# Plane wave causing TE mode in waveguide

I've been struggling with a problem given to us during a exercise-session of waveguides. First part of the exercise was calculating the electric field inside of a waveguide in the form of 2 parallel plates, with it being a TE1 mode. This was simple enough and I won't bother you with the numerical values: $$E_y = C.sin(\beta_x x)e^{-\beta.j.z}$$ With $j=\sqrt{-1}$ and $C,\beta,\beta_x$ constants. Now the part with which I have difficulties is the following task: Find the expression of the electronic field in the frequency domain of the uniform plane wave that causes this TE mode. The solution given to us afterwards is this: $$E_y = \frac{C}{2}e^{-j\beta z+x\beta_x}$$ The constants still have the same values and there is little to no explanation given with these solution. I've tried to convert the sine in the first equation to its exponential form, which takes care of the factor 0.5 and turns it into a similar expression as the second expression. But now I'm just lost, I don't know how I can convert this into a uniform planar wave.

Some extra information: the whole situation is considered to be lossless and the 2 plates are conductors so there is only reflection (though I don't think this is relevant)

• Interesting, but tell us if the question is still valid :).... Mar 28, 2016 at 3:58
• The reflection is extremely relevant, old boy! An open ended rectangular waveguide has a return loss of around 10dB, so a considerable amount of the field is reflected at the interface. The impedance of you wave guide will typically be around 450 ohms (dependant on dimensions) - free space is 377 - Hence the mismatch. Jul 31, 2016 at 14:00