Let's solve the Dutch roll motion step-by-step: --- ### a) **General understanding of Dutch roll motion** **Dutch roll** is an oscillatory motion in aircraft characterized by a coupled yawing and rolling movement. This motion occurs due to the interaction of lateral (side-to-side) and directional (yawing) stability. It is typically lightly damped and has a higher frequency compared to the roll mode. Dutch roll is important in aircraft dynamics as excessive Dutch roll can make the aircraft uncomfortable or unsafe to fly, especially in swept-wing aircraft. --- ### b) **Rewrite the state space equations by substituting numerical values** Given state-space form: \[ \begin{bmatrix} \dot{\Delta \beta} \\ \dot{\Delta r} \end{bmatrix} = \begin{bmatrix} \frac{Y_\beta}{u_0} & 1 - \frac{Y_r}{u_0} \\ N_\beta & N_r \end{bmatrix} \begin{bmatrix} \Delta \beta \\ \Delta r \end{bmatrix} + \begin{bmatrix} \frac{Y_{\delta_r}}{u_0} \\ N_{\delta_r} \end{bmatrix} \Delta \delta_r \] Given values: - \( Y_\beta = -7.8 \) ft/s² - \( Y_r = 2.47 \) ft/s - \( N_\beta = 0.64 \) 1/s² - \( N_r = -0.34 \) 1/s - \( Y_{\delta_r} = -5.236 \) ft/s² - \( N_{\delta_r} = 0.616 \) 1/s² - \( u_0 = 154 \) ft/s Calculate each term: #### For the matrix: - \( \frac{Y_\beta}{u_0} = \frac{-7.8}{154} \approx -0.0506 \) - \( \frac{Y_r}{u_0} = \frac{2.47}{154} \approx 0.0160 \) - \( 1 - \frac{Y_r}{u_0} = 1 - 0.0160 = 0.984 \) - \( N_\beta = 0.64 \) - \( N_r = -0.34 \) - \( \frac{Y_{\delta_r}}{u_0} = \frac{-5.236}{154} \approx -0.0340 \) - \( N_{\delta_r} = 0.616 \) #### Substitute: \[ \begin{bmatrix} \dot{\Delta \beta} \\ \dot{\Delta r} \end{bmatrix} = \begin{bmatrix} -0.0506 & 0.984 \\ 0.64 & -0.34 \end{bmatrix} \begin{bmatrix} \Delta \beta \\ \Delta r \end{bmatrix} + \begin{bmatrix} -0.0340 \\ 0.616 \end{bmatrix} \Delta \delta_r \] --- ### c) **Determine the eigenvalues of the system** Eigenvalues of the matrix \( A \): \[ A = \begin{bmatrix} -0.0506 & 0.984 \\ 0.64 & -0.34 \end{bmatrix} \] Eigenvalues \( \lambda \) satisfy: \[ \text{det}(A - \lambda I) = 0 \] So, \[ \begin{vmatrix} -0.0506 - \lambda & 0.984 \\ 0.64 & -0.34 - \lambda \end{vmatrix} = 0 \] \[ (-0.0506 - \lambda)(-0.34 - \lambda) - (0.984 \times 0.64) = 0 \] \[ (-0.0506 - \lambda)(-0.34 - \lambda) - 0.62976 = 0 \] Expanding, \[ (-0.0506 - \lambda)(-0.34 - \lambda) = 0.0506 \times 0.34 + 0.0506 \lambda + 0.34 \lambda + \lambda^2 = 0.017204 + 0.0506\lambda + 0.34\lambda + \lambda^2 = 0.017204 + 0.3906\lambda + \lambda^2 \] So: \[ 0.017204 + 0.3906\lambda + \lambda^2 - 0.62976 = 0 \] \[ \lambda^2 + 0.3906\lambda - 0.612556 = 0 \] Use the quadratic formula: \[ \lambda = \frac{-0.3906 \pm \sqrt{(0.3906)^2 - 4\times1\times(-0.612556)}}{2} \] \[ = \frac{-0.3906 \pm \sqrt{0.152568 + 2.450224}}{2} \] \[ = \frac{-0.3906 \pm \sqrt{2.602792}}{2} \] \[ \sqrt{2.602792} \approx 1.612 \] So, \[ \lambda_1 = \frac{-0.3906 + 1.612}{2} \approx 0.6107 \] \[ \lambda_2 = \frac{-0.3906 - 1.612}{2} \approx -1.001 \] --- ### d) **Response to unit-step function (using MATLAB or manually)** The system is: \[ \dot{\mathbf{x}} = A\mathbf{x} + B u \] With zero initial conditions (\(\mathbf{x}(0) = 0\)), and \(u(t)\) is a step input (\(u(t) = 1\) for \(t \geq 0\)). The solution: \[ \mathbf{x}(t) = \int_0^t e^{A(t-\tau)} B u(\tau) d\tau \] For a step input, \[ \mathbf{x}(t) = \left[ \int_0^t e^{A(t-\tau)} d\tau \right] B \] Or, in Laplace domain: \[ X(s) = (sI - A)^{-1} B \cdot \frac{1}{s} \] You can use MATLAB with: ```matlab A = [-0.0506, 0.984; 0.64, -0.34]; B = [-0.0340; 0.616]; C = eye(2); % Output both states D = [0; 0]; sys = ss(A, B, C, D); t = 