update week 43
This commit is contained in:
Morten Hjorth-Jensen
@@ -366,225 +366,7 @@ We rather code the convolutions in the minimal memory footprint that they requir
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Does the number of floating point operations change here when we use the commutative property?
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!split
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===== Convolution Examples: Principle of Superposition and Periodic Forces (Fourier Transforms) =====
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For problems with so-called harmonic oscillations, given by for example the following differential equation
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!bt
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\[
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m\frac{d^2x}{dt^2}+\eta\frac{dx}{dt}+x(t)=F(t),
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\]
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!et
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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.
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If one has several driving forces, $F(t)=\sum_n F_n(t)$, one can find
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the particular solution to each $F_n$, $x_{pn}(t)$, and the particular
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solution for the entire driving force is then given by a series like
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!bt
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\begin{equation}
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x_p(t)=\sum_nx_{pn}(t).
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\end{equation}
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!et
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!split
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===== Principle of Superposition =====
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This is known as the principle of superposition. It only applies when
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the homogenous equation is linear. If there were an anharmonic term
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such as $x^3$ in the homogenous equation, then when one summed various
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solutions, $x=(\sum_n x_n)^2$, one would get cross
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terms. Superposition is especially useful when $F(t)$ can be written
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as a sum of sinusoidal terms, because the solutions for each
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sinusoidal (sine or cosine) term is analytic.
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Driving forces are often periodic, even when they are not
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sinusoidal. Periodicity implies that for some time $\tau$
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!bt
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\begin{eqnarray}
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F(t+\tau)=F(t).
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\end{eqnarray}
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!et
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One example of a non-sinusoidal periodic force is a square wave. Many
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components in electric circuits are non-linear, e.g. diodes, which
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makes many wave forms non-sinusoidal even when the circuits are being
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driven by purely sinusoidal sources.
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!split
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===== Simple Code Example =====
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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 $\tau=0.2$.
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!bc pycod
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import numpy as np
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import math
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from scipy import signal
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import matplotlib.pyplot as plt
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# number of points
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n = 500
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# start and final times
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t0 = 0.0
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tn = 1.0
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# Period
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t = np.linspace(t0, tn, n, endpoint=False)
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SqrSignal = np.zeros(n)
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SqrSignal = 1.0+signal.square(2*np.pi*5*t)
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plt.plot(t, SqrSignal)
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plt.ylim(-0.5, 2.5)
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plt.show()
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!ec
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For the sinusoidal example the
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period is $\tau=2\pi/\omega$. However, higher harmonics can also
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satisfy the periodicity requirement. In general, any force that
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satisfies the periodicity requirement can be expressed as a sum over
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harmonics,
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!bt
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\begin{equation}
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F(t)=\frac{f_0}{2}+\sum_{n>0} f_n\cos(2n\pi t/\tau)+g_n\sin(2n\pi t/\tau).
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\end{equation}
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!et
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!split
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===== Wrapping up Fourier transforms =====
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We can write down the answer for
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$x_{pn}(t)$, by substituting $f_n/m$ or $g_n/m$ for $F_0/m$. By
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writing each factor $2n\pi t/\tau$ as $n\omega t$, with $\omega\equiv
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2\pi/\tau$,
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!bt
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\begin{equation}
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label{eq:fourierdef1}
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F(t)=\frac{f_0}{2}+\sum_{n>0}f_n\cos(n\omega t)+g_n\sin(n\omega t).
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\end{equation}
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!et
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The solutions for $x(t)$ then come from replacing $\omega$ with
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$n\omega$ for each term in the particular solution,
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!bt
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\begin{eqnarray}
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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),\\
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\nonumber
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\alpha_n&=&\frac{f_n/m}{\sqrt{((n\omega)^2-\omega_0^2)+4\beta^2n^2\omega^2}},\\
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\nonumber
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\beta_n&=&\frac{g_n/m}{\sqrt{((n\omega)^2-\omega_0^2)+4\beta^2n^2\omega^2}},\\
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\nonumber
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\delta_n&=&\tan^{-1}\left(\frac{2\beta n\omega}{\omega_0^2-n^2\omega^2}\right).
