Inverting With the Second Shifting Theorem
Chapter Seventy-Two
Syllabus topic Module 1, "1.3 Inverse Laplace Transform"
Pages 160 to 161 of 303
In one line
An exponential in s means the answer is delayed, and the delay has to be written with a unit step or the answer is wrong before the delay.
The rule
L⁻¹{e^(-a s) F(s)} = f(t - a) Heaviside(t - a)
The procedure is three steps.
- Set the exponential aside and invert what is left, getting f(t).
- Replace every t in f by (t minus a).
- Multiply by u(t minus a).
Step three is the one that is skipped, and skipping it gives an answer that is wrong for every t less than a.
Worked
L⁻¹{e^(-2s)/s} = Heaviside(t - 2)
L⁻¹{e^(-3s)/s^2} = (t - 3) Heaviside(t - 3)
L⁻¹{e^(-s)/(s + 1)} = e^(-(t - 1)) Heaviside(t - 1)
L⁻¹{e^(-pi s)/(s^2 + 1)} = sin(t - pi) Heaviside(t - pi)
L⁻¹{e^(-2s)/(s - 3)} = e^(3(t - 2)) Heaviside(t - 2)
In each case: invert the non-exponential part, then shift the t, then attach the step.
Take the third line slowly. One over (s plus 1) inverts to e to the minus t. Replacing t by t minus 1 gives e to the minus (t minus 1). Attaching the step gives the answer. Note that e to the minus (t minus 1) is not e to the minus t times e: it is e to the minus t times e to the plus 1, and writing it as e times e to the minus t is correct but obscures the shift.
The mistake, once
Suppose the step is dropped from the second line above.
(t - 3) Heaviside(t - 3) = t - 3
Those two are not the same function. For t = 1 the left side is zero, because the step is zero; the right side is minus 2. So the answer without the step claims the response was minus 2 at a time before anything had happened, which is nonsense.
Worked, where the rest of F needs work too
Most questions combine this with partial fractions or with completing the square.
F(s) = e^(-s)/(s(s + 1))
Invert the non-exponential part first, by partial fractions: one over s(s plus 1) is one over s minus one over (s plus 1), which inverts to 1 minus e to the minus t.
Now shift and attach the step.
L⁻¹{e^(-s)/(s(s + 1))} = (1 - e^(-(t - 1))) Heaviside(t - 1)
And another, with a completed square.
F(s) = e^(-2s)/(s^2 + 4s + 5)
The denominator is (s plus 2) squared plus 1, so the non-exponential part inverts to e to the minus 2t sin t.
L⁻¹{e^(-2s)/(s^2 + 4s + 5)} = e^(-2(t - 2)) sin(t - 2) Heaviside(t - 2)
Every t in the answer has been shifted, including the one inside the exponential. All of them, or none of them.
Inverting With the Second Shifting Theorem
Several exponentials
A sum of terms, each with its own exponential, gives a sum of delayed pieces, each with its own step. That is how the answer to a switched problem naturally comes out.
L⁻¹{e^(-s)/s - e^(-3s)/s} = Heaviside(t - 1) - Heaviside(t - 3)
L⁻¹{1/s^2 - e^(-2s)/s^2} = t - (t - 2) Heaviside(t - 2)
The first is the window of the step-function chapter. The second is a ramp that stops rising at t = 2, which is what the piecewise chapter built from the other direction. Seeing the same function from both sides is worth the minute.
Writing the answer in pieces
An answer full of steps is correct, and it is also what the question usually wants restated as a piecewise function. The conversion is the piecewise chapter read backwards.
For the second line above: before t = 2 the step is zero, so the answer is t. After t = 2 the step is one, so the answer is t minus (t minus 2), which is 2.
So the answer is t for t less than 2 and 2 for t greater than 2, which is the ramp that holds. Writing it out that way is often worth a mark and always worth the understanding.
Check yourself
L⁻¹{e^(-4s)/s} = Heaviside(t - 4)
L⁻¹{e^(-s)/s^3} = (t - 1)^2 Heaviside(t - 1)/2
L⁻¹{e^(-2s) s/(s^2 + 9)} = cos(3(t - 2)) Heaviside(t - 2)
L⁻¹{e^(-s)/(s - 1)^2} = (t - 1) e^(t - 1) Heaviside(t - 1)
L⁻¹{3 e^(-5s)/(s^2 + 9)} = sin(3(t - 5)) Heaviside(t - 5)
The fourth is the one to check most carefully: the non-exponential part inverts to t e to the t, and both the t and the t in the exponent have to be shifted.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.