The Table of Elementary Transforms
Chapter Forty-Six
Syllabus topic Module 1, "1.2 The Laplace Transform"
Pages 102 to 103 of 303
In one line
These fourteen rows, with their conditions on s, are what the rest of the module reads in both directions.
MU's label
"Table of Elementary Laplace Transforms". Every row below was derived from the definition in one of the previous six chapters, not copied from anywhere, and every one was recomputed by machine before it was printed.
The table
| f(t) | F(s) | Valid for |
|---|---|---|
| 1 | 1/s | s greater than 0 |
| t | 1/s^2 | s greater than 0 |
| t^n | n factorial / s^(n+1) | s greater than 0 |
| e^(at) | 1/(s - a) | s greater than a |
| sin at | a/(s^2 + a^2) | s greater than 0 |
| cos at | s/(s^2 + a^2) | s greater than 0 |
| sinh at | a/(s^2 - a^2) | s above the size of a |
| cosh at | s/(s^2 - a^2) | s above the size of a |
| t e^(at) | 1/(s - a)^2 | s greater than a |
| t^n e^(at) | n factorial / (s - a)^(n+1) | s greater than a |
| e^(at) sin bt | b/((s - a)^2 + b^2) | s greater than a |
| e^(at) cos bt | (s - a)/((s - a)^2 + b^2) | s greater than a |
| t sin at | 2 a s/(s^2 + a^2)^2 | s greater than 0 |
| t cos at | (s^2 - a^2)/(s^2 + a^2)^2 | s greater than 0 |
The first eight rows are the ones derived so far. Rows nine to twelve come from the first shifting theorem, and thirteen and fourteen from the rule for multiplying by t, both of which are the next few chapters. They are printed here so that the table is in one place.
Every row, verified:
L{1} = 1/s
L{t} = 1/s^2
L{t^4} = 24/s^5
L{e^(3t)} = 1/(s - 3)
L{sin(2t)} = 2/(s^2 + 4)
L{cos(2t)} = s/(s^2 + 4)
L{sinh(2t)} = 2/(s^2 - 4)
L{cosh(2t)} = s/(s^2 - 4)
L{t e^(3t)} = 1/(s - 3)^2
L{t^3 e^(2t)} = 6/(s - 2)^4
L{e^(2t) sin(3t)} = 3/((s - 2)^2 + 9)
L{e^(2t) cos(3t)} = (s - 2)/((s - 2)^2 + 9)
L{t sin(3t)} = 6 s/(s^2 + 9)^2
L{t cos(3t)} = (s^2 - 9)/(s^2 + 9)^2
How to read it forwards
Match the shape of f, read off the F, and substitute the numbers. Use linearity to break a sum into its terms.
L{2 + 3t - 4 e^(2t)} = 2/s + 3/s^2 - 4/(s - 2)
L{t^2 + sin(t)} = 2/s^3 + 1/(s^2 + 1)
How to read it backwards, which is harder
Going from F to f needs the F to be made to look like a row of the table, and that is where the work is. Three manoeuvres do almost all of it.
Supply a missing constant. The sine row has an a on top. If the numerator is 1, write it as one over a times a.
The Table of Elementary Transforms
L⁻¹{1/(s^2 + 25)} = sin(5t)/5
Make the coefficient of s equal to one. Divide top and bottom.
L⁻¹{1/(2s + 6)} = e^(-3t)/2
Split a numerator across two rows.
L⁻¹{(2s + 5)/(s^2 + 4)} = 2 cos(2t) + 5 sin(2t)/2
Those three, plus completing the square and partial fractions, are the whole of the inverse-transform technique, and each has its own chapter.
The conditions on s
They are in the table and they should be written down whenever a transform is derived from the definition. In ordinary working, when you are reading off the table to solve a differential equation, they are not carried through, and no marks turn on that.
The one place they matter in practice is a question that asks for which values of s a given transform exists. The answer is always the largest exponential growth rate in f, and it is visible as the rightmost point where the denominator of F vanishes.
Two rows that do not exist, and are invented every year
L{f(t) g(t)} = L{f(t)} L{g(t)}
L{1/f(t)} = 1/L{f(t)}
Neither of those is true. The transform is linear, so it respects sums and constant multiples, and it respects nothing else. There is no rule for the transform of a product or of a reciprocal.
What there is instead is the convolution theorem, which says the product of two transforms is the transform of a convolution, which is not a product at all. That is the subject of its own chapter, and it is the single most misused result in the paper.
Check yourself
L{t^5} = 120/s^6
L{e^(-4t)} = 1/(s + 4)
L{sin(6t)} = 6/(s^2 + 36)
L{t^2 e^(-t)} = 2/(s + 1)^3
L{e^(-t) cos(2t)} = (s + 1)/((s + 1)^2 + 4)
L⁻¹{1/s^6} = t^5/120
L⁻¹{1/(s + 4)^2} = t e^(-4t)
L⁻¹{s/(s^2 + 36)} = cos(6t)
L⁻¹{1/((s - 1)^2 + 4)} = e^t sin(2t)/2
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.