State as of 20240501

README.md

@@ -1,3 +1,6 @@
 # LRFTubes mathematical documentation
 
 [![Build Status](https://drone.ascee.nl/api/badges/ASCEE/lrftubes_doc/status.svg)](https://drone.ascee.nl/ASCEE/lrftubes_doc)
+
+This repository is the source code of the LRFTubes mathematical documentation.
+Releases are automatically build and pushed at: 

lrftubes.lyx

@@ -5600,6 +5600,235 @@ literal "true"
 Exponential duct (horn)
 \end_layout
 
+\begin_layout Standard
+hor.py/class ExpHorn(Duct)
+\end_layout
+
+\begin_layout Standard
+An exponential horn can be modelled using ExpHorn.
+ The transfer matrix is derived from Webster's horn equation.
+ The solution for an exponential horn is
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\begin{equation}
+p\left(x\right)=C_{1}e^{E_{1}x}+C_{2}e^{E_{2}x},\qquad U\left(x\right)=i\frac{S\left(x\right)}{\rho\omega}\left(C_{1}E_{1}e^{E_{1}x}+C_{2}E_{2}e^{E_{2}x}\right)
+\end{equation}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+in which 
+\begin_inset Formula $S(x)$
+\end_inset
+
+ is the local cross-sectional area, 
+\begin_inset Formula $\rho$
+\end_inset
+
+ is the air density, 
+\begin_inset Formula $\omega$
+\end_inset
+
+ is the angular frequency.
+ The exponent terms 
+\begin_inset Formula $E_{1}$
+\end_inset
+
+ and 
+\begin_inset Formula $E_{2}$
+\end_inset
+
+ are given by
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\begin{equation}
+E_{1}=\frac{-m+\sqrt{-4k^{2}+m^{2}}}{2},\qquad E_{2}=-\frac{m+\sqrt{-4k^{2}+m^{2}}}{2}
+\end{equation}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+in which 
+\begin_inset Formula $k$
+\end_inset
+
+ is the wave number and 
+\begin_inset Formula $m$
+\end_inset
+
+ is the flare rate, which is defined as
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\begin{equation}
+m=\frac{1}{L}\ln\left(\frac{S_{R}}{S_{L}}\right)
+\end{equation}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+with 
+\begin_inset Formula $L$
+\end_inset
+
+ being the horn length and 
+\begin_inset Formula $S_{R}$
+\end_inset
+
+ and 
+\begin_inset Formula $S_{L}$
+\end_inset
+
+ being the cross-sectional areas on the right-hand and left-hand side of
+ the horn respectively.
+ 
+\begin_inset Newline newline
+\end_inset
+
+
+\begin_inset Newline newline
+\end_inset
+
+By setting the solutions on 
+\begin_inset Formula $x=0$
+\end_inset
+
+ to 
+\begin_inset Formula $p_{L}$
+\end_inset
+
+ and 
+\begin_inset Formula $U_{L}$
+\end_inset
+
+ and those on 
+\begin_inset Formula $x=L$
+\end_inset
+
+ to 
+\begin_inset Formula $p_{R}$
+\end_inset
+
+ and 
+\begin_inset Formula $U_{R}$
+\end_inset
+
+, the following transfer matrix can be derived:
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\begin{equation}
+T_{ExpHorn}=\frac{1}{E_{2}-E_{1}}\left[\begin{array}{cc}
+E_{2}e^{E_{1}L}-E_{1}e^{E_{2}L} & i\frac{\rho\omega}{S_{L}}\left(e^{E_{1}L}-e^{E_{2}L}\right)\\
+i\frac{S_{R}}{\rho\omega}E_{1}E_{2}\left(e^{E_{1}L}-e^{E_{2}L}\right) & -\frac{S_{R}}{S_{L}}\left(E_{1}e^{E_{1}L}-E_{2}e^{E_{2}L}\right)
+\end{array}\right]
+\end{equation}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status collapsed
+
+\begin_layout Plain Layout
+\begin_inset Formula $p_{L}=p\left(0\right)=C_{1}+C_{2}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Formula $U_{L}=U\left(0\right)=i\frac{S_{L}}{\rho\omega}\left(C_{1}E_{1}+C_{2}E_{2}\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Formula $p_{R}=p\left(L\right)=C_{1}e^{E_{1}L}+C_{2}e^{E_{2}L}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Formula $U_{R}=U\left(L\right)=i\frac{S_{L}}{\rho\omega}\left(C_{1}E_{1}e^{E_{1}L}+C_{2}E_{2}e^{E_{2}L}\right)$
+\end_inset
+
+
+\begin_inset Newline newline
+\end_inset
+
+
+\begin_inset Newline newline
+\end_inset
+
+Solve for constants
+\begin_inset Newline newline
+\end_inset
+
+
+\begin_inset Formula $C_{1}=P_{L}-C_{2}\rightarrow U_{L}=i\frac{S_{L}}{\rho\omega}\left[E_{1}p_{L}+\left(E_{2}-E_{1}\right)C_{2}\right]$
+\end_inset
+
+
+\begin_inset Newline newline
+\end_inset
+
+
+\begin_inset Formula $C_{1}=\frac{1}{E_{2}-E_{1}}\left(i\frac{\rho\omega}{S_{L}}U_{L}+E_{2}P_{L}\right),\qquad C_{2}=-\frac{1}{E_{2}-E_{1}}\left(i\frac{\rho\omega}{S_{L}}U_{L}+E_{1}P_{L}\right)$
+\end_inset
+
+
+\begin_inset Newline newline
+\end_inset
+
+
+\begin_inset Newline newline
+\end_inset
+
+Fill in in solution:
+\begin_inset Newline newline
+\end_inset
+
+
+\begin_inset Formula $p_{R}=\frac{1}{E_{2}E_{1}}\left[\left(E_{2}e^{E_{1}L}-E_{1}e^{E_{2}L}\right)P_{L}+i\frac{\rho\omega}{S_{L}}\left(e^{E_{1}L}-e^{E_{2}L}\right)U_{L}\right]$
+\end_inset
+
+
+\begin_inset Newline newline
+\end_inset
+
+
+\begin_inset Formula $U_{R}=\frac{1}{E_{2}E_{1}}\left[i\frac{S_{R}}{\rho\omega}E_{1}E_{2}\left(e^{E_{1}L}-e^{E_{2}L}\right)P_{L}-\frac{S_{R}}{S_{L}}\left(E_{1}e^{E_{1}L}-E_{2}e^{E_{2}L}\right)U_{L}\right]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
 \begin_layout Standard
 \begin_inset Formula 
 \begin{equation}
@@ -5610,7 +5839,7 @@ S_{f}=\exp\left(\alpha x\right)
 
