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<!DOCTYPE HTML>
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<a href="ref.html#guillemain:jasa05">Guillemain et al. (2005)</a> and
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<title>Introduction</title>
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<h1 id="introduction">Introduction</h1>
<p>The purpose of this project was to learn how to implement a waveguide
based clarinet model using methods seen in class. I accomplished my task
by implementing the stable reed filter model from week nine coupled with
the cylindrical model of week 8. The physical values found in this
report were obtained from <a href="ref.html#guillemain:jasa05">Guillemain et al. (2005)</a>
and <a href="ref.html#scavone:thesis">Scavone (1997)</a> .</p>
<p>As my main instrument is the clarinet, I wished to learn about the
physics involved in its sound generation. Thank to this class and my
final project, I was able achieve just that.</p>
<p>Now let's start by going through a quick review of the clarinet modules.</p>
<h2 id="cylindrical-air-column">Cylindrical Air Column</h2>
<p><img src="./images/media/image1.png" class="center" width="50%">
<center><strong>Figure 1:</strong>: A cylindrical pipe in cylindrical polar coordinates <a href="ref.html#scavone:wbpage">[3]</a>.</center>
</p>
<p>The cylindrical air column is a type of resonator that utilizes its
shape as a feedback system to sustain specific frequencies within a
sound oscillating through it. By varying its tube length the air column
can play different frequencies. The length must not vary the timbre of
the sound, in other words the ratio between the $1^{st}$ and consequent
harmonics of the sound. The wave propagation within the column is
planar, and so can simply be represented by a one dimensional wave
equation.</p>
<p><img src="./images/media/image2.png" alt="Shape Description automatically generated with medium
confidence" class="center" width="6%">
</p>
<p>The transverse modes do exist within the waves motion, however the
cylindrical bore only weakly excites them. More can be learned about
this effect in <a href="ref.html#scavone:thesis">Gary's thesis paper</a>.</p>
<p>Hence if we are to represent the cylindrical bore as a waveguide model
it would look like this:</p>
<p><img src="./images/media/image3.png" class="center" width="50%">
<center><strong>Figure 2:</strong> Digital waveguide model of a closed-open cylindrical bore<a href="ref.html#scavone:wbpage">[3]</a>.</center>
</p>
<p>The two delay lines would represent the wave's motion propagating down
the air column and back.</p>
<p><img src="./images/media/image4.png" alt="Shape Description automatically generated with medium
confidence" class="center" width="9%"></p>
<p>The shorter the length, the shorter the delay line. Once the pressure
wave reaches the end of the pipe at $x = L$ , we must calculate
the reflection coefficient of the sound $R_{L}$ with first or second
order filter. Generally, the reflection function is:</p>
<p><img src="./images/media/image5.png" alt="Shape Description automatically generated with medium
confidence" class="center small_equation" width="12%"></p>
<p>Where $Z_{L}$ is the load impedance at the end of the tube and $Z_{c}$
is the characteristic impedance of the cylindrical bore.</p>
<p>The characteristic impedance can be determined by:</p>
<p><img src="./images/media/image6.png" alt="" class="center small_equation" width="36%">
<img src="./images/media/image7.png" alt="Shape Description automatically generated with medium
confidence" class="center small_equation" width="12%">
</p>
<p>Where $P(x,t)$ is the equation of motion of the pressure wave and
$U(x,t)$ it's volume flow equivalent. So, we notice that $Z_{c}$ is
dependent on the mass density of air $\rho$ speed of sound c and the
cross-sectional area of the cylinder.</p>
<p>The Load impedance is determined numerically, however we have some edge
cases we can discuss. When the cylinder is an open end, $Z_{L} = 0$,
$R_{L} = - 1$ which means complete reflectance with inversion. When
$Z_{L} = \infty$ then $R_{L} = 1$, complete reflection of the wave
without inversion.</p>
<p>Once we have determined these values, we can compute the pressure
directly at any point in the model. However, to save on computation
power we only need to calculate the pressure response at the entrance of
the air column at $x = 0$ by the introduction of either a unitary
pressure impulse at the entrance to the model
$\left( Z_{c}\delta\lbrack n\rbrack \right)$ which returns the impulse
response $h(t)$ or a continual pressure input from an excitation source
like a reed model.</p>
<h2 id="reed-model">Reed model.</h2>
<p>The reed can be considered a pressure-controlled excitor that inputs
an amount of pressure into the instrument. It then gets
converted into acoustic energy within the bore and outputs a sound who's
