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 a smooth plucked shape. 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 is a raised cosine bump of height $h$ and half-width $w$ centered at the pluck point $x_p$ :

u_0(x) = \begin{cases} \dfrac{h}{2}\left(1 + \cos\dfrac{\pi (x - x_p)}{w}\right), & |x - x_p| \le w, \\[1ex] 0, & \text{otherwise.} \end{cases}

Note that $u_0$ is continuously differentiable and vanishes at both ends, since $w < x_p < L - w$ .

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
$x_p$	0.1625	m	pluck center ( $L/4$ )
$w$	0.03	m	pluck half-width
$x_m$	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.

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 $a_n'(0) = 0$ , where

A_n = \frac{2}{L} \int_0^L u_0(x) \sin(k_n x)\, dx = \frac{2h}{L}\, \sin(k_n x_p)\, \sin(k_n w)\, \frac{(\pi / w)^2}{k_n \left( (\pi/w)^2 - k_n^2 \right)}.

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 $|A_n| = O(n^{-3})$ and $|a_n'|$ carries an extra factor of $n$ , the terms decay like $n^{-2}$ 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 $s(0) = 0$ .