Part 4 - Differential equation models

Introduction

In this part we will learn about systems that can be modeled with differential equations.

Definition

Some systems whose input, $f(t)$, and output, $y(t)$, are related by linear differential equations, often have the form: $$ \dfrac{d^n y}{dt^n} + a_{n - 1} \dfrac{d^{n - 1}y}{dt^{n - 1}} + \ldots + a_1 \dfrac{dy}{dt} + a_0 y(t) = $$

$$ b_m + \dfrac{d^m f}{dt^m} + b_{m - 1} + \dfrac{d^{m - 1} f}{dt^{m - 1}} + \ldots + b_1 \dfrac{df}{dt} + b_0 f(t) $$

This notation can be quite long and tedious, so let’s say that $D = \dfrac{d}{dt}$ $$ (D^n + a_{n - 1}D^{n - 1} + \ldots + a_1 D + a_0)y(t) = $$

$$ (D^m + b_{m - 1}D^{m - 1} + \ldots + b_1 D + b_0)f(t) = $$

Let’s call these $Q(D)$ and $P(D)$ respectively: $$ Q(D)y(t) = P(D)f(t) $$

A general rule of thumb is that $m \leq n$, this to limit noise.

Data needed to compute system response

For $t \geq 0$, a systems output is the result of two independent causes.

Therefore, we need to know:

1. The initial conditions of the system (also called system state) at $t = 0$.
2. The input, $f(t)$, for $t \geq 0$.
3. If the system is linear

Let’s also quickly refresh on if a system is linear.

Linearity example

Determine if the system described by the equation: $$ \dfrac{d y(t)}{dt} + 4y(t) = f(t) $$

with, $f(t)$, as input and, $y(t)$, as output of the system, is linear.

So let’s check if the superposition principle holds!

$$ \dfrac{d\ \alpha y_1(t)}{dt} + 4 \alpha y_1(t) = f_1(t) $$

$$ \dfrac{d\ \beta y_2(t)}{dt} + 4 \beta y_2(t) = f_2(t) $$

$$ f_1(t) + f_2(t) = \dfrac{d(\ \alpha y_1(t) + \beta y_2(t))}{dt} + 4(\alpha y_1(t) + \beta y_2(t)) $$

Now let’s try with $y(t) = \alpha y_1(t) + \beta y_2(t)$: $$ \dfrac{d(\ \alpha y_1(t) + \beta y_2(t))}{dt} + 4(\alpha y_1(t) + \beta y_2(t)) $$

We can see that $f(t) = f_1(t) + f_2(t)$.

Solving these systems

As we defined, the output of these systems are the result of two independent cases.

So for the zero-input response: $$ Q(D)y_0(t) = 0 $$

The solution for $y_0(t)$ will be: $$ y_0(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t} + \ldots + c_n e^{\lambda_n t} $$

This is if we have no repeated roots, in that case: $$ y_0(t) = (c_1 + c_2t + \ldots + c_r t^{r - 1})e^{\lambda_1 t} + c_{r +1}e^{\lambda_{r + 1} t} + \ldots + c_n e^{\lambda_n t} $$

Also in the case we have complex roots, $\alpha \pm j \beta$. $$ y_0(t) = \dfrac{c}{2} e^{j \theta} e^{(\alpha + j \beta)t} + \dfrac{c}{2} e^{-j \theta} e^{(\alpha - j \beta)t} = \dfrac{c}{2} e^{\alpha t} \left[ e^{j(\beta t + \theta)} + e^{-j(\beta t + \theta)}\right] $$

Using Euler’s formula: $$ y_0(t) = c e^{\alpha t} cos(\beta t + \theta) $$

Example

$$ (D^2 + 3D + 2)y_0(t) = 0 $$

$$ \lambda^2 + 3\lambda + 2 = (\lambda + 1)(\lambda + 2) = 0 $$

$$ \lambda_1 = -1 \newline \lambda_2 = -2 $$

$$ y_0(t) = c_1 e^-t + c_2 e^{-2t} $$