lars/FYS-STK4155
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

update week 43

Browse Source
This commit is contained in:
Morten Hjorth-Jensen
2022-10-27 08:49:01 +02:00
parent 291feb452c
commit 43ba3a8c82
7 changed files with 252 additions and 2086 deletions
Download Patch File Download Diff File Expand all files Collapse all files
doc/pub/week43/html/week43-bs.html
+8 -306
View File
@@ -74,25 +74,6 @@ doconce format html week43.do.txt --html_style=bootstrap --pygments_html_style=d
2,
None,
'a-more-efficient-way-of-coding-the-above-convolution'),
('Convolution Examples: Principle of Superposition and Periodic '
'Forces (Fourier Transforms)',
2,
None,
'convolution-examples-principle-of-superposition-and-periodic-forces-fourier-transforms'),
('Principle of Superposition',
2,
None,
'principle-of-superposition'),
('Simple Code Example', 2, None, 'simple-code-example'),
('Wrapping up Fourier transforms',
2,
None,
'wrapping-up-fourier-transforms'),
('Finding the Coefficients', 2, None, 'finding-the-coefficients'),
('Final words on Fourier Transforms',
2,
None,
'final-words-on-fourier-transforms'),
('Two-dimensional Objects', 2, None, 'two-dimensional-objects'),
('Cross-Correlation', 2, None, 'cross-correlation'),
('More on Dimensionalities', 2, None, 'more-on-dimensionalities'),
@@ -233,12 +214,6 @@ MathJax.Hub.Config({
<!-- navigation toc: --> <li><a href="#convolution-examples-polynomial-multiplication" style="font-size: 80%;">Convolution Examples: Polynomial multiplication</a></li>
<!-- navigation toc: --> <li><a href="#efficient-polynomial-multiplication" style="font-size: 80%;">Efficient Polynomial Multiplication</a></li>
<!-- navigation toc: --> <li><a href="#a-more-efficient-way-of-coding-the-above-convolution" style="font-size: 80%;">A more efficient way of coding the above Convolution</a></li>
<!-- navigation toc: --> <li><a href="#convolution-examples-principle-of-superposition-and-periodic-forces-fourier-transforms" style="font-size: 80%;">Convolution Examples: Principle of Superposition and Periodic Forces (Fourier Transforms)</a></li>
<!-- navigation toc: --> <li><a href="#principle-of-superposition" style="font-size: 80%;">Principle of Superposition</a></li>
<!-- navigation toc: --> <li><a href="#simple-code-example" style="font-size: 80%;">Simple Code Example</a></li>
<!-- navigation toc: --> <li><a href="#wrapping-up-fourier-transforms" style="font-size: 80%;">Wrapping up Fourier transforms</a></li>
<!-- navigation toc: --> <li><a href="#finding-the-coefficients" style="font-size: 80%;">Finding the Coefficients</a></li>
<!-- navigation toc: --> <li><a href="#final-words-on-fourier-transforms" style="font-size: 80%;">Final words on Fourier Transforms</a></li>
<!-- navigation toc: --> <li><a href="#two-dimensional-objects" style="font-size: 80%;">Two-dimensional Objects</a></li>
<!-- navigation toc: --> <li><a href="#cross-correlation" style="font-size: 80%;">Cross-Correlation</a></li>
<!-- navigation toc: --> <li><a href="#more-on-dimensionalities" style="font-size: 80%;">More on Dimensionalities</a></li>
@@ -311,7 +286,7 @@ MathJax.Hub.Config({
</center>
<br>
<center>
<h4>Oct 26, 2022</h4>
<h4>Oct 27, 2022</h4>
</center> <!-- date -->
<br>
@@ -712,279 +687,6 @@ We rather code the convolutions in the minimal memory footprint that they requir
<p>Does the number of floating point operations change here when we use the commutative property?</p>
<!-- !split -->
<h2 id="convolution-examples-principle-of-superposition-and-periodic-forces-fourier-transforms" class="anchor">Convolution Examples: Principle of Superposition and Periodic Forces (Fourier Transforms) </h2>
<p>For problems with so-called harmonic oscillations, given by for example the following differential equation</p>
$$
m\frac{d^2x}{dt^2}+\eta\frac{dx}{dt}+x(t)=F(t),
$$
<p>where \( F(t) \) is an applied external force acting on the system (often called a driving force), one can use the theory of Fourier transformations to find the solutions of this type of equations.</p>
<p>If one has several driving forces, \( F(t)=\sum_n F_n(t) \), one can find
the particular solution to each \( F_n \), \( x_{pn}(t) \), and the particular
solution for the entire driving force is then given by a series like
</p>
$$
\begin{equation}
x_p(t)=\sum_nx_{pn}(t).
\label{_auto1}
\end{equation}
$$
<!-- !split -->
<h2 id="principle-of-superposition" class="anchor">Principle of Superposition </h2>
<p>This is known as the principle of superposition. It only applies when
the homogenous equation is linear. If there were an anharmonic term
such as \( x^3 \) in the homogenous equation, then when one summed various
solutions, \( x=(\sum_n x_n)^2 \), one would get cross
terms. Superposition is especially useful when \( F(t) \) can be written
as a sum of sinusoidal terms, because the solutions for each
sinusoidal (sine or cosine) term is analytic.
</p>
<p>Driving forces are often periodic, even when they are not
sinusoidal. Periodicity implies that for some time \( \tau \)
</p>
$$
\begin{eqnarray}
F(t+\tau)=F(t).
\end{eqnarray}
$$
<p>One example of a non-sinusoidal periodic force is a square wave. Many
components in electric circuits are non-linear, e.g. diodes, which
makes many wave forms non-sinusoidal even when the circuits are being
driven by purely sinusoidal sources.
</p>
<!-- !split -->
<h2 id="simple-code-example" class="anchor">Simple Code Example </h2>
<p>The code here shows a typical example of such a square wave generated using the functionality included in the <b>scipy</b> Python package. We have used a period of \( \tau=0.2 \).</p>
<!-- code=python (!bc pycod) typeset with pygments style "default" -->
<div class="cell border-box-sizing code_cell rendered">
<div class="input">
<div class="inner_cell">
<div class="input_area">
<div class="highlight" style="background: #f8f8f8">
<pre style="line-height: 125%;"><span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">numpy</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">np</span>
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">math</span>
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">scipy</span> <span style="color: #008000; font-weight: bold">import</span> signal
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">matplotlib.pyplot</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">plt</span>
<span style="color: #408080; font-style: italic"># number of points </span>
n <span style="color: #666666">=</span> <span style="color: #666666">500</span>
<span style="color: #408080; font-style: italic"># start and final times </span>
t0 <span style="color: #666666">=</span> <span style="color: #666666">0.0</span>
tn <span style="color: #666666">=</span> <span style="color: #666666">1.0</span>
<span style="color: #408080; font-style: italic"># Period </span>
t <span style="color: #666666">=</span> np<span style="color: #666666">.</span>linspace(t0, tn, n, endpoint<span style="color: #666666">=</span><span style="color: #008000; font-weight: bold">False</span>)
SqrSignal <span style="color: #666666">=</span> np<span style="color: #666666">.</span>zeros(n)
SqrSignal <span style="color: #666666">=</span> <span style="color: #666666">1.0+</span>signal<span style="color: #666666">.</span>square(<span style="color: #666666">2*</span>np<span style="color: #666666">.</span>pi<span style="color: #666666">*5*</span>t)
plt<span style="color: #666666">.</span>plot(t, SqrSignal)
plt<span style="color: #666666">.</span>ylim(<span style="color: #666666">-0.5</span>, <span style="color: #666666">2.5</span>)
plt<span style="color: #666666">.</span>show()
</pre>
</div>
</div>
</div>
</div>
<div class="output_wrapper">
<div class="output">
<div class="output_area">
<div class="output_subarea output_stream output_stdout output_text">
</div>
</div>
</div>
</div>
</div>
<p>For the sinusoidal example the
period is \( \tau=2\pi/\omega \). However, higher harmonics can also
satisfy the periodicity requirement. In general, any force that
satisfies the periodicity requirement can be expressed as a sum over
harmonics,
</p>
$$
\begin{equation}
F(t)=\frac{f_0}{2}+\sum_{n>0} f_n\cos(2n\pi t/\tau)+g_n\sin(2n\pi t/\tau).
\label{_auto2}
\end{equation}
$$
<!-- !split -->
<h2 id="wrapping-up-fourier-transforms" class="anchor">Wrapping up Fourier transforms </h2>
<p>We can write down the answer for
\( x_{pn}(t) \), by substituting \( f_n/m \) or \( g_n/m \) for \( F_0/m \). By
writing each factor \( 2n\pi t/\tau \) as \( n\omega t \), with \( \omega\equiv
2\pi/\tau \),
</p>
$$
\begin{equation}
\label{eq:fourierdef1}
F(t)=\frac{f_0}{2}+\sum_{n>0}f_n\cos(n\omega t)+g_n\sin(n\omega t).
