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The Laplace transform

The integral that carries a function of $t$ to a function of $s$, the short table of pairs it produces, and the linearity that makes that table enough.

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 state the defining integral of the Laplace transform, derive the first two entries of its table, transform any combination of constants, powers, exponentials and sinusoids by linearity, evaluate a transform at a value of $s$, and say for which values of $s$ the defining integral converges.

2. What you already have

You can evaluate an improper integral as a limit, you can integrate by parts, and you can solve a constant-coefficient second-order equation from its characteristic roots. This lesson builds the tool; the next two put it to work, and the reason it is worth the detour is that it will turn an initial value problem into an algebra problem.

3. Words this lesson uses

4. An integral that changes the variable a function lives in

For a function $f$ defined on $t \ge 0$,

$$\mathcal{L}\{f\}(s) = F(s) = \int_0^{\infty} e^{-st} f(t)\,dt.$$

The output is a function of a new variable $s$. Nothing is being simplified and nothing is being approximated: the information in $f$ is carried across intact, and the point is that operations which are hard in $t$ become easy in $s$.

Linearity is what makes a short table enough:

$$\mathcal{L}\{\alpha f + \beta g\} = \alpha F + \beta G.$$

So any combination of the entries below can be transformed on sight.

Every row of that table is one integral, done once, so that it never has to be done again.

5. Where the first two entries come from

Only two integrals are needed to see how the table is built.

For $f = 1$: $\int_0^{\infty} e^{-st}\,dt = \left[-\tfrac{1}{s}e^{-st}\right]_0^{\infty}$. The upper limit gives zero provided $s > 0$, and the lower gives $\tfrac{1}{s}$. If $s \le 0$ the exponential does not decay and there is nothing to evaluate — which is where the convergence condition comes from, rather than being an afterthought.

For $f = t$: integrate by parts with $u = t$ and $dv = e^{-st}dt$. The boundary term vanishes for $s > 0$ and what is left is $\tfrac{1}{s}\int_0^{\infty} e^{-st}dt = \tfrac{1}{s^{2}}$.

Repeating that argument on $t^{n}$ gives $\tfrac{n}{s}$ times the transform of $t^{n-1}$, and the factorial in the table is that recursion run down to the bottom.

6. Where this goes wrong

Confusing $f$ and $F$. They are different functions of different variables. $F(2)$ is the transform evaluated at $s = 2$, and has nothing to do with $f(2)$.

Transforming a product term by term. Linearity is about sums and constant multiples only. $\mathcal{L}\{fg\}$ is not $FG$ — what it actually is needs the convolution theorem, which this course names and does not lean on.

Dropping the region of convergence. A transform without it is a formula that returns numbers where the integral defining it does not exist.

Expecting every function to have one. $e^{t^{2}}$ outgrows every exponential, so no $s$ makes the integral converge and the transform does not exist for any $s$ at all.

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

Transforming a function from the table is a short, fixed routine.

1. Split into terms. Rewrite the function as a sum of pieces that each look like a table row, expanding brackets first: $(1 + t)^{2}$ becomes $1 + 2t + t^{2}$.
2. Pull the constants out. A constant multiplier passes through the transform unchanged.
3. Match each piece to a row, reading off its parameter carefully. In $e^{-3t}$ the table's $a$ is $-3$, so the entry is $\frac{1}{s + 3}$. In $t^{3}$ the table's $n$ is $3$, so the entry is $\frac{3!}{s^{4}} = \frac{6}{s^{4}}$.
4. Add the results.
5. State the region as the strictest of the rows' conditions.

Why linearity is allowed. The transform is an integral, $\int_0^{\infty} e^{-st}f(t)\,dt$, and the integral of a sum is the sum of the integrals, with constants coming outside. That is all linearity says. It does not extend to products: the integral of a product is not the product of the integrals, so $t\sin t$ cannot be transformed by multiplying two table entries.

Why each entry has a region. Every row comes from an integral to infinity, and such an integral exists only when the integrand dies away fast enough. The factor $e^{-st}$ is the thing that makes it die: for $e^{at}$ the integrand is $e^{(a - s)t}$, which decays exactly when $s > a$. A function that grows faster than every exponential, such as $e^{t^{2}}$, has no transform at all.

How to check the answer. Three quick tests catch most slips. First, each piece of $F$ should tend to zero as $s \to \infty$; a transform that grows with $s$ is wrong. Second, $sF(s)$ should tend to $f(0)$ as $s \to \infty$: for $f = 3 + 4t - 2e^{5t}$, $f(0) = 1$, and $s\left(\frac{3}{s} + \frac{4}{s^{2}} - \frac{2}{s - 5}\right) \to 3 + 0 - 2 = 1$. Third, check the sine and cosine rows the right way round: the sine has the frequency on top, the cosine has $s$.

