Two Loop FDEM¶

In [ ]:
from geoscilabs.base import widgetify
import geoscilabs.em.InductionLoop as IND
from ipywidgets import interact, FloatSlider, FloatText


Parameter Descriptions¶

Below are the adjustable parameters for widgets within this notebook:

• $I_p$: Transmitter current amplitude [A]
• $a_{Tx}$: Transmitter loop radius [m]
• $a_{Rx}$: Receiver loop radius [m]
• $x_{Rx}$: Receiver x position [m]
• $z_{Rx}$: Receiver z position [m]
• $\theta$: Receiver normal vector relative to vertical [degrees]
• $R$: Resistance of receiver loop [$\Omega$]
• $L$: Inductance of receiver loop [H]
• $f$: Specific frequency [Hz]
• $t$: Specific time [s]

Background Theory: Induced Currents due to a Harmonic Primary Signal¶

Consider the case in the image above, where a circular loop of wire ($Tx$) caries a harmonic current $I_p (\omega)$. According to the Biot-Savart law, this produces a harmonic primary magnetic field. The harmonic nature of the corresponding magnetic flux which passes through the receiver coil ($Rx$) generates an induced secondary current $I_s (\omega)$, which depends on the coil's resistance ($R$) and inductance ($L$). Here, we will provided final analytic results associated with the app below. Full derivations can be found at the bottom of the page.

Frequency Response¶

The frequency response which characterizes the induced currents in $Rx$ is given by:

$$I_s (\omega) = - \frac{i \omega A \beta_n}{R + i \omega L} I_p(\omega)$$

where $A$ is the area of $Rx$ and $\beta$ contains the geometric information pertaining to the problem. The induced current has both in-phase and quadrature components. These are given by:

\begin{align} I_{Re} (\omega) &= - \frac{i \omega A \beta_n R}{R^2 + (\omega L)^2} I_p(\omega) \\ I_{Im} (\omega) &= - \frac{ \omega^2 A \beta_n L}{R^2 + (\omega L)^2} I_p(\omega) \end{align}

Time-Harmonic Response¶

In the time domain, let us consider a time-harmonic primary current of the form $I_p(t) = I_0 \textrm{cos}(\omega t)$. In this case, the induced currents within $Rx$ are given by:

$$I_s (t) = - \Bigg [ \frac{\omega I_0 A \beta_n}{R \, \textrm{sin} \phi + \omega L \, \textrm{cos} \phi} \Bigg ] \, \textrm{cos} (\omega t -\phi) \;\;\;\;\; \textrm{where} \;\;\;\;\; \phi = \textrm{tan}^{-1} \Bigg ( \frac{\omega L}{R} \Bigg ) \, \in \, [\pi/2, \pi ]$$

The phase-lag between the primary and secondary currents is represented by $\phi$. As a result, there are both in-phase and quadrature components of the induced current, which are given by: \begin{align} \textrm{In phase:} \, I_s (t) &= - \Bigg [ \frac{\omega I_0 A \beta_n}{R \, \textrm{sin} \phi + \omega L \, \textrm{cos} \phi} \Bigg ] \textrm{cos} \phi \, \textrm{cos} (\omega t) \\ \textrm{Quadrature:} \, I_s (t) &= - \Bigg [ \frac{\omega I_0 A \beta_n}{R \, \textrm{sin} \phi + \omega L \, \textrm{cos} \phi} \Bigg ] \textrm{sin} \phi \, \textrm{sin} (\omega t) \end{align}

In [ ]:
# RUN FREQUENCY DOMAIN WIDGET
widgetify(IND.fcn_FDEM_Widget,I=FloatSlider(min=1, max=10., value=1., step=1., continuous_update=False, description = "$I_0$"),\
a1=FloatSlider(min=1., max=20., value=10., step=1., continuous_update=False, description = "$a_{Tx}$"),\
a2=FloatSlider(min=1., max=20.,value=5.,step=1.,continuous_update=False,description = "$a_{Rx}$"),\
xRx=FloatSlider(min=-15., max=15., value=0., step=1., continuous_update=False, description = "$x_{Rx}$"),\
zRx=FloatSlider(min=-15., max=15., value=-8., step=1., continuous_update=False, description = "$z_{Rx}$"),\
azm=FloatSlider(min=-90., max=90., value=0., step=10., continuous_update=False, description = "$\\theta$"),\
logR=FloatSlider(min=0., max=6., value=2., step=1., continuous_update=False, description = "$log_{10}(R)$"),\
logL=FloatSlider(min=-7., max=-2., value=-4., step=1., continuous_update=False, description = "$log_{10}(L)$"),\
logf=FloatSlider(min=0., max=8., value=5., step=1., continuous_update=False, description = "$log_{10}(f)$"))



