State as of 20240501
This commit is contained in:
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# LRFTubes mathematical documentation
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[](https://drone.ascee.nl/ASCEE/lrftubes_doc)
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This repository is the source code of the LRFTubes mathematical documentation.
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Releases are automatically build and pushed at:
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474
lrftubes.lyx
474
lrftubes.lyx
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@ -5600,6 +5600,235 @@ literal "true"
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Exponential duct (horn)
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\end_layout
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\begin_layout Standard
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hor.py/class ExpHorn(Duct)
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\end_layout
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\begin_layout Standard
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An exponential horn can be modelled using ExpHorn.
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The transfer matrix is derived from Webster's horn equation.
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The solution for an exponential horn is
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\begin{equation}
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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)
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\end{equation}
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\end_inset
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\end_layout
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\begin_layout Standard
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in which
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\begin_inset Formula $S(x)$
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\end_inset
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is the local cross-sectional area,
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\begin_inset Formula $\rho$
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\end_inset
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is the air density,
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\begin_inset Formula $\omega$
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\end_inset
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is the angular frequency.
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The exponent terms
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\begin_inset Formula $E_{1}$
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\end_inset
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and
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\begin_inset Formula $E_{2}$
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\end_inset
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are given by
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\begin{equation}
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E_{1}=\frac{-m+\sqrt{-4k^{2}+m^{2}}}{2},\qquad E_{2}=-\frac{m+\sqrt{-4k^{2}+m^{2}}}{2}
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\end{equation}
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\end_inset
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\end_layout
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\begin_layout Standard
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in which
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\begin_inset Formula $k$
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\end_inset
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is the wave number and
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\begin_inset Formula $m$
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\end_inset
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is the flare rate, which is defined as
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\begin{equation}
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m=\frac{1}{L}\ln\left(\frac{S_{R}}{S_{L}}\right)
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\end{equation}
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\end_inset
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\end_layout
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\begin_layout Standard
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with
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\begin_inset Formula $L$
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\end_inset
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being the horn length and
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\begin_inset Formula $S_{R}$
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\end_inset
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and
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\begin_inset Formula $S_{L}$
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\end_inset
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being the cross-sectional areas on the right-hand and left-hand side of
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the horn respectively.
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\begin_inset Newline newline
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\end_inset
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\begin_inset Newline newline
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\end_inset
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By setting the solutions on
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\begin_inset Formula $x=0$
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\end_inset
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to
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\begin_inset Formula $p_{L}$
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\end_inset
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and
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\begin_inset Formula $U_{L}$
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\end_inset
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and those on
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\begin_inset Formula $x=L$
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\end_inset
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to
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\begin_inset Formula $p_{R}$
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\end_inset
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and
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\begin_inset Formula $U_{R}$
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\end_inset
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, the following transfer matrix can be derived:
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\begin{equation}
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T_{ExpHorn}=\frac{1}{E_{2}-E_{1}}\left[\begin{array}{cc}
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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)\\
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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)
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\end{array}\right]
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\end{equation}
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\end_inset
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\end_layout
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\begin_layout Standard
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\begin_inset Note Note
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status collapsed
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\begin_layout Plain Layout
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\begin_inset Formula $p_{L}=p\left(0\right)=C_{1}+C_{2}$
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\end_inset
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\end_layout
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\begin_layout Plain Layout
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\begin_inset Formula $U_{L}=U\left(0\right)=i\frac{S_{L}}{\rho\omega}\left(C_{1}E_{1}+C_{2}E_{2}\right)$
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\end_inset
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\end_layout
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\begin_layout Plain Layout
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\begin_inset Formula $p_{R}=p\left(L\right)=C_{1}e^{E_{1}L}+C_{2}e^{E_{2}L}$
