Laplace Transform Lab

Explore forward Laplace transform pairs for core function families, their poles, region of convergence, and initial/final values.

Worked example — Laplace Transform Lab

Poles	−1
Initial value f(0⁺)	1
Final value f(∞)	0

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Laplace Transform Lab

The Laplace transform trades a function of time for a function of a complex variable s, turning calculus in t into algebra in s. The catalog here is a dictionary of eight such pairs, and beside each one the page prints four facts that are read off F(s) alone, without ever returning to the time domain.

What each symbol means

F(s): the transform. A ratio of polynomials for every entry in this catalog, which is why the next three rows exist at all.
poles: the values of s that make the denominator zero. Everything else on the page is downstream of where they sit.
ROC: the region of convergence — the half-plane where the defining integral actually converges. Step 3 is about how little independent information this row carries.
f(∞): the final value, printed as a number, as ∞, or as a refusal.

Where the formula comes from

Read the pair. At the default exponential, e^(−t) becomes 1/(s+1) — a decay rate in time turns into the position of a single pole.
Find the poles by setting the denominator to zero. s+1=0 gives s = −1, which is what the Poles row shows.
Now compare the Poles row with the ROC row, and keep comparing as you step through the catalog. The ROC is the real part of the rightmost pole. Exponential: pole at −1, ROC Re(s) > −1. Step: pole at 0, ROC Re(s) > 0. Sine: poles at ±j2, whose real part is zero, so ROC Re(s) > 0 again. Eight rows, one rule.
The two value rows come from opposite ends of the same F(s): f(0⁺) = lim(s→∞) sF(s) and f(∞) = lim(s→0) sF(s). That is why step and cosine agree that f(0⁺) = 1 and disagree about the ending.

How to read what you see

Walk the eight entries and watch two columns that look independent behave as one. Step, ramp and power all have their poles at the origin and all read Re(s) > 0. Exponential and both damped waves have a pole with real part −1 and all read Re(s) > −1. Sine and cosine look like the exception — their poles are imaginary — and they are not: the real part of ±j2 is zero, so they read Re(s) > 0 with the ones at the origin. Then look at the endings, which the poles also decide: −1 real part gives 0, a single pole at the origin gives a constant, a repeated pole at the origin gives ∞, and a purely imaginary pair gives no ending at all.

Assumes: That f(t) = 0 before t = 0 — this is the one-sided transform, which is why every entry is written against a unit step. That the amplitude A scales F(s) without moving a pole, so it changes the initial value and never the ROC or the ending. And that every entry is a ratio of polynomials: a transform with a branch cut or an essential singularity has no pole list to read, and three of the four rows would have nothing to say.
Breaks when: The final-value theorem is the one people quote without its conditions, and this catalog contains the counterexample. Applied blindly to the sine, lim(s→0) s·ω/(s²+ω²) is 0 — clean, definite, and wrong: sin(ωt) has no final value. The theorem holds only when the poles of sF(s) lie strictly in the left half-plane, and the sine’s sit on the imaginary axis. The page refuses rather than reporting the limit, which is why that row reads DNE — oscillates instead of the 0 the formula happily produces. The ramp fails the same test from the other side and gets ∞. Two of the eight entries break the rule you were about to generalise from the other six.

The poles row already tells you the ending 🖖

Step through the catalog and compare only two rows of the table: Poles and Final value. Every pair whose poles sit strictly left of the imaginary axis ends at 0 — e^−at at s = −a, the damped sine at s = −a ± jω. Pairs with poles on the axis never settle: sin and cos sit at s = ±jω, and the table prints "DNE — oscillates". The ramp's double pole at s = 0 runs to ∞ instead. Only the step, with a single simple pole at s = 0, lands on a finite non-zero value. The real part of the rightmost pole decides whether a signal dies, holds or diverges — which is why control engineers design in the s-plane and never invert the transform at all.

A dictionary between time and s 🖖

At heart this tool is a translation table. Every function of time — a step, a ramp, a decaying exponential — has a matching partner in the s-domain, and the lab shows both sides of each pair. Solve your problem in the easier s-domain, then translate the answer back using these pairs. A decay e^-at always maps to 1/(s+a), and the number a tells you how fast it fades.