0:0.01:10; [y, t, x] = step(sys, t); plot(t, y(:,1), 'b', t, y(:,2), 'r') xlabel('Time (s)') ylabel('State response') legend('\Delta\beta','\Delta r') title('Step Response of Dutch Roll Mode') ``` --- ### e) **Discuss the stability of the system** - The eigenvalues are \( 0.6107 \) and \( -1.001 \). - The presence of a **positive eigenvalue (\( 0.6107 \))** means the system has an **unstable mode**, and thus, the system is **unstable**. - Normally, for stability, all eigenvalues must have negative real parts. Here, the Dutch roll mode is unstable and will grow without bound in response to disturbance. --- ## **Summary Table** | Step | Result | |------|--------| | a | Dutch roll is a coupled yaw/roll oscillation in aircraft. | | b | State space: \(\begin{bmatrix} -0.0506 & 0.984\\ 0.64 & -0.34 \end{bmatrix}\) | | c | Eigenvalues: \( 0.6107, -1.001 \) | | d | Response: Use MATLAB code above or Laplace solution. | | e | Stability: **Unstable** (positive eigenvalue present). | --- If you need the MATLAB plot or a manual time-domain solution, let me know!

Let's analyze the Dutch roll motion step-by-step, providing detailed calculations and explanations for each part. --- ## **a) General understanding of Dutch roll motion** **Dutch roll** is a dynamic oscillatory mode involving coupled yawing and rolling motions of an aircraft. It is characterized by a lightly damped oscillation where the aircraft yaws and rolls in a coupled manner. This motion results from the interaction of lateral stability derivatives and yawing stability derivatives, leading to a natural oscillation that can be stable or unstable depending on the system parameters. Key points: - It’s a *lateral-directional* oscillation. - It involves a combination of yaw and roll. - Usually lightly damped, with a natural frequency typically higher than the roll mode. - Unstable Dutch roll can cause discomfort and control issues. --- ## **b) Rewrite the state-space equations by substituting the numerical values** Given the general form: \[ \begin{bmatrix} \dot{\Delta \beta} \\ \dot{\Delta r} \end{bmatrix} = A \begin{bmatrix} \Delta \beta \\ \Delta r \end{bmatrix} + B \Delta \delta_r \] where: \[ A = \begin{bmatrix} \frac{Y_\beta}{u_0} & 1 - \frac{Y_r}{u_0} \\ N_\beta & N_r \end{bmatrix} \] \[ B = \begin{bmatrix} \frac{Y_{\delta_r}}{u_0} \\ N_{\delta_r} \end{bmatrix} \] ### Given data: | Parameter | Value | Units | |------------|--------|--------| | \(Y_\beta\) | -7.8 | ft/sec² (from problem) | | \(Y_r\) | 2.47 | ft/sec | | \(N_\beta\) | 0.64 | 1/sec² | | \(N_r\) | -0.34 | 1/sec | | \(Y_{\delta_r}\) | -5.236 | ft/sec² | | \(N_{\delta_r}\) | 0.616 | 1/sec² | | \(u_0\) | 154 | ft/sec | ### Step 1: Calculate each matrix element 1. \( \frac{Y_\beta}{u_0} = \frac{-7.8}{154} \) \[ \frac{-7.8}{154} \approx -0.0506 \] 2. \( 1 - \frac{Y_r}{u_0} = 1 - \frac{2.47}{154} \) \[ \frac{2.47}{154} \approx 0.0160 \] \[ 1 - 0.0160 = 0.984 \] 3. \( N_\beta = 0.64 \) 4. \( N_r = -0.34 \) 5. \( \frac{Y_{\delta_r}}{u_0} = \frac{-5.236}{154} \) \[ \frac{-5.236}{154} \approx -0.0340 \] 6. \( N_{\delta_r} = 0.616 \) ### **Final matrices:** \[ A = \begin{bmatrix} -0.0506 & 0.984 \\ 0.64 & -0.34 \end{bmatrix} ,\quad B = \begin{bmatrix} -0.0340 \\ 0.616 \end{bmatrix} \] --- ## **c) Determine the eigenvalues of the system** Eigenvalues are found by solving: \[ \det(A - \lambda I) = 0 \] \[ \det \begin{bmatrix} -0.0506 - \lambda & 0.984 \\ 0.64 & -0.34 - \lambda \end{bmatrix} = 0 \] Calculate determinant: \[ (-0.0506 - \lambda)(-0.34 - \lambda) - (0.984)(0.64) = 0 \] ### Step 1: Expand the first product: \[ (-0.0506 - \lambda)(-0.34 - \lambda) = (-0.0506)(-0.34) + (-0.0506)(-\lambda) + (-\lambda)(-0.34) + \lambda^2 \] Calculate each term: - \( (-0.0506)(-0.34) = 0.017204 \) - \( (-0.0506)(-\lambda) = 0.0506 \lambda \) - \( (-\lambda)(-0.34) = 0.34 \lambda \) Adding these: \[ 0.017204 + 0.0506 \lambda + 0.34 \lambda + \lambda^2 = 0.017204 + 0.3906 \lambda + \lambda^2 \] ### Step 2: Subtract the product \( 0.984 \times 0.64 = 0.62976 \): \[ 0.017204 + 0.3906 \lambda + \lambda^2 - 0.62976 = 0 \] Simplify: \[ \lambda^2 + 0.3906 \lambda + (0.017204 - 0.62976) = 0 \] \[ \lambda^2 + 0.3906 \lambda - 0.612556 = 0 \] ### Step 3: Use quadratic formula: \[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] with: - \( a = 1 \) - \( b = 0.3906 \) - \( c = -0.612556 \) Compute discriminant: \[ D = b^2 - 4ac = (0.3906)^2 - 4 \times 1 \times (-0.612556) \] \[ D = 0.152568 + 2.450224 = 2.602792 \] Square root: \[ \sqrt{D} \approx 1.612 \] Calculate eigenvalues: \[ \lambda_1 = \frac{-0.3906 + 1.612}{2} = \frac{1.2214}{2} \approx 0.6107 \] \[ \lambda_2 = \frac{-0.3906 - 1.612}{2} = \frac{-2.0026}{2} \approx -1.0013 \] ### **Final eigenvalues:** \[ \boxed{ \lambda_1 \approx 0.611 \quad (\text{unstable mode}) \\ \lambda_2 \approx -1.001 \quad (\text{stable mode}) } \] --- ## **d) Response of the system to a unit-step input** Since the system is second-order, with eigenvalues \(\lambda_1\) and \(\lambda_2\), and zero initial conditions, the response to a step input can be computed using Laplace transforms or MATLAB. ### **Manual approach:** The transfer function for the states: \[ \mathbf{X}(s) = (sI - A)^{-1} B \cdot \frac{1}{s} \] But it's more straightforward to simulate the response using MATLAB or Python. ### **MATLAB code snippet:** ```matlab A = [-0.0506, 0.984; 0.64, -0.34]; B = [-0.0340; 0.616]; C = eye(2); % output both states D = [0; 0]; sys = ss(A, B, C, D); t = 0:0.01:20; % time vector [y, t] = step(sys, t); plot(t, y(:,1), 'b', t, y(:,2), 'r') xlabel('Time (s)') ylabel('States: \Delta\beta (blue), \Delta r (red)') title('Step Response of Dutch Roll Modes') legend('\Delta\beta', '\Delta r') grid on ``` This code will show how the yaw and roll rate respond to a step input. --- ## **e) Discuss the stability of the system** - **Eigenvalues:** \[ \lambda_1 \approx 0.611 \quad (\text{positive real part}) \\ \lambda_2 \approx -1.001 \quad (\text{negative real part}) \] - Since **one eigenvalue is positive**, the system exhibits **unstable behavior**: the mode associated with \(\lambda_1\) will grow exponentially, causing divergence over time. - **Conclusion:** The *Dutch roll* mode in this configuration is **unstable** due to the positive eigenvalue, indicating that without damping or control input, the aircraft's yaw-rolling oscillation will grow unboundedly. --- ## **Summary** | Part | Result | Explanation | |-------|---------|--------------| | **a** | Dutch roll is a coupled yaw/roll oscillation | A natural lateral-directional oscillation in aircraft stability | | **b** | State matrix \(A\): \(\begin{bmatrix} -0.0506 & 0.984 \\ 0.64 & -0.34 \end{bmatrix}\) | Derived by substituting given parameters | | **c** | Eigenvalues: \(\boxed{0.611, -1.001}\) | One positive eigenvalue indicates instability in at least one mode | | **d** | Response: Use MATLAB or Laplace inverse; MATLAB example provided | Response involves oscillations with growth or decay depending on eigenvalues | | **e** | **System is unstable** | Due to the positive eigenvalue, the Dutch roll mode grows over time | --- **Note:** To improve stability, aircraft design or control systems must be adjusted to ensure both eigenvalues have negative real parts.