Question

proof this equation

general wave behaviors along uniform guiding structures

proof in the table equations

OH x(1O- OED xax 10 -=- CX Ox ах су General Wave Behaviors along Uniform Guiding Structures 10-2 523 Similarly, from Eq. (10-4) we have (10-8) We note that each of Eqs. (10-7) and (10-8) differential equations, one for each component of E and H. The exact these component equations depends on the cross-sectional geometry and the bound- ary conditions that a particular field component must satisfy at conductor-dielectric interfaces. We note further that by writing V, for the transversal operator VZ Eqs (10-7) and (10-8) become the governing equations for waveguides with a circular is really three second-order partial solution of cross scction. f course, the is not necessary to solve all six second-order partial differential equations for the six components of E and H. Let us examine the interrelationships among the six com- ponents in Cartesian coordinates by expanding the two source-free curl equations, Eqs. (7-104a) and (7-104b: I are not all i From V x H-jaxE: EJouH (10-9a) ay -7E E jo(10-9b) Note that partial derivatives with respect to z have been replaced by multiplications by (-7) All the component field quantities in the equations above are phasors that depend only on.x and y, the common e factor for z-dependence having been omitted. By manipulating these equations we can express the transverse field com- ponents H, H9, and Eg, and E9 in terms of the two longitudinal compopents E and Ho. For instance, Eqs. (10-9a) and (10-10b) can be combined to eliminate E and obtain H in terms of E and Ho. We have (10-21) (10-22) hay (10-13) (10-14)

Answer 1

iu«n Subettute for O. b 0え

We wnte Ga+ Substitu뇨 と w E lo-loc

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