## Sunday, November 23, 2014

### EM II Notes 2014_11_23: Homework sketches

Just a few notes on how to proceed on the penultimate homework of the semester.

We're to show that the solutions for the 30/60/90 triangular waveguide given in the last homework set will also work for a waveguide that's formed from an equilateral traingle.  The three corners of the equilateral traingle are located at $\left(x,y\right) = \left(0, 0\right)$, $\left(x,y\right) = \left(a, a/\sqrt{3}\right)$, and $\left(x,y\right) = \left(a, -a/\sqrt{3}\right)$.

This falls out immediately from last week's homeowrk.  Because the sine function is peiodic in $\pi$ over the domain from $\left(-\infty, \infty\right)$, the solution given last week in terms of sines will still evaluate to zero on the wall that falls at negative $y$. coordites.  The positive $x$ coordinates of the functions will evaluate to 0 on the wall in the same manner they did before???  There's an issue here.  It's products of the $x$ and $y$ sinusoids that all sum to zero.  These will need to be checked to determine if they still evaluate to zero.  Since the y terms are converted to x terms using the coordinates of the walls, there's very little to do in checking that things are stil OK.  Everyting converts down to a difference of cosines expressed as

$cos\dfrac{m-n}{3a} - cos\dfrac{n-m}{3a}$

However, because the three $y$ sin functions are odd, they will merely wind up with negative signs out in front that can be factored out of the entire expression.  The result will be the same.