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organizing toctree in EM docs
This commit is contained in:
Lindsey Heagy
@@ -121,15 +121,15 @@ For the two formulations, the discretization of the physical properties, fields
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Note that resistivity is the inverse of conductivity, \\(\\rho = \\sigma^{-1}\\).
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E-B Formulation:
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****************
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E-B Formulation
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---------------
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.. math ::
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\mathbf{C} \mathbf{e} + i \omega \mathbf{b} = \mathbf{s_m} \\
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\mathbf{C^T} \mathbf{M^f_{\mu^{-1}}} \mathbf{b} - \mathbf{M^e_\sigma} \mathbf{e} = \mathbf{M^e} \mathbf{s_e}
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H-J Formulation:
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****************
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H-J Formulation
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---------------
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.. math ::
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\mathbf{C^T} \mathbf{M^f_\rho} \mathbf{j} + i \omega \mathbf{M^e_\mu} \mathbf{h} = \mathbf{M^e} \mathbf{s_m} \\
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@@ -48,6 +48,305 @@
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\newcommand{\I}{\vec{I}}
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Time Domain Electromagnetics
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****************************
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.. _api_TDEM_derivation:
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Time-Domain EM Derivation
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=========================
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The following shows the derivation for the TDEM problem. We use the b-formulation below.
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(More to come soon..!)
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Sensitivity Calculation
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-----------------------
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.. math::
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\begin{align}
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\dcurl \e^{(t+1)} + \frac{\b^{(t+1)} - \b^{(t)}}{\delta t} = 0 \\
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\dcurl^\top \MfMui \b^{(t+1)} - \MeSig \e^{(t+1)} = \Me \j_s^{(t+1)}
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\end{align}
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Using Gauss-Newton to solve the inverse problem requires the ability to calculate the product of the
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Jacobian and a vector, as well as the transpose of the Jacobian times a vector.
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The above system can be rewritten as:
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.. math::
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\begin{align}
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\mathbf{A} \u^{(t+1)} + \mathbf{B} \u^{(t)}= \s^{(t+1)}
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\end{align}
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where
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.. math::
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\begin{align}
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\mathbf{A} =
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\left[
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\begin{array}{cc}
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\frac{1}{\delta t} \MfMui & \MfMui\dcurl \\
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\dcurl^\top \MfMui & -\MeSig
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\end{array}
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\right] \\
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\mathbf{B} =
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\left[
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\begin{array}{cc}
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-\frac{1}{\delta t} \MfMui & 0 \\
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0 & 0
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\end{array}
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\right] \\
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\u^{(k)} = \left[
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\begin{array}{c}
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\b^{(k)}\\
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\e^{(k)}
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\end{array}
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\right] \\
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\s^{(k)} = \left[
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\begin{array}{c}
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0\\
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\Me \j^{(k)}_s
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\end{array}
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\right]
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\end{align}
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.. note::
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Here we have multiplied through by \\(\\MfMui\\) to make A and B symmetric!
