Fourier analysis, Laplace transforms, and special functions

Declare a period and its phase

A signal here means a supplied scalar function of time; it need not be a measurement. A positive T is a period when f(t + T) = f(t) for every allowed t. A periodic function can have more than one period. Our Fourier example is supplied for all real t, measured in seconds from a chosen reference, and its values use metres. These assigned values do not specify a moving body's path.

Set T = 4 s and define φ(t) = 2πt/T. This is the dimensionless numerical radian value of the angle θ(t) = φ(t) rad. Sine, cosine and the exponential below take numerical dimensionless arguments. The phase increases by 2π when t increases by T. We use the cycle frequency ν = 1/T = 0.25 Hz; frequency in cycles per second differs from an angular frequency in radians per second.

The nth harmonic, with positive integer n, uses cos(nφ) or sin(nφ) and has cycle frequency nν. The first harmonic has period T, while the second has period T/2 = 2 s; both repeat after T. Harmonic means an integer multiple of the chosen base frequency, not an independently measured physical resonance.

Use full-period orthogonality

A finite Fourier decomposition writes a function as f(t) = c₀ + Σₙ₌₁ᴺ [aₙ cos(nφ) + bₙ sin(nφ)], with a finite positive integer N. The symbol Σ means add those terms. We call c₀ the constant term; this convention differs from writing the constant term as a₀/2. Fourier coefficient normalization must be stated. Here c₀, aₙ and bₙ all have the signal's metre units, and the sine/cosine basis functions are dimensionless.

Orthogonality means the integral of the product of distinct selected basis functions over one full period is zero. For positive integer indices, this applies to two cosines with different indices, two sines with different indices, and any sine–cosine pair. Each sine or cosine also has zero full-period mean. The same-function integrals are ∫₀ᵀ cos²(nφ) dt = ∫₀ᵀ sin²(nφ) dt = T/2; the constant basis has ∫₀ᵀ 1 dt = T.

These identities give c₀ = (1/T)∫₀ᵀ f(t) dt, aₙ = (2/T)∫₀ᵀ f(t)cos(nφ) dt and bₙ = (2/T)∫₀ᵀ f(t)sin(nφ) dt. For a supplied finite sum, distribute the integral through its finitely many terms: orthogonality removes every other basis component and the T/2 normalization cancels the factor 2/T. All integrals exist here because the supplied finite trigonometric sum is continuous. An incomplete period does not generally have the same orthogonality identities.

Check one exact finite decomposition

Supply the original exact function f(t) = 2 m + (3 m)cos φ − (1 m)sin(2φ). Its finite decomposition has c₀ = 2 m, a₁ = 3 m, b₂ = −1 m, and all other sine/cosine coefficients zero. For example, multiplying by sin(2φ) and integrating leaves only −(1 m)(T/2), so b₂ = −1 m. A signed coefficient is not a new frequency or the maximum value of the whole signal.

The selected values in the table follow from this formula, not a measurement or an interpolation rule. At t = 0.5 s, φ = π/4, cos φ = √2/2 and sin(2φ) = 1. Thus f = (1 + 3√2/2) m. The simpler function f₀(t) = 2 m + (3 m)cos φ agrees with f at t = 0, 1, 2 and 3 s, because sin(2φ) is zero there, but differs by 1 m at t = 0.5 s. Those four samples cannot distinguish the two functions.

The supplied finite formula is exact for every real t by construction. Extending Fourier analysis to an arbitrary function requires stating its class, the coefficients' integrability, and the intended meaning and conditions of convergence; a finite truncation need not recover the whole function. This lesson neither infers an arbitrary signal's complete spectrum from samples nor proves an infinite-series reconstruction or inversion theorem.

Define a one-sided Laplace transform

Now use a function g(t) supplied on t ≥ 0. Its one-sided Laplace transform is G(p) = ∫₀∞ e^(−pt)g(t) dt, wherever that improper integral converges. One-sided specifies the interval starting at zero. Here p is a real parameter with inverse-second units, so pt is dimensionless; the usual broader theory can also use a complex parameter. The kernel e^(−pt) is the exponential weighting function, with e the base of the natural exponential.

An improper integral means take lim as R → ∞ of ∫₀ᴿ e^(−pt)g(t) dt, where R is a positive time. Its existence depends on the combined integrand near the endpoints and at large time. A finite formula obtained by algebra is not sufficient if that defining limit fails. For negative p the kernel grows, but a sufficiently fast-decaying g can still make the product integrable.

