The First Shifting Theorem
Chapter Forty-Eight
Syllabus topic Module 1, "1.2 The Laplace Transform"
Pages 106 to 107 of 303
In one line
Multiplying f(t) by e to the at shifts the transform: replace s by s minus a everywhere.
The theorem
MU's label is "First Shifting Theorem".
L{e^(a t) f(t)} = F(s - a)
In words: if you know the transform of f, then the transform of e to the at times f is the same function of s with s replaced by s minus a.
It is also called the shifting property in s, or the first translation theorem.
The proof, which is one line
Write the transform of e to the at f(t) as its integral. The two exponentials combine.
e^(-s t) e^(a t) = e^(-(s - a) t)
So the integral is the defining integral of the transform of f, with s minus a in place of s. That is F(s minus a), and the proof is finished.
The condition on s shifts with it: if F(s) was valid for s greater than c, then F(s minus a) is valid for s greater than c plus a.
What it buys: eight new table rows, with no integration
Every row of the elementary table can be multiplied by an exponential, for free.
| f(t) | F(s) | e^(at) f(t) | F(s - a) |
|---|---|---|---|
| 1 | 1/s | e^(at) | 1/(s - a) |
| t | 1/s^2 | t e^(at) | 1/(s - a)^2 |
| t^n | n! / s^(n+1) | t^n e^(at) | n! / (s - a)^(n+1) |
| sin bt | b/(s^2 + b^2) | e^(at) sin bt | b/((s - a)^2 + b^2) |
| cos bt | s/(s^2 + b^2) | e^(at) cos bt | (s - a)/((s - a)^2 + b^2) |
| sinh bt | b/(s^2 - b^2) | e^(at) sinh bt | b/((s - a)^2 - b^2) |
| cosh bt | s/(s^2 - b^2) | e^(at) cosh bt | (s - a)/((s - a)^2 - b^2) |
Every one of those, verified:
L{t e^(5t)} = 1/(s - 5)^2
L{t^3 e^(-2t)} = 6/(s + 2)^4
L{e^(3t) sin(4t)} = 4/((s - 3)^2 + 16)
L{e^(3t) cos(4t)} = (s - 3)/((s - 3)^2 + 16)
L{e^(-t) sinh(2t)} = 2/((s + 1)^2 - 4)
L{e^(-t) cosh(2t)} = (s + 1)/((s + 1)^2 - 4)
The one place people go wrong
The substitution is everywhere in F, not just in one place.
Take the transform of cos 4t, which is s over (s squared plus 16). Multiplying by e to the 3t means replacing s by s minus 3 in both the numerator and the denominator.
L{e^(3t) cos(4t)} = (s - 3)/((s - 3)^2 + 16)
Replacing it only in the denominator, and leaving an s on top, is the standard error.
(s - 3)/((s - 3)^2 + 16) = s/((s - 3)^2 + 16)
The two are different functions, and the checker proved the equality false, which is exactly what a student's own check should do: put s = 3 into both. The left side is zero; the right side is three sixteenths.
The First Shifting Theorem
Worked
Find the transform of e to the minus 2t times (t squared plus 3 sin t).
Do the inside first, with linearity.
L{t^2 + 3 sin(t)} = 2/s^3 + 3/(s^2 + 1)
Now shift, replacing every s by s plus 2, since a is minus 2.
L{e^(-2t)(t^2 + 3 sin(t))} = 2/(s + 2)^3 + 3/((s + 2)^2 + 1)
That is the answer. Note the order of operations: transform first, then shift. Trying to shift a function of t makes no sense.
Reading it backwards, which is where it matters most
The theorem read in reverse is the main tool for inverting a fraction whose denominator does not factorise: if F(s) can be written as a function of (s minus a), then the inverse is e to the at times the inverse of that function of s.
L⁻¹{1/(s - 4)^3} = t^2 e^(4t)/2
L⁻¹{1/((s + 1)^2 + 9)} = e^(-t) sin(3t)/3
L⁻¹{(s + 2)/((s + 2)^2 + 4)} = e^(-2t) cos(2t)
The manoeuvre that gets a denominator into that shape is completing the square, and it has its own chapter in the inverse-transform section.
Check yourself
L{t e^(-3t)} = 1/(s + 3)^2
L{e^(2t) t^2} = 2/(s - 2)^3
L{e^(-t) sin(t)} = 1/((s + 1)^2 + 1)
L{e^(4t)(1 + t)} = 1/(s - 4) + 1/(s - 4)^2
L⁻¹{1/(s - 1)^2} = t e^t
L⁻¹{2/((s - 3)^2 + 4)} = e^(3t) sin(2t)
L⁻¹{(s - 1)/((s - 1)^2 + 1)} = e^t cos(t)
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.