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The entire time dependent system can be written in a single matrix expression
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.. math::
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\begin{align}
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\hat{\mathbf{A}} \hat{u} = \hat{s}
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\end{align}
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where
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.. math::
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\begin{align}
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\mathbf{\hat{A}} = \left[
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\begin{array}{cccc}
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A & 0 & & \\
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B & A & & \\
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& \ddots & \ddots & \\
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& & B & A
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\end{array}
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\right] \\
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\hat{u} = \left[
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\begin{array}{c}
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\u^{(1)} \\
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\u^{(2)} \\
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\vdots \\
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\u^{(N)}
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\end{array} \right]\\
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\hat{s} = \left[
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\begin{array}{c}
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\s^{(1)} - \mathbf{B} \u^{(0)} \\
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\s^{(2)} \\
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\vdots \\
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\s^{(N)}
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\end{array}
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\right]
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\end{align}
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For the fields \\(\\u\\), the measured data is given by
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.. math::
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\begin{align}
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\vec{d} = \mathbf{Q} \u
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\end{align}
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The sensitivity matrix **J** is then defined as
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.. math::
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\begin{align}
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\mathbf{J} = \mathbf{Q} \frac{\partial \u}{\partial \sigma}
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\end{align}
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Defining the function \\(\\c(m,\\u)\\) to be
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.. math::
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\begin{align}
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\vec{c}(m,\u) = \hat{\mathbf{A}} \vec{u} - \vec{q} = \vec{0}
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\end{align}
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then
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.. math::
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\begin{align}
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\frac{\partial \vec{c}}{\partial m} \partial m
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+ \frac{\partial \vec{c}}{\partial \u} \partial \vec{u} = 0
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\end{align}
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or
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.. math::
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\begin{align}
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\frac{\partial \vec{u}}{\partial m} = -\left(\frac{\partial \vec{c}}{\partial \u} \right)^{-1} \frac{\partial \vec{c}}{\partial m}
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\end{align}
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Differentiating, we find that
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.. math::
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\begin{align}
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\frac{\partial \vec{c}}{\partial \hat{u}} = \hat{\mathbf{A}}
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\end{align}
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and
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.. math::
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\begin{align}
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\frac{\partial \vec{c}}{\partial \sigma} = \mathbf{G}_\sigma =
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\left[
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\begin{array}{c}
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g_\sigma^{(1)}\\
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g_\sigma^{(2)}\\
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\vdots \\
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g_\sigma^{(N)}
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\end{array}
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\right]
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\end{align}
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with
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.. math::
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\begin{align}
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g_\sigma^{(n)} =
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\left[
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\begin{array}{c}
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\mathbf{0} \\
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- \diag{\e^{(n)}} \Ace \diag{\vec{V}}
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\end{array}
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\right]
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\end{align}
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Implementing **J** times a vector
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---------------------------------
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Multiplying **J** onto a vector can be broken into three steps
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* Compute \\(\\vec{p} = \\mathbf{G}m\\)
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* Solve \\(\\hat{\\mathbf{A}} \\vec{y} = \\vec{p}\\)
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* Compute \\(\\vec{w} = -\\mathbf{Q} \\vec{y}\\)
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.. math::
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\begin{align}
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\vec{p}^{(n)} = \left[
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\begin{array}{c}
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\vec{p}_b^{(n)} \\
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\vec{p}_e^{(n)}
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\end{array}
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\right] \\
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\vec{p}_b^{(n)} = 0 \\
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\vec{p}_e^{(n)} = - \diag{\e^{(n)}} \Ace \diag{V} m
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\end{align}
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For all time steps:
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.. math::
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\begin{align}
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\frac{1}{\delta t} \MfMui\vec{y}_{b}^{(t+1)} + \MfMui\dcurl \vec{y}_{e}^{(t+1)}
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- \frac{1}{\delta t} \MfMui \vec{y}_{b}^{(t)}
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= \vec{p}_b^{(t+1)} \\
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\dcurl^\top \MfMui \vec{y}_b^{(t+1)} - \MeSig \vec{y}_e^{(t+1)} = \vec{p}_e^{(t+1)}
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\end{align}
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and
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.. math::
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\begin{align}
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\left( \MfMui \dcurl \MeSig^{-1} \dcurl^\top \MfMui + \frac{1}{\delta t} \MfMui \right) \vec{y}_{b}^{(t+1)} =
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\frac{1}{\delta t} \MfMui \vec{y}_b^{(t)}
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+ \MfMui \dcurl \MeSig^{-1} \vec{p}_e^{(t+1)} + \vec{p}_b^{(t+1)} \\
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\vec{y}_e^{(t+1)} = \MeSig^{-1} \dcurl^\top \MfMui \vec{y}_b^{(t+1)} - \MeSig^{-1} \vec{p}_e^{(t+1)}
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\end{align}
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.. note::
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For the first time step, \\\(t=0\\\), the term: \\\(\\frac{1}{\\delta t} \\MfMui \\vec{y}_b^{(0)}\\\) is zero.