\end{equation}
$$
<p>The solutions for \( x(t) \) then come from replacing \( \omega \) with
\( n\omega \) for each term in the particular solution,
</p>
$$
\begin{eqnarray}
x_p(t)&=&\frac{f_0}{2k}+\sum_{n>0} \alpha_n\cos(n\omega t-\delta_n)+\beta_n\sin(n\omega t-\delta_n),\\
\nonumber
\alpha_n&=&\frac{f_n/m}{\sqrt{((n\omega)^2-\omega_0^2)+4\beta^2n^2\omega^2}},\\
\nonumber
\beta_n&=&\frac{g_n/m}{\sqrt{((n\omega)^2-\omega_0^2)+4\beta^2n^2\omega^2}},\\
\nonumber
\delta_n&=&\tan^{-1}\left(\frac{2\beta n\omega}{\omega_0^2-n^2\omega^2}\right).
\end{eqnarray}
$$
<!-- !split -->
<h2 id="finding-the-coefficients" class="anchor">Finding the Coefficients </h2>
<p>Because the forces have been applied for a long time, any non-zero
damping eliminates the homogenous parts of the solution, so one need
only consider the particular solution for each \( n \).
</p>
<p>The problem is considered solved if one can find expressions for the
coefficients \( f_n \) and \( g_n \), even though the solutions are expressed
as an infinite sum. The coefficients can be extracted from the
function \( F(t) \) by
</p>
$$
\begin{eqnarray}
\label{eq:fourierdef2}
f_n&=&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~F(t)\cos(2n\pi t/\tau),\\
\nonumber
g_n&=&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~F(t)\sin(2n\pi t/\tau).
\end{eqnarray}
$$
<p>To check the consistency of these expressions and to verify
Eq. \eqref{eq:fourierdef2}, one can insert the expansion of \( F(t) \) in
Eq. \eqref{eq:fourierdef1} into the expression for the coefficients in
Eq. \eqref{eq:fourierdef2} and see whether
</p>
$$
\begin{eqnarray}
f_n&=?&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~\left\{
\frac{f_0}{2}+\sum_{m>0}f_m\cos(m\omega t)+g_m\sin(m\omega t)
\right\}\cos(n\omega t).
\end{eqnarray}
$$
<p>Immediately, one can throw away all the terms with \( g_m \) because they
convolute an even and an odd function. The term with \( f_0/2 \)
disappears because \( \cos(n\omega t) \) is equally positive and negative
over the interval and will integrate to zero. For all the terms
\( f_m\cos(m\omega t) \) appearing in the sum, one can use angle addition
formulas to see that \( \cos(m\omega t)\cos(n\omega
t)=(1/2)(\cos[(m+n)\omega t]+\cos[(m-n)\omega t] \). This will integrate
to zero unless \( m=n \). In that case the \( m=n \) term gives
</p>
$$
\begin{equation}
\int_{-\tau/2}^{\tau/2}dt~\cos^2(m\omega t)=\frac{\tau}{2},
\label{_auto3}
\end{equation}
$$
<p>and</p>
$$
\begin{eqnarray}
f_n&=?&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~f_n/2\\
\nonumber
&=&f_n~\checkmark.
\end{eqnarray}
$$
<p>The same method can be used to check for the consistency of \( g_n \).</p>
<!-- !split -->
<h2 id="final-words-on-fourier-transforms" class="anchor">Final words on Fourier Transforms </h2>
<p>The code here uses the Fourier series applied to a
square wave signal. The code here
visualizes the various approximations given by Fourier series compared
with a square wave with period \( T=0.2 \) (dimensionless time), width \( 0.1 \) and max value of the force \( F=2 \). We
see that when we increase the number of components in the Fourier
series, the Fourier series approximation gets closer and closer to the
square wave signal.
</p>
<!-- code=python (!bc pycod) typeset with pygments style "default" -->
<div class="cell border-box-sizing code_cell rendered">
<div class="input">
<div class="inner_cell">
<div class="input_area">
<div class="highlight" style="background: #f8f8f8">
<pre style="line-height: 125%;"><span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">numpy</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">np</span>
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">math</span>
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">scipy</span> <span style="color: #008000; font-weight: bold">import</span> signal
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">matplotlib.pyplot</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">plt</span>
<span style="color: #408080; font-style: italic"># number of points </span>
n <span style="color: #666666">=</span> <span style="color: #666666">500</span>
<span style="color: #408080; font-style: italic"># start and final times </span>
t0 <span style="color: #666666">=</span> <span style="color: #666666">0.0</span>
tn <span style="color: #666666">=</span> <span style="color: #666666">1.0</span>
<span style="color: #408080; font-style: italic"># Period </span>
T <span style="color: #666666">=0.2</span>
<span style="color: #408080; font-style: italic"># Max value of square signal </span>
Fmax<span style="color: #666666">=</span> <span style="color: #666666">2.0</span>
<span style="color: #408080; font-style: italic"># Width of signal </span>
Width <span style="color: #666666">=</span> <span style="color: #666666">0.1</span>
t <span style="color: #666666">=</span> np<span style="color: #666666">.</span>linspace(t0, tn, n, endpoint<span style="color: #666666">=</span><span style="color: #008000; font-weight: bold">False</span>)
SqrSignal <span style="color: #666666">=</span> np<span style="color: #666666">.</span>zeros(n)
FourierSeriesSignal <span style="color: #666666">=</span> np<span style="color: #666666">.</span>zeros(n)
SqrSignal <span style="color: #666666">=</span> <span style="color: #666666">1.0+</span>signal<span style="color: #666666">.</span>square(<span style="color: #666666">2*</span>np<span style="color: #666666">.</span>pi<span style="color: #666666">*5*</span>t<span style="color: #666666">+</span>np<span style="color: #666666">.</span>pi<span style="color: #666666">*</span>Width<span style="color: #666666">/</span>T)
a0 <span style="color: #666666">=</span> Fmax<span style="color: #666666">*</span>Width<span style="color: #666666">/</span>T
FourierSeriesSignal <span style="color: #666666">=</span> a0
Factor <span style="color: #666666">=</span> <span style="color: #666666">2.0*</span>Fmax<span style="color: #666666">/</span>np<span style="color: #666666">.</span>pi
<span style="color: #008000; font-weight: bold">for</span> i <span style="color: #AA22FF; font-weight: bold">in</span> <span style="color: #008000">range</span>(<span style="color: #666666">1</span>,<span style="color: #666666">500</span>):
FourierSeriesSignal <span style="color: #666666">+=</span> Factor<span style="color: #666666">/</span>(i)<span style="color: #666666">*</span>np<span style="color: #666666">.</span>sin(np<span style="color: #666666">.</span>pi<span style="color: #666666">*</span>i<span style="color: #666666">*</span>Width<span style="color: #666666">/</span>T)<span style="color: #666666">*</span>np<span style="color: #666666">.</span>cos(i<span style="color: #666666">*</span>t<span style="color: #666666">*2*</span>np<span style="color: #666666">.</span>pi<span style="color: #666666">/</span>T)
plt<span style="color: #666666">.</span>plot(t, SqrSignal)
plt<span style="color: #666666">.</span>plot(t, FourierSeriesSignal)
plt<span style="color: #666666">.</span>ylim(<span style="color: #666666">-0.5</span>, <span style="color: #666666">2.5</span>)
plt<span style="color: #666666">.</span>show()
</pre>
</div>
</div>
</div>
</div>
<div class="output_wrapper">
<div class="output">
<div class="output_area">
<div class="output_subarea output_stream output_stdout output_text">
</div>
</div>
</div>
</div>
</div>
<!-- !split -->
<h2 id="two-dimensional-objects" class="anchor">Two-dimensional Objects </h2>
@@ -2483,7 +2185,7 @@ samples
$$
\begin{equation}
x = g(z; \theta^{(g)})
\label{_auto4}
\label{_auto1}
\end{equation}
$$
@@ -2500,7 +2202,7 @@ value given by
$$
\begin{equation}
d(x; \theta^{(d)})
\label{_auto5}
\label{_auto2}
\end{equation}
$$
@@ -2513,7 +2215,7 @@ which a function
$$
\begin{equation}
v(\theta^{(g)}, \theta^{(d)})
\label{_auto6}
\label{_auto3}
\end{equation}
$$
@@ -2524,7 +2226,7 @@ conjugate reward
$$
\begin{equation}
-v(\theta^{(g)}, \theta^{(d)})
\label{_auto7}
\label{_auto4}
\end{equation}
$$
@@ -2559,7 +2261,7 @@ $$
\begin{equation}
g^* = \underset{g}{\mathrm{argmin}}\hspace{2pt}
\underset{d}{\mathrm{max}}v(\theta^{(g)}, \theta^{(d)})
\label{_auto8}
\label{_auto5}
\end{equation}
$$
@@ -2569,7 +2271,7 @@ $$
v(\theta^{(g)}, \theta^{(d)}) = \mathbb{E}_{x\sim p_\mathrm{data}}\log d(x)
+ \mathbb{E}_{x\sim p_\mathrm{model}}
\log (1 - d(x))
\label{_auto9}
\label{_auto6}
\end{equation}
$$
@@ -2580,7 +2282,7 @@ approximation of a partition function. In the case where
$$
\begin{equation}
\underset{d}{\mathrm{max}}v(\theta^{(g)}, \theta^{(d)})
\label{_auto10}
\label{_auto7}
\end{equation}
$$
doc/pub/week43/html/week43-reveal.html
+8 -303
View File
@@ -184,7 +184,7 @@ MathJax.Hub.Config({
</center>
<br>
<center>
<h4>Oct 26, 2022</h4>
<h4>Oct 27, 2022</h4>
</center> <!-- date -->
<br>
@@ -623,301 +623,6 @@ We rather code the convolutions in the minimal memory footprint that they requir
<p>Does the number of floating point operations change here when we use the commutative property?</p>
</section>
<section>
<h2 id="convolution-examples-principle-of-superposition-and-periodic-forces-fourier-transforms">Convolution Examples: Principle of Superposition and Periodic Forces (Fourier Transforms) </h2>
<p>For problems with so-called harmonic oscillations, given by for example the following differential equation</p>
<p>&nbsp;<br>
$$
m\frac{d^2x}{dt^2}+\eta\frac{dx}{dt}+x(t)=F(t),
$$
<p>&nbsp;<br>
<p>where \( F(t) \) is an applied external force acting on the system (often called a driving force), one can use the theory of Fourier transformations to find the solutions of this type of equations.</p>
<p>If one has several driving forces, \( F(t)=\sum_n F_n(t) \), one can find
the particular solution to each \( F_n \), \( x_{pn}(t) \), and the particular
solution for the entire driving force is then given by a series like
</p>
<p>&nbsp;<br>
$$
\begin{equation}
x_p(t)=\sum_nx_{pn}(t).
