Partial Fractions: Repeated Factors
Chapter Sixty-Nine
Syllabus topic Module 1, "1.3 Inverse Laplace Transform"
Pages 153 to 154 of 303
In one line
A factor repeated n times needs n terms, one for each power, and the answer carries powers of t.
The shape
If a linear factor appears n times, you need a term for every power from 1 to n.
F(s) = 1/((s - 1)(s - 2)^2) = A/(s - 1) + B/(s - 2) + C/(s - 2)^2
Writing only the highest power, C over (s minus 2) squared, is the standard mistake, and it makes the identity impossible to satisfy.
Why every power is needed
Count the unknowns against the equations. Multiplying up gives an identity between polynomials of degree 2, which has three coefficients, so three equations. Three equations need three unknowns, and the three terms above supply exactly three.
Leave out the middle term and you have two unknowns for three equations, which in general has no solution. That is the algebraic reason, and it is the answer to "why do we need the lower powers".
Finding the constants
Substitution still works for the repeated root's highest power and for every distinct root, but not for the lower powers of the repeated factor, because substituting the repeated root kills them too.
So the routine is:
- Substitute each distinct root to get its constant.
- Substitute the repeated root to get the constant on the highest power.
- Get the remaining constants by comparing coefficients, or by substituting any convenient extra value of s.
Worked, on the example above. Multiplying up: 1 equals A(s minus 2) squared plus B(s minus 1)(s minus 2) plus C(s minus 1).
At s = 1: 1 equals A times 1, so A is 1. At s = 2: 1 equals C times 1, so C is 1. Now put s = 0: 1 equals A times 4 plus B times 2 plus C times minus 1, that is 1 equals 4 plus 2B minus 1, so 2B is minus 2 and B is minus 1.
1/((s - 1)(s - 2)^2) = 1/(s - 1) - 1/(s - 2) + 1/(s - 2)^2
Inverting, and where the t comes from
The row for a repeated factor carries a power of t.
L⁻¹{1/(s - a)^2} = t e^(a t)
L⁻¹{1/(s - a)^3} = t^2 e^(a t)/2
L⁻¹{1/(s - a)^4} = t^3 e^(a t)/6
So the example above inverts as follows.
L⁻¹{1/((s - 1)(s - 2)^2)} = e^t - e^(2t) + t e^(2t)
A repeated factor means a t in the answer. That is the single fact to carry away, and it has a physical meaning: a repeated root is the boundary between oscillation and pure decay, and the t is the system on that boundary.
Worked, a repeated factor at zero
1/(s^2 (s + 1)) = 1/s^2 - 1/s + 1/(s + 1)
Partial Fractions: Repeated Factors
At s = 0 the highest power's constant: 1 equals A times 1, so the constant on one over s squared is 1. At s = minus 1: 1 equals C times 1, so the constant on one over (s plus 1) is 1. Comparing the coefficients of s squared: 0 equals B plus C, so B is minus 1.
L⁻¹{1/(s^2 (s + 1))} = t - 1 + e^(-t)
That answer is worth a moment. It contains a t, from the repeated factor at zero, a constant, and a decaying exponential. In a circuit that is a ramp with an offset and a transient, which is what a constant voltage applied to an inductor and a resistor in series actually does.
Worked, a triple factor
1/(s (s + 1)^3) = 1/s - 1/(s + 1) - 1/(s + 1)^2 - 1/(s + 1)^3
Four terms for four unknowns. Then:
L⁻¹{1/(s (s + 1)^3)} = 1 - e^(-t) - t e^(-t) - t^2 e^(-t)/2
The alternative for the lower powers: differentiate
There is a neater way to get the lower constants, and it is worth knowing because it is quicker for a triple factor.
Multiply F by the full repeated factor, giving a function with no pole at the repeated root. Then the constant on the highest power is its value at the root, the constant on the next power down is its first derivative at the root, the next is half its second derivative, and so on.
That is Taylor's theorem in disguise, and it is also L'Hopital's rule in disguise, which is the third of the three places the indeterminate-forms chapters said this paper needs them.
For the triple-factor example: multiply by (s plus 1) cubed to get one over s. Its value at s = minus 1 is minus 1, which is the constant on the cube. Its derivative is minus one over s squared, whose value at minus 1 is minus 1, the constant on the square. Half its second derivative is one over s cubed, whose value at minus 1 is minus 1, the constant on the single power. All three agree with the answer above.
The check
Same as before: substitute a convenient value of s into both forms.
1/(1 (1 + 1)^3) = 1/8
1/1 - 1/(1 + 1) - 1/(1 + 1)^2 - 1/(1 + 1)^3 = 1/8
Check yourself
1/(s (s + 2)^2) = 1/(4s) - 1/(4(s + 2)) - 1/(2(s + 2)^2)
(s + 3)/(s + 1)^2 = 1/(s + 1) + 2/(s + 1)^2
L⁻¹{1/(s (s + 2)^2)} = 1/4 - e^(-2t)/4 - t e^(-2t)/2
L⁻¹{(s + 3)/(s + 1)^2} = e^(-t) + 2 t e^(-t)
L⁻¹{1/(s^2 (s - 1))} = e^t - t - 1
L⁻¹{1/(s - 3)^3} = t^2 e^(3t)/2
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.