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Implementing **J** transpose times a vector
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-------------------------------------------
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Multiplying \\(\\mathbf{J}^\\top\\) onto a vector can be broken into three steps
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* Compute \\(\\vec{p} = \\mathbf{Q}^\\top \\vec{v}\\)
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* Solve \\(\\hat{\\mathbf{A}}^\\top \\vec{y} = \\vec{p}\\)
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* Compute \\(\\vec{w} = -\\mathbf{G}^\\top y\\)
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.. math::
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\mathbf{\hat{A}}^\top = \left[
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\begin{array}{cccc}
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A & B & & \\
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& \ddots & \ddots & \\
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& & A & B \\
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& & 0 & A
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\end{array}
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\right]
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For the all time-steps (going backwards in time):
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.. math::
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A \vec{y}^{(t)} + B \vec{y}^{(t+1)} = \vec{p}^{(t)}
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.. math::
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\begin{align}
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\frac{1}{\delta t} \MfMui\vec{y}_{b}^{(t)} + \MfMui\dcurl \vec{y}_{e}^{(t)}
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- \frac{1}{\delta t} \MfMui \vec{y}_{b}^{(t+1)}
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= \vec{p}_b^{(t)} \\
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\dcurl^\top \MfMui \vec{y}_b^{(t)} - \MeSig \vec{y}_e^{(t)} = \vec{p}_e^{(t)}
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\end{align}
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and
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.. math::
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\begin{align}
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\left( \MfMui \dcurl \MeSig^{-1} \dcurl^\top \MfMui + \frac{1}{\delta t} \MfMui \right) \vec{y}_{b}^{(t)} =
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\frac{1}{\delta t} \MfMui \vec{y}_b^{(t+1)}
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+ \MfMui \dcurl \MeSig^{-1} \vec{p}_e^{(t)} + \vec{p}_b^{(t)} \\
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\vec{y}_e^{(t)} = \MeSig^{-1} \dcurl^\top \MfMui \vec{y}_b^{(t)} - \MeSig^{-1} \vec{p}_e^{(t)}
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\end{align}
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.. note::
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For the last time step, \\\(t=N\\\), the term: \\\(\\frac{1}{\\delta t} \\MfMui \\vec{y}_b^{(N+1)}\\\) is zero.
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TDEM - B formulation
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====================
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@@ -1,341 +0,0 @@
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.. _api_TDEM_derivation:
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.. math::
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\renewcommand{\div}{\nabla\cdot\,}
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\newcommand{\grad}{\vec \nabla}
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\newcommand{\curl}{{\vec \nabla}\times\,}
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\newcommand {\J}{{\vec J}}
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\renewcommand{\H}{{\vec H}}
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\newcommand {\E}{{\vec E}}
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\newcommand{\dcurl}{{\mathbf C}}
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\newcommand{\dgrad}{{\mathbf G}}
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\newcommand{\Acf}{{\mathbf A_c^f}}
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\newcommand{\Ace}{{\mathbf A_c^e}}
