bmatrix} \\ 1 \end{bmatrix}\), \(C = \begin{bmatrix} 1 & \end{bmatrix}\), \(D = \) So, \[ H(s) = \begin{bmatrix} 1 & \end{bmatrix} \left( \frac{1}{(s+2)(s+3)} \begin{bmatrix} s+5 & 1 \\ -6 & s \end{bmatrix} \right) \begin{bmatrix} \\ 1 \end{bmatrix} \] Only the (1,2) entry is used: \[ H(s) = \frac{1}{(s+2)(s+3)} \cdot 1 = \frac{1}{(s+2)(s+3)} \] ### **Poles and Stability:** Poles at \(s = -2, -3\) (both negative real parts). **Conclusion:** **The system is stable** (all poles in the left half-plane). --- ## **Summary of Answers** ### **i) State Transition Matrix** \[ \boxed{ \Phi(t) = e^{At} = \begin{bmatrix} 3e^{-2t} - 2e^{-3t} & e^{-2t} - e^{-3t} \\ -6e^{-2t} + 6e^{-3t} & -2e^{-2t} + 3e