Supporting Derivation for the Frequency Response¶

Consider a transmitter loop which carries a harmonic primary current $I_p(\omega)$. According to the Biot-Savart law, this results in a primary magnetic field: $$\mathbf{B_p} (\mathbf{r},\omega) = \boldsymbol{\beta} \, I_p(\omega) \;\;\;\; \textrm{where} \;\;\;\;\; \boldsymbol{\beta} = \frac{\mu_0}{4 \pi} \int_C \frac{d \mathbf{l} \times \mathbf{r'}}{|\mathbf{r'}|^2}$$ where $\boldsymbol{\beta}$ contains the problem geometry. Assume the magnetic field is homogeneous through the receiver loop. The primary field generates an EMF within the receiver loop equal to: $$EMF = - i\omega \Phi \;\;\;\;\; \textrm{where} \;\;\;\;\; \Phi = A \beta_n I_p(\omega)$$ where $A$ is the area of the receiver loop and $\beta_n$ is the component of $\boldsymbol{\beta}$ along $\hat n$. The EMF induces a secondary current $I_s(\omega)$ within the receiver loop. The secondary current is defined by the following expression: $$V = - i \omega A \beta_n I_p (\omega) = \big (R + i\omega L \big )I_s(\omega)$$ Rearranging this expression to solve for the secondary current we obtain $$I_s (\omega) = - \frac{i \omega A \beta_n}{R + i \omega L} I_p(\omega)$$ The secondary current has both real and imaginary components. These are given by: $$I_{Re} (\omega) = - \frac{i \omega A \beta_n R}{R^2 + (\omega L)^2} I_p(\omega)$$ and $$I_{Im} (\omega) = - \frac{ \omega^2 A \beta_n L}{R^2 + (\omega L)^2} I_p(\omega)$$

Supporting Derivation for the Time-Harmonic Response¶

Consider a transmitter loop which carries a harmonic primary current of the form: $$I_p(t) = I_0 \textrm{cos} (\omega t)$$ According to the Biot-Savart law, this results in a primary magnetic field: $$\mathbf{B_p} (\mathbf{r},t) = \boldsymbol{\beta} \, I_0 \, \textrm{cos} (\omega t) \;\;\;\; \textrm{where} \;\;\;\;\; \boldsymbol{\beta} = \frac{\mu_0}{4 \pi} \int_C \frac{d \mathbf{l} \times \mathbf{r'}}{|\mathbf{r'}|^2}$$ where $\boldsymbol{\beta}$ contains the problem geometry. If the magnetic field is homogeneous through the receiver loop, the primary field generates an EMF within the receiver loop equal to: $$EMF = - \frac{\partial \Phi}{\partial t} \;\;\;\;\; \textrm{where} \;\;\;\;\; \Phi = A\hat n \cdot \mathbf{B_p} = I_0 A \beta_n \, \textrm{cos} (\omega t)$$ where $A$ is the area of the receiver loop and $\beta_n$ is the component of $\boldsymbol{\beta}$ along $\hat n$. The EMF induces a secondary current $I_s$ within the receiver loop. The secondary current is defined by the following ODE: $$V = \omega I_0 A \beta_n \, \textrm{sin} (\omega t) = I_s R + L \frac{dI_s}{dt}$$ The ODE has a solution of the form: $$I_s (t) = \alpha \, \textrm{cos} (\omega t - \phi)$$ where $\alpha$ is the amplitude of the secondary current and $\phi$ is the phase lag. By solving the ODE, the secondary current induced in the receiver loop is given by: $$I_s (t) = - \Bigg [ \frac{\omega I_0 A \beta_n}{R \, \textrm{sin} \phi + \omega L \, \textrm{cos} \phi} \Bigg ] \, \textrm{cos} (\omega t -\phi) \;\;\;\;\; \textrm{where} \;\;\;\;\; \phi = \textrm{tan}^{-1} \Bigg ( \frac{\omega L}{R} \Bigg ) \, \in \, [\pi/2, \pi ]$$ The secondary current has both in-phase and quadrature components, these are given by: $$\textrm{In phase:} \, I_s (t) = - \Bigg [ \frac{\omega I_0 A \beta_n}{R \, \textrm{sin} \phi + \omega L \, \textrm{cos} \phi} \Bigg ] \textrm{cos} \phi \, \textrm{cos} (\omega t)$$ and $$\textrm{Quadrature:} \, I_s (t) = - \Bigg [ \frac{\omega I_0 A \beta_n}{R \, \textrm{sin} \phi + \omega L \, \textrm{cos} \phi} \Bigg ] \textrm{sin} \phi \, \textrm{sin} (\omega t)$$

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