| 1 |
ahallam |
3003 |
|
| 2 |
|
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
| 3 |
|
|
% |
| 4 |
|
|
% Copyright (c) 2003-2010 by University of Queensland |
| 5 |
|
|
% Earth Systems Science Computational Center (ESSCC) |
| 6 |
|
|
% http://www.uq.edu.au/esscc |
| 7 |
|
|
% |
| 8 |
|
|
% Primary Business: Queensland, Australia |
| 9 |
|
|
% Licensed under the Open Software License version 3.0 |
| 10 |
|
|
% http://www.opensource.org/licenses/osl-3.0.php |
| 11 |
|
|
% |
| 12 |
|
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
| 13 |
|
|
|
| 14 |
|
|
|
| 15 |
|
|
|
| 16 |
|
|
The acoustic wave equation governs the propagation of pressure waves. Wave |
| 17 |
|
|
types that obey this law tend to travel in liquids or gases where shear waves |
| 18 |
ahallam |
3004 |
or longitudinal style wave motion is not possible. An obvious example is sound |
| 19 |
ahallam |
3003 |
waves. |
| 20 |
|
|
|
| 21 |
ahallam |
3004 |
The acoustic wave equation is defined as; |
| 22 |
ahallam |
3003 |
\begin{equation} |
| 23 |
|
|
\nabla ^2 p - \frac{1}{c^2} \frac{\partial ^2 p}{\partial t^2} = 0 |
| 24 |
|
|
\label{eqn:acswave} |
| 25 |
|
|
\end{equation} |
| 26 |
ahallam |
3004 |
where $p$ is the pressure, $t$ is the time and $c$ is the wave velocity. |
| 27 |
ahallam |
3003 |
|
| 28 |
ahallam |
3004 |
\section{The Laplacian in \esc} |
| 29 |
|
|
The Laplacian opperator which can be written as $\Delta$ or $\nabla^2$ is |
| 30 |
|
|
calculated via the divergence of the gradient of the object, which is in this |
| 31 |
|
|
example $p$. Thus we can write; |
| 32 |
|
|
\begin{equation} |
| 33 |
ahallam |
3029 |
\nabla^2 p = \nabla \cdot \nabla p = |
| 34 |
|
|
\sum\hackscore{i}^n |
| 35 |
|
|
\frac{\partial^2 p}{\partial x^2\hackscore{i}} |
| 36 |
ahallam |
3004 |
\label{eqn:laplacian} |
| 37 |
|
|
\end{equation} |
| 38 |
ahallam |
3232 |
For the two dimensional case in Cartesian coordinates \autoref{eqn:laplacian} |
| 39 |
ahallam |
3004 |
becomes; |
| 40 |
|
|
\begin{equation} |
| 41 |
|
|
\nabla^2 p = \frac{\partial^2 p}{\partial x^2} |
| 42 |
|
|
+ \frac{\partial^2 p}{\partial y^2} |
| 43 |
|
|
\end{equation} |
| 44 |
ahallam |
3003 |
|
| 45 |
ahallam |
3004 |
In \esc the Laplacian is calculated using the divergence representation and the |
| 46 |
|
|
intrinsic functions \textit{grad()} and \textit{trace()}. The fucntion |
| 47 |
|
|
\textit{grad{}} will return the spatial gradients of an object. |
| 48 |
|
|
For a rank 0 solution, this is of the form; |
| 49 |
|
|
\begin{equation} |
| 50 |
|
|
\nabla p = \left[ |
| 51 |
|
|
\frac{\partial p}{\partial x \hackscore{0}}, |
| 52 |
|
|
\frac{\partial p}{\partial x \hackscore{1}} |
| 53 |
|
|
\right] |
| 54 |
|
|
\label{eqn:grad} |
| 55 |
|
|
\end{equation} |
| 56 |
|
|
Larger ranked solution objects will return gradient tensors. For example, a |
| 57 |
|
|
pressure field which acts in the directions $p \hackscore{0}$ and $p |
| 58 |
|
|
\hackscore{1}$ would return; |
| 59 |
|
|
\begin{equation} |
| 60 |
|
|
\nabla p = \begin{bmatrix} |
| 61 |
|
|
