Cutoff frequency: Béda antarrépisi

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Baris ka-61:
Substituting and evaluating the time derivative gives
:<math>
\left(\nabla^2 + \frac{\omega^2}{c^2}\right) \psi(x,y,z) = 0.
</math>
The function <math> \psi </math> here refers to whichever field (the electric field or the magnetic field) has no vector component in the longitudinal direction - the "transverse" field. It is a property of all the eigenmodes of the electromagnetic waveguide that at least one of the two fields is transverse. The ''z'' axis is defined to be along the axis of the waveguide.
Baris ka-71:
where <math>k_z</math> is the longitudinal [[wavenumber]], resulting in
:<math>
\left(\nabla_{T}^2 - k_{z}^2 + \frac{\omega^2}{c^2}\right) \psi(x,y,z) = 0,
</math>
where subscript T indicates a 2-dimensional transverse Laplacian. The final step depends on the geometry of the waveguide. The easiest geometry to solve is the rectangular waveguide. In that case the remainder of the Laplacian can be evaluated to its characteristic equation by considering solutions of the form