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\renewcommand{\S}{{\mathbf \Sigma}}
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\newcommand{\St}{{\mathbf \Sigma_\tau}}
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\newcommand{\T}{{\mathbf T}}
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\newcommand{\Tt}{{\mathbf T_\tau}}
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\newcommand{\diag}[1]{\,{\sf diag}\left( #1 \right)}
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\newcommand{\M}{{\mathbf M}}
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\newcommand{\MfMui}{{\M^f_{\mu^{-1}}}}
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\newcommand{\MeSig}{{\M^e_\sigma}}
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\newcommand{\MeSigInf}{{\M^e_{\sigma_\infty}}}
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\newcommand{\MeSigO}{{\M^e_{\sigma_0}}}
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\newcommand{\Me}{{\M^e}}
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\newcommand{\Mes}[1]{{\M^e_{#1}}}
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\newcommand{\Mee}{{\M^e_e}}
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\newcommand{\Mej}{{\M^e_j}}
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\newcommand{\BigO}[1]{\mathcal{O}\bigl(#1\bigr)}
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\newcommand{\bE}{\mathbf{E}}
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\newcommand{\bH}{\mathbf{H}}
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\newcommand{\B}{\vec{B}}
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\newcommand{\D}{\vec{D}}
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\renewcommand{\H}{\vec{H}}
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\newcommand{\s}{\vec{s}}
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\newcommand{\bfJ}{\bf{J}}
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\newcommand{\vecm}{\vec m}
|
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\renewcommand{\Re}{\mathsf{Re}}
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\renewcommand{\Im}{\mathsf{Im}}
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\renewcommand {\j} { {\vec j} }
|
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\newcommand {\h} { {\vec h} }
|
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\renewcommand {\b} { {\vec b} }
|
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\newcommand {\e} { {\vec e} }
|
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\newcommand {\c} { {\vec c} }
|
||||
\renewcommand {\d} { {\vec d} }
|
||||
\renewcommand {\u} { {\vec u} }
|
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\newcommand{\I}{\vec{I}}
|
||||
|
||||
|
||||
Time-Domain EM Derivation
|
||||
*************************
|
||||
|
||||
The following shows the derivation for the TDEM problem. We use the b-formulation below.
|
||||
(More to come soon..!)
|
||||
|
||||
|
||||
Sensitivity Calculation
|
||||
=======================
|
||||
|
||||
.. math::
|
||||
|
||||
\begin{align}
|
||||
\dcurl \e^{(t+1)} + \frac{\b^{(t+1)} - \b^{(t)}}{\delta t} = 0 \\
|
||||
\dcurl^\top \MfMui \b^{(t+1)} - \MeSig \e^{(t+1)} = \Me \j_s^{(t+1)}
|
||||
\end{align}
|
||||
|
||||
Using Gauss-Newton to solve the inverse problem requires the ability to calculate the product of the
|
||||
Jacobian and a vector, as well as the transpose of the Jacobian times a vector.
|
||||
The above system can be rewritten as:
|
||||
|
||||
.. math::
|
||||
|
||||
\begin{align}
|
||||
\mathbf{A} \u^{(t+1)} + \mathbf{B} \u^{(t)}= \s^{(t+1)}
|
||||
\end{align}
|
||||
|
||||
where
|
||||
|
||||
.. math::
|
||||
|
||||
\begin{align}
|
||||
\mathbf{A} =
|
||||
\left[
|
||||
\begin{array}{cc}
|
||||
\frac{1}{\delta t} \MfMui & \MfMui\dcurl \\
|
||||
\dcurl^\top \MfMui & -\MeSig
|
||||
\end{array}
|
||||
\right] \\
|
||||
\mathbf{B} =
|
||||
\left[
|
||||
\begin{array}{cc}
|
||||
-\frac{1}{\delta t} \MfMui & 0 \\
|
||||
0 & 0
|
||||
\end{array}
|
||||
\right] \\
|
||||
\u^{(k)} = \left[
|
||||
\begin{array}{c}
|
||||
\b^{(k)}\\
|
||||
\e^{(k)}
|
||||
\end{array}
|
||||
\right] \\
|
||||
\s^{(k)} = \left[
|
||||
\begin{array}{c}
|
||||
0\\
|
||||
\Me \j^{(k)}_s
|
||||
\end{array}
|
||||
\right]
|
||||
\end{align}
|
||||
|
||||
.. note::
|
||||
|
||||
Here we have multiplied through by \\(\\MfMui\\) to make A and B symmetric!
