1. **Problem Statement:** Given the linear system: $$\dot{x} = \begin{bmatrix} -6 & 4 \\ -2 & 0 \end{bmatrix} x + \begin{bmatrix} 1 \\ 1 \end{bmatrix} u(t)$$ $$Y = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} x$$ with input $u(t) = 1(t)$ (unit step), initial conditions $x_1(0) = 1$, $x_2(0) = 0$. Find: (i) The complete solution for $y(t)$. (ii) The transfer function $G(s) = \frac{Y(s)}{U(s)}$. 2. **Step 1: Write the system in state-space form** State matrix: $$A = \begin{bmatrix} -6 & 4 \\ -2 & 0 \end{bmatrix}$$ Input matrix: $$B = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$ Output matrix: $$C = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$$ 3. **Step 2: Find the homogeneous solution $x_h(t)$** Solve $\dot{x} = A x$ with initial condition $x(0) = \begin{bmatrix}1 \\ 0\end{bmatrix}$. Find eigenvalues of $A$ by solving: $$\det(sI - A) = 0 \Rightarrow \det\begin{bmatrix} s+6 & -4 \\ 2 & s \end{bmatrix} = (s+6)s - (-4)(2) = s^2 + 6s + 8 = 0$$ Solve quadratic: $$s = \frac{-6 \pm \sqrt{36 - 32}}{2} = \frac{-6 \pm 2}{2}$$ Eigenvalues: $$s_1 = -2, \quad s_2 = -4$$ 4. **Step 3: Find eigenvectors** For $s_1 = -2$: $$(A + 2I)v = 0 \Rightarrow \begin{bmatrix} -4 & 4 \\ -2 & 2 \end{bmatrix} v = 0$$ From first row: $-4 v_1 + 4 v_2 = 0 \Rightarrow v_2 = v_1$ Choose $v^{(1)} = \begin{bmatrix}1 \\ 1\end{bmatrix}$. For $s_2 = -4$: $$(A + 4I)v = 0 \Rightarrow \begin{bmatrix} -2 & 4 \\ -2 & 4 \end{bmatrix} v = 0$$ From first row: $-2 v_1 + 4 v_2 = 0 \Rightarrow v_2 = \frac{1}{2} v_1$ Choose $v^{(2)} = \begin{bmatrix}1 \\ \frac{1}{2}\end{bmatrix}$. 5. **Step 4: Write homogeneous solution** $$x_h(t) = c_1 e^{-2t} \begin{bmatrix}1 \\ 1\end{bmatrix} + c_2 e^{-4t} \begin{bmatrix}1 \\ \frac{1}{2}\end{bmatrix}$$ Apply initial condition $x(0) = \begin{bmatrix}1 \\ 0\end{bmatrix}$: $$c_1 \begin{bmatrix}1 \\ 1\end{bmatrix} + c_2 \begin{bmatrix}1 \\ \frac{1}{2}\end{bmatrix} = \begin{bmatrix}1 \\ 0\end{bmatrix}$$ From first component: $$c_1 + c_2 = 1$$ From second component: $$c_1 + \frac{1}{2} c_2 = 0$$ Subtract second from first: $$\cancel{c_1} + c_2 - (\cancel{c_1} + \frac{1}{2} c_2) = 1 - 0 \Rightarrow c_2 - \frac{1}{2} c_2 = 1 \Rightarrow \frac{1}{2} c_2 = 1 \Rightarrow c_2 = 2$$ Then: $$c_1 = 1 - c_2 = 1 - 2 = -1$$ 6. **Step 5: Find particular solution $x_p(t)$ for $u(t) = 1$** Since $u(t)$ is a step, steady state $x_p$ satisfies: $$0 = A x_p + B \Rightarrow A x_p = -B$$ Solve: $$\begin{bmatrix} -6 & 4 \\ -2 & 0 \end{bmatrix} \begin{bmatrix} x_{p1} \\ x_{p2} \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \end{bmatrix}$$ From first row: $$-6 x_{p1} + 4 