\frac{\partial p \hackscore{0}}{\partial x \hackscore{0}} & |
| 62 |
|
|
\frac{\partial p \hackscore{1}}{\partial x \hackscore{0}} \\ |
| 63 |
|
|
\frac{\partial p \hackscore{0}}{\partial x \hackscore{1}} & |
| 64 |
|
|
\frac{\partial p \hackscore{1}}{\partial x \hackscore{1}} |
| 65 |
|
|
\end{bmatrix} |
| 66 |
|
|
\label{eqn:gradrank1} |
| 67 |
|
|
\end{equation} |
| 68 |
ahallam |
3003 |
|
| 69 |
ahallam |
3232 |
\autoref{eqn:grad} corresponds to the Linear PDE general form value |
| 70 |
ahallam |
3029 |
$X$. Notice however that the general form contains the term $X |
| 71 |
|
|
\hackscore{i,j}$\footnote{This is the first derivative in the $j^{th}$ |
| 72 |
|
|
direction for the $i^{th}$ component of the solution.}, |
| 73 |
ahallam |
3004 |
hence for a rank 0 object there is no need to do more than calculate the |
| 74 |
|
|
gradient and submit it to the solver. In the case of the rank 1 or greater |
| 75 |
|
|
object, it is nesscary to calculate the trace also. This is the sum of the |
| 76 |
ahallam |
3232 |
diagonal in \autoref{eqn:gradrank1}. |
| 77 |
ahallam |
3004 |
|
| 78 |
|
|
Thus when solving for equations containing the Laplacian one of two things must |
| 79 |
ahallam |
3063 |
be completed. If the object \verb!p! is less than rank 1 the gradient is |
| 80 |
ahallam |
3004 |
calculated via; |
| 81 |
ahallam |
3025 |
\begin{python} |
| 82 |
ahallam |
3063 |
gradient=grad(p) |
| 83 |
ahallam |
3025 |
\end{python} |
| 84 |
ahallam |
3004 |
and if the object is greater thank or equal to a rank 1 tensor, the trace of |
| 85 |
|
|
the gradient is calculated. |
| 86 |
ahallam |
3025 |
\begin{python} |
| 87 |
ahallam |
3004 |
gradient=trace(grad(p)) |
| 88 |
ahallam |
3025 |
\end{python} |
| 89 |
ahallam |
3004 |
These valuse can then be submitted to the PDE solver via the general form term |
| 90 |
|
|
$X$. The Laplacian is then computed in the solution process by taking the |
| 91 |
|
|
divergence of $X$. |
| 92 |
|
|
|
| 93 |
ahallam |
3025 |
Note, if you are unsure about the rank of your tensor, the \textit{getRank} |
| 94 |
|
|
command will return the rank of the PDE object. |
| 95 |
|
|
\begin{python} |
| 96 |
|
|
rank = p.getRank() |
| 97 |
|
|
\end{python} |
| 98 |
|
|
|
| 99 |
|
|
|
| 100 |
|
|
\section{Numerical Solution Stability} \label{sec:nsstab} |
| 101 |
ahallam |
3004 |
Unfortunately, the wave equation belongs to a class of equations called |
| 102 |
|
|
\textbf{stiff} PDEs. These types of equations can be difficult to solve |
| 103 |
ahallam |
3025 |
numerically as they tend to oscilate about the exact solution which can |
| 104 |
|
|
eventually lead to a catastrophic failure in the solution. To counter this |
| 105 |
|
|
problem, explicitly stable schemes like |
| 106 |
ahallam |
3004 |
the backwards Euler method are required. There are two variables which must be |
| 107 |
|
|
considered for stability when numerically trying to solve the wave equation. |
| 108 |
ahallam |
3025 |
For linear media, the two variables are related via; |
| 109 |
ahallam |
3004 |
