The Unit Step Function
Chapter Fifty-Seven
Syllabus topic Module 1, "1.3 Inverse Laplace Transform"
Pages 125 to 126 of 303
In one line
The unit step is zero before a and one after it, and it is how you write "switches on at time a" in a formula.
Where this chapter sits, and why it is not where MU puts it
MU files the Heaviside unit step function under her 1.3, among the inverse-transform topics. Nothing about it is an inverse transform, and the second shifting theorem, which she puts in 1.2, cannot even be stated without it. So this book teaches it here, in the forward-transform section, and the plan records the move. It stays inside Module 1 either way.
The definition
The unit step function, or Heaviside function, written u(t minus a), is zero for t less than a and one for t greater than a.
Figure 57.1 The unit step on the left. The right-hand picture is the pulse of unit area used to build the impulse function two chapters from here; both are drawn together because the impulse is the derivative of the step.
The value at t = a is a matter of convention and never matters here, because a single point contributes no area to an integral. Some books set it to one half.
The simplest case, a = 0, is written u(t) and is 1 for all positive t. So the constant function 1 and the unit step u(t) are the same thing as far as the Laplace transform is concerned, since the transform never looks at negative t.
Its transform
L{Heaviside(t - 3)} = e^(-3s)/s
L{Heaviside(t - 2)} = e^(-2s)/s
L{Heaviside(t)} = 1/s
The general result:
L{Heaviside(t - a)} = e^(-a s)/s
The derivation is one line. The integrand is zero up to a, so the integral runs from a to infinity instead of from zero. Substituting v = t minus a turns it into the transform of 1 multiplied by e to the minus as.
Note what appears: an exponential in s. That is the signature of a shift in time, and the chapter on the second shifting theorem makes it general. Any time you see an e to the minus as in a transform, something is being switched on at time a.
What it is for: switching
A function multiplied by u(t minus a) is that function with everything before time a erased.
| Expression | What it is |
|---|---|
| u(t - 2) | 0 until t = 2, then 1 |
| 5 u(t - 2) | 0 until t = 2, then 5 |
| u(t - 1) - u(t - 3) | 1 between t = 1 and t = 3, and 0 elsewhere |
| f(t) u(t - a) | f, but switched on at a |
| f(t - a) u(t - a) | f, delayed to start at a |
The Unit Step Function
The last two rows are different and the difference is the whole of the next chapter. The fourth switches f on at time a but f is still being evaluated at t, so you see the middle of f. The fifth delays f, so what you see from time a onwards is f starting from its own beginning.
The third row is the one to learn as a pattern: a difference of two steps is a window. Everything that switches on and later off is built from it.
Writing a piecewise function with steps
This is the skill the second shifting theorem needs, and the chapter after next drills it. In outline: a function given in pieces can always be written as a single expression in steps, by adding, at each break point, a step multiplied by the change in the formula at that point.
The simplest example. Suppose f is 0 before 2 and 3 afterwards. Then f is 3u(t minus 2). Suppose instead f is 1 before 2 and 4 afterwards: the change at t = 2 is plus 3, so f is 1 plus 3u(t minus 2).
L{3 Heaviside(t - 2)} = 3 e^(-2s)/s
L{1 + 3 Heaviside(t - 2)} = 1/s + 3 e^(-2s)/s
Its derivative, in advance
The unit step is flat everywhere except at the single point a, where it jumps. So its derivative is zero everywhere except at a, where it is not defined in any ordinary sense.
That object, zero everywhere except at one point but with total area one, is the Dirac delta, or unit impulse, and it has its own chapter shortly. The relationship is worth knowing now: the impulse is the derivative of the step, and the step is the integral of the impulse.
Check yourself
L{Heaviside(t - 5)} = e^(-5s)/s
L{4 Heaviside(t - 1)} = 4 e^(-s)/s
L{Heaviside(t - 1) - Heaviside(t - 4)} = e^(-s)/s - e^(-4s)/s
L{Heaviside(t)} = 1/s
The third line is the window between t = 1 and t = 4, and its transform is the difference of two step transforms, which is exactly what linearity says it should be.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.