Laplace Transform of cos(2t): F(s), Poles and ROC

Laplace transform pairs, looked up and evaluated

A damped vibration e^−2tsin(5t) becomes 5 / ((s + 2)² + 25), with two poles at s = −2 ± 5j and the value 0.147059 at s = 1. This Laplace transform calculator returns that kind of answer for sixteen standard functions: powers of t (whole or fractional, with or without an exponential factor), e^at, sinh and cosh, sine and cosine, their damped versions, the unit step, the Dirac delta and a constant. Every result lists F(s), the region of convergence, the poles, the inverse pair read backwards, and the numerical value of F at any real s you type.

Quick start. Choose "Trigonometric", then "sin(ωt)", and type 3 for ω. The answer is F(s) = 3 / (s² + 9), converging for Re(s) > 0. Leave s empty and you get an example value, F(2) = 0.230769.

Five moves from f(t) to F(s)

Pick the family of f(t): a power of t, an exponential or hyperbolic function, sine or cosine, a damped oscillation, or one of the special signals (step, impulse, constant).
Pick the exact function from the second list, for example cosh(at) or e^atcos(ωt). Nothing is chosen for you; a missing choice gets a message, not a guess.
Type the parameters: the power n (any number above −1, such as 2 or 0.5), the rate a in e^at (negative for decay), the angular frequency ω in rad/s (positive), or the delay a of a step or an impulse in seconds.
Optionally type s, a real number, to get F(s) as a number. If it lies outside the region of convergence, the result says so instead of printing a meaningless value.
Read the four parts: F(s) in the dark box, the region of convergence, the poles, and the inverse line that turns the pair around.

What the transform does to a function

The one-sided Laplace transform multiplies a function by the decaying weight e^−st and adds it up over all t from 0 to infinity:

F(s) = ∫0∞ f(t) e−st dt

Two things make this useful. First, a derivative turns into multiplication: L{f′(t)} = sF(s) − f(0). A linear differential equation with constant coefficients becomes an algebraic equation in s, which you solve with ordinary algebra and then transform back. Second, the shape of F(s) reads like a summary of the signal. Where F(s) blows up (its poles) tells you whether f(t) grows, decays or oscillates, and how fast. Control engineers, circuit designers and anyone solving a spring-mass problem use the transform for exactly these two reasons.

Because the integral starts at 0, the transform only sees t ≥ 0. That is why a unit step switched on at t = −2 has the same transform as one switched on at t = 0, and why an impulse at t = −1 transforms to zero. The calculator handles both cases on purpose rather than printing a formula with a double minus sign.

Sixteen transform pairs worth keeping

Every row is a choice in the calculator. The last column is the region of convergence: F(s) exists only for s to the right of that line.

f(t), t ≥ 0	F(s) = L{f(t)}	Converges for
1 or u(t)	1 / s	Re(s) > 0
constant c	c / s	Re(s) > 0
t	1 / s²	Re(s) > 0
tⁿ, n = 0, 1, 2, ...	n! / sⁿ⁺¹	Re(s) > 0
t^p, p > −1	Γ(p + 1) / s^p+1	Re(s) > 0
e^at	1 / (s − a)	Re(s) > a
tⁿe^at	n! / (s − a)ⁿ⁺¹	Re(s) > a
sinh(at)	a / (s² − a²)	Re(s) > \|a\|
cosh(at)	s / (s² − a²)	Re(s) > \|a\|
sin(ωt)	ω / (s² + ω²)	Re(s) > 0
cos(ωt)	s / (s² + ω²)	Re(s) > 0
e^atsin(ωt)	ω / ((s − a)² + ω²)	Re(s) > a
e^atcos(ωt)	(s − a) / ((s − a)² + ω²)	Re(s) > a
u(t − a), a ≥ 0	e^−as / s	Re(s) > 0
δ(t)	1	every s
δ(t − a), a ≥ 0	e^−as	every s

Six pairs, each run with real inputs

Pair 1: a vibration that dies out

Input: damped, e^atsin(ωt), a = −2, ω = 5, s = 1.
Result: F(s) = 5 / ((s + 2)² + 25), poles at s = −2 ± 5j, F(1) = 0.147059. The oscillation runs at 0.795775 Hz, one cycle every 1.25664 s.

The real part of the poles, −2, is the decay rate; the imaginary part, 5, is the angular frequency.

Pair 2: a power times an exponential

Input: power of t, n = 3, a = −2, s = 1.
Result: F(s) = 3! / (s + 2)⁴ = 6 / (s + 2)⁴, one pole of order 4 at s = −2, F(1) = 0.0740741, which is 6/81.

A repeated pole of order n + 1 is the signature of a tⁿ factor in the answer.

Pair 3: the square root of t

Input: power of t, n = 0.5, s = 2.
Result: F(s) = Γ(1.5) / s^1.5 with Γ(1.5) ≈ 0.886227 (half the square root of π), and F(2) = 0.313329.