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\end_inset
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\end_layout
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\begin_layout Plain Layout
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\begin_inset Formula $U_{R}=U\left(L\right)=i\frac{S_{L}}{\rho\omega}\left(C_{1}E_{1}e^{E_{1}L}+C_{2}E_{2}e^{E_{2}L}\right)$
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\end_inset
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\begin_inset Newline newline
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\end_inset
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\begin_inset Newline newline
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\end_inset
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Solve for constants
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\begin_inset Newline newline
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\end_inset
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\begin_inset Formula $C_{1}=P_{L}-C_{2}\rightarrow U_{L}=i\frac{S_{L}}{\rho\omega}\left[E_{1}p_{L}+\left(E_{2}-E_{1}\right)C_{2}\right]$
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\end_inset
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\begin_inset Newline newline
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\end_inset
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\begin_inset Formula $C_{1}=\frac{1}{E_{2}-E_{1}}\left(i\frac{\rho\omega}{S_{L}}U_{L}+E_{2}P_{L}\right),\qquad C_{2}=-\frac{1}{E_{2}-E_{1}}\left(i\frac{\rho\omega}{S_{L}}U_{L}+E_{1}P_{L}\right)$
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\end_inset
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\begin_inset Newline newline
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\end_inset
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\begin_inset Newline newline
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\end_inset
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Fill in in solution:
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\begin_inset Newline newline
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\end_inset
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\begin_inset Formula $p_{R}=\frac{1}{E_{2}E_{1}}\left[\left(E_{2}e^{E_{1}L}-E_{1}e^{E_{2}L}\right)P_{L}+i\frac{\rho\omega}{S_{L}}\left(e^{E_{1}L}-e^{E_{2}L}\right)U_{L}\right]$
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\end_inset
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\begin_inset Newline newline
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\end_inset
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\begin_inset Formula $U_{R}=\frac{1}{E_{2}E_{1}}\left[i\frac{S_{R}}{\rho\omega}E_{1}E_{2}\left(e^{E_{1}L}-e^{E_{2}L}\right)P_{L}-\frac{S_{R}}{S_{L}}\left(E_{1}e^{E_{1}L}-E_{2}e^{E_{2}L}\right)U_{L}\right]$
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\end_inset
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\end_layout
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\end_inset
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\begin{equation}
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@ -5610,7 +5839,7 @@ S_{f}=\exp\left(\alpha x\right)
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\begin_inset Note Note
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status collapsed
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status open
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\begin_layout Plain Layout
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\begin_inset Formula $\frac{\mathrm{d}^{2}p}{\mathrm{d}x^{2}}+\alpha\frac{\mathrm{d}p}{\mathrm{d}x}+\Gamma^{2}p=0$
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@ -7922,239 +8151,6 @@ Z_{cpm}\left(f\right)=\frac{1}{i2\pi fC_{dyn}\left(f\right)}
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\end_inset
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\end_layout
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\begin_layout Section
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Exponential horn
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\end_layout
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\begin_layout Standard
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hon.py/class ExpHorn(Duct)
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\end_layout
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\begin_layout Standard
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An exponential horn can be modelled using ExpHorn.
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The transfer matrix is derived from Webster's horn equation.
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The solution for an exponential horn is
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\begin{equation}
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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)
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\end{equation}
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\end_inset
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\end_layout
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\begin_layout Standard
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in which
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\begin_inset Formula $S(x)$
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\end_inset
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|
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is the local cross-sectional area,
|
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\begin_inset Formula $\rho$
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\end_inset
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is the air density,
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\begin_inset Formula $\omega$
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\end_inset
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is the angular frequency.
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The exponent terms
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\begin_inset Formula $E_{1}$
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\end_inset
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and
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\begin_inset Formula $E_{2}$
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\end_inset
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are given by
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\begin{equation}
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E_{1}=\frac{-m+\sqrt{-4k^{2}+m^{2}}}{2},\qquad E_{2}=-\frac{m+\sqrt{-4k^{2}+m^{2}}}{2}
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\end{equation}
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\end_inset
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\end_layout
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\begin_layout Standard
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in which
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\begin_inset Formula $k$
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\end_inset
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is the wave number and
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\begin_inset Formula $m$
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\end_inset
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is the flare rate, which is defined as
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\begin{equation}
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m=\frac{1}{L}\ln\left(\frac{S_{R}}{S_{L}}\right)
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\end{equation}
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\end_inset
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\end_layout
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\begin_layout Standard
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with
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\begin_inset Formula $L$
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\end_inset
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being the horn length and
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\begin_inset Formula $S_{R}$
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\end_inset
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and
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\begin_inset Formula $S_{L}$
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\end_inset
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being the cross-sectional areas on the right-hand and left-hand side of
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the horn respectively.