 
 \begin_inset Note Note
-status collapsed
+status open
 
 \begin_layout Plain Layout
 \begin_inset Formula $\frac{\mathrm{d}^{2}p}{\mathrm{d}x^{2}}+\alpha\frac{\mathrm{d}p}{\mathrm{d}x}+\Gamma^{2}p=0$
@@ -7922,239 +8151,6 @@ Z_{cpm}\left(f\right)=\frac{1}{i2\pi fC_{dyn}\left(f\right)}
 \end_inset
 
 
-\end_layout
-
-\begin_layout Section
-Exponential horn
-\end_layout
-
-\begin_layout Standard
-hon.py/class ExpHorn(Duct)
-\end_layout
-
-\begin_layout Standard
-An exponential horn can be modelled using ExpHorn.
- The transfer matrix is derived from Webster's horn equation.
- The solution for an exponential horn is
-\end_layout
-
-\begin_layout Standard
-\begin_inset Formula 
-\begin{equation}
-p\left(x\right)=C_{1}e^{E_{1}x}+C_{2}e^{E_{2}x},\qquad U\left(x\right)=i\frac{S\left(x\right)}{\rho\omega}\left(C_{1}E_{1}e^{E_{1}x}+C_{2}E_{2}e^{E_{2}x}\right)
-\end{equation}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-in which 
-\begin_inset Formula $S(x)$
-\end_inset
-
- is the local cross-sectional area, 
-\begin_inset Formula $\rho$
-\end_inset
-
- is the air density, 
-\begin_inset Formula $\omega$
-\end_inset
-
- is the angular frequency.
- The exponent terms 
-\begin_inset Formula $E_{1}$
-\end_inset
-
- and 
-\begin_inset Formula $E_{2}$
-\end_inset
-
- are given by
-\end_layout
-
-\begin_layout Standard
-\begin_inset Formula 
-\begin{equation}
-E_{1}=\frac{-m+\sqrt{-4k^{2}+m^{2}}}{2},\qquad E_{2}=-\frac{m+\sqrt{-4k^{2}+m^{2}}}{2}
-\end{equation}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-in which 
-\begin_inset Formula $k$
-\end_inset
-
- is the wave number and 
-\begin_inset Formula $m$
-\end_inset
-
- is the flare rate, which is defined as
-\end_layout
-
-\begin_layout Standard
-\begin_inset Formula 
-\begin{equation}
-m=\frac{1}{L}\ln\left(\frac{S_{R}}{S_{L}}\right)
-\end{equation}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-with 
-\begin_inset Formula $L$
-\end_inset
-
- being the horn length and 
-\begin_inset Formula $S_{R}$
-\end_inset
-
- and 
-\begin_inset Formula $S_{L}$
-\end_inset
-
- being the cross-sectional areas on the right-hand and left-hand side of
- the horn respectively.
- 
-\begin_inset Newline newline
-\end_inset
-
-
-\begin_inset Newline newline
-\end_inset
-
-By setting the solutions on 
-\begin_inset Formula $x=0$
-\end_inset
-
- to 
-\begin_inset Formula $p_{L}$
-\end_inset
-
- and 
-\begin_inset Formula $U_{L}$
-\end_inset
-
- and those on 
-\begin_inset Formula $x=L$
-\end_inset
-
- to 
-\begin_inset Formula $p_{R}$
-\end_inset
-
- and 
-\begin_inset Formula $U_{R}$
-\end_inset
-
-, the following transfer matrix can be derived:
-\end_layout
-
-\begin_layout Standard
-\begin_inset Formula 
-\begin{equation}
-T_{ExpHorn}=\frac{1}{E_{2}-E_{1}}\left[\begin{array}{cc}
-E_{2}e^{E_{1}L}-E_{1}e^{E_{2}L} & i\frac{\rho\omega}{S_{L}}\left(e^{E_{1}L}-e^{E_{2}L}\right)\\
-i\frac{S_{R}}{\rho\omega}E_{1}E_{2}\left(e^{E_{1}L}-e^{E_{2}L}\right) & -\frac{S_{R}}{S_{L}}\left(E_{1}e^{E_{1}L}-E_{2}e^{E_{2}L}\right)