In our convention g uses metres, the kernel is dimensionless and dt uses seconds, so G has m·s units. This differs from the normalized Fourier coefficients above, whose factors 1/T or 2/T cancel the integration's time unit. A transform represents a function in terms of another parameter; p is not an additional measured time or a new spatial coordinate.

Derive the domain before the formula

Supply the exact function g(t) = A e^(−t/τ) for all t ≥ 0, with A = 4 m and τ = 2 s. This is a mathematical input, without a claim about a measured decay process. Its weighted integrand is A e^(−ct), where c = p + 1/τ has inverse-second units. For real p it decays exponentially exactly when c > 0, or p > −0.5 s⁻¹. This is the convergence domain for the selected defining integral.

For c > 0, the finite integral is ∫₀ᴿ A e^(−ct) dt = (A/c)(1 − e^(−cR)). It follows by differentiating the antiderivative −(A/c)e^(−ct). Since e^(−cR) tends to zero, G(p) = A/(p + 1/τ), with the domain retained. At p = 0.5 s⁻¹, c = 1 s⁻¹ and G = 4 m·s; at p = 0, G = 8 m·s.

At c = 0 the integrand is the positive constant A and its integral A R grows without bound. At c < 0 the integrand grows exponentially and also diverges. For instance, substituting p = −1 s⁻¹ into the rational expression gives −8 m·s, but it is outside this transform's convergence domain and is not a value of the defining integral. The expression and its domain belong together.

Keep initial values in transform algebra

A useful scoped rule is L[g′](p) = pG(p) − g(0). Here L means the same one-sided transform. Sufficient conditions for this calculation are: g is continuously differentiable for t ≥ 0, the weighted integrals of g and g′ converge absolutely, and e^(−pR)g(R) → 0 as R → ∞. Absolute convergence means the integral of the absolute value is finite. These are sufficient assumptions for the displayed derivation, not a classification of all transformable functions.

Integration by parts uses ∫ u v′ dt = [uv] − ∫ u′v dt. With u = e^(−pt) and v = g(t), it gives ∫₀ᴿ e^(−pt)g′(t) dt = e^(−pR)g(R) − g(0) + p∫₀ᴿ e^(−pt)g(t) dt. Under the stated assumptions the limit gives the rule. Its initial value remains present: differentiation does not become multiplication by p alone.

For our exponential g, g′ = −g/τ and all these assumptions hold when p > −1/τ. Consequently −G/τ = pG − A, or (p + 1/τ)G = A, consistent with the directly integrated result. The units on both sides of the derivative rule are metres: (1/s)(m·s) = m. Such rules can turn suitable linear differential equations into algebraic relations while retaining initial data. Recovering an unknown original function is a separate step requiring an appropriate inversion method and conditions.

Introduce a defined special function

Mathematics gives standard names and definitions to useful functions beyond the familiar powers, exponentials and trigonometric formulas. A named special function can package a recurring integral and its documented properties. Here use the error function, defined for a real dimensionless z by erf(z) = (2/√π)∫₀ᶻ e^(−u²) du. The integration variable u is a dimensionless number, unrelated to the time variable t. This real definition is sufficient for our example; the reference also treats a broader complex domain.

The integrand is continuous, so this finite integral exists for every real z. For negative z, the integral from 0 to z reverses the orientation. Its definition gives erf(0) = 0, and evenness of e^(−u²) gives erf(−z) = −erf(z). The fundamental theorem of calculus gives d erf(z)/dz = (2/√π)e^(−z²) for real z. These are mathematical properties of a defined function.

The name lets us retain an exact expression involving this integral without assigning an unchecked decimal or replacing it with a familiar elementary formula. In a unit-bearing model the argument must first be formed as a dimensionless quantity, such as t/τ. A probability, error-bar or material interpretation would require its own physical or statistical model; the name alone supplies none.

Express and bound a finite integral

Supply a separate original function q(t) = B e^[−(t/τ)²] only on 0 ≤ t ≤ τ, with B = 3 m and τ = 2 s. Ask for I = ∫₀^τ q(t) dt. This integrates a length-valued signal over time, giving m·s units; it is not automatically a path distance. Define z = t/τ, so the endpoints become 0 and 1 and dt = τ dz.