\begin{equation} \label{eqn:freqvel} |
| 110 |
|
|
f=\frac{v}{\lambda} |
| 111 |
|
|
\end{equation} |
| 112 |
ahallam |
3025 |
The velocity $v$ that a wave travels in a medium is an important variable. For |
| 113 |
|
|
stability the analytical wave must not propagate faster than the numerical wave |
| 114 |
|
|
is able to, and in general, needs to be much slower than the numerical wave. |
| 115 |
ahallam |
3003 |
For example, a line 100m long is discretised into 1m intervals or 101 nodes. If |
| 116 |
|
|
a wave enters with a propagation velocity of 100m/s then the travel time for |
| 117 |
|
|
the wave between each node will be 0.01 seconds. The time step, must therefore |
| 118 |
|
|
be significantly less than this. Of the order $10E-4$ would be appropriate. |
| 119 |
|
|
|
| 120 |
ahallam |
3004 |
The wave frequency content also plays a part in numerical stability. The |
| 121 |
|
|
nyquist-sampling theorem states that a signals bandwidth content will be |
| 122 |
|
|
accurately represented when an equispaced sampling rate $f \hackscore{n}$ is |
| 123 |
|
|
equal to or greater than twice the maximum frequency of the signal |
| 124 |
|
|
$f\hackscore{s}$, or; |
| 125 |
|
|
\begin{equation} \label{eqn:samptheorem} |
| 126 |
|
|
f\hackscore{n} \geqslant f\hackscore{s} |
| 127 |
|
|
\end{equation} |
| 128 |
|
|
For example a 50Hz signal will require a sampling rate greater than 100Hz or |
| 129 |
|
|
one sample every 0.01 seconds. The wave equation relies on a spatial frequency, |
| 130 |
|
|
thus the sampling theorem in this case applies to the solution mesh spacing. In |
| 131 |
|
|
this way, the frequency content of the input signal directly affects the time |
| 132 |
|
|
discretisation of the problem. |
| 133 |
|
|
|
| 134 |
|
|
To accurately model the wave equation with high resolutions and velocities |
| 135 |
|
|
means that very fine spatial and time discretisation is necessary for most |
| 136 |
|
|
problems. |
| 137 |
|
|
This requirement makes the wave equation arduous to |
| 138 |
ahallam |
3003 |
solve numerically due to the large number of time iterations required in each |
| 139 |
ahallam |
3004 |
solution. Models with very high velocities and frequencies will be the worst |
| 140 |
ahallam |
3029 |
affected by this problem. |
| 141 |
ahallam |
3003 |
|
| 142 |
|
|
\section{Displacement Solution} |
| 143 |
|
|
\sslist{example07a.py} |
| 144 |
|
|
|
| 145 |
|
|
We begin the solution to this PDE with the centred difference formula for the |
| 146 |
|
|
second derivative; |
| 147 |
|
|
\begin{equation} |
| 148 |
|
|
f''(x) \approx \frac{f(x+h - 2f(x) + f(x-h)}{h^2} |
| 149 |
|
|
\label{eqn:centdiff} |
| 150 |
|
|
\end{equation} |
| 151 |
ahallam |
3232 |
substituting \autoref{eqn:centdiff} for $\frac{\partial ^2 p }{\partial t ^2}$ |
| 152 |
|
|
in \autoref{eqn:acswave}; |
| 153 |
ahallam |
3003 |
\begin{equation} |
| 154 |
ahallam |
3004 |
\nabla ^2 p - \frac{1}{c^2h^2} \left[p\hackscore{(t+1)} - 2p\hackscore{(t)} + |
| 155 |
|
|
p\hackscore{(t-1)} \right] |
| 156 |
ahallam |
