Transfer Functions for a Motor

Engineers describe a motor by one function of s instead of a differential equation. For each s, the Laplace transform weighs the motor's response by a fading exponential and adds it all up. That single number per s turns calculus into algebra.

How it works

The Laplace transform

In one sentence

The Laplace transform of $f$ is $F(s) = \int_0^\infty e^{-st}f(t)\,dt$ , defined for $s$ large enough that the integral converges; it is linear and satisfies $\mathcal{L}\{f'\} = sF(s) - f(0)$ .

General formula

{\htmlData{sym=0}{F(s)}} = {\htmlData{sym=1}{\mathcal{L}}}\{f\}({\htmlData{sym=4}{s}}) = \int_0^{\infty} {\htmlData{sym=3}{e^{-st}}}{\htmlData{sym=2}{f(t)}}\,dt

$e^{-st}$ is a fading window: for each $s$ , $F(s)$ is a weighted total of the signal that counts the early part most.

What the symbols mean

$F(s)$
Transform
One number for each s: the weighted total of the signal.
$\mathcal{L}$
Laplace transform
The operation that turns f(t) into F(s).
$f(t)$
Signal
The motor's response over time (its step response here).
$e^{-st}$
Weight
A fading window: early times count most, and larger s fades faster.
$s$
Transform variable
Sets how fast the weight fades.

The Laplace transform weighs a signal $f(t)$ by the decaying exponential $e^{-st}$ and adds it all up.
The result $F(s)$ is a function of $s$ : for each $s$ , the area under $e^{-st}f(t)$ .
For a motor's step response $f = 1 - e^{-t/\tau}$ the area is $F(s) = \frac{1}{s} - \frac{1}{s + 1/\tau}$ .
Derivatives become multiplication, $\mathcal{L}\{f'\} = sF(s) - f(0)$ , which turns differential equations into algebra.

Learn the math

The Laplace TransformDiff Eq · Lesson 22, Laplace Transforms Computing Laplace TransformsDiff Eq · Lesson 23, Laplace Transforms