add LaTeX original of tempo doc
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doc/tempo.tex
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doc/tempo.tex
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% vim: tw=80 spl=en
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% Copyright (C) 2018 Julien "_FrnchFrgg_" RIVAUD
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%
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% This program is free software: you can redistribute it and/or modify
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% it under the terms of the GNU General Public License as published by
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% the Free Software Foundation, either version 3 of the License, or
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% (at your option) any later version.
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%
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% This program is distributed in the hope that it will be useful,
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% but WITHOUT ANY WARRANTY; without even the implied warranty of
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% MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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% GNU General Public License for more details.
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%
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% You should have received a copy of the GNU General Public License
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% along with this program. If not, see <http://www.gnu.org/licenses/>.
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\documentclass[10pt]{article}
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%\usepackage{fontspec}
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%\usepackage{polyglossia}
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%\setmainlanguage{english}
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\usepackage[english]{babel}
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%\usepackage{lualatex-math}
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\usepackage{amsmath}
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%\usepackage[math-style=ISO,bold-style=ISO]{unicode-math}
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\usepackage{geometry}
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\geometry{reset,margin=1.5cm,left=6cm,includefoot,marginparwidth=4.5cm}
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\usepackage{cellprops}
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\begin{document}
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\def\define#1{\textbf{#1}}
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\def\d{\mathrm{d}}
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\section{A natural definition for tempo ramping}
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\subsection{Some definitions}
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The \define{time}, often denoted $t$, is the duration in some multiple unit of
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seconds (samples, superclock, etc).
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We define $b$ to be the \define{beats function}: $b$~is the number of beats that
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occurred since $t = 0$ until the current time. It is assumed to be continuous and
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even derivable rather than being a staircase function: $b$~will thus value~$4.5$
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when exactly in the middle of beat~$4$, halfway\footnote{Not necessarily halfway
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in time units, though.} between pulsations $4$~and~$5$.
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The \define{tempo function}, or simply tempo, is the number~$T$ of pulsations
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per unit of time. Thus $T = \dfrac{\d b}{\d t}$.
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\def\DT{\Delta T}
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\def\Dt{\Delta t}
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\def\Db{\Delta b}
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A \define{tempo ramp} is a definition of $T$ for $0 \le t \le \Dt$ such that
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$T_{t=0} = T_0$ and $T_{t=\Dt} = T_{\text{end}} = T_0 + \DT$ where $\Dt$, $T_0$
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and $T_{\text{end}}$~or~$\DT$ are given. Most of the times they are set by the
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user, but $\Dt$ can be sometimes defined in beats, that is be the unique~$\Dt$
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such that $b_{|t=\Dt} = \Db$.
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\subsection{Linear tempo ramping}
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The simplest definition that comes to mind is a tempo ramp where the tempo
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increases from $T_0$~to~$T_0 + \DT$ \emph{linearly} with time.
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Here $T = T_0 + \dfrac{\DT}{\Dt}\times t$, and thus
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$b = \dfrac{\DT}{2\Dt}\times t^2 + T_0 \times t$, assuming $b_{|t=0} = 0$.
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If $\Dt$~is defined implicitly from a duration~$\Db$ in beats, then
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\[ \Db = \dfrac{\DT}{2\Dt}\times \Dt^2 + T_0 \Dt
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= \left( \frac{1}{2}\DT + T_0 \right) \times\Dt \]
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thus
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\[
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\Dt = \dfrac{2\Db}{\DT + 2T_0}
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\]
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\subsection{Exponential tempo ramping}
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\def\e{\mathrm{e}}
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Humans are not very good at keeping track precisely of absolute time for long
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durations, and are better at comparing short durations, that is maintain a
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reasonably stable time span between successive pulsations, and have a precise
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sense of elapsed time \emph{in terms of these pulsations}.
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In particular, when changing the tempo, it is less far-fetched to imagine a
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human feeling the tempo increase as a function of elapsed \emph{beats} rather
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than absolute time in seconds which is a pulsation that is very hard to maintain
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constant and even sense when competing with the (increasing) musical pulsation.
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In other words, a definition more suitable than that of linear ramping would be
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that the tempo increases \emph{linearly} with the value of~$b$. We thus want
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\[ T = T_0 + \dfrac{\DT}{\Db}\times b
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\text{, that is }
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\dfrac{\d b}{\d t} - \dfrac{\DT}{\Db}\times b = T_0 \]
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The solutions of this order 1 linear differential equation are\footnote{The
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exponential is the solution of the associated homogeneous equation and the
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second term is the constant solution of the equation.}
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$b = A \e^{\frac{\DT}{\Db}t} - \dfrac{\Db T_0}{\DT}$, and
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since we want $b_{|t=0} = 0$, we get $A = \dfrac{\Db T_0}{\DT}$ that is
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\[ b = \dfrac{\Db T_0}{\DT} (\e^{\frac{\DT}{\Db}t} - 1) \]
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Denoting $\omega = \dfrac{\DT}{\Db}$ we find
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$b = \frac{T_0}{\omega} (\e^{\omega t} - 1)$.
