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Partial fractions and the inverse transform

Splitting a rational $Y(s)$ into pieces the table already holds, completing the square when it will not factor, and reading the solution back as a function of time.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to split a proper rational transform into partial fractions for distinct roots and for repeated ones, find each numerator by covering up, complete the square when the denominator has no real roots, invert each piece from the table, and check the result against the initial value it came from.

2. What you already have

You can transform an initial value problem and solve for $Y(s)$, which arrives as a rational function. You have also split rational functions into partial fractions in an integration course. This lesson is the return journey, and partial fractions is the whole of the technique.

3. Words this lesson uses

4. Splitting until every piece is a table entry

The transform is one-to-one on continuous functions: if two of them have the same transform, they are the same function. So there is exactly one right answer to what did this come from, and the table may be read in either direction.

The table is short, though, and $Y(s)$ almost never appears in it. The technique is therefore to split $Y$ into a sum of things that are in it, which for a rational function means partial fractions. Three cases cover everything this course meets.

Distinct linear factors. $\dfrac{N(s)}{(s-a)(s-b)}$ splits as $\dfrac{A}{s-a} + \dfrac{B}{s-b}$, and each piece inverts to an exponential.

A repeated factor. $(s-a)^{2}$ needs both $\dfrac{A}{s-a}$ and $\dfrac{B}{(s-a)^{2}}$, and the second inverts to $te^{at}$ — the same factor of $t$ that a repeated characteristic root produced in unit 2.

An irreducible quadratic. Do not force a split. Complete the square to $(s+\alpha)^{2} + \beta^{2}$, which is a sine or cosine entry with $s$ shifted, and a shift in $s$ is multiplication by an exponential in $t$.

The three cases match the three root types of unit 2 exactly, because the denominator of $Y$ is the characteristic polynomial.

5. The cover-up method, and why it works

For $\dfrac{N(s)}{(s-a)(s-b)} = \dfrac{A}{s-a} + \dfrac{B}{s-b}$, multiply everything by $(s-a)$:

$$\frac{N(s)}{s-b} = A + \frac{B(s-a)}{s-b}.$$

Now set $s = a$. The last term has a factor of $(s-a)$ in it and vanishes, leaving $A = \dfrac{N(a)}{a-b}$ immediately. In practice you do not write any of that down: you cover the factor $(s-a)$ with a finger and evaluate what is left at $s = a$.

It works for every simple real root, however many there are, and finds each coefficient independently — no system, no comparison of coefficients, no algebra to slip in. For a repeated root it gives the highest power's coefficient only, and the rest need one of the slower methods.

6. Where this goes wrong

Inverting before splitting. $\mathcal{L}^{-1}$ is linear, so it goes through a sum — but a single compound fraction is not a sum, and $\dfrac{1}{(s-a)(s-b)}$ does not invert to a product of exponentials.

Forcing a factorisation. If the discriminant is negative there are no real roots, and pretending otherwise produces complex coefficients where completing the square would have produced a real answer in two lines.

Giving a repeated factor one term. $(s-a)^{2}$ needs two, one for each power, or the split has too few unknowns to be solvable.

Forgetting the frequency on a sine. The entry is $k/(s^{2}+k^{2})$, so $1/(s^{2}+k^{2})$ inverts to $\tfrac{1}{k}\sin kt$, not $\sin kt$.

7. The method, step by step, and how to check it

Inverting a transform means turning a fraction in $s$ back into a function of $t$, and every inversion in this course follows the same decisions.

1. Make the fraction proper. If the top has degree at least that of the bottom, divide first. Transforms of ordinary functions are always proper, so a top-heavy fraction usually means an error earlier.
2. Factor the denominator into linear factors and irreducible quadratics. Check each quadratic's discriminant: negative means it stays whole.
3. Write the split:

1. Find the coefficients. Cover-up for each simple root; for a repeated root, cover-up gives the top power and comparing coefficients gives the rest.
2. Invert term by term from the table, using linearity. The shift rule says that replacing $s$ by $s - a$ in a transform multiplies the function by $e^{at}$.

Why term by term is allowed. The inverse transform is linear, exactly as the transform is, so the inverse of a sum is the sum of the inverses. It is not multiplicative: $\frac{1}{(s - 1)(s - 4)}$ is not the transform of $e^{t}e^{4t}$. That is the reason for splitting at all.