\tag{1}
\end{equation}
$$
<p>&nbsp;<br>
</section>
<section>
<h2 id="principle-of-superposition">Principle of Superposition </h2>
<p>This is known as the principle of superposition. It only applies when
the homogenous equation is linear. If there were an anharmonic term
such as \( x^3 \) in the homogenous equation, then when one summed various
solutions, \( x=(\sum_n x_n)^2 \), one would get cross
terms. Superposition is especially useful when \( F(t) \) can be written
as a sum of sinusoidal terms, because the solutions for each
sinusoidal (sine or cosine) term is analytic.
</p>
<p>Driving forces are often periodic, even when they are not
sinusoidal. Periodicity implies that for some time \( \tau \)
</p>
<p>&nbsp;<br>
$$
\begin{eqnarray}
F(t+\tau)=F(t).
\end{eqnarray}
$$
<p>&nbsp;<br>
<p>One example of a non-sinusoidal periodic force is a square wave. Many
components in electric circuits are non-linear, e.g. diodes, which
makes many wave forms non-sinusoidal even when the circuits are being
driven by purely sinusoidal sources.
</p>
</section>
<section>
<h2 id="simple-code-example">Simple Code Example </h2>
<p>The code here shows a typical example of such a square wave generated using the functionality included in the <b>scipy</b> Python package. We have used a period of \( \tau=0.2 \).</p>
<!-- code=python (!bc pycod) typeset with pygments style "perldoc" -->
<div class="cell border-box-sizing code_cell rendered">
<div class="input">
<div class="inner_cell">
<div class="input_area">
<div class="highlight" style="background: #eeeedd">
<pre style="font-size: 80%; line-height: 125%;"><span style="color: #8B008B; font-weight: bold">import</span> <span style="color: #008b45; text-decoration: underline">numpy</span> <span style="color: #8B008B; font-weight: bold">as</span> <span style="color: #008b45; text-decoration: underline">np</span>
<span style="color: #8B008B; font-weight: bold">import</span> <span style="color: #008b45; text-decoration: underline">math</span>
<span style="color: #8B008B; font-weight: bold">from</span> <span style="color: #008b45; text-decoration: underline">scipy</span> <span style="color: #8B008B; font-weight: bold">import</span> signal
<span style="color: #8B008B; font-weight: bold">import</span> <span style="color: #008b45; text-decoration: underline">matplotlib.pyplot</span> <span style="color: #8B008B; font-weight: bold">as</span> <span style="color: #008b45; text-decoration: underline">plt</span>
<span style="color: #228B22"># number of points </span>
n = <span style="color: #B452CD">500</span>
<span style="color: #228B22"># start and final times </span>
t0 = <span style="color: #B452CD">0.0</span>
tn = <span style="color: #B452CD">1.0</span>
<span style="color: #228B22"># Period </span>
t = np.linspace(t0, tn, n, endpoint=<span style="color: #8B008B; font-weight: bold">False</span>)
SqrSignal = np.zeros(n)
SqrSignal = <span style="color: #B452CD">1.0</span>+signal.square(<span style="color: #B452CD">2</span>*np.pi*<span style="color: #B452CD">5</span>*t)
plt.plot(t, SqrSignal)
plt.ylim(-<span style="color: #B452CD">0.5</span>, <span style="color: #B452CD">2.5</span>)
plt.show()
</pre>
</div>
</div>
</div>
</div>
<div class="output_wrapper">
<div class="output">
<div class="output_area">
<div class="output_subarea output_stream output_stdout output_text">
</div>
</div>
</div>
</div>
</div>
<p>For the sinusoidal example the
period is \( \tau=2\pi/\omega \). However, higher harmonics can also
satisfy the periodicity requirement. In general, any force that
satisfies the periodicity requirement can be expressed as a sum over
harmonics,
</p>
<p>&nbsp;<br>
$$
\begin{equation}
F(t)=\frac{f_0}{2}+\sum_{n>0} f_n\cos(2n\pi t/\tau)+g_n\sin(2n\pi t/\tau).
\tag{2}
\end{equation}
$$
<p>&nbsp;<br>
</section>
<section>
<h2 id="wrapping-up-fourier-transforms">Wrapping up Fourier transforms </h2>
<p>We can write down the answer for
\( x_{pn}(t) \), by substituting \( f_n/m \) or \( g_n/m \) for \( F_0/m \). By
writing each factor \( 2n\pi t/\tau \) as \( n\omega t \), with \( \omega\equiv
2\pi/\tau \),
</p>
<p>&nbsp;<br>
$$
\begin{equation}
\tag{3}
F(t)=\frac{f_0}{2}+\sum_{n>0}f_n\cos(n\omega t)+g_n\sin(n\omega t).
\end{equation}
$$
<p>&nbsp;<br>
<p>The solutions for \( x(t) \) then come from replacing \( \omega \) with
\( n\omega \) for each term in the particular solution,
</p>
<p>&nbsp;<br>
$$
\begin{eqnarray}
x_p(t)&=&\frac{f_0}{2k}+\sum_{n>0} \alpha_n\cos(n\omega t-\delta_n)+\beta_n\sin(n\omega t-\delta_n),\\
\nonumber
\alpha_n&=&\frac{f_n/m}{\sqrt{((n\omega)^2-\omega_0^2)+4\beta^2n^2\omega^2}},\\
\nonumber
\beta_n&=&\frac{g_n/m}{\sqrt{((n\omega)^2-\omega_0^2)+4\beta^2n^2\omega^2}},\\
\nonumber
\delta_n&=&\tan^{-1}\left(\frac{2\beta n\omega}{\omega_0^2-n^2\omega^2}\right).
\end{eqnarray}
$$
<p>&nbsp;<br>
</section>
<section>
<h2 id="finding-the-coefficients">Finding the Coefficients </h2>
<p>Because the forces have been applied for a long time, any non-zero
damping eliminates the homogenous parts of the solution, so one need
only consider the particular solution for each \( n \).
</p>
<p>The problem is considered solved if one can find expressions for the
coefficients \( f_n \) and \( g_n \), even though the solutions are expressed
as an infinite sum. The coefficients can be extracted from the
function \( F(t) \) by
</p>
<p>&nbsp;<br>
$$
\begin{eqnarray}
\tag{4}
f_n&=&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~F(t)\cos(2n\pi t/\tau),\\
\nonumber
g_n&=&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~F(t)\sin(2n\pi t/\tau).
\end{eqnarray}
$$
<p>&nbsp;<br>
<p>To check the consistency of these expressions and to verify
Eq. <a href="#mjx-eqn-4">(4)</a>, one can insert the expansion of \( F(t) \) in
Eq. <a href="#mjx-eqn-3">(3)</a> into the expression for the coefficients in
Eq. <a href="#mjx-eqn-4">(4)</a> and see whether
</p>
<p>&nbsp;<br>
$$
\begin{eqnarray}
f_n&=?&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~\left\{
\frac{f_0}{2}+\sum_{m>0}f_m\cos(m\omega t)+g_m\sin(m\omega t)
\right\}\cos(n\omega t).