The engineer who scandalized the mathematicians 🖖

Long before the transform was made rigorous, the self-taught engineer Oliver Heaviside solved circuit equations by boldly treating the derivative operator d/dt as an ordinary algebraic symbol — even dividing by it. Pure mathematicians were appalled by the lack of proof; he reportedly retorted, "Shall I refuse my dinner because I do not fully understand digestion?" His step function still bears his name, and the Laplace transform later supplied the rigor he never bothered with.

Problems solved in full

Endpoint values and decay of f(t) = e −1.5t 6 steps
f(t) = e^−1.5t for t ≥ 0. Get F(s) from the definition, read the pole and the region of convergence off the same line of work, and pin both endpoint values without ever inverting a transform. Then answer the question neither endpoint theorem answers: how long does the decay actually take?
1. The transform is a single integral, and everything else on the panel is a by-product of evaluating it. Both exponentials share a base, so they merge into one exponent, −(s + 1.5)t.
  
  $F (s) = \int_{0}^{\infty} e^{- 1.5 t} e^{- s t} d t = [\frac{- e ^{- (s + 1.5) t}}{s + 1.5}]_{0}^{\infty}$
2. The upper limit is where the work is. e^−(s+1.5)t collapses to 0 as t grows only when the real part of s + 1.5 is positive, and that one inequality hands you both remaining entries: the region of convergence Re(s) > −1.5, and the pole at s = −1.5 sitting exactly on its boundary. The pole is not a separate fact to memorise — it is the value of s at which the integral stops existing.
  
  $Re (s + 1.5) > 0 ⟺ Re (s) > - 1.5 ⟹ F (s) = \frac{1}{s + 1.5}$
3. Derive the initial-value theorem rather than quoting it. A large s weights e^−st so heavily towards small t that sF(s) becomes a probe of f near the origin. Here s/(s + 1.5) tends to 1, and e⁰ = 1 confirms it — you have read f(0⁺) off a transform you never inverted.
  
  $f (0^{+}) = s \to \infty lim s F (s) = s \to \infty lim \frac{s}{s + 1.5} = 1 = e^{0}$
4. Push s the other way for the final value. This limit comes with a licence attached: it is only a theorem when every pole of sF(s) lies strictly to the left of the imaginary axis, and the single pole here is at −1.5. The oscillating case below is what the licence is protecting you from.
  
  $f (\infty) = s \to 0 lim s F (s) = \frac{0}{1.5} = 0$
5. Neither theorem carries a clock. One describes t = 0 and the other t = ∞, and between them the curve could take a microsecond or a century. Solve f(τ) = f(0)/e for the missing scale.
  
  $e^{- 1.5 τ} = e^{- 1} ⟹ τ = \frac{1}{1.5} = \frac{2}{3} = 0.667 s$
6. Now leave the panel behind and ask when the signal is down to 1% of where it started.
  
  $e^{- 1.5 t} = 0.01 ⟹ t = \frac{ln 100}{1.5} = \frac{4.6052}{1.5} = 3.07 s$
Answer

F(s) = 1/(s + 1.5), a pole at −1.5, and τ = 0.667 s — the panel prints the pole and the time constant, and you derived both from one integral. Keep the last line instead. Because a cancels out of ln(100)/a divided by 1/a, reaching 1% always takes 4.61 time constants, whatever the decay rate is: 4.61 is a property of the number 100, not of this signal. All the pole controls is how long one τ lasts. Double a to 3 and every duration on this page halves. Slide it the other way, towards the imaginary axis, and τ runs off to infinity — and once the pole reaches the axis the final value stops existing altogether, which is the next problem.

$τ = \frac{1}{a}, t_{1%} = \frac{ln 100}{a} = 4.61 τ (a > 0)$
The final-value limit for f(t) = sin(4t) that returns 0 5 steps
f(t) = sin(4t) for t ≥ 0. The final-value limit runs perfectly smoothly here and returns 0. Nothing blows up and nothing is indeterminate. Work out why 0 is nevertheless the wrong answer, and what the theorem was really asking you to check.
1. Build this transform out of the previous problem's result rather than a lookup table. Euler splits a sine into a pair of complex exponentials, and each one obeys the rule you already derived from the integral.
  