note and timbre is dependent on the dimensions of the pipe.</p>
<p>The flow and reed movement are controlled by the difference in pressures
in the mouthpiece and the resulting air columns pressure,
$p\Delta = p_{m} - p(0,t)$,</p>
<p><img src="./images/media/image8.png" alt="Diagram Description automatically
generated" class="center" width="25%">
<center><strong>Figure 3:</strong> An approximate reed orifice geometry<a href="ref.html#scavone:wbpage">[3]</a>.</center>
</p>
<p>The volume flow $u$ through the reed is derived from the Bernoulli
equation,</p>
<p><img src="./images/media/image9.png" alt="" class="center" width="24%"></p>
<p>If we were to assume that there is no change in height within each
cavity, $y_{1} = y_{2} = 0$ then,</p>
<p><img src="./images/media/image10.png" alt="" class="center" width="26%"></p>
<p>Then if in the big cavity $u_{u} = 0$ then we can say that it's velocity is
$v_{1} = 0$ and thus,</p>
<p><img src="./images/media/image11.png" alt="Shape Description automatically generated with medium
confidence" class="center" width="14%"></p>
<p><img src="./images/media/image12.png" alt="Shape Description automatically generated with medium
confidence" class="center" width="23%"></p>
<p>And assuming that y is balanced along a point of equilibrium $H$</p>
<p><img src="./images/media/image13.png" alt="Shape Description automatically generated with medium
confidence" class="center" width="20%">.</p>
<p>The reed's motion, represented in the volume flow as $y$, is composed by
its interaction with the rest of the instrument by the difference in
pressure $p\Delta$ which forces the reed opening either shut or open,
causing it to oscillate. As such we can view it as a mass spring damper
system.</p>
<p><img src="./images/media/image14.png" class="center" width="50%">
<center><strong>Figure 4:</strong> The single-reed as a mechanical spring blown closed<a href="ref.html#scavone:wbpage">[3]</a>.</center>
</p>
<p>Its system of equation is:</p>
<p><img src="./images/media/image15.png" alt="Shape Description automatically generated with medium
confidence" class="center" width="15%"></p>
<p>To solve the approximate interaction at the junction between the reed
and the instruments air column $p(0,t)$, we must start by solving the
mass spring dampers system of equation.</p>
<p>First by deriving its Laplace transform and then directly applying
the bilinear transform, we get</p>
<p><img src="./images/media/image16.png" alt="" class="center " width="35%"></p>
<p><img src="./images/media/image17.png" alt="Shape Description automatically generated with medium
confidence" class="center" width="7%"></p>
<p>Where $\alpha$ is the bilinear transform's constant.</p>
<p>Now, in the implementation of the filter model, there is the use of a feed
forward coefficient which when fed into the volume flow model would make
the reed model non-explicit. As a solution we do not need to consider
the feedforward values, leaving us with:</p>
<p><img src="./images/media/image18.png" alt="" class="center " width="30%"></p>
<p>Since the zeros are at $z = \pm 1$ as we can see here,</p>
<p><img src="./images/media/image19.png" alt="" class="center" width="20%"></p>
<p>The numerator values only affect the phase offset and the decaying
oscillation term and thus we can modify them without affecting the
desirable qualities of the filter like its stability or behavior.</p>
<p>Now given that the volume flow solved above does not immediately depend
on $p\Delta$, we can create an explicit solution:</p>
<p><img src="./images/media/image20.png" alt="" class="center" width="35%"></p>
<p><img src="./images/media/image21.png" alt="Shape Description automatically generated with medium
confidence" class="center" width="20%"></p>
<p>
<img src="./images/media/image22.png" alt="" class="center" width="15%">
</p>
<p><img src="./images/media/image23.png" alt="Shape Description automatically generated with medium
confidence" class="center" width="12%"></p>
<p><img src="./images/media/image24.png" alt="Shape Description automatically generated with medium
confidence" class="center" width="12%">
<p></p>
<img src="./images/media/image25.png" alt="" class="center" width="15%"></p>
<p></p>
<p></p>
<p><img src="./images/media/image26.png" alt="Shape Description automatically generated with medium
confidence" class="center" height="40"></p>
<p></p>
<p><img src="./images/media/image27.png" alt="" class="center" width="35%">
<img src="./images/media/image28.png" alt="" class="center" width="20%">
</p>
<p><img src="./images/media/image29.png" alt="Shape Description automatically generated with medium
confidence" class="center " width="25%">
<p>
If $y + H < 0$ then $u_{0} = 0$.</p>
</p>
<p>This has been a summary of the section that composes the clarinet model
seen in this paper.</p>
<P>
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Made by Maxwell Gentili-Morin.
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