What the transform is for. Nothing in this lesson has solved an equation yet. The next lesson uses one more property, that differentiating $f$ multiplies $F$ by $s$ and subtracts the starting value, to turn a differential equation into algebra. Fluency with the table is what makes that step fast.

8. In the world: how engineers describe a system by its transform

Control engineers rarely write the differential equation of a system down twice. They transform it once and describe the system by its transfer function, the ratio of the output's transform to the input's. A heater whose temperature excess $y$ obeys $\tau y' + y = Ku$, with $u$ the power setting, transforms (from rest) to $\tau sY + Y = KU$, so $\frac{Y}{U} = \frac{K}{\tau s + 1}$. Everything about the heater is in that fraction: its gain $K$ and its time constant $\tau$.

The reason this works is the rule that makes the transform worth learning: differentiating in time becomes multiplying by $s$. A system built from derivatives becomes a system built from multiplications, and connecting two systems one after another multiplies their transfer functions. Engineers draw block diagrams of heaters, motors and sensors and combine them by algebra, with the table of this lesson as their dictionary between time and $s$.

The standard test input is a step, the power switched from $0$ to a constant $u_0$ at $t = 0$, whose transform is $\frac{u_0}{s}$ from the first row of the table. The heater's response then has transform $\frac{Ku_0}{s(\tau s + 1)}$, which the lesson after next turns back into $Ku_0\left(1 - e^{-t/\tau}\right)$.

9. In the world: reading a signal's start and end from its transform

Two quick facts let an engineer read a signal's first and last values from its transform without inverting it. For a well-behaved $f$,

$$f(0) = \lim_{s \to \infty} sF(s), \qquad \lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s),$$

the second only when $f$ really does settle. Take a sensor whose output is $f(t) = 5 + 3e^{-2t}$ volts. Its transform, term by term from the table, is $F(s) = \frac{5}{s} + \frac{3}{s + 2}$. Then $sF(s) = 5 + \frac{3s}{s + 2}$, which tends to $5 + 3 = 8$ as $s \to \infty$ and to $5 + 0 = 5$ as $s \to 0$: the sensor starts at $8$ volts and settles at $5$, exactly what $f$ says.

The value of the two limits is that they work on a transform that has not been, or cannot easily be, inverted: the transform of a whole control loop, built by multiplying block transfer functions, can have a long denominator, and the engineer's first question, where does it settle and does it settle at the right value, is answered by one limit. The second fact fails for a signal that oscillates for ever or grows, such as $\sin t$ or $e^{t}$, and a designer checks that first, which the next two lessons show how to do.

10. In the world: why the table stops where it does

The table of this lesson is short because the signals engineers switch on are short in kind: constants, ramps, decaying and growing exponentials, and oscillations. Almost every input to a mechanical or electrical system is built from those by adding and delaying, which is why the table, linearity and the delay rule of lesson 16 together cover nearly everything met in practice.

When a signal is not on the table, it is usually built from entries that are. A motor current that rises linearly for two seconds and then holds is a ramp minus a delayed ramp; a pulse is a step minus a delayed step; a signal that swings and fades is a cosine times a decaying exponential, whose transform is the cosine's entry with $s$ shifted. An engineer faced with a new input first asks which entries it is made of, transforms each, and adds, which is the method of this lesson applied to the world rather than to an exercise. The row that is genuinely missing, a function like $e^{t^{2}}$ that grows faster than any exponential, does not arise from any physical device, because no real quantity grows that fast for long.

11. The transform is not a substitution, and the table is not a dictionary of symbols

Because the table is used by pattern-matching, it is easy to start treating the transform as a symbol-for-symbol swap — $t$ becomes $s$, and a product of functions becomes a product of transforms. It is neither. Each row of the table is the outcome of an integral, and the only structural rules that hold in general are the linear ones: sums split and constants come out. Multiplication does not survive, and neither does composition; $\mathcal{L}\{f(2t)\}$ is not $F(2s)$ but $\tfrac{1}{2}F(s/2)$, and the factor in front is exactly what a symbol swap would lose. When a rule is needed for something other than a sum, it is a theorem with a proof, not a habit of notation.

12. A sum, done term by term

1. Transform $f(t) = 3 + 4t - 2e^{5t}$. Split it by linearity and pull the constants out.

   $\mathcal{L}\{f\} = 3\mathcal{L}\{1\} + 4\mathcal{L}\{t\} - 2\mathcal{L}\{e^{5t}\}$

   Three separate table entries, with the constants pulled out.

