Step by step
The question
Question: find the Laplace transform of 4sin(2t) + cos(2t).
ℒ{4sin(2t) + cos(2t)} =
1 The idea
The Laplace transform turns a function of t into a function of s.
It is an integral, but we almost never work it out: we use a table made from it.
ℒ{f(t)} = ∫ from 0 to ∞ of e^(−st) f(t) dt
The table rows we need.
sin(bt) → bs² + b²
cos(bt) → ss² + b²
2 Split it up
The transform of a sum is the sum of the transforms.
A number in front waits outside.
= 4ℒ{sin(2t)} + ℒ{cos(2t)}
ℒ{af + bg} = aℒ{f} + bℒ{g}
3 Use the table
Each part by its row: sin(2t) gives 2/(s² + 4), cos(2t) gives s/(s² + 4).
= 4 × 2s2 + 4 + ss2 + 4
Careful!
sin has b on top: 2, not 1.
Careful: not ℒ{sin(2t)} = 1s² + 4 ✗
4 Tidy up
Multiply the numbers: 8/(s² + 4) + s/(s² + 4).
= 8s2 + 4 + ss2 + 4
5 Check
Let's check: read the table backwards.
8/(s² + 4) + s/(s² + 4) comes from 4sin(2t) + cos(2t): the question.
Right!
backwards: 8s2 + 4 + ss2 + 4 → 4sin(2t) + cos(2t)
Let's look back.
First: the idea.
Then: split it up.
Then: use the table.
Then: tidy up.
Last: check.
Answer
= 8s2 + 4 + ss2 + 4