And now for a totally non-relationship oriented post (let's see how many comments I get).
Taking the curl of Faraday's Law says:
Applying Ampere’s Law and a vector identity give:
Assuming no sources and losses inside the waveguide requires that:
So:
None of this is new; we worked out all this math for plane waves. But for plane waves, we said that for a wave propagating in the *z*-direction, there was no *x* and *y* dependence. In other words, E = Ae^{-γz}, where A was a complex constant. Now we are instead going to assume that E *does* have an *x* and *y* dependence: **E = **X(x)Y(y)e^{-γz}, where X(x) and Y(y) are functions of *x* and *y*.
It follows that the derivatives with respect to *x* and *y* contained in the vector Laplacian have non-zero values:
Consider only the vectors in the *z* direction. Rearranging terms gives:
Likewise:
Now, given the form of our expected solution, E_{z}** = **X(x)Y(y)e^{-γz}, it follows that
Dividing through by X(x)Y(y) e^{-γz} gives:
Note that –(γ^{2} + ω^{2}με)
must be a constant. Because
X”/X and Y”/Y are each only dependent on *x* and *y*,
respectively, it follows that both X”/X and Y”/Y must also be constants.
Therefore let’s assign the constants –M^{2} = X”/X and –N^{2}
= Y”/Y so that:
M^{2}–
N^{2} = –(γ^{2} + ω^{2}με)
and γ = √(M^{2} + N^{2} – ω^{2}με).
Note: this is the first equation we have
developed for γ in this context.
You may recall from plane waves that, for a dielectric, γ = jβ = jω√(με). But now have some additional factors to
consider.
Note
that a propagating wave requires that γ have an imaginary component; if
γ is real, we simply have an attenuating field. It follows that
M^{2} + N^{2} < ω^{2}με.
Now
let’s assume that X(x) takes the form Ae^{λx}. It follows that
X” = λ^{2}Ae^{λx} =
–M^{2} Ae^{λx}.
So
λ = ±jM and X(x) = Ae^{jMx}+
Be^{-jMx} = X_{1}sin(Mx) + X_{2}cos(Mx), where
X_{1} and X_{2} are complex. Likewise, Y(y) = Y_{1}sin(Ny) + Y_{2}cos(Ny). Therefore:
Similarly,
for different functions X(x) and
Y(y). We could give them
subscripts, but we never use them at the same time, so why bother.
Armed with these new equations, let’s return to Faraday’s Law and expand
the curl of E as described on the back cover of the text:
Remembering
our *z* dependency, and separating
the vector components, we can write:
Likewise,
from Ampere’s Law
Substituting
equation (5) into equation (1) yields:
Similarly:
Point: If we know the E_{z}
and H_{z} components, we can find the transverse components!
Question: Is there such a thing as a TEM mode
(i.e. E_{z} = H_{z} = 0) in a
waveguide?

Answer: Let H_{z} = 0; i.e.
Gauss’ Law requires that
i.e. there are no magnetic sources.
Law requires that
It follows that if the transverse components of H are
non-zero, the z component of E must also be non-zero!

Likewise, by Faraday’s Law, if E_{z}_{ }= 0 and the transverse components
of E are non-zero, then H_{z} ≠ 0.

Therefore,
a waveguide *cannot* support TEM mode.

__Boundary Conditions__

_{z}

_{, hollow part}= E

_{z}

_{, metal part}. But inside the metal, σ → ∞, so σE = J = a finite value requires E

_{z}

_{, inside metal}= 0. Therefore E

_{z}

_{, hollow part}= 0 at the inside surface of the waveguide. Furthermore, for a wave guide of cross-sectional dimensions a and b, E

_{x}= 0 when y = {0, b} and E

_{y}= 0 when x = {0, a}.

** TM Mode**: H

_{z}= 0 everywhere inside the waveguide.

_{z}we developed earlier: For the boundary x = 0, For the boundary y = 0, So: For the boundary x = a: For the boundary y = b: Plugging in these values for N and M into the equation we developed for Remember that in order to propagate , so the effective cutoff frequency occurs when where v

_{p}= propagation velocity in the dielectric. Note that, because E

_{z}uses the sine function, and sin(0) = 0, both m and n must be greater than zero; otherwise, E

_{z}= 0 and we have no fields for the reason explained earlier.

** TE Mode**: E

_{z}= 0 everywhere inside the waveguide.

Note: without knowing the surface currents, we
cannot directly make any conclusions about H_{x}
and H_{y} at the boundary. However, we can use the equations we
developed earlier,
and
to say that at boundaries x = {0, a},
At the boundaries y = {0, b}:
Given the format of:
we can say that:
At x = 0,
At y = 0,
So:
At x = a,
At y = b,
So
Notice: because H_{z} in the TE mode
uses the cosine function instead of the sine function, and cos(0) = 1, we can get a
wave if either m or n is equal to zero, but not if both are equal to zero. This means that our formula for the
cutoff frequency yields
as the lowest possible propagation frequency for both the TM
and TE cases when a is the larger of the two dimensions,
m = 1 and n = 0. We therefore refer
to TE_{10} as the **dominant mode**.

## 1 comment:

Nice try. There you go with

anotherattempt to reduce propagation, magnetism and dominance to an equation.Post a Comment