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\begin_inset Newline newline
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\end_inset
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\begin_inset Newline newline
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\end_inset
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|
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By setting the solutions on
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\begin_inset Formula $x=0$
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\end_inset
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to
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\begin_inset Formula $p_{L}$
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\end_inset
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and
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\begin_inset Formula $U_{L}$
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\end_inset
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and those on
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\begin_inset Formula $x=L$
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\end_inset
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to
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\begin_inset Formula $p_{R}$
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\end_inset
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and
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\begin_inset Formula $U_{R}$
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\end_inset
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, the following transfer matrix can be derived:
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\begin{equation}
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T_{ExpHorn}=\frac{1}{E_{2}-E_{1}}\left[\begin{array}{cc}
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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)\\
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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)
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\end{array}\right]
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\end{equation}
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\end_inset
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\end_layout
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\begin_layout Standard
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\begin_inset Note Note
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status open
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\begin_layout Plain Layout
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\begin_inset Formula $p_{L}=p\left(0\right)=C_{1}+C_{2}$
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\end_inset
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\end_layout
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\begin_layout Plain Layout
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\begin_inset Formula $U_{L}=U\left(0\right)=i\frac{S_{L}}{\rho\omega}\left(C_{1}E_{1}+C_{2}E_{2}\right)$
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\end_inset
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\end_layout
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\begin_layout Plain Layout
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\begin_inset Formula $p_{R}=p\left(L\right)=C_{1}e^{E_{1}L}+C_{2}e^{E_{2}L}$
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\end_inset
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\end_layout
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\begin_layout Plain Layout
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\begin_inset Formula $U_{R}=U\left(L\right)=i\frac{S_{L}}{\rho\omega}\left(C_{1}E_{1}e^{E_{1}L}+C_{2}E_{2}e^{E_{2}L}\right)$
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\end_inset
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\begin_inset Newline newline
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\end_inset
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\begin_inset Newline newline
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\end_inset
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Solve for constants
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\begin_inset Newline newline
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\end_inset
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\begin_inset Formula $C_{1}=P_{L}-C_{2}\rightarrow U_{L}=i\frac{S_{L}}{\rho\omega}\left[E_{1}p_{L}+\left(E_{2}-E_{1}\right)C_{2}\right]$
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\end_inset
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\begin_inset Newline newline
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\end_inset
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\begin_inset Formula $C_{1}=\frac{1}{E_{2}-E_{1}}\left(i\frac{\rho\omega}{S_{L}}U_{L}+E_{2}P_{L}\right),\qquad C_{2}=-\frac{1}{E_{2}-E_{1}}\left(i\frac{\rho\omega}{S_{L}}U_{L}+E_{1}P_{L}\right)$
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\end_inset
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||||
\begin_inset Newline newline
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||||
\end_inset
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||||
|
||||
|
||||
\begin_inset Newline newline
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||||
\end_inset
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||||
|
||||
Fill in in solution:
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\begin_inset Newline newline
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||||
\end_inset
|
||||
|
||||
|
||||
\begin_inset Formula $p_{R}=\frac{1}{E_{2}E_{1}}\left[\left(E_{2}e^{E_{1}L}-E_{1}e^{E_{2}L}\right)P_{L}+i\frac{\rho\omega}{S_{L}}\left(e^{E_{1}L}-e^{E_{2}L}\right)U_{L}\right]$
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\end_inset
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||||
\begin_inset Newline newline
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\end_inset
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||||
\begin_inset Formula $U_{R}=\frac{1}{E_{2}E_{1}}\left[i\frac{S_{R}}{\rho\omega}E_{1}E_{2}\left(e^{E_{1}L}-e^{E_{2}L}\right)P_{L}-\frac{S_{R}}{S_{L}}\left(E_{1}e^{E_{1}L}-E_{2}e^{E_{2}L}\right)U_{L}\right]$
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\end_inset
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\end_layout
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\end_inset
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\end_layout
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\begin_layout Section
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@ -8686,6 +8682,7 @@ The wavelength is much larger than the thermal penetration depth (
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).
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\end_layout
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\begin_deeper
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\begin_layout Standard
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We can derive the following impedance boundary condition
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\begin_inset CommandInset citation
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|
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@ -8698,7 +8695,7 @@ literal "true"
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|||
|
||||
:
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||||
\begin_inset Note Note
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||||
status collapsed
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||||
status open
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||||
|
||||
\begin_layout Plain Layout
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||||
Delta EC User guide:
|
||||
|
|
@ -8770,6 +8767,7 @@ Hence the impedance of a hard wall scales with
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|||
.
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||||
\end_layout
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||||
|
||||
\end_deeper
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||||
\begin_layout Standard
|
||||
\begin_inset Float figure
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wide false
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||||
|
|
@ -9727,7 +9725,7 @@ V_{\mathrm{in}}-V_{\mathrm{bemf}}=Z_{\mathrm{el}}I,
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|
||||
\end_inset
|
||||
|
||||
where
|
||||
where
|
||||
\begin_inset Formula $Z_{\mathrm{el}}$
|
||||
\end_inset
|
||||
|
||||
|
|
@ -15015,8 +15013,8 @@ literal "true"
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\end_inset
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|
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.
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Assuming axial symmetrySuch that the acoustic pressure in for example tube
|
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|
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Assuming axial symmetry, such that the acoustic pressure in for example
|
||||
tube
|
||||
\begin_inset Formula $B$
|
||||
\end_inset
|
||||
|
||||
|
|
|
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Loading…
Reference in New Issue