1# string-pluck-v1: plucked damped string, pickup velocity
3This is the hello-world challenge of the collection. The model is a linear damped string with fixed ends, released from rest from its static equilibrium under a finger load of finite width. The deliverable is a two second mono wav file containing the transverse velocity of the string at a fixed pickup point. Every quantity below is defined in the continuum, so the true solution is a specific function of time that does not depend on any discretization. A modal series solution is given in the appendix; it converges to the true solution and can be evaluated to machine precision, so this challenge is mainly a test of the submission and scoring pipeline.
5The machine-readable version of this specification is [challenge.json](challenge.json).
7## Model
9Let $u(x, t)$ denote the transverse displacement in meters of the string at position $x \in [0, L]$ and time $t \ge 0$. The string satisfies
11$$
12\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2} - 2\sigma_0 \frac{\partial u}{\partial t} + \kappa \frac{\partial^3 u}{\partial x^2 \partial t}, \qquad 0 < x < L, \quad t > 0,
13$$
15with fixed ends
17$$
18u(0, t) = u(L, t) = 0, \qquad t \ge 0,
19$$
21and initial conditions
23$$
24u(x, 0) = u_0(x), \qquad \frac{\partial u}{\partial t}(x, 0) = 0, \qquad 0 \le x \le L.
25$$
27Here $c$ is the wave speed, $\sigma_0$ is a frequency-independent loss rate, and $\kappa$ is a Kelvin-Voigt viscosity that damps high frequencies faster than low ones.
29The initial shape $u_0$ is the static equilibrium of the string under a transverse load of finite width centered at the pluck point $x_p$, which is how a finger or plectrum deforms the string before release. The load profile is a raised cosine of half-width $w$,
31$$
32g(x) = \begin{cases} \dfrac{Q}{2w}\left(1 + \cos\dfrac{\pi (x - x_p)}{w}\right), & |x - x_p| \le w, \\[1ex] 0, & \text{otherwise,} \end{cases}
33$$
35and $u_0$ solves the static problem $u_0'' = -g$ on $(0, L)$ with $u_0(0) = u_0(L) = 0$. Here $Q = \int_0^L g\,dx$ is the total load expressed as a slope (force divided by tension), and it is fixed by requiring the pluck height $u_0(x_p) = h$. In closed form,
37$$
38u_0(x) = Q\,\tau(x) + Q\,\kappa(x - x_p), \qquad
39\tau(x) = \begin{cases} x (L - x_p) / L, & x \le x_p, \\ x_p (L - x) / L, & x \ge x_p, \end{cases}
40$$
42$$
43\kappa(y) = \begin{cases} y\,H(y) - \dfrac{1}{2w}\left[ \dfrac{(y + w)^2}{2} - \dfrac{w^2}{\pi^2}\left(1 + \cos\dfrac{\pi y}{w}\right) \right], & |y| \le w, \\[1ex] 0, & \text{otherwise,} \end{cases}
44\qquad
45Q = \frac{h}{x_p (L - x_p) / L - w\left(\tfrac{1}{4} - \tfrac{1}{\pi^2}\right)},
46$$
48where $H$ is the Heaviside step ($H(y) = 1$ for $y > 0$, else $0$). $\tau$ is the triangle produced by a point load, and $\kappa$ is the correction from spreading the load over the contact region; it vanishes outside $|y| \le w$. Note that $u_0$ is three times continuously differentiable, since $g$ is continuously differentiable, and that $w < x_p < L - w$ so the fixed-end conditions hold.
50## Parameters
52| Symbol | Value | Unit | Meaning |
53| --- | --- | --- | --- |
54| $L$ | 0.65 | m | string length |
55| $c$ | 143 | m/s | wave speed (fundamental $c / 2L = 110$ Hz) |
56| $\sigma_0$ | 3 | 1/s | frequency-independent loss rate |
57| $\kappa$ | 0.004 | m^2/s | Kelvin-Voigt viscosity |
58| $h$ | 0.002 | m | pluck height $u_0(x_p)$ |
59| $x_p$ | 0.1625 | m | pluck center ($L/4$) |
60| $w$ | 0.006 | m | half-width of the finger load |
61| $x_m$ | 0.54 | m | pickup position |
63All values are exact decimal numbers, not rounded measurements.
65## Output signal
67The audio signal is the transverse velocity of the string at the pickup point,
69$$
70s(t) = \frac{\partial u}{\partial t}(x_m, t), \qquad \text{in m/s}.
71$$
73The deliverable is a wav file with the following properties.
75| Property | Value |
76| --- | --- |
77| channels | 1 |
78| sample rate $f_s$ | 48000 Hz |
79| number of samples $N$ | 96000 (2.0 s) |
80| sample format | 32-bit IEEE float (WAVE_FORMAT_IEEE_FLOAT) |
81| sample $k$, for $k = 0, \dots, N-1$ | $s(k / f_s) / s_\text{ref}$ |
82| $s_\text{ref}$ | see `output.signal_scale` in challenge.json (m/s) |
84The first sample is at $t = 0$, where $s(0) = 0$. Samples are pointwise values of the continuum signal. No anti-aliasing filter is applied, even though $s(t)$ contains frequency content above the Nyquist frequency; that content is heavily damped by the Kelvin-Voigt term and is part of the exact answer. The scale $s_\text{ref}$ is a fixed constant chosen so that the peak sample magnitude is roughly 0.5. Do not normalize the signal; absolute amplitude is compared. The signal has a large crest factor (a loud transient followed by a much quieter ring), so it sounds faint when played raw; `playback.gain_db` in challenge.json is a suggested listening gain, which the results sites apply at playback together with a limiter, and it is not part of the deliverable.
