What the Inverse Transform Is
Chapter Sixty-Five
Syllabus topic Module 1, "1.3 Inverse Laplace Transform"
Pages 145 to 146 of 303
In one line
The inverse transform answers the question "whose transform is this", and it is the half of the work that actually produces the answer.
The definition
If F(s) is the Laplace transform of f(t), then f(t) is the inverse Laplace transform of F(s), written with an inverse superscript on the L.
L⁻¹{1/s} = 1
L⁻¹{1/(s - 3)} = e^(3t)
L⁻¹{1/(s^2 + 4)} = sin(2t)/2
That is all the definition says. There is no new integral to learn for this paper: the inverse is found by recognising F as a transform you already know, not by computing anything.
There is a formula for the inverse as a contour integral in the complex plane, called the Bromwich integral, and it is beyond this syllabus. MU does not name it and no question needs it.
Why the question has only one answer
Because of the uniqueness result quoted in the existence chapter: two continuous functions with the same transform are the same function.
So if you find any f whose transform is the F in front of you, it is the answer. You do not have to worry that some other function has the same transform, and you do not have to justify your method. Guessing, checking, and stopping is a complete and rigorous procedure.
That licence is worth knowing, because it means the whole of the inverse-transform technique is a set of recognition patterns rather than a calculation.
Linearity, again
The inverse is linear, for the same reason the forward transform is.
L⁻¹{2/s + 3/(s - 1)} = 2 + 3 e^t
L⁻¹{1/s^2 - 1/(s^2 + 1)} = t - sin(t)
So the plan for any F is always: break it into pieces each of which is a table row, and invert each piece. Every chapter in this section is a technique for doing that breaking.
The five techniques, and where each applies
| F(s) looks like | Technique | Chapter |
|---|---|---|
| a table row, up to constants | read the table backwards | the next one |
| a quadratic denominator that does not factorise | complete the square, then shift | third from here |
| a fraction with a factorisable denominator | partial fractions | four chapters |
| an exponential in s times something | the second shifting theorem | later |
| a product of two recognisable pieces | convolution | later |
That table is the whole of this section, and reading it is how you decide what to do with an F you have never seen.
The first thing to do with any F
Look at the denominator, and in this order.
- Does it factorise into linear factors? Then partial fractions, and the answer is exponentials.
- Is it an irreducible quadratic? Then complete the square, and the answer oscillates.
- Is there a factor of s on its own? Then there is a constant term in the answer, or an integration.
- Is there an exponential in s anywhere? Then something is delayed, and there will be a unit step in the answer.
- Is a factor raised to a power? Then there will be a power of t multiplying something.
What the Inverse Transform Is
Those five readings, made before any algebra, will tell you the shape of the answer, and knowing the shape before you start is what stops a wrong turn costing five minutes of an hour.
Why this is the half that matters
The forward transform of a differential equation is mechanical: read the table, term by term. What comes out is an algebraic equation, and solving it is school algebra.
Then you are left with a function of s and a question: what function of t is this? Everything hard about the method is there. So this section is longer than the forward section, and its techniques are worth more practice.
A warning about the notation
The inverse is written with an inverse superscript on the L, and it is not a reciprocal. The inverse transform of one over F(s) has nothing to do with the inverse transform of F(s), and writing the operator as a fraction is a mistake that leads straight to nonsense.
L⁻¹{F(s)} = 1/L{F(s)}
That statement is false and meaningless. The superscript means "the operation that undoes L", exactly as it does on a function.
Check yourself
L⁻¹{5/s} = 5
L⁻¹{1/(s + 2)} = e^(-2t)
L⁻¹{1/s^3} = t^2/2
L⁻¹{s/(s^2 + 9)} = cos(3t)
L⁻¹{2/(s^2 - 4)} = sinh(2t)
L⁻¹{1/(s - 1)^2} = t e^t
All six are table rows read backwards, with one adjustment each at most. If any of them took more than a few seconds, the next chapter is the one to work through slowly.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.