|
||||
|
||||
The entire time dependent system can be written in a single matrix expression
|
||||
|
||||
.. math::
|
||||
|
||||
\begin{align}
|
||||
\hat{\mathbf{A}} \hat{u} = \hat{s}
|
||||
\end{align}
|
||||
|
||||
where
|
||||
|
||||
.. math::
|
||||
|
||||
\begin{align}
|
||||
\mathbf{\hat{A}} = \left[
|
||||
\begin{array}{cccc}
|
||||
A & 0 & & \\
|
||||
B & A & & \\
|
||||
& \ddots & \ddots & \\
|
||||
& & B & A
|
||||
\end{array}
|
||||
\right] \\
|
||||
\hat{u} = \left[
|
||||
\begin{array}{c}
|
||||
\u^{(1)} \\
|
||||
\u^{(2)} \\
|
||||
\vdots \\
|
||||
\u^{(N)}
|
||||
\end{array} \right]\\
|
||||
\hat{s} = \left[
|
||||
\begin{array}{c}
|
||||
\s^{(1)} - \mathbf{B} \u^{(0)} \\
|
||||
\s^{(2)} \\
|
||||
\vdots \\
|
||||
\s^{(N)}
|
||||
\end{array}
|
||||
\right]
|
||||
\end{align}
|
||||
|
||||
For the fields \\(\\u\\), the measured data is given by
|
||||
|
||||
.. math::
|
||||
|
||||
\begin{align}
|
||||
\vec{d} = \mathbf{Q} \u
|
||||
\end{align}
|
||||
|
||||
The sensitivity matrix **J** is then defined as
|
||||
|
||||
.. math::
|
||||
|
||||
\begin{align}
|
||||
\mathbf{J} = \mathbf{Q} \frac{\partial \u}{\partial \sigma}
|
||||
\end{align}
|
||||
|
||||
|
||||
Defining the function \\(\\c(m,\\u)\\) to be
|
||||
|
||||
.. math::
|
||||
|
||||
\begin{align}
|
||||
\vec{c}(m,\u) = \hat{\mathbf{A}} \vec{u} - \vec{q} = \vec{0}
|
||||
\end{align}
|
||||
|
||||
then
|
||||
|
||||
.. math::
|
||||
|
||||
\begin{align}
|
||||
\frac{\partial \vec{c}}{\partial m} \partial m
|
||||
+ \frac{\partial \vec{c}}{\partial \u} \partial \vec{u} = 0
|
||||
\end{align}
|
||||
|
||||
or
|
||||
|
||||
.. math::
|
||||
|
||||
\begin{align}
|
||||
\frac{\partial \vec{u}}{\partial m} = -\left(\frac{\partial \vec{c}}{\partial \u} \right)^{-1} \frac{\partial \vec{c}}{\partial m}
|
||||
\end{align}
|
||||
|
||||
|
||||
Differentiating, we find that
|
||||
|
||||
.. math::
|
||||
|
||||
\begin{align}
|
||||
\frac{\partial \vec{c}}{\partial \hat{u}} = \hat{\mathbf{A}}
|
||||
\end{align}
|
||||
|
||||
and
|
||||
|
||||
.. math::
|
||||
|
||||
\begin{align}
|
||||
\frac{\partial \vec{c}}{\partial \sigma} = \mathbf{G}_\sigma =
|
||||
\left[
|
||||
\begin{array}{c}
|
||||
g_\sigma^{(1)}\\
|
||||
g_\sigma^{(2)}\\
|
||||
\vdots \\
|
||||
g_\sigma^{(N)}
|
||||
\end{array}
|
||||
\right]
|
||||
\end{align}
|
||||
|
||||
with
|
||||
|
||||
.. math::
|
||||
|
||||
\begin{align}
|
||||
g_\sigma^{(n)} =
|
||||
\left[
|
||||
\begin{array}{c}
|
||||
\mathbf{0} \\
|
||||
- \diag{\e^{(n)}} \Ace \diag{\vec{V}}
|
||||
\end{array}
|
||||
\right]
|
||||
\end{align}
|
||||
|
||||
|
||||
Implementing **J** times a vector
|
||||
=================================
|
||||
|
||||
Multiplying **J** onto a vector can be broken into three steps