How to check the answer. Recombine the partial fractions over a common denominator and confirm you get the original numerator. Then compare $y(0)$ with $\lim_{s \to \infty} sY$, which is the ratio of the leading coefficients when the top has degree one less than the bottom, and zero when it has less. These two checks take a line each and catch a wrong coefficient, a wrong sign in an exponent and a forgotten frequency in a sine.

Reading the answer before finding it. The denominator's roots are the exponents that will appear in $y$: a root $a$ gives $e^{at}$, a repeated root adds $te^{at}$, and a complex pair $-\alpha \pm \beta i$ gives $e^{-\alpha t}$ times a cosine and a sine of frequency $\beta$. So the shape of the answer is known before a single coefficient is computed, and an answer with an exponent that is not a root of the denominator is certainly wrong.

8. In the world: a machine's response to being switched on

When a motor, a heater or an amplifier is switched on, its output's transform typically arrives as a fraction such as

$$Y(s) = \frac{6}{s(s + 2)(s + 3)},$$

the step input $\frac{1}{s}$ times a system with two decay rates. Partial fractions turn it into a time response an engineer can plot. Cover up $s$ and put $s = 0$: $A = \frac{6}{2 \times 3} = 1$. Cover up $s + 2$ and put $s = -2$: $B = \frac{6}{(-2)(1)} = -3$. Cover up $s + 3$ and put $s = -3$: $C = \frac{6}{(-3)(-1)} = 2$. So

$$y(t) = 1 - 3e^{-2t} + 2e^{-3t}.$$

Each term has a job. The $1$ is where the output settles; the two exponentials are the transient, and the slower one, $e^{-2t}$, sets how long settling takes. Check the start: $y(0) = 1 - 3 + 2 = 0$, as a system starting from rest must. Check the slope: $y'(0) = 6 - 6 = 0$, so the output leaves zero smoothly, which is the gentle start a motor with inertia shows.

9. In the world: poles, and reading stability from the denominator

Engineers call the roots of the denominator of a transfer function its poles, and they judge a design by where the poles lie before inverting anything, because each pole becomes one term of the answer. A real pole at $s = -a$ gives $e^{-at}$; a pair $-\alpha \pm \beta i$ gives $e^{-\alpha t}$ times an oscillation at $\beta$.

So a pole with a positive real part is a warning. A feedback loop whose denominator is $(s + 3)(s - 0.5)$ contains $e^{0.5t}$: its output doubles every $\frac{\ln 2}{0.5} \approx 1.4$ seconds and the system is unstable, whatever the numerators say. A denominator $s^{2} + 2s + 26 = (s + 1)^{2} + 25$ has poles at $-1 \pm 5i$: the output rings at $5$ radians a second and dies with time constant one second, stable but lively. Moving poles further left makes a system settle faster; moving them away from the real axis makes it ring more.

This is how the partial fractions of this lesson become design: an engineer adjusting a controller's gains is moving poles, and the inverse transform is what tells them what the machine will then do.

10. In the world: why a repeated pole is the boundary

When two poles meet on the real axis the partial fraction needs the $te^{-at}$ term, and the response is the fastest that does not overshoot: the critical damping of lesson 8, now seen as two poles colliding. Instrument makers design meters and scale balances to sit just there, so the needle arrives quickly and does not swing.

A bathroom scale shows the idea. Step on it and the needle, a mass on a spring with some damping, has a transform whose denominator is the characteristic polynomial. If its poles are a complex pair, the needle overshoots your weight and swings back and forth before settling, and you wait. If they are two separated real poles, the needle creeps up slowly. The designer tunes the damping until the two poles meet, at a repeated root: the partial fraction then has a term $\frac{B}{(s + a)^{2}}$, the needle's motion contains $te^{-at}$, and it reaches the reading as fast as it can without passing it. Every step of that design, from choosing the damping to predicting the needle's motion, is a partial fraction split and an inversion. The same reasoning sets the damping of a car's speedometer, a galvanometer and the arm of a record player, anywhere a reading must arrive quickly and then stay put. In each case the number the designer adjusts is the one that decides whether two poles meet, and the partial fraction shows what happens on either side of that choice.