3003 |
= 0 |
| 157 |
|
|
\label{eqn:waveu} |
| 158 |
|
|
\end{equation} |
| 159 |
|
|
Rearranging for $p_{(t+1)}$; |
| 160 |
|
|
\begin{equation} |
| 161 |
ahallam |
3004 |
p\hackscore{(t+1)} = c^2 h^2 \nabla ^2 p\hackscore{(t)} +2p\hackscore{(t)} - |
| 162 |
|
|
p\hackscore{(t-1)} |
| 163 |
ahallam |
3003 |
\end{equation} |
| 164 |
|
|
this can be compared with the general form of the \modLPDE module and it |
| 165 |
ahallam |
3004 |
becomes clear that $D=1$, $X\hackscore{i,j}=-c^2 h^2 \nabla ^2 p_{(t)}$ and |
| 166 |
|
|
$Y=2p_{(t)} - p_{(t-1)}$. |
| 167 |
ahallam |
3003 |
|
| 168 |
ahallam |
3025 |
The solution script is similar to others that we have created in previous |
| 169 |
ahallam |
3004 |
chapters. The general steps are; |
| 170 |
|
|
\begin{enumerate} |
| 171 |
|
|
\item The necessary libraries must be imported. |
| 172 |
|
|
\item The domain needs to be defined. |
| 173 |
|
|
\item The time iteration and control parameters need to be defined. |
| 174 |
|
|
\item The PDE is initialised with source and boundary conditions. |
| 175 |
|
|
\item The time loop is started and the PDE is solved at consecutive time steps. |
| 176 |
|
|
\item All or select solutions are saved to file for visualisation lated on. |
| 177 |
|
|
\end{enumerate} |
| 178 |
|
|
|
| 179 |
|
|
Parts of the script which warrant more attention are the definition of the |
| 180 |
|
|
source, visualising the source, the solution time loop and the VTK data export. |
| 181 |
|
|
|
| 182 |
|
|
\subsection{Pressure Sources} |
| 183 |
|
|
As the pressure is a scalar, one need only define the pressure for two |
| 184 |
|
|
time steps prior to the start of the solution loop. Two known solutions are |
| 185 |
|
|
required because the wave equation contains a double partial derivative with |
| 186 |
|
|
respect to time. This is often a good opportunity to introduce a source to the |
| 187 |
|
|
solution. This model has the source located at it's centre. The source should |
| 188 |
|
|
be smooth and cover a number of samples to satisfy the frequency stability |
| 189 |
|
|
criterion. Small sources will generate high frequency signals. Here, the source |
| 190 |
|
|
is defined by a cosine function. |
| 191 |
ahallam |
3025 |
\begin{python} |
| 192 |
ahallam |
3004 |
U0=0.01 # amplitude of point source |
| 193 |
|
|
xc=[500,500] #location of point source |
| 194 |
|
|
# define small radius around point xc |
| 195 |
|
|
src_radius = 30 |
| 196 |
|
|
# for first two time steps |
| 197 |
ahallam |
3025 |
u=U0*(cos(length(x-xc)*3.1415/src_radius)+1)*\ |
| 198 |
|
|
whereNegative(length(x-xc)-src_radius) |
| 199 |
ahallam |
3004 |
u_m1=u |
| 200 |
ahallam |
3025 |
\end{python} |
| 201 |
ahallam |
3004 |
When using a rectangular domain |
| 202 |
|
|
|
| 203 |
ahallam |
3025 |
\subsection{Visualising the Source} |
| 204 |
|
|
There are two options for visualising the source. The first is to export the |
| 205 |
|
|
initial conditions of the model to VTK, which can be interpreted as a scalar |
| 206 |
|
|
