string-pluck-v1: plucked damped string, pickup velocity#

This 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.

The machine-readable version of this specification is challenge.json.

Model#

Let $u(x, t)$ denote the transverse displacement in meters of the string at position $x \in [0, L]$ and time $t \ge 0$ . The string satisfies

\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,

with fixed ends

u(0, t) = u(L, t) = 0, \qquad t \ge 0,

and initial conditions

u(x, 0) = u_0(x), \qquad \frac{\partial u}{\partial t}(x, 0) = 0, \qquad 0 \le x \le L.

Here $c$ 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.

The 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 $w$ ,

g(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}

and $u_0$ 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,

u_0(x) = Q\,\tau(x) + Q\,\kappa(x - x_p), \qquad \tau(x) = \begin{cases} x (L - x_p) / L, & x \le x_p, \\ x_p (L - x) / L, & x \ge x_p, \end{cases}

\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} \qquad Q = \frac{h}{x_p (L - x_p) / L - w\left(\tfrac{1}{4} - \tfrac{1}{\pi^2}\right)},

where $H$ 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.

Parameters#

Symbol	Value	Unit	Meaning
$L$	0.65	m	string length
$c$	143	m/s	wave speed (fundamental $c / 2L = 110$ Hz)
$\sigma_0$	3	1/s	frequency-independent loss rate
$\kappa$	0.004	m^2/s	Kelvin-Voigt viscosity
$h$	0.002	m	pluck height $u_0(x_p)$
$x_p$	0.1625	m	pluck center ( $L/4$ )
$w$	0.006	m	half-width of the finger load
$x_m$	0.54	m	pickup position

All values are exact decimal numbers, not rounded measurements.

Output signal#

The audio signal is the transverse velocity of the string at the pickup point,

s(t) = \frac{\partial u}{\partial t}(x_m, t), \qquad \text{in m/s}.

The deliverable is a wav file with the following properties.

Property	Value
channels	1
sample rate $f_s$	48000 Hz
number of samples $N$	96000 (2.0 s)
sample format	32-bit IEEE float (WAVE_FORMAT_IEEE_FLOAT)
sample $k$ , for $k = 0, \dots, N-1$	$s(k / f_s) / s_\text{ref}$
$s_\text{ref}$	see `output.signal_scale` in challenge.json (m/s)

The 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.

Error measures#

Let $a_k$ 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.

Waveform error (primary):

E_\text{wave} = \frac{\left( \sum_k (a_k - r_k)^2 \right)^{1/2}}{\left( \sum_k r_k^2 \right)^{1/2}}.

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

E_\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}}.

The 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.

Reference solution#

There is no externally supplied ground truth. The reference is chosen automatically from the submitted solutions by the procedure described in the collection README, 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.

Optional field export#

A 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.

This appendix is provided for convenience and verification; it is not part of the challenge definition, and a submission may use any method.

Expanding $u(x, t) = \sum_{n \ge 1} a_n(t) \sin(k_n x)$ with $k_n = n\pi / L$ , each coefficient satisfies

a_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,

with $a_n(0) = A_n$ and $a_n'(0) = 0$ . Integrating by parts twice, using $u_0'' = -g$ and the fixed ends,

A_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}.

With 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 $\omega_n = \sqrt{c^2 k_n^2 - \alpha_n^2}$ ,

a_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.

The pickup velocity is then

s(t) = \sum_{n \ge 1} a_n'(t) \sin(k_n x_m).

Since $|A_n| = O(n^{-5})$ 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 $s(0) = 0$ .