|
||||
|
||||
|
||||
* Compute \\(\\vec{p} = \\mathbf{G}m\\)
|
||||
* Solve \\(\\hat{\\mathbf{A}} \\vec{y} = \\vec{p}\\)
|
||||
* Compute \\(\\vec{w} = -\\mathbf{Q} \\vec{y}\\)
|
||||
|
||||
.. math::
|
||||
|
||||
\begin{align}
|
||||
\vec{p}^{(n)} = \left[
|
||||
\begin{array}{c}
|
||||
\vec{p}_b^{(n)} \\
|
||||
\vec{p}_e^{(n)}
|
||||
\end{array}
|
||||
\right] \\
|
||||
\vec{p}_b^{(n)} = 0 \\
|
||||
\vec{p}_e^{(n)} = - \diag{\e^{(n)}} \Ace \diag{V} m
|
||||
\end{align}
|
||||
|
||||
|
||||
For all time steps:
|
||||
|
||||
.. math::
|
||||
|
||||
\begin{align}
|
||||
\frac{1}{\delta t} \MfMui\vec{y}_{b}^{(t+1)} + \MfMui\dcurl \vec{y}_{e}^{(t+1)}
|
||||
- \frac{1}{\delta t} \MfMui \vec{y}_{b}^{(t)}
|
||||
= \vec{p}_b^{(t+1)} \\
|
||||
\dcurl^\top \MfMui \vec{y}_b^{(t+1)} - \MeSig \vec{y}_e^{(t+1)} = \vec{p}_e^{(t+1)}
|
||||
\end{align}
|
||||
|
||||
and
|
||||
|
||||
.. math::
|
||||
|
||||
\begin{align}
|
||||
\left( \MfMui \dcurl \MeSig^{-1} \dcurl^\top \MfMui + \frac{1}{\delta t} \MfMui \right) \vec{y}_{b}^{(t+1)} =
|
||||
\frac{1}{\delta t} \MfMui \vec{y}_b^{(t)}
|
||||
+ \MfMui \dcurl \MeSig^{-1} \vec{p}_e^{(t+1)} + \vec{p}_b^{(t+1)} \\
|
||||
\vec{y}_e^{(t+1)} = \MeSig^{-1} \dcurl^\top \MfMui \vec{y}_b^{(t+1)} - \MeSig^{-1} \vec{p}_e^{(t+1)}
|
||||
\end{align}
|
||||
|
||||
.. note::
|
||||
|
||||
For the first time step, \\\(t=0\\\), the term: \\\(\\frac{1}{\\delta t} \\MfMui \\vec{y}_b^{(0)}\\\) is zero.
|
||||
|
||||
|
||||
|
||||
|
||||
Implementing **J** transpose times a vector
|
||||
===========================================
|
||||
|
||||
Multiplying \\(\\mathbf{J}^\\top\\) onto a vector can be broken into three steps
|
||||
|
||||
|
||||
* Compute \\(\\vec{p} = \\mathbf{Q}^\\top \\vec{v}\\)
|
||||
* Solve \\(\\hat{\\mathbf{A}}^\\top \\vec{y} = \\vec{p}\\)
|
||||
* Compute \\(\\vec{w} = -\\mathbf{G}^\\top y\\)
|
||||
|
||||
|
||||
.. math::
|
||||
|
||||
\mathbf{\hat{A}}^\top = \left[
|
||||
\begin{array}{cccc}
|
||||
A & B & & \\
|
||||
& \ddots & \ddots & \\
|
||||
& & A & B \\
|
||||
& & 0 & A
|
||||
\end{array}
|
||||
\right]
|
||||
|
||||
For the all time-steps (going backwards in time):
|
||||
|
||||
|
||||
.. math::
|
||||
|
||||
A \vec{y}^{(t)} + B \vec{y}^{(t+1)} = \vec{p}^{(t)}
|
||||
|
||||
|
||||
.. math::
|
||||
|
||||
\begin{align}
|
||||
\frac{1}{\delta t} \MfMui\vec{y}_{b}^{(t)} + \MfMui\dcurl \vec{y}_{e}^{(t)}
|
||||
- \frac{1}{\delta t} \MfMui \vec{y}_{b}^{(t+1)}
|
||||
= \vec{p}_b^{(t)} \\
|
||||
\dcurl^\top \MfMui \vec{y}_b^{(t)} - \MeSig \vec{y}_e^{(t)} = \vec{p}_e^{(t)}
|
||||
\end{align}
|
||||
|
||||
and
|
||||
|
||||
.. math::
|
||||
|
||||
\begin{align}
|
||||
\left( \MfMui \dcurl \MeSig^{-1} \dcurl^\top \MfMui + \frac{1}{\delta t} \MfMui \right) \vec{y}_{b}^{(t)} =