suface in mayavi. The second is to take a cross section of the model. |
| 207 |
|
|
|
| 208 |
|
|
For the later, we will require the \textit{Locator} function. |
| 209 |
ahallam |
3063 |
First \verb!Locator! must be imported; |
| 210 |
ahallam |
3025 |
\begin{python} |
| 211 |
|
|
from esys.escript.pdetools import Locator |
| 212 |
|
|
\end{python} |
| 213 |
|
|
The function can then be used on the domain to locate the nearest domain node |
| 214 |
|
|
to the point or points of interest. |
| 215 |
|
|
|
| 216 |
|
|
It is now necessary to build a list of $(x,y)$ locations that specify where are |
| 217 |
|
|
model slice will go. This is easily implemeted with a loop; |
| 218 |
|
|
\begin{python} |
| 219 |
|
|
cut_loc=[] |
| 220 |
|
|
src_cut=[] |
| 221 |
|
|
for i in range(ndx/2-ndx/10,ndx/2+ndx/10): |
| 222 |
|
|
cut_loc.append(xstep*i) |
| 223 |
|
|
src_cut.append([xstep*i,xc[1]]) |
| 224 |
|
|
\end{python} |
| 225 |
ahallam |
3063 |
We then submit the output to \verb!Locator! and finally return the appropriate |
| 226 |
|
|
values using the \verb!getValue! function. |
| 227 |
ahallam |
3025 |
\begin{python} |
| 228 |
|
|
src=Locator(mydomain,src_cut) |
| 229 |
|
|
src_cut=src.getValue(u) |
| 230 |
|
|
\end{python} |
| 231 |
|
|
It is then a trivial task to plot and save the output using \mpl. |
| 232 |
|
|
\begin{python} |
| 233 |
|
|
pl.plot(cut_loc,src_cut) |
| 234 |
|
|
pl.axis([xc[0]-src_radius*3,xc[0]+src_radius*3,0.,2*U0]) |
| 235 |
|
|
pl.savefig(os.path.join(savepath,"source_line.png")) |
| 236 |
|
|
\end{python} |
| 237 |
|
|
\begin{figure}[h] |
| 238 |
ahallam |
3029 |
\centering |
| 239 |
ahallam |
3063 |
\includegraphics[width=6in]{figures/sourceline.png} |
| 240 |
ahallam |
3025 |
\caption{Cross section of the source function.} |
| 241 |
|
|
\label{fig:cxsource} |
| 242 |
|
|
\end{figure} |
| 243 |
|
|
|
| 244 |
|
|
|
| 245 |
|
|
\subsection{Point Monitoring} |
| 246 |
|
|
In the more general case where the solution mesh is irregular or specific |
| 247 |
|
|
locations need to be monitored, it is simple enough to use the \textit{Locator} |
| 248 |
|
|
function. |
| 249 |
|
|
\begin{python} |
| 250 |
|
|
rec=Locator(mydomain,[250.,250.]) |
| 251 |
|
|
\end{python} |
| 252 |
ahallam |
3029 |
When the solution \verb u is updated we can extract the value at that point |
| 253 |
ahallam |
3025 |
via; |
| 254 |
|
|
\begin{python} |
| 255 |
|
|
u_rec=rec.getValue(u) |
| 256 |
|
|
\end{python} |
| 257 |
ahallam |
3063 |
For consecutive time steps one can record the values from \verb!u_rec! in an |
| 258 |
|
|
array initialised as \verb!u_rec0=[]! with; |
| 259 |
ahallam |
3025 |
\begin{python} |
| 260 |
|
|
u_rec0.append(rec.getValue(u)) |
| 261 |
|
|
\end{python} |
| 262 |
|
|
|
| 263 |
|
|
It can be useful to monitor the value at a single or multiple individual points |
| 264 |
|
|
in the model during the modelling process. This is done using |
| 265 |
ahallam |
3063 |
the \verb!Locator! function. |
| 266 |
ahallam |
3025 |
|
| 267 |
|
|
|
| 268 |
ahallam |
3003 |
\section{Acceleration Solution} |
| 269 |
|