\end{eqnarray}
$$
<p>&nbsp;<br>
<p>Immediately, one can throw away all the terms with \( g_m \) because they
convolute an even and an odd function. The term with \( f_0/2 \)
disappears because \( \cos(n\omega t) \) is equally positive and negative
over the interval and will integrate to zero. For all the terms
\( f_m\cos(m\omega t) \) appearing in the sum, one can use angle addition
formulas to see that \( \cos(m\omega t)\cos(n\omega
t)=(1/2)(\cos[(m+n)\omega t]+\cos[(m-n)\omega t] \). This will integrate
to zero unless \( m=n \). In that case the \( m=n \) term gives
</p>
<p>&nbsp;<br>
$$
\begin{equation}
\int_{-\tau/2}^{\tau/2}dt~\cos^2(m\omega t)=\frac{\tau}{2},
\tag{5}
\end{equation}
$$
<p>&nbsp;<br>
<p>and</p>
<p>&nbsp;<br>
$$
\begin{eqnarray}
f_n&=?&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~f_n/2\\
\nonumber
&=&f_n~\checkmark.
\end{eqnarray}
$$
<p>&nbsp;<br>
<p>The same method can be used to check for the consistency of \( g_n \).</p>
</section>
<section>
<h2 id="final-words-on-fourier-transforms">Final words on Fourier Transforms </h2>
<p>The code here uses the Fourier series applied to a
square wave signal. The code here
visualizes the various approximations given by Fourier series compared
with a square wave with period \( T=0.2 \) (dimensionless time), width \( 0.1 \) and max value of the force \( F=2 \). We
see that when we increase the number of components in the Fourier
series, the Fourier series approximation gets closer and closer to the
square wave signal.
</p>
<!-- code=python (!bc pycod) typeset with pygments style "perldoc" -->
<div class="cell border-box-sizing code_cell rendered">
<div class="input">
<div class="inner_cell">
<div class="input_area">
<div class="highlight" style="background: #eeeedd">
<pre style="font-size: 80%; line-height: 125%;"><span style="color: #8B008B; font-weight: bold">import</span> <span style="color: #008b45; text-decoration: underline">numpy</span> <span style="color: #8B008B; font-weight: bold">as</span> <span style="color: #008b45; text-decoration: underline">np</span>
<span style="color: #8B008B; font-weight: bold">import</span> <span style="color: #008b45; text-decoration: underline">math</span>
<span style="color: #8B008B; font-weight: bold">from</span> <span style="color: #008b45; text-decoration: underline">scipy</span> <span style="color: #8B008B; font-weight: bold">import</span> signal
<span style="color: #8B008B; font-weight: bold">import</span> <span style="color: #008b45; text-decoration: underline">matplotlib.pyplot</span> <span style="color: #8B008B; font-weight: bold">as</span> <span style="color: #008b45; text-decoration: underline">plt</span>
<span style="color: #228B22"># number of points </span>
n = <span style="color: #B452CD">500</span>
<span style="color: #228B22"># start and final times </span>
t0 = <span style="color: #B452CD">0.0</span>
tn = <span style="color: #B452CD">1.0</span>
<span style="color: #228B22"># Period </span>
T =<span style="color: #B452CD">0.2</span>
<span style="color: #228B22"># Max value of square signal </span>
Fmax= <span style="color: #B452CD">2.0</span>
<span style="color: #228B22"># Width of signal </span>
Width = <span style="color: #B452CD">0.1</span>
t = np.linspace(t0, tn, n, endpoint=<span style="color: #8B008B; font-weight: bold">False</span>)
SqrSignal = np.zeros(n)
FourierSeriesSignal = np.zeros(n)
SqrSignal = <span style="color: #B452CD">1.0</span>+signal.square(<span style="color: #B452CD">2</span>*np.pi*<span style="color: #B452CD">5</span>*t+np.pi*Width/T)
a0 = Fmax*Width/T
FourierSeriesSignal = a0
Factor = <span style="color: #B452CD">2.0</span>*Fmax/np.pi
<span style="color: #8B008B; font-weight: bold">for</span> i <span style="color: #8B008B">in</span> <span style="color: #658b00">range</span>(<span style="color: #B452CD">1</span>,<span style="color: #B452CD">500</span>):
FourierSeriesSignal += Factor/(i)*np.sin(np.pi*i*Width/T)*np.cos(i*t*<span style="color: #B452CD">2</span>*np.pi/T)
plt.plot(t, SqrSignal)
plt.plot(t, FourierSeriesSignal)
plt.ylim(-<span style="color: #B452CD">0.5</span>, <span style="color: #B452CD">2.5</span>)
plt.show()
</pre>
</div>
</div>
</div>
</div>
<div class="output_wrapper">
<div class="output">
<div class="output_area">
<div class="output_subarea output_stream output_stdout output_text">
</div>
</div>
</div>
</div>
</div>
</section>
<section>
<h2 id="two-dimensional-objects">Two-dimensional Objects </h2>
@@ -2439,7 +2144,7 @@ samples
$$
\begin{equation}
x = g(z; \theta^{(g)})
\tag{6}
\tag{1}
\end{equation}
$$
<p>&nbsp;<br>
@@ -2458,7 +2163,7 @@ value given by
$$
\begin{equation}
d(x; \theta^{(d)})
\tag{7}
\tag{2}
\end{equation}
$$
<p>&nbsp;<br>
@@ -2473,7 +2178,7 @@ which a function
$$
\begin{equation}
v(\theta^{(g)}, \theta^{(d)})
\tag{8}
\tag{3}
\end{equation}
$$
<p>&nbsp;<br>
@@ -2486,7 +2191,7 @@ conjugate reward
$$
\begin{equation}
-v(\theta^{(g)}, \theta^{(d)})
\tag{9}
\tag{4}
\end{equation}
$$
<p>&nbsp;<br>
@@ -2524,7 +2229,7 @@ $$
\begin{equation}
g^* = \underset{g}{\mathrm{argmin}}\hspace{2pt}
\underset{d}{\mathrm{max}}v(\theta^{(g)}, \theta^{(d)})
\tag{10}
\tag{5}
\end{equation}
$$
<p>&nbsp;<br>
@@ -2536,7 +2241,7 @@ $$
v(\theta^{(g)}, \theta^{(d)}) = \mathbb{E}_{x\sim p_\mathrm{data}}\log d(x)
+ \mathbb{E}_{x\sim p_\mathrm{model}}
\log (1 - d(x))
\tag{11}
\tag{6}
\end{equation}
$$
<p>&nbsp;<br>
@@ -2549,7 +2254,7 @@ approximation of a partition function. In the case where
$$
\begin{equation}
\underset{d}{\mathrm{max}}v(\theta^{(g)}, \theta^{(d)})
\tag{12}
\tag{7}
\end{equation}
$$
<p>&nbsp;<br>
doc/pub/week43/html/week43-solarized.html
+8 -300
View File
@@ -101,25 +101,6 @@ div.toc p,a {
2,
None,
'a-more-efficient-way-of-coding-the-above-convolution'),
('Convolution Examples: Principle of Superposition and Periodic '
'Forces (Fourier Transforms)',
2,
None,
'convolution-examples-principle-of-superposition-and-periodic-forces-fourier-transforms'),
('Principle of Superposition',
2,
None,
'principle-of-superposition'),
('Simple Code Example', 2, None, 'simple-code-example'),
('Wrapping up Fourier transforms',
2,
None,
'wrapping-up-fourier-transforms'),
('Finding the Coefficients', 2, None, 'finding-the-coefficients'),
('Final words on Fourier Transforms',
2,
None,
'final-words-on-fourier-transforms'),
('Two-dimensional Objects', 2, None, 'two-dimensional-objects'),
('Cross-Correlation', 2, None, 'cross-correlation'),
('More on Dimensionalities', 2, None, 'more-on-dimensionalities'),
@@ -247,7 +228,7 @@ MathJax.Hub.Config({
</center>
<br>
<center>
<h4>Oct 26, 2022</h4>
<h4>Oct 27, 2022</h4>
</center> <!-- date -->
<br>
@@ -641,279 +622,6 @@ We rather code the convolutions in the minimal memory footprint that they requir
<p>Does the number of floating point operations change here when we use the commutative property?</p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="convolution-examples-principle-of-superposition-and-periodic-forces-fourier-transforms">Convolution Examples: Principle of Superposition and Periodic Forces (Fourier Transforms) </h2>
<p>For problems with so-called harmonic oscillations, given by for example the following differential equation</p>
$$
m\frac{d^2x}{dt^2}+\eta\frac{dx}{dt}+x(t)=F(t),
$$
<p>where \( F(t) \) is an applied external force acting on the system (often called a driving force), one can use the theory of Fourier transformations to find the solutions of this type of equations.</p>
<p>If one has several driving forces, \( F(t)=\sum_n F_n(t) \), one can find
the particular solution to each \( F_n \), \( x_{pn}(t) \), and the particular
solution for the entire driving force is then given by a series like
</p>
$$
\begin{equation}
x_p(t)=\sum_nx_{pn}(t).
\label{_auto1}
\end{equation}
$$
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="principle-of-superposition">Principle of Superposition </h2>
<p>This is known as the principle of superposition. It only applies when
the homogenous equation is linear. If there were an anharmonic term
such as \( x^3 \) in the homogenous equation, then when one summed various
solutions, \( x=(\sum_n x_n)^2 \), one would get cross
terms. Superposition is especially useful when \( F(t) \) can be written
as a sum of sinusoidal terms, because the solutions for each
sinusoidal (sine or cosine) term is analytic.
