Let's work through the problem 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 \( y(t) \) for \( x(0) = 0 \) and \( u(t) = 10 \) (step input of magnitude 10)** #### **Step 1: General Solution for \( x(t) \) with Zero Initial Condition** The general solution for a state-space system: \[ \dot{x} = Ax + Bu \] with \( x(0) = 0 \), \( u(t) = U_0 \) (constant, here \( U_0 = 10 \)), is: \[ x(t) = \int_0^t e^{A(t-\tau)} B u(\tau) d\tau \] For a step input \( u(\tau) = 10 \) for all \( \tau \), this becomes: \[ x(t) = 10 \int_0^t e^{A(t-\tau)} B d\tau \] #### **Step 2: Compute \( A \) and \( B \) Structure** Given \( A \) is upper-triangular with -4, -5, -2 on the diagonal: \[ A = \begin{bmatrix} -4 & 1 & 0 \\ 0 & -5 & 1 \\ 0 & 0 & -2 \end{bmatrix} \] \( B = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \) #### **Step 3: Find \( x(t) \) Coordinates** Notice \( B \) only affects the third state, and \( A \) is in "companion" form (upper shift structure). Let \( x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \) So, \[ \dot{x}_1 = -4 x_1 + x_2 \\ \dot{x}_2 = -5 x_2 + x_3 \\ \dot{x}_3 = -2 x_3 + 10 \] Start with \( x_3 \): ##### **Solve for \( x_3(t) \):** \[ \dot{x}_3 + 2 x_3 = 10 \] This is a first order ODE. The solution (zero initial condition) \[ x_3(t) = 10 \int_0^t e^{-2 (t-\tau)} d\tau = 10 \cdot \frac{1 - e^{-2t}}{2} \] So, \[ x_3(t) = 5 (1 - e^{-2t}) \] ##### **Solve for \( x_2(t) \):** \[ \dot{x}_2 + 5 x_2 = x_3(t) \] This is a non-homogeneous linear ODE, can be solved by integrating factor. The integrating factor is \( e^{5t} \): \[ \frac{d}{dt}[e^{5t} x_2] = e^{5t} x_3(t) \] \[ x_2(t) = e^{-5t} \int_0^t e^{5\tau} x_3(\tau) d\tau \] \[ x_3(\tau) = 5(1 - e^{-2\tau}) \] So, \[ x_2(t) = e^{-5t} \int_0^t e^{5\tau} 5(1 - e^{-2\tau}) d\tau \] \[ = 5 e^{-5t} \int_0^t [e^{5\tau} - e^{3\tau}] d\tau \] \[ = 5 e^{-5t} \left[ \int_0^t e^{5\tau} d\tau - \int_0^t e^{3\tau} d\tau \right] \] \[ = 5 e^{-5t} \left[ \frac{1}{5}(e^{5t} - 1) - \frac{1}{3}(e^{3t} - 1) \right] \] \[ = e^{-5t} (e^{5t} - 1 - \frac{5}{3} (e^{3t} - 1)) \] \[ = (1 - e^{-5t}) - \frac{5}{3}(e^{-2t} - e^{-5t}) \] So, \[ x_2(t) = (1 - e^{-5t}) - \frac{5}{3}(e^{-2t} - e^{-5t}) \] \[ = 1 - e^{-5t} - \frac{5}{3}e^{-2t} + \frac{5}{3}e^{-5t} \] \[ = 1 - \frac{5}{3}e^{-2t} - e^{-5t} + \frac{5}{3}e^{-5t} \] \[ = 1 - \frac{5}{3}e^{-2t} + \frac{2}{3}e^{-5t} \] ##### **Solve for \( x_1(t) \):** \[ \dot{x}_1 + 4x_1 = x_2(t) \] Integrating factor is \( e^{4t} \): \[ \frac{d}{dt}[e^{4t} x_1] = e^{4t} x_2(t) \] \[ x_1(t) = e^{-4t} \int_0^t e^{4\tau} x_2(\tau) d\tau \] Let’s plug in \( x_2(\tau) \): \[ x_2(\tau) = 1 - \frac{5}{3} e^{-2\tau} + \frac{2}{3} e^{-5\tau} \] So, \[ x_1(t) = e^{-4t} \int_0^t e^{4\tau} \left[1 - \frac{5}{3} e^{-2\tau} + \frac{2}{3} e^{-5\tau}\right] d\tau \] \[ = e^{-4t} \left[ \int_0^t e^{4\tau} d\tau - \frac{5}{3} \int_0^t e^{4\tau} e^{-2\tau} d\tau + \frac{2}{3} \int_0^t e^{4\tau} e^{-5\tau} d\tau \right] \] \[ = e^{-4t} \left[ \int_0^t e^{4\tau} d\tau - \frac{5}{3} \int_0^t e^{2\tau} d\tau + \frac{2}{3} \int_0^t e^{-\tau} d\tau \right] \] Compute each integral: - \( \int_0^t e^{4\tau} d\tau = \frac{1}{4}(e^{4t} - 1) \) - \( \int_0^t e^{2\tau} d\tau = \frac{1}{2}(e^{2t} - 1) \) - \( \int_0^t e^{-\tau} d\tau = - (e^{-t} - 1) = 1 - e^{-t} \) So, \[ x_1(t) = e^{-4t} \left[ \frac{1}{4} (e^{4t} - 1) - \frac{5}{3} \cdot \frac{1}{2}(e^{2t} - 1) + \frac{2}{3}(1 - e^{-t}) \right] \] \[ = e^{-4t} \left[ \frac{1}{4}(e^{4t} - 1) - \frac{5}{6}(e^{2t} - 1) + \frac{2}{3}(1 - e^{-t}) \right] \] Expand and group terms: \[ = \frac{1}{4} e^{-4t} e^{4t} - \frac{1}{4} e^{-4t} - \frac{5}{6} e^{-4t} e^{2t} + \frac{5}{6} e^{-4t} + \frac{2}{3} e^{-4t} - \frac{2}{3} e^{-4t} e^{-t} \] \[ = \frac{1}{4} - \frac{1}{4} e^{-4t} - \frac{5}{6} e^{-2t} + \frac{5}{6} e^{-4t} + \frac{2}{3} e^{-4t} - \frac{2}{3} e^{-5t} \] Now, combine like terms for \( e^{-4t} \): - \( -\frac{1}{4} + \frac{5}{6} + \frac{2}{3} = -\frac{3}{12} + \frac{10}{12} + \frac{8}{12} = \frac{15}{12} = \frac{5}{4} \) So, \[ x_1(t) = \frac{1}{4} - \frac{5}{6} e^{-2t} + \frac{5}{4} e^{-4t} - \frac{2}{3} e^{-5t} \] #### **Step 4: Find \( y(t) \):** \[ y(t) = [1 \quad 1 \quad 0] x(t) = x_1(t) + x_2(t) \] \[ = \left( \frac{1}{4} - \frac{5}{6} e^{-2t} + \frac{5}{4} e^{-4t} - \frac{2}{3} e^{-5t} \right ) + \left( 1 - \frac{5}{3} e^{-2t} + \frac{2}{3} e^{-5t} \right ) \] \[ = \left( \frac{1}{4} + 1 \right ) - \left( \frac{5}{6} + \frac{5}{3} \right ) e^{-2t} + \frac{5}{4} e^{-4t} + \left( -\frac{2}{3} + \frac{2}{3} \right ) e^{-5t} \] \[ = \frac{5}{4} - \frac{5}{6} e^{-2t} - \frac{10}{6} e^{-2t} + \frac{5}{4} e^{-4t} \] \[ = \frac{5}{4} - \frac{15}{6} e^{-2t} + \frac{5}{4} e^{-4t} \] \[ = \frac{5}{4} - \frac{5}{2} e^{-2t} + \frac{5}{4} e^{-4t} \] ## **Final answer for part (a):** \[ \boxed{ y(t) = \frac{5}{4} - \frac{5}{2} e^{-2t} + \frac{5}{4} e^{-4t} } \] --- ## b) **Convert the state space model to the diagonal form** ### **What is the transformation matrix? Write the new state space model.** #### **Step 1: The eigenvalues of \( A \) are its diagonal entries (since \( A \) is upper-triangular):** \[ \lambda_1 = -4, \quad \lambda_2 = -5, \quad \lambda_3 = -2 \] #### **Step 2: Find the eigenvectors** Let’s find eigenvectors \( v_i \) for \( \lambda_i \): ##### For \( \lambda_1 = -4 \): \[ (A + 4I)v_1 = 0 \] \[ \begin{bmatrix} 0 & 1 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = 0 \] From row 1: \( 0 \cdot a + 1 \cdot b + 0 \cdot c = 0 \implies b = 0 \) Row 2: \( 0 \cdot a - 1 \cdot b + 1 \cdot c = 0 \implies -b + c = 0 \implies c = b = 0 \) Row 3: \( 0 \cdot a + 0 \cdot b + 2 \cdot c = 0 \implies 2c = 0 \implies c = 0 \) Therefore, \( v_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end