2. Look up the constant and the power of $t$.

   $3 \cdot \dfrac{1}{s} = \dfrac{3}{s}, \qquad 4 \cdot \dfrac{1}{s^{2}} = \dfrac{4}{s^{2}}, \qquad s > 0$

   $\mathcal{L}\{1\} = \frac{1}{s}$ and $\mathcal{L}\{t\} = \frac{1!}{s^{2}}$.

3. Look up the exponential, which has $a = 5$.

   $-2 \cdot \dfrac{1}{s - 5} = -\dfrac{2}{s - 5}, \qquad s > 5$

   $\mathcal{L}\{e^{at}\} = \frac{1}{s - a}$, and the integral converges only while $e^{-st}$ beats $e^{5t}$.

4. Add the three results.

   $F(s) = \dfrac{3}{s} + \dfrac{4}{s^{2}} - \dfrac{2}{s - 5}$

   Constants came straight through, signs included.

5. State the region of convergence.

   $s > 0, \quad s > 0, \quad s > 5 \quad\Rightarrow\quad s > 5$

   The region is the strictest of the three, because every piece must converge for the sum to.

13. A function with no transform

1. For $f(t) = e^{t^{2}}$, write the defining integral and combine the exponents in the integrand.

   $\displaystyle\int_0^{\infty} e^{-st}e^{t^{2}}\,dt = \int_0^{\infty} e^{t^{2} - st}\,dt$

   $e^{A}e^{B} = e^{A + B}$.

2. Factor the exponent to see its sign for large $t$.

   $t^{2} - st = t(t - s)$

   A product is easier to read than a difference: both factors are positive once $t > s$.

3. Bound the integrand below beyond $t = s$ (take $s > 0$; smaller $s$ only makes it larger).

   $t \ge s + 1: \quad t(t - s) \ge (s + 1) \cdot 1 > 0 \quad\Rightarrow\quad e^{t^{2} - st} > 1$

   The exponent is positive, and $e$ to a positive power exceeds $1$.

4. Compare the tail of the integral with the integral of $1$.

   $\displaystyle\int_{s + 1}^{\infty} e^{t^{2} - st}\,dt > \int_{s + 1}^{\infty} 1\,dt = \infty$

   An integrand that stays above $1$ over an infinite interval has an infinite integral.

5. Conclude that no transform exists.

   $\mathcal{L}\{e^{t^{2}}\} \text{ exists for no } s$

   No choice of $s$ rescues it: growth faster than every exponential is exactly the hypothesis that fails.

14. A table entry derived from the definition

1. Find $\mathcal{L}\{\sin kt\}$ from the integral, for $s > 0$. Name it $I$ and integrate by parts with $u = \sin kt$, $dv = e^{-st}\,dt$.

   $\displaystyle I = \int_0^{\infty} e^{-st}\sin kt\,dt = \left[-\frac{e^{-st}}{s}\sin kt\right]_0^{\infty} + \frac{k}{s}\int_0^{\infty} e^{-st}\cos kt\,dt$

   $v = -\frac{e^{-st}}{s}$ and $du = k\cos kt\,dt$; the minus signs combine to a plus.

2. Evaluate the boundary term.

   $\left[-\dfrac{e^{-st}}{s}\sin kt\right]_0^{\infty} = 0 - 0 = 0 \quad\Rightarrow\quad I = \dfrac{k}{s}J, \quad J = \displaystyle\int_0^{\infty} e^{-st}\cos kt\,dt$

   At infinity $e^{-st} \to 0$ with $\sin$ bounded; at $0$, $\sin 0 = 0$.

3. Integrate $J$ by parts the same way, with $u = \cos kt$.

   $J = \left[-\dfrac{e^{-st}}{s}\cos kt\right]_0^{\infty} - \dfrac{k}{s}\displaystyle\int_0^{\infty} e^{-st}\sin kt\,dt = \dfrac{1}{s} - \dfrac{k}{s}I$

   At $0$ the boundary term is $-\left(-\frac{1}{s}\cos 0\right) = \frac{1}{s}$, and $du = -k\sin kt\,dt$.

4. Substitute $J$ back into $I = \frac{k}{s}J$.

   $I = \dfrac{k}{s}\left(\dfrac{1}{s} - \dfrac{k}{s}I\right) = \dfrac{k}{s^{2}} - \dfrac{k^{2}}{s^{2}}I$

   The original integral has come back, which is what makes the trick work.