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\end{eqnarray}
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!et
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!split
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===== Finding the Coefficients =====
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Because the forces have been applied for a long time, any non-zero
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damping eliminates the homogenous parts of the solution, so one need
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only consider the particular solution for each $n$.
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The problem is considered solved if one can find expressions for the
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coefficients $f_n$ and $g_n$, even though the solutions are expressed
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as an infinite sum. The coefficients can be extracted from the
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function $F(t)$ by
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!bt
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\begin{eqnarray}
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label{eq:fourierdef2}
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f_n&=&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~F(t)\cos(2n\pi t/\tau),\\
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\nonumber
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g_n&=&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~F(t)\sin(2n\pi t/\tau).
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\end{eqnarray}
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!et
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To check the consistency of these expressions and to verify
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Eq. (ref{eq:fourierdef2}), one can insert the expansion of $F(t)$ in
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Eq. (ref{eq:fourierdef1}) into the expression for the coefficients in
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Eq. (ref{eq:fourierdef2}) and see whether
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!bt
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\begin{eqnarray}
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f_n&=?&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~\left\{
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\frac{f_0}{2}+\sum_{m>0}f_m\cos(m\omega t)+g_m\sin(m\omega t)
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\right\}\cos(n\omega t).
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\end{eqnarray}
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!et
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Immediately, one can throw away all the terms with $g_m$ because they
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convolute an even and an odd function. The term with $f_0/2$
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disappears because $\cos(n\omega t)$ is equally positive and negative
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over the interval and will integrate to zero. For all the terms
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$f_m\cos(m\omega t)$ appearing in the sum, one can use angle addition
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formulas to see that $\cos(m\omega t)\cos(n\omega
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t)=(1/2)(\cos[(m+n)\omega t]+\cos[(m-n)\omega t]$. This will integrate
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to zero unless $m=n$. In that case the $m=n$ term gives
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!bt
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\begin{equation}
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\int_{-\tau/2}^{\tau/2}dt~\cos^2(m\omega t)=\frac{\tau}{2},
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\end{equation}
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!et
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and
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!bt
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\begin{eqnarray}
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f_n&=?&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~f_n/2\\
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\nonumber
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&=&f_n~\checkmark.
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\end{eqnarray}
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!et
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The same method can be used to check for the consistency of $g_n$.
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!split
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===== Final words on Fourier Transforms =====
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The code here uses the Fourier series applied to a
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square wave signal. The code here
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visualizes the various approximations given by Fourier series compared
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with a square wave with period $T=0.2$ (dimensionless time), width $0.1$ and max value of the force $F=2$. We
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see that when we increase the number of components in the Fourier
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series, the Fourier series approximation gets closer and closer to the
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square wave signal.
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!bc pycod
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import numpy as np
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import math
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from scipy import signal
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import matplotlib.pyplot as plt
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# number of points
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n = 500
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# start and final times
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t0 = 0.0
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tn = 1.0
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# Period
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T =0.2
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# Max value of square signal
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Fmax= 2.0
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# Width of signal
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Width = 0.1
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t = np.linspace(t0, tn, n, endpoint=False)
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SqrSignal = np.zeros(n)
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FourierSeriesSignal = np.zeros(n)
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SqrSignal = 1.0+signal.square(2*np.pi*5*t+np.pi*Width/T)
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a0 = Fmax*Width/T
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FourierSeriesSignal = a0
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Factor = 2.0*Fmax/np.pi
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for i in range(1,500):
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FourierSeriesSignal += Factor/(i)*np.sin(np.pi*i*Width/T)*np.cos(i*t*2*np.pi/T)
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plt.plot(t, SqrSignal)
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plt.plot(t, FourierSeriesSignal)
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plt.ylim(-0.5, 2.5)
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plt.show()
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!ec
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!split
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