Completing the Square, and the First Shifting Theorem Backwards
Chapter Sixty-Seven
Syllabus topic Module 1, "1.3 Inverse Laplace Transform"
Pages 149 to 150 of 303
In one line
If the quadratic denominator does not factorise, complete the square, and the shift comes out as an exponential.
When to reach for it
Look at the discriminant of the quadratic in the denominator. If it is negative, the quadratic does not factorise over the reals, partial fractions cannot help, and completing the square is the method.
| Denominator | Discriminant | Method |
|---|---|---|
| s^2 + 4s + 3 | 16 - 12 = 4, positive | partial fractions |
| s^2 + 4s + 4 | 16 - 16 = 0 | a repeated factor |
| s^2 + 4s + 8 | 16 - 32 = -16, negative | complete the square |
Checking the discriminant first takes ten seconds and saves the two minutes spent trying to factorise something that will not.
Completing the square
For s squared plus bs plus c, take half of b, square it, add and subtract.
s^2 + 6s + 13 = (s + 3)^2 + 4
s^2 - 4s + 13 = (s - 2)^2 + 9
s^2 + 2s + 5 = (s + 1)^2 + 4
s^2 + 8s + 25 = (s + 4)^2 + 9
In each case the result is (s minus a) squared plus b squared, and the two table rows that fit are the shifted sine and the shifted cosine.
The two rows
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)
Note what a and b do. a comes out as the exponential and b as the frequency, and a's sign is the sign inside the bracket reversed: (s plus 3) squared means a is minus 3, so the answer decays.
Worked, numerator constant
L⁻¹{1/(s^2 + 6s + 13)} = e^(-3t) sin(2t)/2
L⁻¹{1/(s^2 - 4s + 13)} = e^(2t) sin(3t)/3
L⁻¹{5/(s^2 + 2s + 5)} = 5 e^(-t) sin(2t)/2
Each is the same three steps: complete the square, read a and b, divide by b.
Worked, numerator containing s
This is the case that needs the extra step, and it is the commoner one in practice, because a second-order differential equation nearly always produces it.
Invert the function below.
F(s) = (s + 5)/(s^2 + 4s + 8)
Complete the square: the denominator is (s plus 2) squared plus 4, so a is minus 2 and b is 2.
Now the numerator must be written in terms of s plus 2, because that is what the cosine row needs on top. Write s plus 5 as (s plus 2) plus 3.
s + 5 = (s + 2) + 3
Now split into two rows.
L⁻¹{(s + 2)/((s + 2)^2 + 4)} = e^(-2t) cos(2t)
L⁻¹{3/((s + 2)^2 + 4)} = 3 e^(-2t) sin(2t)/2
Completing the Square, and the First Shifting Theorem Backwards
And the answer is the sum.
L⁻¹{(s + 5)/(s^2 + 4s + 8)} = e^(-2t) cos(2t) + 3 e^(-2t) sin(2t)/2
The procedure, which is worth learning as five steps
- Check the discriminant. Negative means this method.
- Complete the square, giving (s minus a) squared plus b squared.
- Rewrite the numerator so that every s appears as (s minus a), splitting off a constant.
- Invert the (s minus a) part with the cosine row.
- Invert the constant part with the sine row, dividing by b.
Step three is the one that is skipped, and skipping it leaves an s on top of a shifted denominator, which matches no table row at all.
Two more, worked quickly
L⁻¹{s/(s^2 + 2s + 2)} = e^(-t) cos(t) - e^(-t) sin(t)
L⁻¹{(2s - 1)/(s^2 - 2s + 10)} = 2 e^t cos(3t) + e^t sin(3t)/3
For the first: the denominator is (s plus 1) squared plus 1, and s is (s plus 1) minus 1, so the answer is the cosine row minus the sine row, with b = 1.
For the second: the denominator is (s minus 1) squared plus 9, and 2s minus 1 is 2(s minus 1) plus 1, so the answer is twice the cosine row plus one third of the sine row.
When the denominator has a factor of s as well
Then there are two things going on: complete the square on the quadratic and use partial fractions on the s. The next four chapters handle partial fractions; the combination looks like this.
L⁻¹{1/(s(s^2 + 2s + 2))} = 1/2 - e^(-t) cos(t)/2 - e^(-t) sin(t)/2
Partial fractions splits off the one over s term, and what is left is a quadratic over a quadratic, which completes the square.
Check yourself
L⁻¹{1/(s^2 + 4s + 5)} = e^(-2t) sin(t)
L⁻¹{1/(s^2 - 6s + 10)} = e^(3t) sin(t)
L⁻¹{(s + 1)/(s^2 + 2s + 10)} = e^(-t) cos(3t)
L⁻¹{s/(s^2 + 4s + 13)} = e^(-2t) cos(3t) - 2 e^(-2t) sin(3t)/3
The third is the pleasant case: the numerator is already exactly (s minus a), so no splitting is needed and the answer is one term. Recognising that saves a step, and it happens more often than you would expect because a well-set question is often built that way.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.