- #1

- 343

- 0

## Homework Statement

A parallel plate waveguide has perfectly conducting plates at y = 0 and y = b for 0 ≤ x < ∞ and -∞ < z < ∞. Inside that bound, the waveguide is filled with a dielectric with k as a propagation constant.

The Green's function to be satisfied is

[itex] \nabla^2G + k^2G = -\delta(x-x')\delta(y-y')[/itex]

where 0 ≤ x' < ∞ and 0 ≤ y' ≤ b. G = 0 at the walls and there is no dependence on z. Find the Green's function everywhere inside the waveguide.

## Homework Equations

[itex] \nabla^2G + k^2G = -\delta(x-x')\delta(y-y')[/itex]

[itex]G = f(x)g(y)[/itex]

[itex] f(0) = 0, f(∞) = 0, g(0) = 0, g(b) = 0[/itex]

also

limit as [itex]x→x'^- f(x) =[/itex] limit as [itex]x→x'^+ f(x), [/itex]limit as [itex]x→x'^- f'(x) = [/itex]limit as [itex]x→x'^+ f'(x)[/itex]

## The Attempt at a Solution

We first look for solutions for the homogenous equation:

[itex] \nabla^2G + k^2G = 0 [/itex]

Using separation of variables, it is possible to form an ODE from this PDE. If k^2 = k_x^2+k_y^2 then the PDE looks like

[itex] \frac{∂^2G}{∂x^2} + \frac{∂^2G}{∂y^2} + (k_x^2+k_y^2)G = 0 [/itex]

Let G = f(x)g(y) and the two ODE that form are the following:

[itex]f''(x)+k_x^2f(x) = 0[/itex]

[itex]g''(y)+k_y^2g(y) = 0[/itex]

Now some basic general solutions to f(x) are:

[itex]Ae^{ik_xx}+Be^{-ik_xx}[/itex]

or

[itex]Acos(k_xx)-Bsin(k_xx)[/itex]

and actually, I believe the most general solution to f(x) would be a summation of one of the two solutions above, like the following:

\begin{align}\overset{∞}{\underset{n=1}{\sum}} A_ne^{ik_xx}+B_ne^{-ik_xx}\end{align}

g(y) has similar solutions to f(x) and here are the two basic general solutions:

[itex]Ce^{ik_yy}+De^{-ik_yy}[/itex]

or

[itex]Ccos(k_yy)-Dsin(k_yy)[/itex]

I just need to choose 2 of the 4 basic equations here and work with the boundary conditions. Lets say I choose the following:

[itex]f(x) = Ae^{ik_xx}+Be^{-ik_xx}[/itex]

[itex]g(y) = Ccos(k_yy)-Dsin(k_yy)[/itex]

The boundary conditions that I see are as follows:

[itex] f(0) = 0, f(∞) = 0, g(0) = 0, g(b) = 0[/itex]

also

limit as [itex]x→x'^- f(x) =[/itex] limit as [itex]x→x'^+ f(x), [/itex]limit as [itex]x→x'^- f'(x) = [/itex]limit as [itex]x→x'^+ f'(x)[/itex]

This is what I have been able to determine so far. When I start to apply boundary conditions is when I feel like I don't know what I am doing. Lets apply the BC for g(y):

[itex]g(0) = 0 = Ccos(k_y*0)-Dsin(k_y*0)[/itex]

This tells us C = 0.

[itex]g(b) = 0 = -Dsin(k_y*b)[/itex]

Now for the right side to be 0, [itex]k_y*b=n\pi[/itex] where n are integers. This means that [itex]k_y=n\pi/b[/itex]. So if I did stuff right here, g(y) is the following:

[itex]g(y) = -Dsin(n\pi*y/b)[/itex]

Lets look at the BC for f(x):

[itex] f(0) = 0 = A + B[/itex]

So this says that A = -B which I don't have too much of a problem with. I think that this only applies for when 0 ≤ x ≤ x'. The f(x) for this region would then be:

[itex]f(x) = A(e^{+ik_xx}-e^{-ik_xx}) = i*2A*sin(k_xx)[/itex]

Where my problem arises is for the following BC when x' ≤ x < ∞:

[itex]f(∞) = 0 = Ae^{ik_xx} + Be^{-ik_xx}[/itex]

With this equation, I can't seem to figure out how to make this equation true without it being the trivial solution (ie, A=B=0). If A = 0 then we have:

[itex]f(∞) = 0 = Be^{-ik_xx}[/itex]

[itex]f(∞) = 0 = Bcos(k_xx)-iBsin(k_xx)[/itex]

but I still don't know how to handle that. I don't know what happens when a complex exponential goes to -∞. Another idea would be to work with the sine of the complex exponential and make it so it equals zero by using k_x. The problem here is that I can't do that without making k_x dependent on x which isn't allowed.

[itex] k_x = n\pi/x[/itex]

The only other thing to do would be to set k_x = 0 but that just gives the trivial solution. Perhaps the answer to my problem is just so obvious that I am oblivious to it but can someone point me in the right direction?

Thanks!