86## Error measures
88Let $a_k$ and $r_k$, $k = 0, \dots, N-1$, be the samples of a submitted solution and of the reference solution. Two error measures are computed. Both are relative and dimensionless, and both are reported for every solution.
90**Waveform error** (primary):
92$$
93E_\text{wave} = \frac{\left( \sum_k (a_k - r_k)^2 \right)^{1/2}}{\left( \sum_k r_k^2 \right)^{1/2}}.
94$$
96**Spectrogram error.** Let $M = 2048$ be the frame length and $H = 512$ the hop. The periodic Hann window is $g_j = \tfrac{1}{2}\left(1 - \cos(2\pi j / M)\right)$ for $j = 0, \dots, M-1$. Frame $m$ of a signal $x$ is $x_{mH + j}\, g_j$, for $m = 0, \dots, \lfloor (N - M) / H \rfloor$, with no zero padding at either end. Let $X_m[q]$, $q = 0, \dots, M/2$, be its discrete Fourier transform, $X_m[q] = \sum_{j=0}^{M-1} x_{mH+j}\, g_j\, e^{-2\pi i q j / M}$. Then
98$$
99E_\text{spec} = \frac{\left( \sum_{m, q} \left( |A_m[q]| - |R_m[q]| \right)^2 \right)^{1/2}}{\left( \sum_{m, q} |R_m[q]|^2 \right)^{1/2}}.
100$$
102The spectrogram error ignores phase within a frame and is less sensitive than the waveform error to small frequency errors that accumulate into phase drift over the two seconds.
104## Reference solution
106There is no externally supplied ground truth. The reference is chosen automatically from the submitted solutions by the procedure described in the [collection README](../../README.md), unless `reference_solution` in challenge.json names one explicitly. For this particular challenge the modal series in the appendix gives an essentially exact answer, which is expected to be selected.
108## Optional field export
110A submission may additionally provide the displacement field for visualization. This is not scored. The file is HDF5 with three datasets: `x`, float64, shape (201,), the positions $x_i = i L / 200$ in meters; `t`, float64, shape (4001,), the times $t_j = j / 2000$ in seconds; and `u`, float32, shape (4001, 201), with `u[j, i]` $= u(x_i, t_j)$ in meters. Root attributes `challenge_id`, `solution_id`, `quantity` (`"transverse displacement"`) and `units` (`"m"`) are recommended. The grid is stated under `fields` in challenge.json. A file of this size is about 3 MB uncompressed and can live in the solution repository; gzip compression inside the file is welcome.
112## Appendix: modal series solution
114This appendix is provided for convenience and verification; it is not part of the challenge definition, and a submission may use any method.
116Expanding $u(x, t) = \sum_{n \ge 1} a_n(t) \sin(k_n x)$ with $k_n = n\pi / L$, each coefficient satisfies
118$$
119a_n'' + 2\alpha_n a_n' + c^2 k_n^2 a_n = 0, \qquad \alpha_n = \sigma_0 + \tfrac{1}{2}\kappa k_n^2,
120$$
122with $a_n(0) = A_n$ and $a_n'(0) = 0$. Integrating by parts twice, using $u_0'' = -g$ and the fixed ends,
124$$
125A_n = \frac{2}{L} \int_0^L u_0(x) \sin(k_n x)\, dx = \frac{2}{L k_n^2} \int_0^L g(x) \sin(k_n x)\, dx = \frac{2Q}{L\, w\, k_n^3}\, \sin(k_n x_p)\, \sin(k_n w)\, \frac{(\pi / w)^2}{(\pi/w)^2 - k_n^2}.
126$$
128With the parameters above, $(\pi/w)^2 - k_n^2 \ne 0$ for every integer $n$, since $L / w$ is not an integer. For every mode that matters $\alpha_n < c k_n$ (the modes are underdamped up to roughly $n = 14800$), and in that case, with $\omega_n = \sqrt{c^2 k_n^2 - \alpha_n^2}$,
130$$
131a_n(t) = A_n e^{-\alpha_n t} \left( \cos \omega_n t + \frac{\alpha_n}{\omega_n} \sin \omega_n t \right), \qquad a_n'(t) = -A_n \frac{c^2 k_n^2}{\omega_n} e^{-\alpha_n t} \sin \omega_n t.
132$$
134The pickup velocity is then
136$$
137s(t) = \sum_{n \ge 1} a_n'(t) \sin(k_n x_m).
138$$
140Since $|A_n| = O(n^{-5})$ and $|a_n'|$ carries an extra factor of $n$, the terms decay like $n^{-4}$ before damping is taken into account, and for $t > 0$ the exponential factor makes the series converge rapidly. In practice a few hundred modes give the signal to single precision for all $t \ge 1 / f_s$; at $t = 0$ the exact value is $s(0) = 0$.