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(Note: $\omega$ is called $c$ in Nick's text, but I like $\omega$ better since
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it is a frequency).
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The \emph{tempo} is $T = \dfrac{\d b}{\d t}
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= T_0 \e^{\frac{\DT}{\Db}t} = T_0 \e^{\omega t}$, and this is where the name
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exponential tempo ramp comes from.
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Recall that $\Db$, $\DT$ and $T_0$ are parameters of the ramp, while $t$~is the
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arbitrary time within the ramp at which we want to know $b$~and~$T$.
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Let us compute the reciprocal of $b$:
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\[
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b = \dfrac{\Db T_0}{\DT} (\e^{\frac{\DT}{\Db}t} - 1)
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\iff
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\dfrac{\DT}{\Db T_0} b = \e^{\frac{\DT}{\Db}t} - 1
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\iff
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1 + \dfrac{\DT}{\Db T_0} b = \e^{\frac{\DT}{\Db}t}
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\]
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\[
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\iff
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\log \left( 1 + \dfrac{\DT}{\Db T_0} b \right) = \frac{\DT}{\Db}t
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\iff
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t = \frac{\Db}{\DT} \log \left( 1 + \dfrac{\DT}{\Db T_0} b \right)
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\]
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If $\Dt$~is defined directly instead of implicitly from a duration~$\Db$ in
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beats, then we have to find out the value of~$\Db$. At $t=\Dt$, we have
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$b = \Db$ thus
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\[
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\Dt = \frac{\Db}{\DT} \log \left( 1 + \dfrac{\DT}{\Db T_0} \Db \right)
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\iff
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\Dt = \frac{\Db}{\DT} \log \left( 1 + \dfrac{\DT}{T_0} \right)
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\iff
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\Db = \frac{\DT\Dt}{\log \left( 1 + \dfrac{\DT}{T_0} \right) }
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\]
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A more useful expression is
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\[
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\omega = \frac{\DT}{\Db}
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= \frac{1}{\Dt}\log\left(1+ \frac{\DT}{T_0}\right)
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\]
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which gives
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\[
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b = \frac{\Dt T_0}{\log\left(1+ \frac{\DT}{T_0}\right)}
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\e^{\frac{t}{\Dt}\log\left(1+ \frac{\DT}{T_0}\right)}
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= \frac{\Dt T_0}{\log\left(1+ \frac{\DT}{T_0}\right)}
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\left(1+ \frac{\DT}{T_0}\right)^{\frac{t}{\Dt}}
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\]
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and
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\[
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T
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= T_0 \e^{\frac{t}{\Dt}\log\left(1+ \frac{\DT}{T_0}\right)}
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= T_0 \left(1+ \frac{\DT}{T_0}\right)^{\frac{t}{\Dt}}
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\]
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We at last compute the reciprocals:
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\[
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b = \frac{\Dt T_0}{\log\left(1+ \frac{\DT}{T_0}\right)}
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\e^{\frac{t}{\Dt}\log\left(1+ \frac{\DT}{T_0}\right)}
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\iff
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\frac{\log\left(1+ \frac{\DT}{T_0}\right)}{\Dt T_0} b
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= \e^{\frac{t}{\Dt}\log\left(1+ \frac{\DT}{T_0}\right)}
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\]
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\[
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\iff
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\log\left( \frac{\log\left(1+ \frac{\DT}{T_0}\right)}{\Dt T_0} b \right)
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= \frac{t}{\Dt}\log\left(1+ \frac{\DT}{T_0}\right)
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\iff
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t = \frac{
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\Dt
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\log\left( \frac{\log\left(1+ \frac{\DT}{T_0}\right)}{\Dt T_0} b \right)
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}{
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\log\left(1+ \frac{\DT}{T_0}\right)
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}
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\]
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\subsection{Summary with $T$ as main tempo representation}
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The following table gathers all results. For code factorization we might want
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to only consider the red formulas, that is compute (and maybe cache) $\omega$
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beforehand. Note that in the formulas, $\omega$ is \emph{always} the inverse of
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a time, never in per-beats.