|
\sslist{example07b.py} |
| 270 |
|
|
|
| 271 |
|
|
An alternative method is to solve for the acceleration $\frac{\partial ^2 |
| 272 |
ahallam |
3029 |
p}{\partial t^2}$ directly, and derive the displacement solution from the |
| 273 |
ahallam |
3232 |
PDE solution. \autoref{eqn:waveu} is thus modified; |
| 274 |
ahallam |
3003 |
\begin{equation} |
| 275 |
|
|
\nabla ^2 p - \frac{1}{c^2} a = 0 |
| 276 |
|
|
\label{eqn:wavea} |
| 277 |
|
|
\end{equation} |
| 278 |
ahallam |
3004 |
and can be solved directly with $Y=0$ and $X=-c^2 \nabla ^2 p\hackscore{(t)}$. |
| 279 |
ahallam |
3003 |
After each iteration the displacement is re-evaluated via; |
| 280 |
|
|
\begin{equation} |
| 281 |
ahallam |
3004 |
p\hackscore{(t+1)}=2p\hackscore{(t)} - p\hackscore{(t-1)} + h^2a |
| 282 |
ahallam |
3003 |
\end{equation} |
| 283 |
|
|
|
| 284 |
ahallam |
3029 |
\subsection{Lumping} |
| 285 |
ahallam |
3004 |
For \esc, the acceleration solution is prefered as it allows the use of matrix |
| 286 |
|
|
lumping. Lumping or mass lumping as it is sometimes known, is the process of |
| 287 |
|
|
aggressively approximating the density elements of a mass matrix into the main |
| 288 |
|
|
diagonal. The use of Lumping is motivaed by the simplicity of diagonal matrix |
| 289 |
|
|
inversion. As a result, Lumping can significantly reduce the computational |
| 290 |
ahallam |
3029 |
requirements of a problem. Care should be taken however, as this |
| 291 |
|
|
function can only be used when the $A$, $B$ and $C$ coefficients of the |
| 292 |
|
|
general form are zero. |
| 293 |
ahallam |
3003 |
|
| 294 |
ahallam |
3004 |
To turn lumping on in \esc one can use the command; |
| 295 |
ahallam |
3025 |
\begin{python} |
| 296 |
ahallam |
3004 |
mypde.getSolverOptions().setSolverMethod(mypde.getSolverOptions().LUMPING) |
| 297 |
ahallam |
3025 |
\end{python} |
| 298 |
ahallam |
3004 |
It is also possible to check if lumping is set using; |
| 299 |
ahallam |
3025 |
\begin{python} |
| 300 |
ahallam |
3004 |
print mypde.isUsingLumping() |
| 301 |
ahallam |
3025 |
\end{python} |
| 302 |
ahallam |
3004 |
|
| 303 |
|
|
\section{Stability Investigation} |
| 304 |
|
|
It is now prudent to investigate the stability limitations of this problem. |
| 305 |
ahallam |
3025 |
First, we let the frequency content of the source be very small. If we define |
| 306 |
|
|
the source as a cosine input, than the wavlength of the input is equal to the |
| 307 |
|
|
radius of the source. Let this value be 5 meters. Now, if the maximum velocity |
| 308 |
|
|
of the model is $c=380.0ms^{-1}$ then the source |
| 309 |
|
|
frequency is $f\hackscore{r} = \frac{380.0}{5} = 76.0 Hz$. This is a worst case |
| 310 |
|
|
scenario with a small source and the models maximum velocity. |
| 311 |
|
|
|
| 312 |
ahallam |
3232 |
Furthermore, we know from \autoref{sec:nsstab}, that the spatial sampling |
| 313 |
ahallam |
3025 |
frequency must be at least twice this value to ensure stability. If we assume |
| 314 |
|
|
the model mesh is a square equispaced grid, |