</p>
<p>Driving forces are often periodic, even when they are not
sinusoidal. Periodicity implies that for some time \( \tau \)
</p>
$$
\begin{eqnarray}
F(t+\tau)=F(t).
\end{eqnarray}
$$
<p>One example of a non-sinusoidal periodic force is a square wave. Many
components in electric circuits are non-linear, e.g. diodes, which
makes many wave forms non-sinusoidal even when the circuits are being
driven by purely sinusoidal sources.
</p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="simple-code-example">Simple Code Example </h2>
<p>The code here shows a typical example of such a square wave generated using the functionality included in the <b>scipy</b> Python package. We have used a period of \( \tau=0.2 \).</p>
<!-- code=python (!bc pycod) typeset with pygments style "perldoc" -->
<div class="cell border-box-sizing code_cell rendered">
<div class="input">
<div class="inner_cell">
<div class="input_area">
<div class="highlight" style="background: #eeeedd">
<pre style="line-height: 125%;"><span style="color: #8B008B; font-weight: bold">import</span> <span style="color: #008b45; text-decoration: underline">numpy</span> <span style="color: #8B008B; font-weight: bold">as</span> <span style="color: #008b45; text-decoration: underline">np</span>
<span style="color: #8B008B; font-weight: bold">import</span> <span style="color: #008b45; text-decoration: underline">math</span>
<span style="color: #8B008B; font-weight: bold">from</span> <span style="color: #008b45; text-decoration: underline">scipy</span> <span style="color: #8B008B; font-weight: bold">import</span> signal
<span style="color: #8B008B; font-weight: bold">import</span> <span style="color: #008b45; text-decoration: underline">matplotlib.pyplot</span> <span style="color: #8B008B; font-weight: bold">as</span> <span style="color: #008b45; text-decoration: underline">plt</span>
<span style="color: #228B22"># number of points </span>
n = <span style="color: #B452CD">500</span>
<span style="color: #228B22"># start and final times </span>
t0 = <span style="color: #B452CD">0.0</span>
tn = <span style="color: #B452CD">1.0</span>
<span style="color: #228B22"># Period </span>
t = np.linspace(t0, tn, n, endpoint=<span style="color: #8B008B; font-weight: bold">False</span>)
SqrSignal = np.zeros(n)
SqrSignal = <span style="color: #B452CD">1.0</span>+signal.square(<span style="color: #B452CD">2</span>*np.pi*<span style="color: #B452CD">5</span>*t)
plt.plot(t, SqrSignal)
plt.ylim(-<span style="color: #B452CD">0.5</span>, <span style="color: #B452CD">2.5</span>)
plt.show()
</pre>
</div>
</div>
</div>
</div>
<div class="output_wrapper">
<div class="output">
<div class="output_area">
<div class="output_subarea output_stream output_stdout output_text">
</div>
</div>
</div>
</div>
</div>
<p>For the sinusoidal example the
period is \( \tau=2\pi/\omega \). However, higher harmonics can also
satisfy the periodicity requirement. In general, any force that
satisfies the periodicity requirement can be expressed as a sum over
harmonics,
</p>
$$
\begin{equation}
F(t)=\frac{f_0}{2}+\sum_{n>0} f_n\cos(2n\pi t/\tau)+g_n\sin(2n\pi t/\tau).
\label{_auto2}
\end{equation}
$$
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="wrapping-up-fourier-transforms">Wrapping up Fourier transforms </h2>
<p>We can write down the answer for
\( x_{pn}(t) \), by substituting \( f_n/m \) or \( g_n/m \) for \( F_0/m \). By
writing each factor \( 2n\pi t/\tau \) as \( n\omega t \), with \( \omega\equiv
2\pi/\tau \),
</p>
$$
\begin{equation}
\label{eq:fourierdef1}
F(t)=\frac{f_0}{2}+\sum_{n>0}f_n\cos(n\omega t)+g_n\sin(n\omega t).
\end{equation}
$$
<p>The solutions for \( x(t) \) then come from replacing \( \omega \) with
\( n\omega \) for each term in the particular solution,
</p>
$$
\begin{eqnarray}
x_p(t)&=&\frac{f_0}{2k}+\sum_{n>0} \alpha_n\cos(n\omega t-\delta_n)+\beta_n\sin(n\omega t-\delta_n),\\
\nonumber
\alpha_n&=&\frac{f_n/m}{\sqrt{((n\omega)^2-\omega_0^2)+4\beta^2n^2\omega^2}},\\
\nonumber
\beta_n&=&\frac{g_n/m}{\sqrt{((n\omega)^2-\omega_0^2)+4\beta^2n^2\omega^2}},\\
\nonumber
\delta_n&=&\tan^{-1}\left(\frac{2\beta n\omega}{\omega_0^2-n^2\omega^2}\right).
\end{eqnarray}
$$
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="finding-the-coefficients">Finding the Coefficients </h2>
<p>Because the forces have been applied for a long time, any non-zero
damping eliminates the homogenous parts of the solution, so one need
only consider the particular solution for each \( n \).
</p>
<p>The problem is considered solved if one can find expressions for the
coefficients \( f_n \) and \( g_n \), even though the solutions are expressed
as an infinite sum. The coefficients can be extracted from the
function \( F(t) \) by
</p>
$$
\begin{eqnarray}
\label{eq:fourierdef2}
f_n&=&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~F(t)\cos(2n\pi t/\tau),\\
\nonumber
g_n&=&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~F(t)\sin(2n\pi t/\tau).
\end{eqnarray}
$$
<p>To check the consistency of these expressions and to verify
Eq. \eqref{eq:fourierdef2}, one can insert the expansion of \( F(t) \) in
Eq. \eqref{eq:fourierdef1} into the expression for the coefficients in
Eq. \eqref{eq:fourierdef2} and see whether
</p>
$$
\begin{eqnarray}
f_n&=?&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~\left\{
\frac{f_0}{2}+\sum_{m>0}f_m\cos(m\omega t)+g_m\sin(m\omega t)
\right\}\cos(n\omega t).
\end{eqnarray}
$$
<p>Immediately, one can throw away all the terms with \( g_m \) because they
convolute an even and an odd function. The term with \( f_0/2 \)
disappears because \( \cos(n\omega t) \) is equally positive and negative
over the interval and will integrate to zero. For all the terms
\( f_m\cos(m\omega t) \) appearing in the sum, one can use angle addition
formulas to see that \( \cos(m\omega t)\cos(n\omega
t)=(1/2)(\cos[(m+n)\omega t]+\cos[(m-n)\omega t] \). This will integrate
to zero unless \( m=n \). In that case the \( m=n \) term gives
</p>
$$
\begin{equation}
\int_{-\tau/2}^{\tau/2}dt~\cos^2(m\omega t)=\frac{\tau}{2},
\label{_auto3}
\end{equation}
$$
<p>and</p>
$$
\begin{eqnarray}
f_n&=?&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~f_n/2\\
\nonumber
&=&f_n~\checkmark.
\end{eqnarray}
$$
<p>The same method can be used to check for the consistency of \( g_n \).</p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="final-words-on-fourier-transforms">Final words on Fourier Transforms </h2>
<p>The code here uses the Fourier series applied to a
square wave signal. The code here
visualizes the various approximations given by Fourier series compared
with a square wave with period \( T=0.2 \) (dimensionless time), width \( 0.1 \) and max value of the force \( F=2 \). We
see that when we increase the number of components in the Fourier
series, the Fourier series approximation gets closer and closer to the
square wave signal.
