The First Shifting Theorem at Work
Chapter Forty-Nine
Syllabus topic Module 1, "1.2 The Laplace Transform"
Pages 108 to 109 of 303
In one line
Every transform of the form e to the at times something is this theorem, and every inverse of a completed square is it read backwards.
Forward, a graded set
Each of these is the same two steps: transform the function, then replace s by s minus a.
L{e^(2t) t^4} = 24/(s - 2)^5
L{e^(-3t) t} = 1/(s + 3)^2
L{e^(t/2) sin(t)} = 1/((s - 1/2)^2 + 1)
L{e^(-4t) cos(3t)} = (s + 4)/((s + 4)^2 + 9)
L{e^(2t)(3 + t^2)} = 3/(s - 2) + 2/(s - 2)^3
L{e^(-t)(cos(2t) - sin(2t))} = (s + 1)/((s + 1)^2 + 4) - 2/((s + 1)^2 + 4)
The fifth and sixth show the order that matters: linearity first, inside the bracket, and the shift applied to the whole result.
A product of two exponentials, which is not a special case
L{e^(2t) e^(3t)} = 1/(s - 5)
Combine the exponentials before doing anything else. Applying the shifting theorem twice would also work and gives the same answer, but combining is quicker and less error-prone.
A product of an exponential and a hyperbolic function
L{e^(t) sinh(2t)} = 2/((s - 1)^2 - 4)
L{e^(t) cosh(2t)} = (s - 1)/((s - 1)^2 - 4)
Alternatively, write the hyperbolic function as two exponentials and combine each with the first, which avoids the theorem altogether.
L{e^(t) sinh(2t)} = 1/(2(s - 3)) - 1/(2(s + 1))
Both answers are correct and they are equal. Which form you leave it in depends on what comes next: the second is already in partial fractions, which is convenient if it is going to be inverted.
Backwards: the recognition that makes it work
The theorem read in reverse says: if every s in F(s) appears as (s minus a), pull out an e to the at.
The three shapes to recognise:
L⁻¹{1/(s - a)^n} = t^(n - 1) e^(a t)/factorial(n - 1)
with n a positive whole number; and, with the quadratic completed,
L⁻¹{1/((s - a)^2 + b^2)} = e^(a t) sin(b t)/b
L⁻¹{(s - a)/((s - a)^2 + b^2)} = e^(a t) cos(b t)
Those three rows cover nearly every inverse in this paper that is not a plain partial fraction.
Worked inverse, with the completing of the square shown
Invert the function below.
F(s) = 1/(s^2 + 6s + 13)
The denominator does not factorise over the reals, because its discriminant, 36 minus 52, is negative. So complete the square.
s^2 + 6s + 13 = (s + 3)^2 + 4
Now every s appears as s plus 3, so a is minus 3 and b is 2.
L⁻¹{1/((s + 3)^2 + 4)} = e^(-3t) sin(2t)/2
The one over two comes from the sine row needing a b on top, and b is 2.
The First Shifting Theorem at Work
Worked inverse, where the numerator needs adjusting too
Invert the function below.
F(s) = (s + 1)/(s^2 + 4s + 8)
Complete the square in the denominator: s squared plus 4s plus 8 is (s plus 2) squared plus 4.
Now the numerator must be written in terms of s plus 2 as well, because the table's cosine row has exactly (s minus a) on top. So write s plus 1 as (s plus 2) minus 1.
s + 1 = (s + 2) - 1
Now split into two table entries.
L⁻¹{(s + 2)/((s + 2)^2 + 4)} = e^(-2t) cos(2t)
L⁻¹{1/((s + 2)^2 + 4)} = e^(-2t) sin(2t)/2
So the answer is the first minus the second.
L⁻¹{(s + 1)/(s^2 + 4s + 8)} = e^(-2t) cos(2t) - e^(-2t) sin(2t)/2
That three-step routine, complete the square, rewrite the numerator to match, split into two rows, is the single most useful procedure in the inverse half of this module. It appears in almost every second-order differential equation whose solution oscillates.
The procedure
Forwards. Identify a, transform the rest, replace every s by s minus a.
Backwards. Complete the square if the quadratic will not factorise. Rewrite the numerator in terms of the same shifted variable. Split into the sine row and the cosine row. Supply the missing constant on the sine.
Check yourself
L{e^(5t) t^2} = 2/(s - 5)^3
L{e^(-2t) sin(3t)} = 3/((s + 2)^2 + 9)
L⁻¹{1/(s^2 + 2s + 5)} = e^(-t) sin(2t)/2
L⁻¹{s/(s^2 + 2s + 5)} = e^(-t) cos(2t) - e^(-t) sin(2t)/2
L⁻¹{1/(s^2 - 4s + 3)} = e^(3t)/2 - e^t/2
The last one is different in kind: that denominator does factorise, into (s minus 1)(s minus 3), so partial fractions is the route and no shifting is needed. Always check the discriminant before completing the square.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.