5. Collect the $I$ terms on the left.

   $I + \dfrac{k^{2}}{s^{2}}I = \dfrac{k}{s^{2}} \quad\Rightarrow\quad I \cdot \dfrac{s^{2} + k^{2}}{s^{2}} = \dfrac{k}{s^{2}}$

   Add $\frac{k^{2}}{s^{2}}I$ to both sides and factor.

6. Multiply both sides by $\frac{s^{2}}{s^{2} + k^{2}}$.

   $I = \mathcal{L}\{\sin kt\} = \dfrac{k}{s^{2} + k^{2}}, \qquad s > 0$

   This is the table entry. The same calculation gives $J = \frac{s}{s^{2} + k^{2}}$ for the cosine.

7. Check it with a limit.

   $sF(s) = \dfrac{ks}{s^{2} + k^{2}} \to 0 = \sin 0 \text{ as } s \to \infty$

   For a well-behaved $f$, $sF(s)$ tends to $f(0)$ for large $s$, so the formula passes a quick test.

15. Your turn: transform $f(t) = 6t^{2} + \cos 4t$, and say where it converges

1. Transform the $t^{2}$ term.

   $6\mathcal{L}\{t^{2}\} = 6 \cdot \dfrac{2!}{s^{3}} = \dfrac{12}{s^{3}}$

   The coefficient $6$ and the factorial $2$ multiply.

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

   Transform the cosine.

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

   Add them and state the region.

16. Guided practice

Match each function of $t$ to its Laplace transform.

17. Guided practice

Complete the worked solution: the Laplace transform of $f(t) = 7t^{2} + 5t^{3}$.

1. Split the transform by linearity, pulling each coefficient out.

   $\mathcal{L}\{f\} = 7\mathcal{L}\{t^{2}\} + 5\mathcal{L}\{t^{3}\}$

   The transform of a sum is the sum of the transforms, and constants pass through.

2. Look up each power of $t$ in the table.

   $\mathcal{L}\{t^{2}\} = \dfrac{2!}{s^{3}}, \qquad \mathcal{L}\{t^{3}\} = \dfrac{3!}{s^{4}}$

   $\mathcal{L}\{t^{n}\} = \frac{n!}{s^{n + 1}}$.

3. Multiply the first coefficient by $2!$ to get the numerator over $s^{3}$.

   $7 \times 2! = 7 \times 2 =$ _p

   $2! = 2 \times 1$.

4. Multiply the second coefficient by $3!$ to get the numerator over $s^{4}$.

   $5 \times 3! = 5 \times 3 \times 2 =$ _q

   $3! = 3 \times 2 \times 1$.

5. Assemble the transform from the two numerators, and state the region.

   $F(s) = \dfrac{7 \cdot 2!}{s^{3}} + \dfrac{5 \cdot 3!}{s^{4}}, \qquad s > 0$

   Both table entries converge for $s > 0$.

18. Guided practice

Which is the Laplace transform of $4 + 3t$?

• $\dfrac{4}{s} + \dfrac{3}{s^{2}}$
• $\dfrac{4 + 3}{s^{3}}$
• $\dfrac{4}{s} + \dfrac{3}{s}$
• $\dfrac{4s + 3s^{2}}{1}$

19. Practice

The transform of $f(t) = 4 + 2t + 4t^{2}$ is a sum of three terms, each with the power of $s$ shown. Fill in the numerator of each.

20. Practice

Let $f(t) = 5 + 5t$. What is $F(4)$, the value of its Laplace transform at $s = 4$?

Answer:

21. Practice

A sensor's output voltage $t$ seconds after it is switched on is the signal below, and the control system that reads it works with its Laplace transform. Find the Laplace transform of $f(t) = 6t + 6e^{-4t} + 2\cos 2t$. Write $F(s)$ as a formula in $s$.

Answer:

22. Somewhere new

For $f(t) = e^{5t}$ the defining integral is $\int_0^{\infty} e^{-st}e^{5t}\,dt$. For which values of $s$ does it converge?

This task has no paper form; do it on a device.

23. Lesson test

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

24. Test question

A sensor's output voltage $t$ seconds after it is switched on is the signal below, and the control system that reads it works with its Laplace transform. Find the Laplace transform of $f(t) = 5t + 4e^{-4t} + 5\cos 2t$. Write $F(s)$ as a formula in $s$.

Answer:

25. What you can do now

You can transform a sum of standard functions term by term and say where the result converges. Say in your own words why the transform of a product is not the product of the transforms.

Working for the steps left to you

15. Your turn: transform $f(t) = 6t^{2} + \cos 4t$, and say where it converges, step 2

15. Your turn: transform $f(t) = 6t^{2} + \cos 4t$, and say where it converges, step 3