</p>
<!-- code=python (!bc pycod) typeset with pygments style "perldoc" -->
<div class="cell border-box-sizing code_cell rendered">
<div class="input">
<div class="inner_cell">
<div class="input_area">
<div class="highlight" style="background: #eeeedd">
<pre style="line-height: 125%;"><span style="color: #8B008B; font-weight: bold">import</span> <span style="color: #008b45; text-decoration: underline">numpy</span> <span style="color: #8B008B; font-weight: bold">as</span> <span style="color: #008b45; text-decoration: underline">np</span>
<span style="color: #8B008B; font-weight: bold">import</span> <span style="color: #008b45; text-decoration: underline">math</span>
<span style="color: #8B008B; font-weight: bold">from</span> <span style="color: #008b45; text-decoration: underline">scipy</span> <span style="color: #8B008B; font-weight: bold">import</span> signal
<span style="color: #8B008B; font-weight: bold">import</span> <span style="color: #008b45; text-decoration: underline">matplotlib.pyplot</span> <span style="color: #8B008B; font-weight: bold">as</span> <span style="color: #008b45; text-decoration: underline">plt</span>
<span style="color: #228B22"># number of points </span>
n = <span style="color: #B452CD">500</span>
<span style="color: #228B22"># start and final times </span>
t0 = <span style="color: #B452CD">0.0</span>
tn = <span style="color: #B452CD">1.0</span>
<span style="color: #228B22"># Period </span>
T =<span style="color: #B452CD">0.2</span>
<span style="color: #228B22"># Max value of square signal </span>
Fmax= <span style="color: #B452CD">2.0</span>
<span style="color: #228B22"># Width of signal </span>
Width = <span style="color: #B452CD">0.1</span>
t = np.linspace(t0, tn, n, endpoint=<span style="color: #8B008B; font-weight: bold">False</span>)
SqrSignal = np.zeros(n)
FourierSeriesSignal = np.zeros(n)
SqrSignal = <span style="color: #B452CD">1.0</span>+signal.square(<span style="color: #B452CD">2</span>*np.pi*<span style="color: #B452CD">5</span>*t+np.pi*Width/T)
a0 = Fmax*Width/T
FourierSeriesSignal = a0
Factor = <span style="color: #B452CD">2.0</span>*Fmax/np.pi
<span style="color: #8B008B; font-weight: bold">for</span> i <span style="color: #8B008B">in</span> <span style="color: #658b00">range</span>(<span style="color: #B452CD">1</span>,<span style="color: #B452CD">500</span>):
FourierSeriesSignal += Factor/(i)*np.sin(np.pi*i*Width/T)*np.cos(i*t*<span style="color: #B452CD">2</span>*np.pi/T)
plt.plot(t, SqrSignal)
plt.plot(t, FourierSeriesSignal)
plt.ylim(-<span style="color: #B452CD">0.5</span>, <span style="color: #B452CD">2.5</span>)
plt.show()
</pre>
</div>
</div>
</div>
</div>
<div class="output_wrapper">
<div class="output">
<div class="output_area">
<div class="output_subarea output_stream output_stdout output_text">
</div>
</div>
</div>
</div>
</div>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="two-dimensional-objects">Two-dimensional Objects </h2>
@@ -2412,7 +2120,7 @@ samples
$$
\begin{equation}
x = g(z; \theta^{(g)})
\label{_auto4}
\label{_auto1}
\end{equation}
$$
@@ -2429,7 +2137,7 @@ value given by
$$
\begin{equation}
d(x; \theta^{(d)})
\label{_auto5}
\label{_auto2}
\end{equation}
$$
@@ -2442,7 +2150,7 @@ which a function
$$
\begin{equation}
v(\theta^{(g)}, \theta^{(d)})
\label{_auto6}
\label{_auto3}
\end{equation}
$$
@@ -2453,7 +2161,7 @@ conjugate reward
$$
\begin{equation}
-v(\theta^{(g)}, \theta^{(d)})
\label{_auto7}
\label{_auto4}
\end{equation}
$$
@@ -2488,7 +2196,7 @@ $$
\begin{equation}
g^* = \underset{g}{\mathrm{argmin}}\hspace{2pt}
\underset{d}{\mathrm{max}}v(\theta^{(g)}, \theta^{(d)})
\label{_auto8}
\label{_auto5}
\end{equation}
$$
@@ -2498,7 +2206,7 @@ $$
v(\theta^{(g)}, \theta^{(d)}) = \mathbb{E}_{x\sim p_\mathrm{data}}\log d(x)
+ \mathbb{E}_{x\sim p_\mathrm{model}}
\log (1 - d(x))
\label{_auto9}
\label{_auto6}
\end{equation}
$$
@@ -2509,7 +2217,7 @@ approximation of a partition function. In the case where
$$
\begin{equation}
\underset{d}{\mathrm{max}}v(\theta^{(g)}, \theta^{(d)})
\label{_auto10}
\label{_auto7}
\end{equation}
$$
doc/pub/week43/html/week43.html
+8 -300
View File
@@ -178,25 +178,6 @@ div.toc p,a {
2,
None,
'a-more-efficient-way-of-coding-the-above-convolution'),
('Convolution Examples: Principle of Superposition and Periodic '
'Forces (Fourier Transforms)',
2,
None,
'convolution-examples-principle-of-superposition-and-periodic-forces-fourier-transforms'),
('Principle of Superposition',
2,
None,
'principle-of-superposition'),
('Simple Code Example', 2, None, 'simple-code-example'),
('Wrapping up Fourier transforms',
2,
None,
'wrapping-up-fourier-transforms'),
('Finding the Coefficients', 2, None, 'finding-the-coefficients'),
('Final words on Fourier Transforms',
2,
None,
'final-words-on-fourier-transforms'),
('Two-dimensional Objects', 2, None, 'two-dimensional-objects'),
('Cross-Correlation', 2, None, 'cross-correlation'),
('More on Dimensionalities', 2, None, 'more-on-dimensionalities'),
@@ -324,7 +305,7 @@ MathJax.Hub.Config({
</center>
<br>
<center>
<h4>Oct 26, 2022</h4>
<h4>Oct 27, 2022</h4>
</center> <!-- date -->
<br>
@@ -718,279 +699,6 @@ We rather code the convolutions in the minimal memory footprint that they requir
<p>Does the number of floating point operations change here when we use the commutative property?</p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="convolution-examples-principle-of-superposition-and-periodic-forces-fourier-transforms">Convolution Examples: Principle of Superposition and Periodic Forces (Fourier Transforms) </h2>
<p>For problems with so-called harmonic oscillations, given by for example the following differential equation</p>
$$
m\frac{d^2x}{dt^2}+\eta\frac{dx}{dt}+x(t)=F(t),
$$
<p>where \( F(t) \) is an applied external force acting on the system (often called a driving force), one can use the theory of Fourier transformations to find the solutions of this type of equations.</p>
<p>If one has several driving forces, \( F(t)=\sum_n F_n(t) \), one can find
the particular solution to each \( F_n \), \( x_{pn}(t) \), and the particular
solution for the entire driving force is then given by a series like
</p>
$$
\begin{equation}
x_p(t)=\sum_nx_{pn}(t).
\label{_auto1}
\end{equation}
$$
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="principle-of-superposition">Principle of Superposition </h2>
<p>This is known as the principle of superposition. It only applies when
the homogenous equation is linear. If there were an anharmonic term
such as \( x^3 \) in the homogenous equation, then when one summed various
solutions, \( x=(\sum_n x_n)^2 \), one would get cross
terms. Superposition is especially useful when \( F(t) \) can be written
as a sum of sinusoidal terms, because the solutions for each
sinusoidal (sine or cosine) term is analytic.
</p>
<p>Driving forces are often periodic, even when they are not
sinusoidal. Periodicity implies that for some time \( \tau \)
</p>
$$
\begin{eqnarray}
F(t+\tau)=F(t).
\end{eqnarray}
$$
<p>One example of a non-sinusoidal periodic force is a square wave. Many
components in electric circuits are non-linear, e.g. diodes, which
makes many wave forms non-sinusoidal even when the circuits are being
driven by purely sinusoidal sources.
</p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="simple-code-example">Simple Code Example </h2>
<p>The code here shows a typical example of such a square wave generated using the functionality included in the <b>scipy</b> Python package. We have used a period of \( \tau=0.2 \).</p>
<!-- code=python (!bc pycod) typeset with pygments style "default" -->
<div class="cell border-box-sizing code_cell rendered">
<div class="input">
<div class="inner_cell">
<div class="input_area">
<div class="highlight" style="background: #f8f8f8">
<pre style="line-height: 125%;"><span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">numpy</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">np</span>
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">math</span>
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">scipy</span> <span style="color: #008000; font-weight: bold">import</span> signal
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">matplotlib.pyplot</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">plt</span>
<span style="color: #408080; font-style: italic"># number of points </span>
n <span style="color: #666666">=</span> <span style="color: #666666">500</span>
<span style="color: #408080; font-style: italic"># start and final times </span>
t0 <span style="color: #666666">=</span> <span style="color: #666666">0.0</span>
tn <span style="color: #666666">=</span> <span style="color: #666666">1.0</span>
<span style="color: #408080; font-style: italic"># Period </span>
t <span style="color: #666666">=</span> np<span style="color: #666666">.</span>linspace(t0, tn, n, endpoint<span style="color: #666666">=</span><span style="color: #008000; font-weight: bold">False</span>)
SqrSignal <span style="color: #666666">=</span> np<span style="color: #666666">.</span>zeros(n)
SqrSignal <span style="color: #666666">=</span> <span style="color: #666666">1.0+</span>signal<span style="color: #666666">.</span>square(<span style="color: #666666">2*</span>np<span style="color: #666666">.</span>pi<span style="color: #666666">*5*</span>t)
plt<span style="color: #666666">.</span>plot(t, SqrSignal)
plt<span style="color: #666666">.</span>ylim(<span style="color: #666666">-0.5</span>, <span style="color: #666666">2.5</span>)
plt<span style="color: #666666">.</span>show()
</pre>
</div>
</div>
</div>
</div>
<div class="output_wrapper">
<div class="output">
<div class="output_area">
<div class="output_subarea output_stream output_stdout output_text">
</div>
</div>
</div>
</div>
</div>
<p>For the sinusoidal example the
period is \( \tau=2\pi/\omega \). However, higher harmonics can also
satisfy the periodicity requirement. In general, any force that
satisfies the periodicity requirement can be expressed as a sum over
harmonics,
</p>
$$
\begin{equation}
F(t)=\frac{f_0}{2}+\sum_{n>0} f_n\cos(2n\pi t/\tau)+g_n\sin(2n\pi t/\tau).