Let's analyze the problem step-by-step carefully. --- ### **Given Data:** **State-space model:** \[ \dot{x} = A x + B u \] where \[ A = \begin{bmatrix} -4 & 1 & 0 \\ 0 & -5 & 1 \\ 0 & 0 & -2 \end{bmatrix}, \quad B = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \] and the output: \[ y = C x, \quad \text{with} \quad C = \begin{bmatrix} 1 & 1 & 0 \end{bmatrix} \] --- ### **Part (a): Find \( y(t) \) assuming \( x(0) = 0 \), \( u(t) = 10 \) (step input of magnitude 10).** --- #### **Step 1:** Recognize the structure of the system The matrix \(A\) is upper triangular, with eigenvalues: \[ \lambda_1 = -4,\quad \lambda_2 = -5,\quad \lambda_3 = -2 \] The input \( u(t) = 10 \) is a step of magnitude 10 applied to the third state directly via \( B \). --- #### **Step 2:** Solve the differential equations sequentially The state equations: \[ \begin{cases} \dot{x}_1 = -4 x_1 + x_2 \\ \dot{x}_2 = -5 x_2 + x_3 \\ \dot{x}_3 = -2 x_3 + 10 \end{cases} \] with initial conditions \( x(0) = 0 \). --- #### **Step 3:** Solve for \( x_3(t) \) \[ \dot{x}_3 + 2 x_3 = 10 \] This is a first-order linear ODE. The integrating factor: \[ \mu(t) = e^{2 t} \] Solution: \[ x_3(t) = e^{-2 t} \int_0^t e^{2 \tau} \times 10 \, d\tau = 10 e^{-2t} \left[ \frac{1}{2} e^{2 \tau} \right]_0^t = 10 e^{-2t} \times \frac{1}{2} (e^{2t} - 1) = 5(1 - e^{-2t}) \] --- #### **Step 4:** Solve for \( x_2(t) \) \[ \dot{x}_2 + 5 x_2 = x_3(t) = 5(1 - e^{-2 t}) \] This is a linear ODE with integrating factor \( e^{5 t} \): \[ \frac{d}{dt} \left( e^{5 t} x_2 \right) = e^{5 t} \times 5(1 - e^{-2 t}) \] Compute RHS: \[ e^{5 t} \times 5(1 - e^{-2 t}) = 5 e^{5 t} - 5 e^{5 t} e^{-2 t} = 5 e^{5 t} - 5 e^{3 t} \] Integrate: \[ e^{5 t} x_2 = \int_0^t \left( 5 e^{5 \tau} - 5 e^{3 \tau} \right) d \tau \] \[ = 5 \int_0^t e^{5 \tau} d \tau - 5 \int_0^t e^{3 \tau} d \tau = 5 \times \frac{1}{5} (e^{5 t} - 1) - 5 \times \frac{1}{3} (e^{3 t} - 1) = (e^{5 t} - 1) - \frac{5}{3} (e^{3 t} - 1) \] Thus, \[ x_2(t) = e^{-5 t} \left[ e^{5 t} - 1 - \frac{5}{3} e^{3 t} + \frac{5}{3} \right] \] \[ = 1 - e^{-5 t} - \frac{5}{3} e^{-5 t} e^{3 t} + \frac{5}{3} e^{-5 t} \] Note \( e^{-5 t} e^{3 t} = e^{-2 t} \). So, \[ x_2(t) = 1 - e^{-5 t} - \frac{5}{3} e^{-2 t} + \frac{5}{3} e^{-5 t} \] Group terms: \[ x_2(t) = 1 - \left( e^{-5 t} - \frac{5}{3} e^{-5 t} \right) - \frac{5}{3} e^{-2 t} = 1 - \left( 1 - \frac{5}{3} \right) e^{-5 t} - \frac{5}{3} e^{-2 t} \] \[ = 1 - \left( - \frac{2}{3} \right) e^{-5 t} - \frac{5}{3} e^{-2 t} = 1 + \frac{2}{3} e^{-5 t} - \frac{5}{3} e^{-2 t} \] --- #### **Step 5:** Solve for \( x_1(t) \) \[ \dot{x}_1 + 4 x_1 = x_2(t) \] Using integrating factor \( e^{4 t} \): \[ \frac{d}{dt} \left( e^{4 t} x_1 \right) = e^{4 t} x_2(t) \] \[ x_1(t) = e^{-4 t} \int_0^t e^{4 \tau} x_2(\tau) d \tau \] Substitute \( x_2(\tau) \): \[ x_2(\tau) = 1 + \frac{2}{3} e^{-5 \tau} - \frac{5}{3} e^{-2 \tau} \] So: \[ x_1(t) = e^{-4 t} \left[ \int_0^t e^{4 \tau} d \tau + \frac{2}{3} \int_0^t e^{4 \tau} e^{-5 \tau} d \tau - \frac{5}{3} \int_0^t e^{4 \tau} e^{-2 \tau} d \tau \right] \] Compute each integral: - \(\int_0^t e^{4 \tau} d \tau = \frac{1}{4} (e^{4 t} - 1)\) - \(\int_0^t e^{(4 - 5) \tau} d \tau = \int_0^t e^{- \tau} d \tau = 1 - e^{- t}\) - \(\int_0^t e^{(4 - 2) \tau} d \tau = \int_0^t e^{2 \tau} d \tau = \frac{1}{2} (e^{2 t} - 1)\) Putting it all together: \[ x_1(t) = e^{-4 t} \left[ \frac{1}{4} (e^{4 t} - 1) + \frac{2}{3} (1 - e^{- t}) - \frac{5}{3} \times \frac{1}{2} (e^{2 t} - 1) \right] \] Simplify: \[ x_1(t) = e^{-4 t} \left[ \frac{1}{4} e^{4 t} - \frac{1}{4} + \frac{2}{3} - \frac{5}{6} e^{2 t} + \frac{5}{6} \right] \] Further: \[ x_1(t) = e^{-4 t} \left( \frac{1}{4} e^{4 t} - \frac{1}{4} + \frac{2}{3} + \frac{5}{6} - \frac{5}{6} e^{2 t} \right) \] Note: \[ e^{-4 t} \times \frac{1}{4} e^{4 t} = \frac{1}{4} \] and \[ e^{-4 t} \times e^{2 t} = e^{-2 t} \] Thus, \[ x_1(t) = \frac{1}{4} - e^{-4 t} \left( \frac{1}{4} - \frac{2}{3} - \frac{5}{6} + \frac{5}{6} e^{2 t} \right) \] Calculate constants: \[ - \frac{1}{4} + \frac{2}{3} + \frac{5}{6} = - \frac{3}{12} + \frac{8}{12} + \frac{10}{12} = \frac{15}{12} = \frac{5}{4} \] So, \[ x_1(t) = \frac{1}{4} - e^{-4 t} \left( - \frac{5}{4} + \frac{5}{6} e^{2 t} \right) \] Distribute: \[ x_1(t) = \frac{1}{4} + \frac{5}{4} e^{-4 t} - \frac{5}{6} e^{-4 t} e^{2 t} = \frac{1}{4} + \frac{5}{4} e^{-4 t} - \frac{5}{6} e^{-2 t} \] --- #### **Step 6:** Find the output \( y(t) \) \[ y(t) = C x(t) = x_1(t) + x_2(t) \] \[ = \left( \frac{1}{4} + \frac{5}{4} e^{-4 t} - \frac{5}{6} e^{-2 t} \right) + \left( 1 + \frac{2}{3} e^{-5 t} - \frac{5}{3} e^{-2 t} \right) \] Combine constants: \[ \frac{1}{4} + 1 = \frac{5}{4} \] Combine the \( e^{-2 t} \) terms: \[ - \frac{5}{6} e^{-2 t} - \frac{5}{3} e^{-2 t} = - \frac{5}{6} e^{-2 t} - \frac{10}{6} e^{-2 t} = - \frac{15}{6} e^{-2 t} = - \frac{5}{2} e^{-2 t} \] Final expression: \[ \boxed{ \boxed{ y(t) = \frac{5}{4} + \frac{5}{4} e^{-4 t} - \frac{5}{2} e^{-2 t} + \frac{2}{3} e^{-5 t} } } \] --- ### **Part (a) Final Answer:** \[ \boxed{ \boxed{ y(t) = \frac{5}{4} + \frac{5}{4} e^{-4 t} - \frac{5}{2} e^{-2 t} + \frac{2}{3} e^{-5 t} } } \] --- ### **Part (b): Convert the state space to diagonal form. What is the transformation matrix? Write the new model.** --- #### **Step 1:** Eigenvalues of \(A\) From the upper triangular matrix: \[ \lambda_1 = -4, \quad \lambda_2 = -5, \quad \lambda_3 = -2 \] Eigenvalues are on the diagonal. --- #### **Step 2:** Find eigenvectors - For \( \lambda_1 = -4 \): \[ (A + 4 I) v_1 = 0 \] \[ A + 4 I = \begin{bmatrix} 0 & 1 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & 2 \end{bmatrix} \] Solve: \[ \begin{cases} 0 \cdot a + 1 \cdot b + 0 \cdot c = 0 \Rightarrow b=0 \\ 0 \cdot a -1 \cdot b + 1 \cdot c=0 \Rightarrow c=0 \\ 0 \cdot a + 0 \cdot b + 2 c=0 \Rightarrow c=0 \end{cases} \] Eigenvector: \[ v_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \] - For \( \lambda_2 = -5 \): \[ (A + 5 I) v_2 = 0 \] \[ A + 5 I = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 3 \end{bmatrix} \] Solve: \[ \begin{cases} 1 \cdot a + 1 \cdot b= 0 \Rightarrow a = -b \\ 0 \cdot a + 0 \cdot b + 1 \cdot c= 0 \Rightarrow c=