11. The inverse is not a formula applied to the whole fraction

Because the forward direction feels mechanical — look up, write down — it is easy to expect the reverse to be mechanical in the same way, and to try to invert $Y(s)$ as it stands. Linearity is what makes the table usable, and linearity applies to a sum; a single fraction with a compound denominator is not a sum until partial fractions makes it one. The related error is to invert numerator and denominator separately, which has no justification of any kind and produces an answer that fails the $y(0)$ check immediately. The reliable discipline is that nothing is inverted until every piece in front of you is, character for character, a row of the table.

12. Two distinct roots

1. Invert $Y = \dfrac{6}{(s - 1)(s - 4)}$. Write the split.

   $\dfrac{6}{(s - 1)(s - 4)} = \dfrac{A}{s - 1} + \dfrac{B}{s - 4}$

   Two distinct linear factors, one term each.

2. Cover up $s - 1$ and set $s = 1$.

   $A = \dfrac{6}{1 - 4} = \dfrac{6}{-3} = -2$

   Every other term vanishes at the root.

3. Cover up $s - 4$ and set $s = 4$.

   $B = \dfrac{6}{4 - 1} = \dfrac{6}{3} = 2$

   The same move for the other factor.

4. Check the split by recombining it.

   $\dfrac{-2}{s - 1} + \dfrac{2}{s - 4} = \dfrac{-2(s - 4) + 2(s - 1)}{(s - 1)(s - 4)} = \dfrac{6}{(s - 1)(s - 4)} \quad \checkmark$

   The $s$ terms cancel and $8 - 2 = 6$.

5. Invert each piece and check at $t = 0$.

   $y = -2e^{t} + 2e^{4t}, \qquad y(0) = -2 + 2 = 0$

   With a constant numerator the coefficients must cancel, because $sY \to 0$: a free check.

13. A quadratic that will not factor

1. Invert $Y = \dfrac{s + 3}{s^{2} + 2s + 5}$. Check the discriminant.

   $\Delta = 2^{2} - 4 \times 5 = 4 - 20 = -16 < 0$

   Check before splitting: there is nothing to split.

2. Complete the square in the denominator.

   $s^{2} + 2s + 5 = (s + 1)^{2} + 4$

   Half of $2$ is $1$, and $5 - 1 = 4$ is left over.

3. Rewrite the numerator in the same shifted variable, $s + 1$.

   $s + 3 = (s + 1) + 2$

   The table entries are in $s + 1$, so the numerator must be too.

4. Split into a shifted cosine and a shifted sine.

   $Y = \dfrac{s + 1}{(s + 1)^{2} + 2^{2}} + \dfrac{2}{(s + 1)^{2} + 2^{2}}$

   $s + 1$ on top is the cosine entry; the frequency $2$ on top is the sine entry.

5. Invert both pieces with the shift rule.

   $y = e^{-t}\cos 2t + e^{-t}\sin 2t$

   Replacing $s$ by $s + 1$ multiplies the function by $e^{-t}$. The shift became the decay and the $4$ became the frequency $2$.

6. Check the starting value.

   $y(0) = 1 \cdot 1 + 1 \cdot 0 = 1, \qquad \lim_{s \to \infty} sY = \lim \dfrac{s^{2} + 3s}{s^{2} + 2s + 5} = 1$

   Both give $1$, so the numerator was shifted correctly.

14. A repeated root

1. Invert $Y = \dfrac{2s + 3}{(s + 1)^{2}}$. A squared factor needs one term for each power.

   $\dfrac{2s + 3}{(s + 1)^{2}} = \dfrac{A}{s + 1} + \dfrac{B}{(s + 1)^{2}}$

   With one term the split would have too few unknowns to match a linear numerator.

2. Multiply both sides by $(s + 1)^{2}$.

   $2s + 3 = A(s + 1) + B$

   Clearing the denominator turns the split into an identity between polynomials.

3. Put $s = -1$ to find $B$.

   $2(-1) + 3 = A \cdot 0 + B \quad\Rightarrow\quad B = 1$

   This is the cover-up value, and it gives only the highest power's coefficient.

4. Compare the coefficients of $s$ to find $A$.

   $2s = As \quad\Rightarrow\quad A = 2$

   The identity holds for every $s$, so the $s$ terms on each side must match.

5. Invert both terms.

   $y = 2e^{-t} + te^{-t}$

   $\frac{1}{s + 1} \to e^{-t}$, and $\frac{1}{(s + 1)^{2}} \to te^{-t}$ from the table row $te^{at}$.