A fractional power gives a branch point at s = 0 rather than a pole. The factorial table cannot handle it; the Gamma function can.

Pair 4: a switch closed after 2 seconds

Input: special, unit step, delay a = 2, s = 1.
Result: F(s) = e^−2s / s, F(1) = 0.135335, which is e⁻².

Any delay of a seconds multiplies the transform by e^−as. That is the second shift theorem, and it is how piecewise inputs are built from steps.

Pair 5: cosh with both poles on the real axis

Input: exponential, cosh(at), a = 2, s = 3.
Result: F(s) = s / (s² − 4), poles at s = 2 and s = −2, F(3) = 0.6.

The function cosh(2t) is half of e^2t plus half of e^−2t, so it carries one pole from each.

Pair 6: an s the integral never reaches

Input: exponential, e^at, a = 2, s = 1.
Result: F(s) = 1 / (s − 2), but s = 1 lies outside Re(s) > 2, so the calculator gives no value.

Plugging s = 1 into 1/(s − 2) gives −1, yet the integral of e^2te^−t grows without limit. The formula outlives the integral; the calculator does not pretend otherwise.

The properties that stretch the table

Sixteen pairs cover far more than sixteen functions once you combine them with a few rules. The transform is linear, so sums and constant multiples go straight through.

Rule	In t	In s
Linearity	a f(t) + b g(t)	a F(s) + b G(s)
First shift	e^at f(t)	F(s − a)
Second shift	f(t − a) u(t − a)	e^−as F(s)
Derivative	f′(t)	sF(s) − f(0)
Second derivative	f″(t)	s²F(s) − s f(0) − f′(0)
Integral	∫₀^t f(τ) dτ	F(s) / s
Multiply by t	t f(t)	−F′(s)
Scaling	f(kt), k > 0	F(s/k) / k

A worked combination: 4 + 3e^−t has the transform 4/s + 3/(s + 1). Run the constant 4 and the exponential with a = −1 separately, then add the two F(s) values at the same s. At s = 1 that is 4 + 1.5 = 5.5.

Poles, and why the region of convergence moves

The region of convergence is a half-plane Re(s) > σ, and σ is set by the rightmost pole. For e^−3t the only pole is at −3, so every s above −3 works, including s = 0. For e^2t the pole sits at +2, and anything at or left of 2 makes the integral diverge.

Pole positions also predict behavior. Poles in the left half-plane mean a response that settles; poles on the imaginary axis mean a steady oscillation; any pole in the right half-plane means growth without bound. A pair at −2 ± 5j is a vibration that fades at the rate e^−2t. A single real pole of order 4 at −2 is a hump, t³e^−2t, that rises and then decays.

The delta function and a zero function have no poles at all, so their transforms exist for every s. A fractional power such as t^0.5 has no pole either, but a branch point at 0: F(s) = Γ(1.5)/s^1.5 is finite for every s > 0 and still undefined at 0.

Reading the four parts of the result

The dark box holds F(s) as a textbook writes it: (s + 2) instead of (s − −2), s² + 4 instead of (s − 0)² + 4. For whole powers the factorial is shown next to its value, so 3! / (s + 2)⁴ = 6 / (s + 2)⁴. The region tile gives Re(s) > σ or "every s". The poles tile lists each pole with its order, or a complex pair as σ ± ωj. The value tile gives F at your s, or at an example two units inside the region when you leave s empty. Under the tiles, the inverse line reads the pair backwards, which is how you use it after solving an equation in s.

Laplace questions from coursework

How do I find an inverse Laplace transform with this tool?

Break F(s) into partial fractions until each piece matches a row of the pair table, then read each row from right to left. For 5 / ((s + 2)² + 25), match ω = 5 and a = −2 in the damped sine row; the inverse line of the result confirms e^−2tsin(5t).

Why does t⁻¹ have no Laplace transform?

Near t = 0 the integral of 1/t behaves like the logarithm, which runs off to minus infinity. Any power p ≤ −1 fails the same way. Powers above −1, such as t^−0.5, are fine: its transform is Γ(0.5)/s^0.5, the square root of π/s.

Is s a real number or a complex one?

In theory s is complex, and the region of convergence is stated for its real part. The value tile takes a real s because that is what homework usually asks for, and because the sign test for convergence is then a simple comparison with σ.

What is the Laplace transform used for?

Mostly for linear differential equations with constant coefficients: circuit currents, spring-mass motion, heat flow in simple geometries, and the transfer functions of control systems. The transform turns the equation into algebra and builds the initial conditions in through the derivative rule.

How is this different from a Fourier transform?

The Fourier transform integrates over all t with e^−jωt and needs a signal that does not grow. The Laplace transform adds the damping factor e^−σt and starts at 0, so growing signals like e^2t still have a transform on part of the plane. Setting s = jω recovers the Fourier transform when the imaginary axis lies inside the region.

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What is the Laplace transform of cos(2t)?

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