User:Egm6321.f10.team5.abc/3

Homework 3
All problems except for 8

Part 1
Derive the following equations.

Part 2
Write the above equations in the form of $$\dot x(t) = A(t) x(t) + B(t) u(t)$$. Find A, B, and u.

Part 1
We use Newtons's second law for Rotation which is $$\Sigma \tau = I \alpha$$ for the derivation. We note that there are three torques applied on the particle 1; gravitational torque, controlling torque, and spring torque. Gravitational torque is Controlling torque is Spring torque is Moment of inertia for the point mass is Angular acceleration for the point mass is We set up the equation of motion for the first particle. Apply the small-angle approximation which states that $$\sin x \approx x$$ and $$\cos x \approx 1$$ to simplify equation as We use the same techniques to derive the equation of motion for second particle. Gravitational torque is Controlling torque is Spring torque is Moment of inertia for the point mass is Angular acceleration for the point mass is We set up the equation of motion for the second particle. Apply the small-angle approximation which states that $$\sin x \approx x$$ and $$\cos x \approx 1$$ to simplify equation as

Part 2
We write $$\dot x(t) = A(t) x(t) + B(t) u(t)$$ in matrix form. We divide the both side of $$m_1 l^2 \ddot \theta_1\ = - k a^2 (\theta_1 - \theta_2) - l \theta_1 m_1 g + l u_1$$ by $$m_1 l^2$$ to get Similarly, we divide the both side of $$m_2 l^2 \ddot \theta_2\ = - k a^2 (\theta_2 - \theta_1) - l \theta_2 m_2 g + l u_2$$ by $$m_2 l^2$$ to get We can easily determine that the all the components on row 1 and row 3 of matrix A and matrix B are null, so we write that Now we plug in the expressions of $$\ddot \theta_1$$ and $$\ddot \theta_2$$ into the matrix.

Problem 3
Solve $$\dot x(t) = a(t) x(t) + b(t) u(t)$$ for the case where $$a(t)$$ and $$b(t)$$ are constant coefficients while u(t) is prescribed.

Solution 3
We state that $$a(t) = a$$ and $$b(t) = b$$ for this case. So we have the equation in the form of $$hx' + h'x = hk$$. Note that we cannot use the equation of $$y(x) = \frac 1 {h(x)} \int^x h(s) d(s) ds$$ since the lower limit is bounded as $$t_0$$. Deriving an equation for the lower bound Apply the modified equation given above to our problem. It is possible to simplify the equation further as

Problem 6
Use $$\frac d {dt} \underline \phi (t,t_0) = \underline A \underline \phi (t,t_0)$$ and $$\underline \phi (t_0,t_0) = \underline 1$$ along with $$\dot {\underline x}(t) = \underline A \underline x(t) + \underline B \underline u(t)$$ to obtain $$\underline x(t) = [\exp \underline A (t - t_0)] \underline x (t_0) + \int_{t_0}^t [\exp \underline A (t - \tau)] \underline B \underline u (\tau) d\tau$$.

Solution 6
We can derive from $$\frac d {dt} \underline \phi (t,t_0) = \underline A \underline \phi (t,t_0)$$ that $$\underline A = \frac {\frac d {dt} \underline \phi (t,t_0)} {\underline \phi (t,t_0)}$$. Because $$\phi (t_0,t_0) = \underline 1$$ which is an identity matrix.

Problem 10
Assume $$h_y y' = 2 y^3$$ and solve $$(6 x y^2) y' + 2 y^3 = h_y y' + h_x$$.

Solution 10
We derive the following procedures according to the given assumption. Note from the equation of $$(6 x y^2) y' + 2 y^3 = h_y y' + h_x$$ that $$h_x = (6 x y^2) y'$$. Take a derivative of $$h(x,y)$$ with respect to $$y$$ to find that Multiply by $$y'$$ to determine that $$k_1(y)$$ is a constant. Since $${k_1}'(y) = 0$$, then $$k_1 (y) = k_1$$.