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\[\everymath{\displaystyle}
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\cellprops{ td:nth-child(1) { math-mode: text; } }
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\begin{array}{lccc}
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\hline
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When you know & \Dt & \Db & \omega \\
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\hline
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Compute $\omega$ &
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\color{red}
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\frac{1}{\Dt}\log\left(1+ \frac{\DT}{T_0}\right)
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&
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\color{red}
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\frac{\DT}{\Db}
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&
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\\
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Compute $b$ from $t$ &
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\frac{\Dt T_0}{\log\left(1+ \frac{\DT}{T_0}\right)}
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\left(1+ \frac{\DT}{T_0}\right)^{\frac{t}{\Dt}}
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&
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\frac{\Db T_0}{\DT} (\e^{\frac{\DT}{\Db}t} - 1)
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&
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\color{red}
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\frac{T_0}{\omega} (\e^{\omega t} - 1)
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\\
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Compute $T$ from $t$ &
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T_0 \left(1+ \frac{\DT}{T_0}\right)^{\frac{t}{\Dt}}
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&
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T_0 \e^{\frac{\DT}{\Db}t}
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&
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\color{red}
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T_0 \e^{\omega t}
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\\
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Compute $t$ from $b$ &
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\frac{
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\Dt
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\log\left( \frac{\log\left(1+ \frac{\DT}{T_0}\right)}
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{\Dt T_0} b \right)
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}{
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\log\left(1+ \frac{\DT}{T_0}\right)
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}
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&
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\frac{\Db}{\DT} \log \left( 1 + \frac{\DT}{\Db T_0} b \right)
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&
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\color{red}
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\frac{1}{\omega} \log \left( 1 + \frac{\omega b}{T_0} \right)
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\\
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Compute $T$ from $b$ &
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T_0 + \frac{1}{\Dt}\log\left(1+ \frac{\DT}{T_0}\right) \times b
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&
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T_0 + \frac{\DT}{\Db}\times b
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&
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\color{red}
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T_0 + \omega b
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\\
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\hline
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\end{array}\]
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Note that there are often library functions to compute $\log(1+x)$ directly
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from~$x$ more precise than computing $1+x$ then its logarithm. The main reason
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is that when adding 1to some very small number you loose a lot of precision
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because the exponent is now tailored to the representation of~$1$, and that
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the Taylor-McLaurin series for $\log(1+x)$ is a very efficient mean to compute
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an approximate value for it.
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\subsection{Using $S = \frac{1}{T}$ to represent tempo}
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\def\DS{\Delta S}
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\everymath{\displaystyle}
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Since the base time unit in ardour is not seconds but far smaller than
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that\footnote{It was samples in previous versions of Ardour, and is even more
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precise in later versions.}, any value of $T$~or~$\DT$ will be very small. A
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better value to store, that can easily be rounded to integer without a huge
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loss of precision, is $S = 1/T$.
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Instead of having $T_0$~and~$T_1 = T_0 + \DT$, we thus assume that when
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computing the formulas we have access to $S_0$~and~$S_1$.
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Then from $\Dt$,
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$\omega = \frac{1}{\Dt}\log\left(1 + \frac{\DT}{T_0}\right)
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= \frac{1}{\Dt}\log\left(\frac{T_1}{T_0}\right)
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= \frac{1}{\Dt}\log\left(\frac{S_0}{S_1}\right)$.
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From beats, $\omega = \frac{\DT}{\Db} = \frac{S_0 - S_1}{S_0S_1\Db}$ (or just
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use $\omega = \left( \frac{1}{S_1} - \frac{1}{S_0} \right) / \Db$).
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To compute $b$~from~$t$ we can use
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$b = \frac{\e^{\omega t}-1}{S_0 \omega}$.
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And for $S$~from~$t$ we have
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$S = \frac{S_0}{\e^{\omega t}} = S_0\e^{-\omega t}$.
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As for $t$~from~$b$ the formula becomes
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$t = \frac{1}{\omega} \log \left( 1 + S_0 \omega b \right)$.
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Lastly, for $S$~from~$b$ we can see:
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\[ S = 1/T = \frac{1}{1/S_0 + \omega b} = \frac{S_0}{1+S_0\omega b}\]
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Maybe we can also to compute and cache
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$\omega' = S_0 \omega = \frac{S_0}{\Dt}\log\left(\frac{S_0}{S_1}\right)
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= \frac{S_0 - S_1}{S_1\Db}$
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Which is in « per-beats » and occurs often in the formulas (but not
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exclusively). Note that the fact that $\omega$~or~$\omega'$ is relevant is
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\emph{not} depending on whether we compute from time or from beats, or even if
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the ramp length is defined in time or in beats.
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\end{document}
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