-\end{array}\right]
-\end{equation}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset Note Note
-status open
-
-\begin_layout Plain Layout
-\begin_inset Formula $p_{L}=p\left(0\right)=C_{1}+C_{2}$
-\end_inset
-
-
-\end_layout
-
-\begin_layout Plain Layout
-\begin_inset Formula $U_{L}=U\left(0\right)=i\frac{S_{L}}{\rho\omega}\left(C_{1}E_{1}+C_{2}E_{2}\right)$
-\end_inset
-
-
-\end_layout
-
-\begin_layout Plain Layout
-\begin_inset Formula $p_{R}=p\left(L\right)=C_{1}e^{E_{1}L}+C_{2}e^{E_{2}L}$
-\end_inset
-
-
-\end_layout
-
-\begin_layout Plain Layout
-\begin_inset Formula $U_{R}=U\left(L\right)=i\frac{S_{L}}{\rho\omega}\left(C_{1}E_{1}e^{E_{1}L}+C_{2}E_{2}e^{E_{2}L}\right)$
-\end_inset
-
-
-\begin_inset Newline newline
-\end_inset
-
-
-\begin_inset Newline newline
-\end_inset
-
-Solve for constants
-\begin_inset Newline newline
-\end_inset
-
-
-\begin_inset Formula $C_{1}=P_{L}-C_{2}\rightarrow U_{L}=i\frac{S_{L}}{\rho\omega}\left[E_{1}p_{L}+\left(E_{2}-E_{1}\right)C_{2}\right]$
-\end_inset
-
-
-\begin_inset Newline newline
-\end_inset
-
-
-\begin_inset Formula $C_{1}=\frac{1}{E_{2}-E_{1}}\left(i\frac{\rho\omega}{S_{L}}U_{L}+E_{2}P_{L}\right),\qquad C_{2}=-\frac{1}{E_{2}-E_{1}}\left(i\frac{\rho\omega}{S_{L}}U_{L}+E_{1}P_{L}\right)$
-\end_inset
-
-
-\begin_inset Newline newline
-\end_inset
-
-
-\begin_inset Newline newline
-\end_inset
-
-Fill in in solution:
-\begin_inset Newline newline
-\end_inset
-
-
-\begin_inset Formula $p_{R}=\frac{1}{E_{2}E_{1}}\left[\left(E_{2}e^{E_{1}L}-E_{1}e^{E_{2}L}\right)P_{L}+i\frac{\rho\omega}{S_{L}}\left(e^{E_{1}L}-e^{E_{2}L}\right)U_{L}\right]$
-\end_inset
-
-
-\begin_inset Newline newline
-\end_inset
-
-
-\begin_inset Formula $U_{R}=\frac{1}{E_{2}E_{1}}\left[i\frac{S_{R}}{\rho\omega}E_{1}E_{2}\left(e^{E_{1}L}-e^{E_{2}L}\right)P_{L}-\frac{S_{R}}{S_{L}}\left(E_{1}e^{E_{1}L}-E_{2}e^{E_{2}L}\right)U_{L}\right]$
-\end_inset
-
-
-\end_layout
-
-\end_inset
-
-
 \end_layout
 
 \begin_layout Section
@@ -8686,6 +8682,7 @@ The wavelength is much larger than the thermal penetration depth (
 ).
 \end_layout
 
+\begin_deeper
 \begin_layout Standard
 We can derive the following impedance boundary condition 
 \begin_inset CommandInset citation
@@ -8698,7 +8695,7 @@ literal "true"
 
 :
 \begin_inset Note Note
-status collapsed
+status open
 
 \begin_layout Plain Layout
 Delta EC User guide:
@@ -8770,6 +8767,7 @@ Hence the impedance of a hard wall scales with
 .
 \end_layout
 
+\end_deeper
 \begin_layout Standard
 \begin_inset Float figure
 wide false
@@ -9727,7 +9725,7 @@ V_{\mathrm{in}}-V_{\mathrm{bemf}}=Z_{\mathrm{el}}I,
 
 \end_inset
 
-where 
+ where 
 \begin_inset Formula $Z_{\mathrm{el}}$
 \end_inset
 
@@ -15015,8 +15013,8 @@ literal "true"
 \end_inset
 
 .
- Assuming axial symmetrySuch that the acoustic pressure in for example tube
+ Assuming axial symmetry, such that the acoustic pressure in for example
- 
+ tube 
 \begin_inset Formula $B$
 \end_inset