\label{_auto2}
\end{equation}
$$
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="wrapping-up-fourier-transforms">Wrapping up Fourier transforms </h2>
<p>We can write down the answer for
\( x_{pn}(t) \), by substituting \( f_n/m \) or \( g_n/m \) for \( F_0/m \). By
writing each factor \( 2n\pi t/\tau \) as \( n\omega t \), with \( \omega\equiv
2\pi/\tau \),
</p>
$$
\begin{equation}
\label{eq:fourierdef1}
F(t)=\frac{f_0}{2}+\sum_{n>0}f_n\cos(n\omega t)+g_n\sin(n\omega t).
\end{equation}
$$
<p>The solutions for \( x(t) \) then come from replacing \( \omega \) with
\( n\omega \) for each term in the particular solution,
</p>
$$
\begin{eqnarray}
x_p(t)&=&\frac{f_0}{2k}+\sum_{n>0} \alpha_n\cos(n\omega t-\delta_n)+\beta_n\sin(n\omega t-\delta_n),\\
\nonumber
\alpha_n&=&\frac{f_n/m}{\sqrt{((n\omega)^2-\omega_0^2)+4\beta^2n^2\omega^2}},\\
\nonumber
\beta_n&=&\frac{g_n/m}{\sqrt{((n\omega)^2-\omega_0^2)+4\beta^2n^2\omega^2}},\\
\nonumber
\delta_n&=&\tan^{-1}\left(\frac{2\beta n\omega}{\omega_0^2-n^2\omega^2}\right).
\end{eqnarray}
$$
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="finding-the-coefficients">Finding the Coefficients </h2>
<p>Because the forces have been applied for a long time, any non-zero
damping eliminates the homogenous parts of the solution, so one need
only consider the particular solution for each \( n \).
</p>
<p>The problem is considered solved if one can find expressions for the
coefficients \( f_n \) and \( g_n \), even though the solutions are expressed
as an infinite sum. The coefficients can be extracted from the
function \( F(t) \) by
</p>
$$
\begin{eqnarray}
\label{eq:fourierdef2}
f_n&=&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~F(t)\cos(2n\pi t/\tau),\\
\nonumber
g_n&=&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~F(t)\sin(2n\pi t/\tau).
\end{eqnarray}
$$
<p>To check the consistency of these expressions and to verify
Eq. \eqref{eq:fourierdef2}, one can insert the expansion of \( F(t) \) in
Eq. \eqref{eq:fourierdef1} into the expression for the coefficients in
Eq. \eqref{eq:fourierdef2} and see whether
</p>
$$
\begin{eqnarray}
f_n&=?&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~\left\{
\frac{f_0}{2}+\sum_{m>0}f_m\cos(m\omega t)+g_m\sin(m\omega t)
\right\}\cos(n\omega t).
\end{eqnarray}
$$
<p>Immediately, one can throw away all the terms with \( g_m \) because they
convolute an even and an odd function. The term with \( f_0/2 \)
disappears because \( \cos(n\omega t) \) is equally positive and negative
over the interval and will integrate to zero. For all the terms
\( f_m\cos(m\omega t) \) appearing in the sum, one can use angle addition
formulas to see that \( \cos(m\omega t)\cos(n\omega
t)=(1/2)(\cos[(m+n)\omega t]+\cos[(m-n)\omega t] \). This will integrate
to zero unless \( m=n \). In that case the \( m=n \) term gives
</p>
$$
\begin{equation}
\int_{-\tau/2}^{\tau/2}dt~\cos^2(m\omega t)=\frac{\tau}{2},
\label{_auto3}
\end{equation}
$$
<p>and</p>
$$
\begin{eqnarray}
f_n&=?&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~f_n/2\\
\nonumber
&=&f_n~\checkmark.
\end{eqnarray}
$$
<p>The same method can be used to check for the consistency of \( g_n \).</p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="final-words-on-fourier-transforms">Final words on Fourier Transforms </h2>
<p>The code here uses the Fourier series applied to a
square wave signal. The code here
visualizes the various approximations given by Fourier series compared
with a square wave with period \( T=0.2 \) (dimensionless time), width \( 0.1 \) and max value of the force \( F=2 \). We
see that when we increase the number of components in the Fourier
series, the Fourier series approximation gets closer and closer to the
square wave signal.
</p>
<!-- code=python (!bc pycod) typeset with pygments style "default" -->
<div class="cell border-box-sizing code_cell rendered">
<div class="input">
<div class="inner_cell">
<div class="input_area">
<div class="highlight" style="background: #f8f8f8">
<pre style="line-height: 125%;"><span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">numpy</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">np</span>
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">math</span>
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">scipy</span> <span style="color: #008000; font-weight: bold">import</span> signal
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">matplotlib.pyplot</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">plt</span>
<span style="color: #408080; font-style: italic"># number of points </span>
n <span style="color: #666666">=</span> <span style="color: #666666">500</span>
<span style="color: #408080; font-style: italic"># start and final times </span>
t0 <span style="color: #666666">=</span> <span style="color: #666666">0.0</span>
tn <span style="color: #666666">=</span> <span style="color: #666666">1.0</span>
<span style="color: #408080; font-style: italic"># Period </span>
T <span style="color: #666666">=0.2</span>
<span style="color: #408080; font-style: italic"># Max value of square signal </span>
Fmax<span style="color: #666666">=</span> <span style="color: #666666">2.0</span>
<span style="color: #408080; font-style: italic"># Width of signal </span>
Width <span style="color: #666666">=</span> <span style="color: #666666">0.1</span>
t <span style="color: #666666">=</span> np<span style="color: #666666">.</span>linspace(t0, tn, n, endpoint<span style="color: #666666">=</span><span style="color: #008000; font-weight: bold">False</span>)
SqrSignal <span style="color: #666666">=</span> np<span style="color: #666666">.</span>zeros(n)
FourierSeriesSignal <span style="color: #666666">=</span> np<span style="color: #666666">.</span>zeros(n)
SqrSignal <span style="color: #666666">=</span> <span style="color: #666666">1.0+</span>signal<span style="color: #666666">.</span>square(<span style="color: #666666">2*</span>np<span style="color: #666666">.</span>pi<span style="color: #666666">*5*</span>t<span style="color: #666666">+</span>np<span style="color: #666666">.</span>pi<span style="color: #666666">*</span>Width<span style="color: #666666">/</span>T)
a0 <span style="color: #666666">=</span> Fmax<span style="color: #666666">*</span>Width<span style="color: #666666">/</span>T
FourierSeriesSignal <span style="color: #666666">=</span> a0
Factor <span style="color: #666666">=</span> <span style="color: #666666">2.0*</span>Fmax<span style="color: #666666">/</span>np<span style="color: #666666">.</span>pi
<span style="color: #008000; font-weight: bold">for</span> i <span style="color: #AA22FF; font-weight: bold">in</span> <span style="color: #008000">range</span>(<span style="color: #666666">1</span>,<span style="color: #666666">500</span>):
FourierSeriesSignal <span style="color: #666666">+=</span> Factor<span style="color: #666666">/</span>(i)<span style="color: #666666">*</span>np<span style="color: #666666">.</span>sin(np<span style="color: #666666">.</span>pi<span style="color: #666666">*</span>i<span style="color: #666666">*</span>Width<span style="color: #666666">/</span>T)<span style="color: #666666">*</span>np<span style="color: #666666">.</span>cos(i<span style="color: #666666">*</span>t<span style="color: #666666">*2*</span>np<span style="color: #666666">.</span>pi<span style="color: #666666">/</span>T)
plt<span style="color: #666666">.</span>plot(t, SqrSignal)
plt<span style="color: #666666">.</span>plot(t, FourierSeriesSignal)
plt<span style="color: #666666">.</span>ylim(<span style="color: #666666">-0.5</span>, <span style="color: #666666">2.5</span>)
plt<span style="color: #666666">.</span>show()
</pre>
</div>
</div>
</div>
</div>
<div class="output_wrapper">
<div class="output">
<div class="output_area">
<div class="output_subarea output_stream output_stdout output_text">
</div>
</div>
</div>
</div>
</div>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="two-dimensional-objects">Two-dimensional Objects </h2>
@@ -2489,7 +2197,7 @@ samples
$$
\begin{equation}
x = g(z; \theta^{(g)})
\label{_auto4}
\label{_auto1}
\end{equation}
$$
@@ -2506,7 +2214,7 @@ value given by
$$
\begin{equation}
d(x; \theta^{(d)})
\label{_auto5}
\label{_auto2}
\end{equation}
$$
@@ -2519,7 +2227,7 @@ which a function
$$
\begin{equation}
v(\theta^{(g)}, \theta^{(d)})
\label{_auto6}
\label{_auto3}
\end{equation}
$$
@@ -2530,7 +2238,7 @@ conjugate reward
$$
\begin{equation}
-v(\theta^{(g)}, \theta^{(d)})
\label{_auto7}
\label{_auto4}
\end{equation}
$$
@@ -2565,7 +2273,7 @@ $$
\begin{equation}
g^* = \underset{g}{\mathrm{argmin}}\hspace{2pt}
\underset{d}{\mathrm{max}}v(\theta^{(g)}, \theta^{(d)})
\label{_auto8}
\label{_auto5}
\end{equation}
$$
@@ -2575,7 +2283,7 @@ $$
v(\theta^{(g)}, \theta^{(d)}) = \mathbb{E}_{x\sim p_\mathrm{data}}\log d(x)
+ \mathbb{E}_{x\sim p_\mathrm{model}}
\log (1 - d(x))
\label{_auto9}
\label{_auto6}
\end{equation}
$$
@@ -2586,7 +2294,7 @@ approximation of a partition function. In the case where
$$
\begin{equation}
\underset{d}{\mathrm{max}}v(\theta^{(g)}, \theta^{(d)})
\label{_auto10}
\label{_auto7}
\end{equation}
$$
doc/pub/week43/ipynb/ipynb-week43-src.tar.gz
BIN
View File
Binary file not shown.
