Changes

*2D Solution Plane

*Quasi-Static Analysis

==Getting Started==

==Creating the Concentric Metallic Cylinders==

Create two PEC groups called PEC_1 and PEC_2 in the Navigation Tree and draw two cylinder objects with the following properties:

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Define a voltage source called VS_1 with a default voltage of 1V and associate it with the PEC_1 group. Your inner conductor is held at a voltage of 1V, while your outer conductor is held at zero potential. Also, define a horizontal (Z-directed) field sensor and place it at (0, 0, 5mm).

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==Running a Quasi-Static Simulation and Visualizing the Results==

Run a quasi-static analysis of your air-filled coaxial line. At the end of the simulation, the output message window reports the computed values of the characteristic impedance and effective permittivity of the transmission line:

Z0: 48.2803 Ohms

Epsilon_Effective: 1

The effective permittivity of 1 is expected as you have an air-filled two-conductor TEM transmission line. You can also view the electric field and electric potential results.

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To validate your simulation results, let's take a look at the analytical formulas for a coaxial cable:

<math> Z_0 = \frac{60}{\sqrt{\epsilon_{r}}} ln(b/a) = \frac{60}{\sqrt{1.0}} ln(2.3) \approx 50\Omega </math>

<math> \Phi(r) = V_0 \left[ 1 - \frac{ln(r/a)}{ln(b/a)} \right] </math>

<math> E_r(r) = \frac{1}{r} \frac{V_0}{ln(b/a)} </math>

where a and b are the radius of the inner and outer conductors, respectively, and &epsilon;_r is the relative permittivity of the dielectric filling. Substituting a = 1mm and b = 2.3mm, we obtain the following values for the electric field and potential, which agree quite well with [[EM.Cube]]'s computed values:

<math> \Phi(r = 1mm) = V_0 = 1V, \quad \quad \Phi(r = 2.3mm) = 0V </math>

<math> E_r(r = 1mm) = \frac{1}{0.001} \frac{1}{ln(2.3)} = 1200V/m, \quad \quad E_r(r = 2.3mm) = \frac{1}{0.0023} \frac{1}{ln(2.3)} = 522V/m</math>

From the above figures, you can see the 1/r decay of the electric field along the radial direction from the inner conductor to the outer conductor.

==Analyzing a Coaxial Line with Dielectric Filling==

In the this part of the tutorial lesson, you will analyze a coaxial cable with a dielectric core made of "Mica" with &epsilon;_r = 5.4. Keep the same inner conductor cylinder of the previous part. But increase the radius of the outer conductor cylinder to 6.93mm. Also, define a new dielectric material group in the Navigation Tree and select Mica as the material from [[EM.Cube]]'s Material List. Under Dielectric_1, draw a new Cylinder of radius 6.93mm with both ends capped. The following table summarizes the geometrical [[parameters]]:

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Keep the same voltage source, field sensor and mesh density as in the previous part. Note that you used a solid dielectric cylinder instead of a hollow dielectric cylinder to model the dielectric core of your coaxial cable. This is because in [[EM.Cube]]'s Static and FDTD Modules, a material hierarchy system is in effect, in which a PEC has precedence over a dielectric material. This means that if a PEC object and a dielectric object overlap each other over some region in the space, the common overlapped regions is assumed to be PEC rather than dielectric.

{{Note| In [[EM.Cube]]'s Static Module, overlapped regions between PEC and dielectric objects are always assumed to be PEC rather than dielectric. This hierarchical rule make construction of concentric structures much easier.}}

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Run a quasi-static analysis of your dielectric-filled coaxial line. At the end of the simulation, the output message window reports the computed values of the characteristic impedance and effective permittivity of the transmission line:

Z0: 49.1214 Ohms

Epsilon_Effective: 5.3994

An effective permittivity of &epsilon;_eff = &epsilon;_r = 5.4 is expected as you have an homogenous two-conductor TEM transmission line. Visualize the electric field and electric potential distributions and plot the 2D graphs of the total electric field along the X-axis.

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You can validate your simulation results using the analytical formulas for a dielectric-filled coaxial cable:

<math> Z_0 = \frac{60}{\sqrt{\epsilon_{r}}} ln(b/a) = \frac{60}{\sqrt{5.4}} ln(6.93) \approx 50\Omega </math>

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