| 315 |
|
|
then the sampling interval is the side length divided by the number of samples, |
| 316 |
|
|
given by $\Delta x = \frac{1000.0m}{400} = 2.5m$ and the maximum sampling |
| 317 |
|
|
frequency capable at this interval is |
| 318 |
|
|
$f\hackscore{s}=\frac{380.0ms^{-1}}{2.5m}=152Hz$ this is just equal to the |
| 319 |
ahallam |
3232 |
required rate satisfying \autoref{eqn:samptheorem}. |
| 320 |
ahallam |
3004 |
|
| 321 |
ahallam |
3232 |
\autoref{fig:ex07sampth} depicts three examples where the grid has been |
| 322 |
ahallam |
3025 |
undersampled, sampled correctly, and over sampled. The grids used had |
| 323 |
|
|
200, 400 and 800 nodes per side respectively. Obviously, the oversampled grid |
| 324 |
|
|
retains the best resolution of the modelled wave. |
| 325 |
ahallam |
3004 |
|
| 326 |
ahallam |
3025 |
The time step required for each of these examples is simply calculated from |
| 327 |
|
|
the propagation requirement. For a maximum velocity of $380.0ms^{-1}$, |
| 328 |
|
|
\begin{subequations} |
| 329 |
|
|
\begin{equation} |
| 330 |
|
|
\Delta t \leq \frac{1000.0m}{200} \frac{1}{380.0} = 0.013s |
| 331 |
|
|
\end{equation} |
| 332 |
|
|
\begin{equation} |
| 333 |
|
|
\Delta t \leq \frac{1000.0m}{400} \frac{1}{380.0} = 0.0065s |
| 334 |
|
|
\end{equation} |
| 335 |
|
|
\begin{equation} |
| 336 |
|
|
\Delta t \leq \frac{1000.0m}{800} \frac{1}{380.0} = 0.0032s |
| 337 |
|
|
\end{equation} |
| 338 |
|
|
\end{subequations} |
| 339 |
|
|
We can see, that for each doubling of the number of nodes in the mesh, we halve |
| 340 |
|
|
the timestep. To illustrate the impact this has, consider our model. If the |
| 341 |
|
|
source is placed at the center, it is $500m$ from the nearest boundary. With a |
| 342 |
|
|
velocity of $380.0ms^{-1}$ it will take $\approx1.3s$ for the wavefront to |
| 343 |
|
|
reach that boundary. In each case, this equates to $100$, $200$ and $400$ time |
| 344 |
|
|
steps. This is again, only a best case scenario, for true stability these time |
| 345 |
|
|
values may need to be halved and possibly havled again. |
| 346 |
ahallam |
3004 |
|
| 347 |
ahallam |
3025 |
\begin{figure}[ht] |
| 348 |
|
|
\centering |
| 349 |
|
|
\subfigure[Undersampled Example]{ |
| 350 |
ahallam |
3054 |
\includegraphics[width=3in]{figures/ex07usamp.png} |
| 351 |
ahallam |
3025 |
\label{fig:ex07usamp} |
| 352 |
|
|
} |
| 353 |
|
|
\subfigure[Just sampled Example]{ |
| 354 |
ahallam |
3054 |
\includegraphics[width=3in]{figures/ex07jsamp.png} |
| 355 |
ahallam |
3025 |
\label{fig:ex07jsamp} |
| 356 |
|
|
} |
| 357 |
|
|
\subfigure[Over sampled Example]{ |
| 358 |
ahallam |
3054 |
\includegraphics[width=3in]{figures/ex07nsamp.png} |
| 359 |
ahallam |
3025 |
\label{fig:ex07nsamp} |
| 360 |
|
|
} |
| 361 |
|
|
\label{fig:ex07sampth} |
| 362 |
|
|
\caption{Sampling Theorem example for stability |
| 363 |
|
|
investigation.} |
| 364 |
|
|
\end{figure} |
| 365 |
ahallam |
3004 |
|
| 366 |
ahallam |
3025 |
|
| 367 |
|
|
|
| 368 |
|
|
|
| 369 |
|
|
|
Parent Directory
| 
Revision Log