doc/pub/week43/ipynb/week43.ipynb
+220 -659
View File
File diff suppressed because it is too large Load Diff
doc/src/week43/week43.do.txt
-218
View File
@@ -366,225 +366,7 @@ We rather code the convolutions in the minimal memory footprint that they requir
Does the number of floating point operations change here when we use the commutative property?
!split
===== Convolution Examples: Principle of Superposition and Periodic Forces (Fourier Transforms) =====
For problems with so-called harmonic oscillations, given by for example the following differential equation
!bt
\[
m\frac{d^2x}{dt^2}+\eta\frac{dx}{dt}+x(t)=F(t),
\]
!et
where $F(t)$ is an applied external force acting on the system (often called a driving force), one can use the theory of Fourier transformations to find the solutions of this type of equations.
If one has several driving forces, $F(t)=\sum_n F_n(t)$, one can find
the particular solution to each $F_n$, $x_{pn}(t)$, and the particular
solution for the entire driving force is then given by a series like
!bt
\begin{equation}
x_p(t)=\sum_nx_{pn}(t).
\end{equation}
!et
!split
===== Principle of Superposition =====
This is known as the principle of superposition. It only applies when
the homogenous equation is linear. If there were an anharmonic term
such as $x^3$ in the homogenous equation, then when one summed various
solutions, $x=(\sum_n x_n)^2$, one would get cross
terms. Superposition is especially useful when $F(t)$ can be written
as a sum of sinusoidal terms, because the solutions for each
sinusoidal (sine or cosine) term is analytic.
Driving forces are often periodic, even when they are not
sinusoidal. Periodicity implies that for some time $\tau$
!bt
\begin{eqnarray}
F(t+\tau)=F(t).
\end{eqnarray}
!et
One example of a non-sinusoidal periodic force is a square wave. Many
components in electric circuits are non-linear, e.g. diodes, which
makes many wave forms non-sinusoidal even when the circuits are being
driven by purely sinusoidal sources.
!split
===== Simple Code Example =====
The code here shows a typical example of such a square wave generated using the functionality included in the _scipy_ Python package. We have used a period of $\tau=0.2$.
!bc pycod
import numpy as np
import math
from scipy import signal
import matplotlib.pyplot as plt
# number of points
n = 500
# start and final times
t0 = 0.0
tn = 1.0
# Period
t = np.linspace(t0, tn, n, endpoint=False)
SqrSignal = np.zeros(n)
SqrSignal = 1.0+signal.square(2*np.pi*5*t)
plt.plot(t, SqrSignal)
plt.ylim(-0.5, 2.5)
plt.show()
!ec
For the sinusoidal example the
period is $\tau=2\pi/\omega$. However, higher harmonics can also
satisfy the periodicity requirement. In general, any force that
satisfies the periodicity requirement can be expressed as a sum over
harmonics,
!bt
\begin{equation}
F(t)=\frac{f_0}{2}+\sum_{n>0} f_n\cos(2n\pi t/\tau)+g_n\sin(2n\pi t/\tau).
\end{equation}
!et
!split
===== Wrapping up Fourier transforms =====
We can write down the answer for
$x_{pn}(t)$, by substituting $f_n/m$ or $g_n/m$ for $F_0/m$. By
writing each factor $2n\pi t/\tau$ as $n\omega t$, with $\omega\equiv
2\pi/\tau$,
!bt
\begin{equation}
label{eq:fourierdef1}
F(t)=\frac{f_0}{2}+\sum_{n>0}f_n\cos(n\omega t)+g_n\sin(n\omega t).
\end{equation}
!et
The solutions for $x(t)$ then come from replacing $\omega$ with
$n\omega$ for each term in the particular solution,
!bt
\begin{eqnarray}
x_p(t)&=&\frac{f_0}{2k}+\sum_{n>0} \alpha_n\cos(n\omega t-\delta_n)+\beta_n\sin(n\omega t-\delta_n),\\
\nonumber
\alpha_n&=&\frac{f_n/m}{\sqrt{((n\omega)^2-\omega_0^2)+4\beta^2n^2\omega^2}},\\
\nonumber
\beta_n&=&\frac{g_n/m}{\sqrt{((n\omega)^2-\omega_0^2)+4\beta^2n^2\omega^2}},\\
\nonumber
\delta_n&=&\tan^{-1}\left(\frac{2\beta n\omega}{\omega_0^2-n^2\omega^2}\right).
\end{eqnarray}
!et
!split
===== Finding the Coefficients =====
Because the forces have been applied for a long time, any non-zero
damping eliminates the homogenous parts of the solution, so one need
only consider the particular solution for each $n$.
The problem is considered solved if one can find expressions for the
coefficients $f_n$ and $g_n$, even though the solutions are expressed
as an infinite sum. The coefficients can be extracted from the
function $F(t)$ by
!bt
\begin{eqnarray}
label{eq:fourierdef2}
f_n&=&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~F(t)\cos(2n\pi t/\tau),\\
\nonumber
g_n&=&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~F(t)\sin(2n\pi t/\tau).
\end{eqnarray}
!et
To check the consistency of these expressions and to verify
Eq. (ref{eq:fourierdef2}), one can insert the expansion of $F(t)$ in
Eq. (ref{eq:fourierdef1}) into the expression for the coefficients in
Eq. (ref{eq:fourierdef2}) and see whether
!bt
\begin{eqnarray}
f_n&=?&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~\left\{
\frac{f_0}{2}+\sum_{m>0}f_m\cos(m\omega t)+g_m\sin(m\omega t)
\right\}\cos(n\omega t).
\end{eqnarray}
!et
Immediately, one can throw away all the terms with $g_m$ because they
convolute an even and an odd function. The term with $f_0/2$
disappears because $\cos(n\omega t)$ is equally positive and negative
over the interval and will integrate to zero. For all the terms
$f_m\cos(m\omega t)$ appearing in the sum, one can use angle addition
formulas to see that $\cos(m\omega t)\cos(n\omega
t)=(1/2)(\cos[(m+n)\omega t]+\cos[(m-n)\omega t]$. This will integrate
to zero unless $m=n$. In that case the $m=n$ term gives
!bt
\begin{equation}
\int_{-\tau/2}^{\tau/2}dt~\cos^2(m\omega t)=\frac{\tau}{2},
\end{equation}
!et
and
!bt
\begin{eqnarray}
f_n&=?&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~f_n/2\\
\nonumber
&=&f_n~\checkmark.
\end{eqnarray}
!et
The same method can be used to check for the consistency of $g_n$.
!split
===== Final words on Fourier Transforms =====
The code here uses the Fourier series applied to a
square wave signal. The code here
visualizes the various approximations given by Fourier series compared
with a square wave with period $T=0.2$ (dimensionless time), width $0.1$ and max value of the force $F=2$. We
see that when we increase the number of components in the Fourier
series, the Fourier series approximation gets closer and closer to the
square wave signal.
!bc pycod
import numpy as np
import math
from scipy import signal
import matplotlib.pyplot as plt
# number of points
n = 500
# start and final times
t0 = 0.0
tn = 1.0
# Period
T =0.2
# Max value of square signal
Fmax= 2.0
# Width of signal
Width = 0.1
t = np.linspace(t0, tn, n, endpoint=False)
SqrSignal = np.zeros(n)
FourierSeriesSignal = np.zeros(n)
SqrSignal = 1.0+signal.square(2*np.pi*5*t+np.pi*Width/T)
a0 = Fmax*Width/T
FourierSeriesSignal = a0
Factor = 2.0*Fmax/np.pi
for i in range(1,500):
FourierSeriesSignal += Factor/(i)*np.sin(np.pi*i*Width/T)*np.cos(i*t*2*np.pi/T)
plt.plot(t, SqrSignal)
plt.plot(t, FourierSeriesSignal)
plt.ylim(-0.5, 2.5)
plt.show()
!ec
!split