Let's solve the problem step by step. --- ## **Given:** \[ \dot{x} = \begin{bmatrix} -4 & 1 & 0 \\ 0 & -5 & 1 \\ 0 & 0 & -2 \end{bmatrix} x + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} u \] \[ y = \begin{bmatrix} 1 & 1 & 0 \end{bmatrix} x \] Initial condition: \[ x(0) = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \] Input: Step of magnitude 10, i.e., \( u(t) = 10 \cdot 1(t) \) --- ## **(a) Solve for \( y(t) \) for given input and initial condition** ### **Step 1: State-Space Response** The general solution: \[ x(t) = e^{At}x(0) + \int_0^t e^{A(t-\tau)}Bu(\tau) d\tau \] Given \( x(0) = 0 \), so: \[ x(t) = \int_0^t e^{A(t-\tau)} B u(\tau) d\tau \] For a step input \( u(\tau) = 10 \): \[ x(t) = 10 \int_0^t e^{A(t-\tau)} B d\tau \] Let \( \theta = t - \tau \Rightarrow d\tau = -d\theta \), bounds: when \( \tau=0, \theta=t \); when \( \tau = t, \theta=0 \): \[ x(t) = 10 \int_{t}^{0} e^{A\theta} B (-d\theta) = 10 \int_0^{t} e^{A\theta} B d\theta \] So, \[ x(t) = 10 \int_0^t e^{A\theta} B d\theta \] ### **Step 2: Compute \( e^{A\theta} B \) ** Since \( A \) is upper triangular, and \( B = [0\ 0\ 1]^T \), this is a classic "chain integrator" with distinct real eigenvalues. Let us denote the elements as \( x = [x_1, x_2, x_3]^T \). Write out the system: \[ \begin{cases} \dot{x}_1 = -4x_1 + x_2 \\ \dot{x}_2 = -5x_2 + x_3 \\ \dot{x}_3 = -2x_3 + u \end{cases} \] We can solve backwards, starting from \( x_3 \). #### **Step 2a: Solve for \( x_3(t) \)** \[ \dot{x}_3 = -2x_3 + 10, \quad x_3(0) = 0 \] This is a standard first-order ODE: Solution: \[ x_3(t) = [x_3(0) - \frac{10}{2}]e^{-2t} + \frac{10}{2} = -5e^{-2t} + 5 \] #### **Step 2b: Solve for \( x_2(t) \)** \[ \dot{x}_2 = -5x_2 + x_3(t), \quad x_2(0) = 0 \] Use integrating factor \( e^{5t} \): \[ \frac{d}{dt}[e^{5t} x_2] = e^{5t} x_3(t) \] Integrate: \[ x_2(t) = e^{-5t} \int_0^t e^{5\tau} x_3(\tau) d\tau \] Plug in \( x_3(\tau) = -5e^{-2\tau} + 5 \): \[ x_2(t) = e^{-5t} \int_0^t e^{5\tau} (-5e^{-2\tau} + 5) d\tau \] \[ = e^{-5t} \int_0^t \left( -5e^{3\tau} + 5e^{5\tau} \right) d\tau \] \[ = e^{-5t} \left[ -5 \int_0^t e^{3\tau} d\tau + 5 \int_0^t e^{5\tau} d\tau \right] \] Calculate the integrals: \[ \int_0^t e^{3\tau} d\tau = \frac{1}{3}(e^{3t} - 1) \] \[ \int_0^t e^{5\tau} d\tau = \frac{1}{5}(e^{5t} - 1) \] So, \[ x_2(t) = e^{-5t} \left[ -5 \cdot \frac{1}{3}(e^{3t} - 1) + 5 \cdot \frac{1}{5}(e^{5t} - 1) \right] \] \[ = e^{-5t} \left[ -\frac{5}{3}(e^{3t} - 1) + (e^{5t} - 1) \right] \] \[ = e^{-5t} \left[ -\frac{5}{3}e^{3t} + \frac{5}{3} + e^{5t} - 1 \right] \] \[ = e^{-5t} (e^{5t} - \frac{5}{3}e^{3t} + \frac{5}{3} - 1) \] \[ = 1 - \frac{5}{3}e^{-2t} + \frac{5}{3}e^{-5t} - e^{-5t} \] \[ = 1 - \frac{5}{3}e^{-2t} + \left(\frac{5}{3} - 1\right) e^{-5t} \] \[ = 1 - \frac{5}{3}e^{-2t} + \frac{2}{3}e^{-5t} \] #### **Step 2c: Solve