The substitution gives I = Bτ∫₀¹ e^(−z²) dz = Bτ(√π/2)erf(1). With the supplied B and τ, I = 3√π m·s erf(1). The special-function argument is the dimensionless upper limit τ/τ = 1. The constant factors retain the integral's physical units.

For 0 ≤ z ≤ 1, 1/e ≤ e^(−z²) ≤ 1. Integrating these bounds and multiplying by Bτ = 6 m·s gives 6/e m·s ≤ I ≤ 6 m·s. These are mathematical bounds within the exact supplied model, not statistical confidence limits or measured uncertainty. Fourier decomposition, Laplace weighting and a named special function have different definitions; each becomes useful when its normalization, domain, conditions and units are kept explicit.

Worked examples

Try it yourself

1. With T = 4 s and φ = 2πt/T, find the base cycle frequency, the second harmonic's cycle frequency, and φ at t = 1 s. Does φ carry a second unit?

   Show the answer and reasoning

   The frequencies are 0.25 Hz and 0.5 Hz; φ = π/2, dimensionless.

   ν = 1/T and the second harmonic has frequency 2ν. The time ratio t/T = 1/4 is dimensionless. φ is the numerical radian value of the associated angle, so it is suitable as the trigonometric argument.

2. For the supplied f, the full-period integral ∫₀ᵀ f(t)sin(2φ) dt is −2 m·s. With T = 4 s and our normalization, find b₂ and its units.

   Show the answer and reasoning

   b₂ = (2/T)(−2 m·s) = −1 m.

   2/T = 0.5 s⁻¹ cancels the integral's time unit. The negative sign belongs to this sine coefficient. A same-basis square integral is T/2, which explains this normalization.

3. Compare the supplied f with f₀ = 2 m + (3 m)cos φ. Do their values at t = 0, 1, 2 and 3 s distinguish them? What is f − f₀ at t = 0.5 s?

   Show the answer and reasoning

   Those four values agree. At t = 0.5 s, f − f₀ = −1 m.

   At the four integer-second times, sin(2φ) is zero. At 0.5 s, 2φ = π/2 and the sine is 1. These two explicit functions demonstrate that those few samples do not fix the full periodic function.

4. For the selected g with A = 4 m and τ = 2 s, is p = 0 inside the convergence domain? If so, find G(0) and its units.

   Show the answer and reasoning

   Yes. G(0) = Aτ = 8 m·s.

   0 > −0.5 s⁻¹, so the combined exponential still decays. Dividing 4 m by 0.5 s⁻¹ gives 8 m·s. At p = 0 the weighting kernel is 1, and the decay of g itself makes the integral finite.

5. The rational expression 4 m/(p + 0.5 s⁻¹) equals −8 m·s at p = −1 s⁻¹. Is this a value of the defining one-sided transform of the selected g?

   Show the answer and reasoning

   No. The defining integral diverges at that p.

   The combined coefficient c = −0.5 s⁻¹ is negative, so the positive integrand is (4 m)e^[(0.5 s⁻¹)t] and grows exponentially. A finite algebraic expression outside the convergence domain does not supply a value of this integral.

6. For the selected q with τ = 2 s, express the integral from 0 to τ of q(t) dt using erf and bound it as above. Why is the upper argument erf(1) rather than erf(2 s)?

   Show the answer and reasoning

   The integral is 3√π m·s erf(1), between 6/e m·s and 6 m·s. The upper dimensionless ratio is τ/τ = 1.

   Substitution z = t/τ gives dt = τ dz and endpoints 0 and 1. erf was defined for a dimensionless real argument. The integrand on this interval lies between 1/e and 1; the factor Bτ = 6 m·s retains the original units.

What to carry forward

• A finite Fourier decomposition uses a declared period, harmonic basis and coefficient normalization.
• Full-period orthogonality recovers coefficients of the supplied finite sum; a few sample values need not identify a whole function.
• A one-sided Laplace transform is an improper weighted integral, with a convergence domain as well as an algebraic expression.
• Transform units depend on the definition; in our Laplace convention a metre-valued input gives m·s.
• A derivative transform rule retains initial values and requires the stated integral and limiting conditions.
• A special function has a precise definition; keep its argument dimensionless and retain units in the model's prefactors.