|
||||
\frac{1}{\delta t} \MfMui \vec{y}_b^{(t+1)}
|
||||
+ \MfMui \dcurl \MeSig^{-1} \vec{p}_e^{(t)} + \vec{p}_b^{(t)} \\
|
||||
\vec{y}_e^{(t)} = \MeSig^{-1} \dcurl^\top \MfMui \vec{y}_b^{(t)} - \MeSig^{-1} \vec{p}_e^{(t)}
|
||||
\end{align}
|
||||
|
||||
|
||||
.. note::
|
||||
|
||||
For the last time step, \\\(t=N\\\), the term: \\\(\\frac{1}{\\delta t} \\MfMui \\vec{y}_b^{(N+1)}\\\) is zero.
|
||||
+10
-9
@@ -4,6 +4,16 @@ simpegEM Utilities
|
||||
SimPEG for EM provides a few EM specific utility codes,
|
||||
sources, and analytic functions.
|
||||
|
||||
Utilities for Electromagnetics
|
||||
==============================
|
||||
|
||||
.. automodule:: SimPEG.EM.Utils
|
||||
:show-inheritance:
|
||||
:members:
|
||||
:undoc-members:
|
||||
:inherited-members:
|
||||
|
||||
|
||||
Analytic Functions - Time
|
||||
=========================
|
||||
|
||||
@@ -22,12 +32,3 @@ Analytic Functions - Frequency
|
||||
:members:
|
||||
:undoc-members:
|
||||
:inherited-members:
|
||||
|
||||
|
||||
Sources
|
||||
=======
|
||||
|
||||
.. autoclass:: SimPEG.EM.FDEM.SrcFDEM.MagDipole
|
||||
:show-inheritance:
|
||||
:members:
|
||||
:undoc-members:
|
||||
|
||||
+9
-27
@@ -3,42 +3,24 @@ Electromagnetics
|
||||
================
|
||||
|
||||
`SimPEG.EM` uses SimPEG as the framework for the forward and inverse
|
||||
electromagnetics geophysical problems.
|
||||
electromagnetics geophysical problems.
|
||||
|
||||
Time Domian Electromagnetics
|
||||
----------------------------
|
||||
|
||||
.. toctree::
|
||||
:maxdepth: 2
|
||||
|
||||
api_TDEM_derivation
|
||||
To solve for predicted data, we follow the framework shown below. The model is
|
||||
what we invert for. This is mapped to a physical property on the simulation
|
||||
mesh. A source which is used to excite the system is specified. Having a model
|
||||
and a source, we can solve Maxwell's equations for fields. We sample these
|
||||
fields with recievers to give us predicted data.
|
||||
|
||||
|
||||
Code for Time Domian Electromagnetics
|
||||
-------------------------------------
|
||||
.. image:: ../images/simpegEM_noMath.png
|
||||
:scale: 50%
|
||||
|
||||
.. toctree::
|
||||
:maxdepth: 2
|
||||
|
||||
api_TDEM
|
||||
|
||||
Frequency Domian Electromagnetics
|
||||
---------------------------------
|
||||
|
||||
.. toctree::
|
||||
:maxdepth: 2
|
||||
|
||||
api_FDEM
|
||||
|
||||
|
||||
Utility Codes
|
||||
-------------
|
||||
|
||||
.. toctree::
|
||||
:maxdepth: 2
|
||||
|
||||
api_TDEM
|
||||
api_Utils
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
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|
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Reference in New Issue
Block a user