University Physics V · topic UPV-01.08
Fourier series, Fourier transforms, and conjugate variables
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Fourier series, Fourier transforms, and conjugate variables: predict a finite change
A Fourier expansion writes a function as amplitudes along the eigenfunctions of −i d/dx — a discrete sum of e^(ikₙx) on a finite interval, a delta-normalised integral over e^(ikx) on the line — and the map between f(x) and f̃(k) is a unitary change of basis on L². Use the supplied topic model Y_out = 1/(2*max(sig,0.01)), where Y_out is Δk RMS WIDTH (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. sig is Packet rms width Δx, measured in nm. Keep x0 (Packet centre x₀) = 0 nm; k0 (Carrier wavenumber k₀) = 0 nm⁻¹ fixed. For this exercise use only 0.57 nm ≤ sig ≤ 1.83 nm on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at sig = 0.8472 nm and 1.0236 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δsig across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
Δk RMS WIDTH against sig
Predictions from the supplied Fourier series, Fourier transforms, and conjugate variables model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): Δk RMS WIDTH / nm⁻¹
Horizontal axis (x): sig: Packet rms width Δx / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| sig: Packet rms width Δx / nm | Δk RMS WIDTH / nm⁻¹ |
|---|---|
| 0.5700000000000001 | 0.8771929824561403 |
| 0.885 | 0.5649717514124294 |
| 1.2000000000000002 | 0.41666666666666663 |
| 1.5150000000000001 | 0.33003300330033003 |
| 1.83 | 0.27322404371584696 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of sig separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At sig = 0.8472 nm, substitute into Y_out = 1/(2*max(sig,0.01)): Y₀ = 0.590179 nm⁻¹.
- At sig = 1.0236 nm, the same fixed-parameter model gives Y₁ = 0.488472 nm⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.101707 nm⁻¹; the input change is Δsig = 0.1764 nm.
- Average sensitivity is ΔY_out/Δsig = -0.576572 nm⁻¹ per nm. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 0.590179 nm⁻¹, Y₁ = 0.488472 nm⁻¹; ΔY_out = -0.101707 nm⁻¹; average sensitivity = -0.576572 nm⁻¹ per nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Fourier series, Fourier transforms, and conjugate variables: infer an input with uncertainty
A Fourier expansion writes a function as amplitudes along the eigenfunctions of −i d/dx — a discrete sum of e^(ikₙx) on a finite interval, a delta-normalised integral over e^(ikx) on the line — and the map between f(x) and f̃(k) is a unitary change of basis on L². Use the supplied topic model Y_out = 1/(2*max(sig,0.01)), where Y_out is Δk RMS WIDTH (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. sig is Packet rms width Δx, measured in nm. Keep x0 (Packet centre x₀) = 0 nm; k0 (Carrier wavenumber k₀) = 0 nm⁻¹ fixed. For this exercise use only 0.57 nm ≤ sig ≤ 1.83 nm on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.37352458 nm⁻¹, with a bounded reading uncertainty ±0.015045 nm⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown sig by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000126 nm, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
Δk RMS WIDTH against sig
Predictions from the supplied Fourier series, Fourier transforms, and conjugate variables model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): Δk RMS WIDTH / nm⁻¹
Horizontal axis (x): sig: Packet rms width Δx / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| sig: Packet rms width Δx / nm | Δk RMS WIDTH / nm⁻¹ |
|---|---|
| 0.5700000000000001 | 0.8771929824561403 |
| 0.885 | 0.5649717514124294 |
| 1.2000000000000002 | 0.41666666666666663 |
| 1.5150000000000001 | 0.33003300330033003 |
| 1.83 | 0.27322404371584696 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(sig) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.57) = 0.877193 nm⁻¹ and Y_out(1.83) = 0.273224 nm⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives sig_line = 0.57 + (0.373525 − 0.877193)(1.83 − 0.57)/(0.273224 − 0.877193) = 1.62075 nm. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives sig = 1.3386 nm; substituting back gives Y_out = 0.37352458 nm⁻¹.
- The interpolation error in the input is |sig_line − sig| = 0.282153 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.35847958 nm⁻¹ and 0.38856958 nm⁻¹. Sort the roots even if the response decreases: 1.28677 nm ≤ sig ≤ 1.39478 nm.
- The lower and upper input deviations are 0.0518292 nm and 0.0561796 nm. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer sig ≈ 1.3386 nm; compatible interval [1.28677, 1.39478] nm. Endpoint interpolation gives 1.62075 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.