6. Check the starting value.

   $y(0) = 2 + 0 = 2, \qquad \lim_{s \to \infty} sY = \lim \dfrac{2s^{2} + 3s}{s^{2} + 2s + 1} = 2$

   The factor of $t$ is the transform's version of the repeated-root $xe^{rx}$.

15. Your turn: invert $Y = \dfrac{2s + 1}{s^{2} + s}$

1. Factor the denominator and write the split.

   $s^{2} + s = s(s + 1), \qquad \dfrac{2s + 1}{s(s + 1)} = \dfrac{A}{s} + \dfrac{B}{s + 1}$

   Factor first, always.

2. Your turn: work this step out. Its working is at the end of the packet.

   Cover up twice.

3. Your turn: work this step out. Its working is at the end of the packet.

   Invert each piece.

16. Guided practice

Put the steps of inverting $\dfrac{1}{(s - 2)(s - 9)}$ into the order they must be done.

Number the steps in order (write the number in the box):

• Factor the denominator
• Write the split with unknown numerators
• Solve for the unknown numerators
• Invert each piece from the table
• Add the pieces

17. Guided practice

Complete the worked solution: invert $\dfrac{24}{(s - 8)(s - 4)}$.

1. Write the split with unknown numerators.

   $\dfrac{24}{(s - 8)(s - 4)} = \dfrac{A}{s - 8} + \dfrac{B}{s - 4}$

   Each distinct linear factor gets one term.

2. Cover up $s - 8$, set $s = 8$, and simplify.

   $A = \dfrac{24}{8 - 4} = \dfrac{24}{4} =$ _a

   Multiplying by $s - 8$ and setting $s$ to its root removes every term but $A$.

3. Cover up $s - 4$, set $s = 4$, and simplify.

   $B = \dfrac{24}{4 - 8} = \dfrac{24}{-4} =$ _c

   The same move for the other factor; this time the subtraction is negative.

4. Invert each piece from the table.

   $f(t) = Ae^{8t} + Be^{4t}$

   $\mathcal{L}^{-1}\left\{\frac{1}{s - r}\right\} = e^{rt}$, and the transform is linear.

5. Check at $t = 0$.

   $f(0) = A + B =$ _z

   A constant over a quadratic decays like $1/s^{2}$, so the coefficients must cancel.

18. Guided practice

Match each transform to the function of $t$ it came from. (One of them is $\dfrac{s}{s^{2} + 25}$, which came from $\cos 5t$ because the same denominator with $s$ on top is the cosine instead.)

19. Practice

Split $\dfrac{12}{(s - 7)(s - 3)}$ as $\dfrac{A}{s - 7} + \dfrac{B}{s - 3}$, then fill in the table.

20. Practice

In the split $\dfrac{12}{(s - 8)(s - 4)} = \dfrac{A}{s - 8} + \dfrac{B}{s - 4}$, what is $A$?

Answer:

21. Practice

The output of a machine switched on at $t = 0$ has the Laplace transform below, found from its equations; to plot what it actually does, it has to be turned back into a function of time. Invert $Y(s) = \dfrac{8s^{2} + 5s - 4}{s(s - 1)(s + 2)}$. Write $y(t)$ as a formula in $t$ (type each exponential as e^(...)).

Answer:

22. Somewhere new

What does $\dfrac{1}{s^{2} + 8s + 25}$ invert to?

• A decaying exponential times a sine, $e^{-4t}\sin 3t$, up to a constant
• A sum of two exponentials, one for each root of the denominator
• A pure sine of frequency $3$, with no exponential factor
• A single exponential $e^{-4t}$, with no oscillation

23. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

24. Test question

The output of a machine switched on at $t = 0$ has the Laplace transform below, found from its equations; to plot what it actually does, it has to be turned back into a function of time. Invert $Y(s) = \dfrac{4s^{2} + 7s - 3}{s(s - 1)(s + 3)}$. Write $y(t)$ as a formula in $t$ (type each exponential as e^(...)).

Answer:

25. What you can do now

You can invert a rational transform by splitting it into table entries, and you can tell from the discriminant whether to split or to complete the square. Say in your own words why a repeated factor needs two terms rather than one.

Working for the steps left to you

15. Your turn: invert $Y = \dfrac{2s + 1}{s^{2} + s}$, step 2

15. Your turn: invert $Y = \dfrac{2s + 1}{s^{2} + s}$, step 3