for \( x_1(t) \)** \[ \dot{x}_1 = -4x_1 + x_2(t), \quad x_1(0) = 0 \] Integrating factor \( e^{4t} \): \[ x_1(t) = e^{-4t} \int_0^t e^{4\tau} x_2(\tau) d\tau \] Plug in \( x_2(\tau) \): \[ x_2(\tau) = 1 - \frac{5}{3}e^{-2\tau} + \frac{2}{3}e^{-5\tau} \] So, \[ x_1(t) = e^{-4t} \int_0^t e^{4\tau} \left[ 1 - \frac{5}{3}e^{-2\tau} + \frac{2}{3}e^{-5\tau} \right] d\tau \] \[ = e^{-4t} \left[ \int_0^t e^{4\tau} d\tau - \frac{5}{3} \int_0^t e^{4\tau} e^{-2\tau} d\tau + \frac{2}{3} \int_0^t e^{4\tau} e^{-5\tau} d\tau \right] \] \[ = e^{-4t} \left[ \int_0^t e^{4\tau} d\tau - \frac{5}{3} \int_0^t e^{2\tau} d\tau + \frac{2}{3} \int_0^t e^{-1\tau} d\tau \right] \] Calculate each term: - \( \int_0^t e^{4\tau} d\tau = \frac{1}{4}(e^{4t} - 1) \) - \( \int_0^t e^{2\tau} d\tau = \frac{1}{2}(e^{2t} - 1) \) - \( \int_0^t e^{-1\tau} d\tau = - (e^{-t} - 1) \) So, \[ x_1(t) = e^{-4t} \left[ \frac{1}{4}(e^{4t} - 1) - \frac{5}{3} \cdot \frac{1}{2}(e^{2t} - 1) + \frac{2}{3} \cdot (1 - e^{-t}) \right] \] \[ = e^{-4t} \left[ \frac{1}{4}(e^{4t} - 1) - \frac{5}{6}(e^{2t} - 1) + \frac{2}{3}(1 - e^{-t}) \right] \] \[ = e^{-4t} \left[ \frac{1}{4}e^{4t} - \frac{1}{4} - \frac{5}{6}e^{2t} + \frac{5}{6} + \frac{2}{3} - \frac{2}{3}e^{-t} \right] \] \[ = \frac{1}{4} - \frac{1}{4}e^{-4t} - \frac{5}{6}e^{-2t} + \frac{5}{6}e^{-4t} + \frac{2}{3}e^{-4t} - \frac{2}{3}e^{-5t} \] But let's expand step by step for clarity: First, distribute \( e^{-4t} \): - \( \frac{1}{4}e^{4t} \cdot e^{-4t} = \frac{1}{4} \) - \( -\frac{1}{4} \cdot e^{-4t} \) - \( -\frac{5}{6}e^{2t} \cdot e^{-4t} = -\frac{5}{6} e^{-2t} \) - \( +\frac{5}{6} \cdot e^{-4t} \) - \( +\frac{2}{3} e^{-4t} \) - \( -\frac{2}{3}e^{-t} \cdot e^{-4t} = -\frac{2}{3}e^{-5t} \) So, collect terms: \[ x_1(t) = \frac{1}{4} - \frac{1}{4}e^{-4t} - \frac{5}{6}e^{-2t} + \frac{5}{6}e^{-4t} + \frac{2}{3}e^{-4t} - \frac{2}{3}e^{-5t} \] Now, combine like terms in \( e^{-4t} \): \( -\frac{1}{4} + \frac{5}{6} + \frac{2}{3} \): First, convert to common denominator (12): - \( -\frac{1}{4} = -\frac{3}{12} \) - \( \frac{5}{6} = \frac{10}{12} \) - \( \frac{2}{3} = \frac{8}{12} \) Sum: \( -\frac{3}{12} + \frac{10}{12} + \frac{8}{12} = \frac{15}{12} = \frac{5}{4} \) So, \[ x_1(t) = \frac{1}{4} - \frac{5}{6}e^{-2t} + \frac{5}{4}e^{-4t} - \frac{2}{3}e^{-5t} \] --- ### **Step 3: Compute Output \( y(t) \)** Recall: \[ y(t) = [1\ 1\ 0] x(t) = x_1(t) + x_2(t) \] Plug in \( x_1(t) \) and \( x_2(t) \): \[ y(t) = \left( \frac{1}{4} - \frac{5}{6}e^{-2t} + \frac{5}{4}e^{-4t} - \frac{2}{3}e^{-5t} \right) + \left( 1 - \frac{5}{3}e^{-2t} + \frac{2}{3}e^{-5t} \right) \] \[ = \frac{1}{4} + 1 - \frac{5}{6}e^{-2t} - \frac{5}{3}e^{-2t} + \frac{5}{4}e^{-4t} - \frac{2}{3}e^{-5t} + \frac{2}{3}e^{-5t} \] \[ = \frac{5}{4} - \left( \frac{5}{6} + \frac{5}{3} \right) e^{-2t} + \frac{5}{4}e^{-4t} + \left( -\frac{2}{3} +