Let's work through each part step by step. --- ### **Given State Space Model** \[ \dot{x} = \begin{bmatrix} -4 & 1 & 0 \\ 0 & -5 & 1 \\ 0 & 0 & -2 \end{bmatrix} x + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} u \] \[ y = \begin{bmatrix} 1 & 1 & 0 \end{bmatrix} x \] --- ## **(a) Solve for \( y(t) \) with \( x(0) = [0~0~0]^T \) and \( u(t) = 10 \cdot 1(t) \) (step input)** ### **Step 1: Find the general solution for \( x(t) \)** The solution to the state space equation with zero initial condition: \[ x(t) = \int_0^t e^{A(t-\tau)} B u(\tau) d\tau \] Since \( u(t) = 10 \) for all \( t \geq 0 \): \[ x(t) = \int_0^t e^{A(t-\tau)} B \cdot 10 d\tau = 10 \int_0^t e^{A(t-\tau)} B d\tau \] Let \( s = t-\tau \), \( d\tau = -ds \), when \( \tau = 0, s = t \), when \( \tau = t, s = 0 \): \[ x(t) = 10 \int_{s = t}^{s = 0} e^{As} B (-ds) = 10 \int_{s = 0}^{s = t} e^{As} B ds \] So, \[ x(t) = 10 \int_0^t e^{A s} B ds \] ### **Step 2: Compute \( e^{A s} B \)** Since \( A \) is upper-triangular, its eigenvalues are on the diagonal: \( -4, -5, -2 \). Let's compute \( e^{A s} B \). Let’s observe that \( B = [0~0~1]^T \). So we only need the third column of \( e^{A s} \). For an upper-triangular (companion) matrix, the third column of \( e^{A s} \) can be computed recursively. #### For \( x_3 \): \[ \dot{x}_3 = -2 x_3 + u \] With zero initial condition and input \( u(t) = 10 \): This is a first order ODE: \[ x_3(t) = \int_0^t e^{-2(t-\tau)} 10 d\tau = 10 \int_0^t e^{-2(t-\tau)} d\tau = 10 \left[ -\frac{1}{2} e^{-2(t-\tau)} \right]_0^t \] \[ = 10 \left( -\frac{1}{2} (e^{-2(t-t)} - e^{-2(t-0)}) \right) = 10 \left( -\frac{1}{2} (1 - e^{-2t}) \right) = 5 (1 - e^{-2t}) \] So, \[ x_3(t) = 5 (1 - e^{-2t}) \] #### For \( x_2 \): \[ \dot{x}_2 = -5 x_2 + x_3 \] With zero initial condition: This is a first-order ODE with a forcing term \( x_3(t) \): Let’s use the integrating factor method. The integrating factor is \( e^{5t} \): \[ \frac{d}{dt}(x_2 e^{5t}) = x_3 e^{5t} \] \[ x_2(t) = e^{-5t} \int_0^t x_3(\tau) e^{5\tau} d\tau \] Plug in \( x_3(\tau) = 5(1-e^{-2\tau}) \): \[ x_2(t) = e^{-5t} \int_0^t 5(1 - e^{-2\tau}) e^{5\tau} d\tau \] \[ = 5 e^{-5t} \int_0^t (e^{5\tau} - e^{3\tau}) d\tau \] \[ = 5 e^{-5t} \left[ \int_0^t e^{5\tau} d\tau - \int_0^t e^{3\tau} d\tau \right] \] \[ = 5 e^{-5t} \left[ \frac{1}{5} (e^{5t} - 1) - \frac{1}{3} (e^{3t} - 1) \right] \] \[ = e^{-5t} \left[ (e^{5t} - 1) - \frac{5}{3} (e^{3t} - 1) \right] \] \[ = e^{-5t} (e^{5t} - 1) - \frac{5}{3} e^{-5t} (e^{3t} - 1) \] \[ = (1 - e^{-5t}) - \frac{5}{3} (e^{-2t} - e^{-5t}) \] \[ = 1 - e^{-5t} - \frac{5}{3} e^{-2t} + \frac{5}{3} e^{-5t} \] \[ = 1 - e^{-5t} - \frac{5}{3} e^{-2t} + \frac{5}{3} e^{-5t} \] \[ = 1 - e^{-5t} - \frac{5}{3} e^{-2t} + \frac{5}{3} e^{-5t} \] \[ = 1 - e^{-5t} - \frac{5}{3} e^{-2t} + \frac{5}{3} e^{-5t} \] \[ = 1 - \left( e^{-5t} - \frac{5}{3} e^{-5t} \right) - \frac{5}{3} e^{-2t} \] \[ = 1 - \left( \frac{3}{3} - \frac{5}{3} \right) e^{-5t} - \frac{5}{3} e^{-2t} \] \[ = 1 + \frac{2}{3} e^{-5t} - \frac{5}{3} e^{-2t} \] #### For \( x_1 \): \[ \dot{x}_1 = -4 x_1 + x_2 \] Zero initial conditions. Integrating factor is \( e^{4t} \): \[ \frac{d}{dt}(x_1 e^{4t}) = x_2 e^{4t} \] \[ x_1(t) = e^{-4t} \int_0^t x_2(\tau) e^{4\tau} d\tau \] Plug in \( x_2(\tau) \) from above: \[ x_2(\tau) = 1 + \frac{2}{3} e^{-5\tau} - \frac{5}{3} e^{-2\tau} \] So, \[ x_1(t) = e^{-4t} \int_0^t \left[ 1 + \frac{2}{3} e^{-5\tau} - \frac{5}{3} e^{-2\tau} \right] e^{4\tau} d\tau \] \[ = e^{-4t} \int_0^t \left[ e^{4\tau} + \frac{2}{3} e^{-5\tau} e^{4\tau} - \frac{5}{3} e^{-2\tau} e^{4\tau} \right] d\tau \] \[ = e^{-4t} \int_0^t \left[ e^{4\tau} + \frac{2}{3} e^{- \tau} - \frac{5}{3} e^{2\tau} \right] d\tau \] Now compute each term: - \( \int_0^t e^{4\tau} d\tau = \frac{1}{4} (e^{4t} - 1) \) - \( \int_0^t e^{-\tau} d\tau = - (e^{-\tau})|_0^t = -(e^{-t} - 1) = 1 - e^{-t} \) - \( \int_0^t e^{2\tau} d\tau = \frac{1}{2} (e^{2t} - 1) \) So, \[ x_1(t) = e^{-4t} \left[ \frac{1}{4}(e^{4t}-1) + \frac{2}{3}(1 - e^{-t}) - \frac{5}{3} \cdot \frac{1}{2}(e^{2t}-1) \right] \] \[ = e^{-4t} \left[ \frac{1}{4}(e^{4t}-1) + \frac{2}{3}(1 - e^{-t}) - \frac{5}{6}(e^{2t}-1) \right] \] Now expand: \[ = e^{-4t} \left[ \frac{1}{4} e^{4t} - \frac{1}{4} + \frac{2}{3} - \frac{2}{3} e^{-t} - \frac{5}{6} e^{2t} + \frac{5}{6} \right] \] Multiply \( e^{-4t} \): - \( \frac{1}{4} e^{4t} \cdot e^{-4t} = \frac{1}{4} \) - \( -\frac{1}{4} e^{-4t} \) - \( \frac{2}{3} e^{-4t} \) - \( -\frac{2}{3} e^{-4t} e^{-t} = -\frac{2}{3} e^{-5t} \) - \( -\frac{5}{6} e^{2t} e^{-4t} = -\frac{5}{6} e^{-2t} \) - \( \frac{5}{6} e^{-4t} \) So sum up: \[ x_1(t) = \frac{1}{4} - \frac{1}{4} e^{-4t} + \frac{2}{3} e^{-4t} - \frac{2}{3} e^{-5t} - \frac{5}{6} e^{-2t} + \frac{5}{6} e^{-4t} \] Now collect \( e^{-4t} \) terms: \[ -\frac{1}{4} + \frac{2}{3} + \frac{5}{6} \] First, let's get a common denominator, which is 12: - \( -\frac{1}{4} = -\frac{3}{12} \) - \( \frac{2}{3} = \frac{8}{12} \) - \( \frac{5}{6} = \frac{10}{12} \) Sum: \( -\frac{3}{12} + \frac{8}{12} + \frac{10}{12} = \frac{15}{12} \) So, \[ x_1(t) = \frac{1}{4} + \frac{15}{12} e^{-4t} - \frac{2}{3} e^{-5t} - \frac{5}{6} e^{-2t} \] Simplify \( \frac{15}{12} = \frac{5}{4} \): \[ x_1(t) = \frac{1}{4} + \frac{5}{4} e^{-4t} - \frac{2}{3} e^{-5t} - \frac{5}{6} e^{-2t} \] --- ### **Step 3: Output \( y(t) \)** Recall: \[ y(t) = [1~1~0] x(t) = x_1(t) + x_2(t) \] Recall: \[ x_1(t) = \frac{1}{4} + \frac{5}{4} e^{-4t} - \frac{2}{3} e^{-5t} - \frac{5}{6} e^{-2t} \] \[ x_2(t) = 1 + \frac{2}{3} e^{-5t} - \frac{5}{3} e^{-2t} \] So, \[ y(t) = x_1(t) + x_2(t) = \left( \frac{1}{4} + 1 \right) + \frac{5}{4} e^{-4t} - \frac{2}{3} e^{-5t} + \frac{2}{3} e^{-5t} - \frac{5}{6} e^{-2t} - \frac{5}{3} e^{-2t} \] Now combine like terms: - Constants: \( \