  $sin 4 t = \frac{e ^{j 4 t} - e ^{- j 4 t}}{2 j}, L {e^{α t}} = \frac{1}{s - α}$
2. Combine over a common denominator. The 2j cancels against the j4 − (−j4) in the numerator, so a complex construction leaves a real function of s. The panel writes the denominator as s² + 4², which is the same thing.
  
  $F (s) = \frac{1}{2 j} (\frac{1}{s - j 4} - \frac{1}{s + j 4}) = \frac{4}{s ^{2} + 16}$
3. Locate the poles. The denominator vanishes at s = ±j4, and their real part is 0 — not negative, not positive, but exactly on the imaginary axis. That is the case the licence above was written to exclude.
  
  $s^{2} + 16 = 0 ⟹ s = \pm j 4, Re (s) = 0$
4. Run the final-value limit anyway, because nothing stops you. The numerator goes to 0, the denominator goes to 16, and the quotient is 0. Arithmetic will not warn you that you had no right to it.
  
  $s \to 0 lim s F (s) = s \to 0 lim \frac{4 s}{s ^{2} + 16} = \frac{0}{16} = 0$
5. Compare that against the function itself. sin(4t) has period 2π/4 = 1.5708 s, which the panel rounds to 1.57 s, and a quarter of the way into every one of those periods it is back at +1. A function that keeps returning to 1 forever cannot converge to 0, or to anything. The panel prints no number in the final-value row and says the signal oscillates instead.
  
  $T = \frac{2 π}{4} = 1.5708 s, f (\frac{T}{4} + k T) = 1$
Answer

Poles at ±j4, period 1.57 s, and no final value at all — even though the limit evaluates cleanly to 0. The lesson is not that the arithmetic is fragile; it is that the arithmetic is not the theorem. Compare the two sF(s) side by side and the difference is one pole location. What makes 0 interesting rather than merely wrong is that it is the correct answer to every neighbouring question: damp the sine by any a > 0 and the poles shift to −a ± j4, the theorem applies, and the final value genuinely is 0 — for every a, however tiny. The limit returns the answer to all the nearby problems and to none of this one. And a provable 0 can still be useless: at a = 0.1, problem 1's clock gives ln(100)/0.1 = 46 s to fall to 1%, which is about 29 full cycles of ringing at this frequency.

$s F (s) = \frac{4 s}{s ^{2} + 16} \Rightarrow s = \pm j 4 s F (s) = \frac{s}{s + 1.5} \Rightarrow s = - 1.5$

References (3)

When the final-value theorem applies, and when the table must say it does not: E. Gluskin, "Let us teach this generalization of the final-value theorem." European Journal of Physics 24(6), 591–597, 2003.
Pole location and stability, at length: Katsuhiko Ogata, Modern Control Engineering, 5th edition. Prentice Hall, 2010. ISBN 978-0-13-615673-4.
Where the region of convergence is derived from the pole positions: A. V. Oppenheim & A. S. Willsky, Signals and Systems, 2nd edition. Prentice Hall, 1997 — chapter 9, on the Laplace transform and its region of convergence. ISBN 978-0-13-814757-0.

By the 42Math team · Updated 2026-07-29

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

step function - A step of height 2 transforms to 2/s, one simple pole at the origin — and it is the only entry in the catalog whose final value is finite and non-zero, at 2.
Ramp - A ramp 2t — transforms to 2/s², a double pole at the origin
Power tⁿ - t³ — transforms to 6/s⁴: the n! = 6 comes from integrating by parts three times, and the pole order is n + 1 = 4
exponential decay - e^(-1.5t) transforms to 1/(s + 1.5): a single real pole at s = -1.5, so the region of convergence is Re(s) > -1.5 and the final value is 0.
sine wave - sin(4t) transforms to 4/(s² + 16), poles at ±j4. They sit on the axis, so nothing decays and the final value does not exist.
Cosine - cos(4t) — poles sit on the imaginary axis at ±j4, so there is no final value: it oscillates forever
damped sine - Damped sine: 3/((s + 1)² + 9), poles at -1 ± j3. Same frequency as plain sine, but the real part -1 is the whole difference between fading and ringing forever.
Damped cosine - A damped cosine — the poles move off the axis to −1 ± j3, the ROC becomes Re(s) > −1, and the final value settles at 0