3.5 The functions e^x and e^jθ: second-order differential and

With our understanding of the functions $e^x$ , $e^{j\Theta }$ , and the quadratic equation $z^2+\frac{b}{a}z+\frac{c}{a}=0$ , we can undertake a rudimentary study of differential and difference equations.

Differential Equations. In your study of circuits and systems you will encounter the homogeneous differential equation

Because the function $e^{st}$ reproduces itself under differentiation, it is plausible to assume that $x\left(t\right)=e^{st}$ is a solution to the differential equation. Let's try it:

If this equation is to be satisfied for all $t$ , then the polynomial in $s$ must be zero. Therefore we require

As we know from our study of this quadratic equation, the solutions are

This means that our assumed solution works, provided $s=s_1$ or $s_2$ . It is a fundamental result from the theory of differential equations that the most general solution for $x\left(t\right)$ is a linear combination of these assumed solutions:

If $a_1^2-4a_2$ is less than zero, then the roots $s_1$ and $s_2$ are complex:

Let's rewrite this solution as

where σ and ω are the constants

With this notation, the solution for $x\left(t\right)$ is

If this solution is to be real, then the two terms on the right-hand side must be complex conjugates. This means that $A_2=A_1^{*}$ and the solution for $x\left(t\right)$ is

The constant $A_1$ may be written as $A_1=\left|A\right|e^{j\phi }$ . Then the solution for $x\left(t\right)$ is

This “damped cosinusoidal solution” is illustrated in [link] .

Difference Equations. In your study of digital filters you will encounter homogeneous difference equations of the form

What this means is that the sequence $\left\{x_n\right\}$ obeys a homogeneous recursion :

A plausible guess at a solution is the geometric sequence $x_n=z^n$ . With this guess, the difference equation produces the result

If this guess is to work, then the second-order polynomial on the left-hand side must equal zero:

$1+a_1{z}^{-1}+a_2{z}^{-2}=0$

The solutions are

The general solution to the difference equation is a linear combination of the assumed solutions:

This general solution is illustrated in [link] .