The following animation shows how the temperature changes on the bar with time (considering only the first 100 terms for the Fourier series for the square wave).
Let’s solve the below diffusion PDE with the given Neumann BCs.
As can be seen from above, the initial condition can be represented as a 2periodic triangle wave function (using even periodic extension), i.e.,
The next animation shows how the different points on the tube arrive at the steadystate solution over time.
The next figure shows the time taken for each point inside the tube, to approach within 1% the steadystate solution.
with the following
The following R code implements the numerical method:
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#Solve heat equation using forward time, centered space scheme #Define simulation parameters x = seq (0,5,5/19) #Spatial grid dt = 0.06 #Time step tMax = 20 #Simulation time nu = 0.5 #Constant of proportionality t = seq (0, tMax, dt) #Time vector n = length (x) m = length (t) fLeft = function (t) 0 #Left boundary condition fRight = function (t) sin (2* pi *t/5) #Right boundary condition fInitial = function (x) matrix ( rep (0,m*n), nrow=m) #Initial condition #Run simulation dx = x[2]x[1] r = nu*dt/dx^2 #Create tridiagonal matrix A with entries 12r and r 
The tridiagonal matrix A is shown in the following figure
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#Impose inital conditions u = fInitial (x) u[1,1] = fLeft (0) u[1,n] = fRight (0) for (i in 1:( length (t)1)) { u[i+1,] = A%*%(u[i,]) # Find solution at next time step u[i+1,1] = fLeft (t[i]) # Impose left B.C. u[i+1,n] = fRight (t[i]) # Impose right B.C. } 
The following animation shows the solution obtained to the heat equation using the numerical method (in R) described above,
The next figure shows how can a numerical method be used to solve the wave PDE
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#Define simulation parameters x = seq (0,1,1/99) #Spatial grid dt = 0.5 #Time step tMax = 200 #Simulation time c = 0.015 #Wave speed fLeft &amp;amp;lt; function (t) 0 #Left boundary condition fRight &amp;amp;lt; function (t) 0.1* sin (t/10) #Right boundary condition fPosInitial &amp;amp;lt; function (x) exp (200*(x0.5)^2) #Initial position fVelInitial &amp;amp;lt; function (x) 0*x #Initial velocity #Run simulation dx = x[2]x[1] r = c*dt/dx n &amp;amp;lt; length (x) t &amp;amp;lt; seq (0,tMax,dt) #Time vector m &amp;amp;lt; length (t) + 1 
Now iteratively compute u(x,t) by imposing the following boundary conditions
1. u(0,t) = 0
2. u(L,t) = (1/10).sin(t/10)
along with the following initial conditions
1. u(x,0) = exp(500.(x1/2)^{2})
2. ∂u(x,0)/∂t = 0.x
as defined in the above code snippet.
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#Create tridiagonal matrix A here, as described in the algorithm #Impose initial condition u &amp;amp;lt; matrix ( rep (0, n*m), nrow=m) u[1,] = fPosInitial (x) u[1,1] = fLeft (0) u[1,n] = fRight (0) for (i in 1: length (t)) { #Find solution at next time step if (i == 1) { u[2,] = 1/2*(A%*%u[1,]) + dt* fVelInitial (x) } else { u[i+1,] = (A%*%u[i,])u[i1,] } u[i+1,1] = fLeft (t[i]) #Impose left B.C. u[i+1,n] = fRight (t[i]) #Impose right B.C. } 
The following animation shows the output of the above implementation of the solution of wave PDE using R, it shows how the waves propagate, given a set of BCs and ICs.
Let’s denoise an input noisy audio file (part of the theme music from Satyajit Ray’s famous movie পথের পাঁচালী, Songs of the road) using python scipy.fftpack module’s fft() implementation.
The noisy input file was generated and uploaded here.
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from scipy.io import wavfile import scipy.fftpack as fp import numpy as np import matplotlib.pylab as plt # first fft to see the pattern then we can edit to see signal and inverse # back to a better sound Fs, y = wavfile.read( 'pather_panchali_noisy.wav' ) # can be found here: "https://github.com/sandipan/edx/blob/courses/pather_panchali_noisy.wav" datamcehref="https://github.com/sandipan/edx/blob/courses/pather_panchali_noisy.wav". # you may want to scale y, in this case, y was already scaled in between [1,1] Y = fp.fftshift(fp.fft(y)) #Take the Fourier series and take a symmetric shift fshift = np.arange(  n / / 2 , n / / 2 ) * (Fs / n) plt.plot(fshift, Y.real) #plot(t, y); #plot the sound signal plt.show() 
You will obtain a figure like the following:
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L = len (fshift) #Find the length of frequency values Yfilt = Y # find the frequencies corresponding to the noise here print (np.argsort(Y.real)[::  1 ][: 2 ]) # 495296 639296 Yfilt[ 495296 ] = Yfilt[ 639296 ] = 0 # find the frequencies corresponding to the spikes plt.plot(fshift, Yfilt.real) plt.show() soundFiltered = fp.ifft(fp.ifftshift(Yfilt)).real wavfile.write( 'pather_panchali_denoised.wav' , Fs, soundNoisy) 
Posted 10 May 2021
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