Review Homework

Express the following differential equation as a system of first-order differential equations in matrix form, by defining $x_1 = y$, $x_2 = y^{(1)}$ and $x_3 = y^{(2)}$:

$y^{(3)} + 5y^{(2)} + 5y^{(1)} - 4y = 0.$

Find the Jordan form for the following matrices:

a) $A_1 = \begin{bmatrix} 5 & -6 \\ 4 & -5 \end{bmatrix}$

b) $A_2 = \begin{bmatrix} -1 & 0 & 1 \\ 0 & 2 & -3 \\ 0 & 0 & -1 \end{bmatrix}$

Hint: You will have to find a generalized eigenvector for part b.

Show that for square matrices $A$ and $B$, which commute, that is

$AB = BA,$

it follows that

$e^{(A+B)} = e^A e^B.$

Hint: Recall that $e^A = \lim_{T\to\infty} \sum_{k=0}^{T} \frac{A^k}{k!} = \sum_{k=0}^{\infty} \frac{A^k}{k!}$. Complete the following:

$e^A e^B = \sum_{k=0}^{\infty} \sum_{l=0}^{\infty} \frac{A^k}{k!} \frac{B^l}{l!}$

Also, show and use the fact that $(A + B)^k = \sum_{m=0}^{k} \binom{k}{m} A^m B^{k-m}$, when $AB = BA$.

Let $A$ be a square matrix. Show that $\frac{d}{dt}e^{At} = Ae^{At}$.

Compute $e^{A_1 t}$, where $A_1$ is defined as in Problem 2.

Hint: The eigenvalues are 1 and -1. The corresponding eigenvectors are $[3 \quad 2]^T$ and $[1 \quad 1]^T$.

Let

$A = \begin{bmatrix} \frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & \frac{4}{5} \end{bmatrix}.$

It can be shown that $A$ has its eigenvalues as 1 and 0 and the eigenvectors are $[1 \quad 2]^T$ and $[-2 \quad 1]^T$.

Now, with $A$ as given, solve the following differential equation:

$x' = Ax + \begin{bmatrix} e^t \\ 0 \end{bmatrix},$

with the initial condition $x(0) = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$.

Let

$A = \begin{bmatrix} a & b \\ -b & a \end{bmatrix}.$

Compute $e^{At}$.

Consider now an equation of the form $x' = Ax$ (with an arbitrary initial condition). Draw the direction field qualitatively for each of the following cases: $a > 0$, $a < 0$ and $a = 0$.

Consider

$\frac{dy}{dt} = ay(t) + u(t) + r(t)$

where $a \in \mathbb{R}$ is a scalar, $u(t)$ is the control input and $r(t)$ is some disturbance/noise acting on the system. The disturbance is external, that is, the controller has nothing to do with it.

If $a > 0$, then in the absence of control, the solution to the system is given with

$y(t) = e^{at}y(0) + \int_0^t e^{a(t-s)} r(s) ds$

In particular, if $r(s) = M$ for $s \ge 0$ for some constant $M > 0$, with $y(0) = 0$, we have that $\lim_{t\to\infty} |y(t)| = \infty$.

On the other hand, if we apply the control input, using the feedback from the state of the system, by the relation $u(t) = -Ky(t)$, this leads to

$\frac{dy}{dt} = (a - K)y(t) + r(t)$

For what values of $K$ can we guarantee that the output, $y(t)$, will remain bounded for every bounded signal $r(t)$? (By bounded we mean that $\|r\|_\infty = \sup_{t \in \mathbb{R}_+} |r(t)| < \infty$).

A very useful case study for understanding non-linear systems is the inverted pendulum on a cart system with masses of the pendulum and cart given with $m$ and $M$, respectively.

The dynamics of the inverted pendulum can be expressed with the following:

$u = M\frac{d^2y}{dt^2} + m\frac{d^2y}{dt^2} + ml\cos(\theta)\frac{d^2\theta}{dt^2} - ml\left(\frac{d\theta}{dt}\right)^2\sin(\theta)$

$ml^2\frac{d^2\theta}{dt^2} = mg\sin(\theta)l - m\frac{d^2y}{dt^2}\cos(\theta)l$

Around the point $(\theta = 0, \frac{d\theta}{dt} = 0)$, we apply the linear approximations $\sin(\theta) \approx \theta$ and $\cos(\theta) \approx 1$, and $\left(\frac{d\theta}{dt}\right)^2 \approx 0$ to arrive at the linearized model. Write this in state space form with $x_1 = y, x_2 = \frac{dy}{dt}, x_3 = \theta, x_4 = \frac{d\theta}{dt}$.