Fourier analysis and wave packets

Start with one localized shape

At the supplied start time, a small transverse disturbance has the shape f(x) = A exp[−(x/ℓ)²] cos(k₀x), with A = 0.02 m, ℓ = 1 m and k₀ = 4 m⁻¹. The fixed longitudinal coordinate x increases to the right on the whole real line; positive transverse displacement is upward. These are stipulated model values, not measurements.

The positive envelope E(x) = A exp[−(x/ℓ)²] makes the oscillations small far from the origin. The cosine is the carrier: it provides the repeated oscillation inside that envelope. Because |cos(k₀x)| ≤ 1, −E(x) ≤ f(x) ≤ E(x). This is one localized wave-packet example. The Gaussian envelope is positive at every finite x.

For preparation, use Algebra, functions, graphs, and logarithms and Geometry, trigonometry, and coordinate systems. Complex numbers and complex exponentials explains the notation used below. Begin with the shape, figure and exact tables; Derivatives, Taylor series, and local approximation and Integrals and multivariable calculus provide optional return paths for the derivative and integral checks. Fourier analysis, Laplace transforms, and special functions gives the earlier finite-decomposition introduction.

Here k and k₀ are angular spatial wavenumbers in m⁻¹. The products kx and k₀x are the dimensionless numerical radian phases: an angle with numerical phase φ is φ rad. The ratios x/ℓ and the exponential arguments are dimensionless as well. A wave pattern and a material point will have different roles when time is introduced.

Recover a finite periodic pattern

First use a separate exact periodic pattern h(x) = a₀ + a₁ cos(2πx/L) + b₂ sin(4πx/L), where L = 4 m, a₀ = 0.003 m, a₁ = 0.008 m and b₂ = −0.004 m. It is supplied for all real x and repeats after L. The constant term in this convention is a₀ itself.

On the full interval 0 ≤ x ≤ L, distinct positive-integer sine or cosine harmonics are orthogonal: the integral of their product is zero. A sine times a cosine also integrates to zero. The integral of the square of one such sine or cosine is L/2. Product-to-sum identities establish these facts by reducing the products to complete oscillations.

Multiplying h by cos(2πx/L) and integrating therefore selects a₁: a₁ = (2/L) ∫₀ᴸ h(x) cos(2πx/L) dx. Similarly b₂ = (2/L) ∫₀ᴸ h(x) sin(4πx/L) dx, and a₀ = (1/L) ∫₀ᴸ h(x) dx. Each integral has unit m²; division by L returns a coefficient in metres.

This finite formula is exact because the three terms were supplied. Orthogonality determines their coefficients without establishing that an arbitrary function equals an infinite Fourier series. Changing the integration interval can remove the orthogonality used here.

State the continuous convention

The localized profile f is not periodic, so we use a continuous spatial variable k rather than integer harmonics of one period. Define its Fourier amplitude by f̂(k) = ∫₋∞^∞ f(x) exp(−ikx) dx. The imaginary unit satisfies i² = −1, and exp(iφ) = cos φ + i sin φ for real numerical phase φ.

Our inverse convention is f(x) = (1/(2π)) ∫₋∞^∞ f̂(k) exp(ikx) dk when the required inversion conditions hold. The location of the factor 1/(2π) is part of the convention. A source that includes this factor in its forward amplitude uses a different amplitude from f̂.

Since f and x both carry metres, f̂ has unit m². In the inverse, dk has unit m⁻¹ and returns a displacement in metres. The variable k labels simultaneous spatial components; it is not time or a material position. Both signs of k are retained.

For this supplied Gaussian-modulated cosine, f and the amplitude derived next are smooth and absolutely integrable. We check the inverse for this particular pair. No finite diagram or convergence of a sampled sum establishes inversion for every possible signal.

Derive the two Gaussian components

A useful established Gaussian integral identity is ∫₋∞^∞ exp(−u²) exp(−iau) du = √π exp(−a²/4) for real dimensionless a. One justification defines this integral as G(a), differentiates under its integrable Gaussian bound and integrates by parts to obtain G′(a) = −aG(a)/2. The Gaussian integral gives G(0) = √π, fixing the stated result.

Use cos(k₀x) = [exp(ik₀x) + exp(−ik₀x)]/2 and substitute u = x/ℓ in the forward integral. Each exponential shifts the Gaussian argument by one carrier wavenumber. The result is f̂(k) = (Aℓ√π/2){exp[−ℓ²(k−k₀)²/4] + exp[−ℓ²(k+k₀)²/4]}.

The inverse integral of each of these two terms is (A/2) exp[−(x/ℓ)²] exp(±ik₀x). Adding them gives the original real cosine profile. The Gaussian identity justifies this exact reconstruction over the whole real k line; no contour integration or infinite-series equality is needed for this example.

For the main values, write C = (√π/100) m². Then f̂(k) is the sum of two Gaussians with prefactor C. It is real, even and positive at every finite k. The carrier coordinates ±4 m⁻¹ locate the centers of the separate terms; overlap means they are not the exact maxima of their sum.

Compare the profile and its amplitude

The upper figure shows f and the two bounding envelopes at t = 0 s. Filled circles select x = −π/4 m, 0 m and π/4 m. The profile can be negative while its positive envelope remains positive; the envelope is a bound on oscillation size, not a second displacement history.

The lower figure shows the real amplitude f̂ against k. Hollow diamonds select k = −4 m⁻¹, 0 m⁻¹ and 4 m⁻¹. The vertical quantity there is measured in m², not the metres of the upper panel. These panels have different horizontal and vertical scales and do not show a geometric particle trajectory.