x_{p2} = -1$$ From second row: $$-2 x_{p1} = -1 \Rightarrow x_{p1} = \frac{1}{2}$$ Substitute $x_{p1}$ into first row: $$-6 \times \frac{1}{2} + 4 x_{p2} = -1 \Rightarrow -3 + 4 x_{p2} = -1 \Rightarrow 4 x_{p2} = 2 \Rightarrow x_{p2} = \frac{1}{2}$$ So, $$x_p = \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \end{bmatrix}$$ 7. **Step 6: Complete solution for $x(t)$** $$x(t) = x_h(t) + x_p = - e^{-2t} \begin{bmatrix}1 \\ 1\end{bmatrix} + 2 e^{-4t} \begin{bmatrix}1 \\ \frac{1}{2}\end{bmatrix} + \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \end{bmatrix}$$ 8. **Step 7: Find output $y(t)$** Recall: $$Y = C x = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} x$$ Calculate: $$y(t) = C x(t) = C x_h(t) + C x_p$$ Calculate $C x_h(t)$: $$C \begin{bmatrix}1 \\ 1\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix} \begin{bmatrix}1 \\ 1\end{bmatrix} = \begin{bmatrix}1 \\ 2\end{bmatrix}$$ $$C \begin{bmatrix}1 \\ \frac{1}{2}\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix} \begin{bmatrix}1 \\ \frac{1}{2}\end{bmatrix} = \begin{bmatrix}1 \\ 1 + \frac{1}{2} \end{bmatrix} = \begin{bmatrix}1 \\ \frac{3}{2} \end{bmatrix}$$ Calculate $C x_p$: $$C \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} + \frac{1}{2} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \\ 1 \end{bmatrix}$$ So, $$y(t) = - e^{-2t} \begin{bmatrix}1 \\ 2\end{bmatrix} + 2 e^{-4t} \begin{bmatrix}1 \\ \frac{3}{2}\end{bmatrix} + \begin{bmatrix} \frac{1}{2} \\ 1 \end{bmatrix}$$ 9. **Step 8: Transfer function $G(s)$** Recall: $$G(s) = C (sI - A)^{-1} B$$ Calculate $sI - A$: $$sI - A = \begin{bmatrix} s+6 & -4 \\ 2 & s \end{bmatrix}$$ Calculate its inverse: $$\det = (s+6)s - (-4)(2) = s^2 + 6s + 8$$ $$ (sI - A)^{-1} = \frac{1}{s^2 + 6s + 8} \begin{bmatrix} s & 4 \\ -2 & s+6 \end{bmatrix}$$ Multiply by $B$: $$(sI - A)^{-1} B = \frac{1}{s^2 + 6s + 8} \begin{bmatrix} s & 4 \\ -2 & s+6 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \frac{1}{s^2 + 6s + 8} \begin{bmatrix} s + 4 \\ -2 + s + 6 \end{bmatrix} = \frac{1}{s^2 + 6s + 8} \begin{bmatrix} s + 4 \\ s + 4 \end{bmatrix}$$ Finally multiply by $C$: $$G(s) = C (sI - A)^{-1} B = \frac{1}{s^2 + 6s + 8} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} s + 4 \\ s + 4 \end{bmatrix} = \frac{1}{s^2 + 6s + 8} \begin{bmatrix} s + 4 \\ (s + 4) + (s + 4) \end{bmatrix} = \frac{1}{s^2 + 6s + 8} \begin{bmatrix} s + 4 \\ 2s + 8 \end{bmatrix}$$ **Final answers:** $$y(t) = \begin{bmatrix} - e^{-2t} + 2 e^{-4t} + \frac{1}{2} \\ -2 e^{-2t} + 3 e^{-4t} + 1 \end{bmatrix}$$ $$G(s) = \frac{1}{s^2 + 6s + 8} \begin{bmatrix} s + 4 \\ 2s + 8 \end{bmatrix}$$