The Second Shifting Theorem
Chapter Fifty-Eight
Syllabus topic Module 1, "1.2 The Laplace Transform"
Pages 127 to 128 of 303
In one line
Delaying f by a multiplies its transform by e to the minus as, provided the delayed function is switched on with a unit step.
The theorem
MU's label is "Second Shifting Theorem".
L{f(t - a) Heaviside(t - a)} = e^(-a s) F(s)
Compare the two shifting theorems, because the pairing is the point.
| Theorem | What happens to f | What happens to F |
|---|---|---|
| First | multiplied by e^(at) | s replaced by s - a |
| Second | delayed by a, and switched on | multiplied by e^(-as) |
The first shifts in s; the second shifts in t. That is why they are named in that order.
The proof
Write the defining integral. The unit step makes the integrand zero up to t = a, so the integral runs from a to infinity.
Substitute v = t minus a, so t is v plus a and dt is dv. The kernel becomes e to the minus s(v plus a), which splits into e to the minus as times e to the minus sv, and the e to the minus as is a constant that comes out in front.
What is left is the defining integral of F in the variable v. So the answer is e to the minus as times F(s), and the proof is three lines.
Why the unit step has to be there
This is the part students skip and then get wrong, so it is worth being blunt.
The theorem as stated is about the function that is zero before a and equals f(t minus a) after it. Without the step factor, f(t minus a) would be some non-zero thing for t less than a, and the integral from 0 to a would contribute something the theorem does not account for.
So "delayed by a" always means "delayed by a and zero before that". If your function is not zero before a, the theorem does not apply and you must write the function out as a sum of pieces first.
Worked
L{Heaviside(t - 2)} = e^(-2s)/s
Here f is the constant 1, whose transform is one over s, delayed by 2. The theorem gives e to the minus 2s over s, which is the result the previous chapter derived directly.
L{(t - 3) Heaviside(t - 3)} = e^(-3s)/s^2
L{(t - 1)^2 Heaviside(t - 1)} = 2 e^(-s)/s^3
L{sin(t - pi) Heaviside(t - pi)} = e^(-pi s)/(s^2 + 1)
L{e^(2(t - 4)) Heaviside(t - 4)} = e^(-4s)/(s - 2)
In each case: identify f from the shape, look up F, multiply by e to the minus as. The pattern to spot is that the same a appears inside the function and inside the step, which is what tells you the theorem applies directly.
The Second Shifting Theorem
When the a does not match: the function is switched on, not delayed
Here is the case that needs work, and it is the one examinations set.
L{t Heaviside(t - 2)}
That is t, switched on at 2, and it is not of the form f(t minus 2) times u(t minus 2), because the t is not written as t minus 2. So the theorem does not apply as it stands.
The fix is to force it into shape by writing t as (t minus 2) plus 2.
t = (t - 2) + 2
Now the function is ((t minus 2) plus 2) times u(t minus 2), which is (t minus 2)u(t minus 2) plus 2u(t minus 2), and both terms are in the right shape.
L{t Heaviside(t - 2)} = e^(-2s)/s^2 + 2 e^(-2s)/s
That manoeuvre, rewrite everything in terms of (t minus a), is the whole technique for this theorem. A harder one, with a square:
t^2 = (t - 1)^2 + 2(t - 1) + 1
L{t^2 Heaviside(t - 1)} = 2 e^(-s)/s^3 + 2 e^(-s)/s^2 + e^(-s)/s
And with a sine, where the compound-angle formula does the rewriting:
sin(t) = sin((t - pi/2) + pi/2)
sin((t - pi/2) + pi/2) = cos(t - pi/2)
L{sin(t) Heaviside(t - pi/2)} = s e^(-pi s/2)/(s^2 + 1)
Reading it backwards
The reverse reading is the one used when solving an equation with a switched input: an e to the minus as in F(s) means the answer is delayed by a and multiplied by a step.
L⁻¹{e^(-2s)/s} = Heaviside(t - 2)
L⁻¹{e^(-3s)/s^2} = (t - 3) Heaviside(t - 3)
L⁻¹{e^(-s)/(s^2 + 4)} = sin(2(t - 1)) Heaviside(t - 1)/2
L⁻¹{e^(-4s)/(s - 2)} = e^(2(t - 4)) Heaviside(t - 4)
The procedure: put the exponential aside, invert what is left, then in the answer replace every t by t minus a and multiply by u(t minus a). Forgetting the step is the standard error, and the answer is then wrong for all t less than a.
Check yourself
L{(t - 5) Heaviside(t - 5)} = e^(-5s)/s^2
L{cos(t - 1) Heaviside(t - 1)} = s e^(-s)/(s^2 + 1)
L{t Heaviside(t - 1)} = e^(-s)/s^2 + e^(-s)/s
L⁻¹{e^(-s)/s^3} = (t - 1)^2 Heaviside(t - 1)/2
L⁻¹{e^(-2s)/(s + 1)} = e^(-(t - 2)) Heaviside(t - 2)
L⁻¹{e^(-pi s) s/(s^2 + 1)} = cos(t - pi) Heaviside(t - pi)
The third line of the first block is the t-switched-on case with a = 1, and it has two terms for exactly the reason the worked example gave.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.