All entries in the nearby tables are exact model expressions. The connected finite curve segments approximate the analytic functions. Colours, solid versus dashed lines and circle versus diamond marks distinguish their roles.

Retain the whole-line integral

The spectral table uses C = (√π/100) m² from the preceding formula. Its equal opposite-k entries reflect this real even profile. A Fourier amplitude is not automatically a probability density, energy density or power spectrum; those interpretations require additional definitions and physical information.

Only −3 ≤ x ≤ 3 m and −10 ≤ k ≤ 10 m⁻¹ are drawn. These are viewing windows, not physical ends or support boundaries. Both the Gaussian envelope and the spectral amplitude have tails outside them.

Replacing the inverse integral by one restricted to the displayed k window changes the problem. At x = 0 the omitted integrand is positive, so omitting those tails gives a strictly smaller value than the full inverse. A drawing or a finite quadrature may be useful, but neither is an exact whole-line reconstruction by itself.

Follow a supplied nondispersive packet

Now supply y(x,t) = f(x−ct), with c = 2 m/s, all real x and 0 ≤ t ≤ 1 s. This is a chosen unforced linear homogeneous wave model with constant c and equation ∂²y/∂t² = c² ∂²y/∂x². A transverse-string interpretation is limited to the small-slope linear regime; no tension, density or measured accuracy is inferred from these supplied numbers.

Let z = x−ct and let f′ and f″ denote derivatives with respect to its length argument. The chain rule gives ∂y/∂t = −cf′(z), ∂²y/∂t² = c²f″(z), and ∂²y/∂x² = f″(z). Thus the supplied history satisfies the equation. Both sides have unit m/s². Its initial data include y(x,0) = f(x) and ∂y/∂t(x,0) = −cf′(x).

Equivalently, each spatial component receives the factor exp[ik(x−ct)]. This common shift preserves the whole shape: in this stipulated nondispersive case the envelope and carrier translate together. Damping, forcing, reflecting ends and a nonlinear wavenumber-frequency relation are outside this case.

At a fixed longitudinal label, y changes transversely with time. The pattern moving to the right at c does not establish a material point moving longitudinally at c. A Fourier description also does not by itself show that a chosen model fits a real medium.

Change one width with a defined meaning

The parameter ℓ is the positive distance at which E(x)/E(0) reaches exp(−1). For either separate shifted spectral Gaussian, an offset |k∓k₀| = 2/ℓ reduces that term by exp(−1). These are defined distances and offsets, not assumed standard deviations of the summed spectrum.

Holding A and k₀ fixed while changing ℓ from 1 m to 0.5 m halves the envelope distance and doubles each individual spectral offset from 2 m⁻¹ to 4 m⁻¹. The prefactor Aℓ√π/2 also halves. This gives a precise example of a shorter spatial envelope requiring a broader range of spatial components.

The comparison is a mathematical property of this specified family and convention. It is not a universal width product, a quantum uncertainty claim or a conclusion about finite experimental resolution. Return to the integral checks when ready; the shape, signs, units and window distinctions already explain the main comparison.

Worked examples

Try it yourself

1. Using the supplied finite h on 0 ≤ x ≤ L, recover a₁ and the coefficient of cos(6πx/L).

   Show the answer and reasoning

   a₁ = 0.008 m; the third cosine coefficient is 0 m.

   Full-period orthogonality selects the supplied first cosine. The constant, first cosine and second sine each give zero projection on the absent third cosine.

2. If f is a displacement in metres and x is in metres, what units do f̂ and the inverse integral have in the stated convention?

   Show the answer and reasoning

   f̂ has unit m²; the inverse integral has unit m.

   The forward factor f dx contributes m². In the inverse, dk contributes m⁻¹ and the exponential and 1/(2π) are dimensionless.

3. At x = π/4 m, does a positive envelope imply positive profile displacement? Give the exact profile value.

   Show the answer and reasoning

   No. f(π/4 m) = −0.02 exp(−π²/16) m.

   The envelope is positive, but the numerical carrier phase is π and cos π = −1. A bound on magnitude does not fix the displacement sign.

4. Does keeping only −10 ≤ k ≤ 10 m⁻¹ in the inverse integral establish exact reconstruction at x = 0?

   Show the answer and reasoning

   No; it omits a strictly positive contribution.

   At x = 0 the inverse exponential is 1 and f̂(k) is positive for every finite k. Its nonzero tails outside the displayed interval belong to the exact whole-line integral.

5. Find y(1 m, 0.5 s). Does this show that the material at that label has travelled longitudinally at 2 m/s?

   Show the answer and reasoning

   y(1 m, 0.5 s) = 0.02 m. No longitudinal material motion is established.

   The argument x−ct is zero, giving f(0). The formula supplies transverse displacement at fixed longitudinal labels; c is the pattern translation speed.

6. With A and k₀ fixed, replace ℓ = 1 m by ℓ = 0.5 m. How do the envelope exp(−1) distance and each separate spectral Gaussian’s exp(−1) offset change?

   Show the answer and reasoning

   The distance halves from 1 m to 0.5 m; the offset doubles from 2 m⁻¹ to 4 m⁻¹.

   The defining conditions are |x| = ℓ and |k∓k₀| = 2/ℓ. They describe this family and each separate spectral term, not a universal width statistic for their sum.

What to carry forward

• Full-period orthogonality recovers coefficients of the supplied finite pattern.
• State the Fourier convention and retain f̂’s unit m² when f and x carry metres.
• This Gaussian-modulated cosine has an exactly checked two-sided Gaussian amplitude and inverse.
• A finite viewing window and a finite polyline do not replace a whole-line integral.
• The stipulated nondispersive packet translates as a pattern; material displacement is a separate quantity.