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450fall09hw12_Sol

# 450fall09hw12_Sol - ECE 450 Fall 2009 Homework 12 Solutions...

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ECE 450 Fall 2009 Homework 12 Solutions Due: Tuesday, Nov 17 th , 2009 1. X -band: 8.2 GHz GHz 12.4 f ≤≤ 22 mc nc ab ⎛⎞ =+ ⎜⎟ ⎝⎠ c f , TE mn Æ m =0 or n =0, but not both; TM mn Æ m 0 and n 0 TE mn Table: f c (GHz) (DNE Æ Does Not Exist) Propgation c ff >→ f c (GHz) n 0 1 2 m 0 DNE 14.71 29.42 1 6.55 16.10 30.13 2 13.10 19.69 32.20 3 19.65 24.54 35.37 TM mn Table: f c (GHz) (DNE Æ Does Not Exist) f c (GHz) n 0 1 2 m 0 DNE DNE DNE 1 DNE 16.10 30.13 2 DNE 19.69 32.20 3 DNE 24.54 35.37 *From the table, we see that only the TE 10 mode propagates because the X-band frequencies (8 GHz ) is greater than the TE 10 mode cutoff frequency (6.55 .2 GHz 12.4 f 1

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GHz). All other modes have cutoff frequencies which are greater than the band, in turn making them evanescent. This is the way it should be, typically one designs a waveguide to only support one dominate, single mode. 2. Given: 0 z x z y j y j H kE ωμ + = E (1) and 0 z y z x H j H x E k j ω + = ε (2), take (2) and solve for H x : 0 0 z z zx y yx z E E j H H x jj k H H xj k −+ == ε ε then substitute in (1) and solve for E y : () 0 0 00 22 0 0 0 z z zy z zz y yz z z y y y z y E E E E H jk E j y x jk j HE y j k j k E xy k E EH Ek k j k j jk j E kk + = ⎛⎞ ∂∂ + ⎜⎟ ⎝⎠ −= += + =− ε ε ε k In the a similar fashion, take (1) and solve for E y : 0 0 x z z y xz y z j E E y k E y H jk = = 2
then substitute in (2) and solve for H x : () 0 0 00 22 0 0 0 x x x x xx z z z z zz xz z z x z H H HH H H j jk H H H y x jk j EH j j jk jk yx jj k k HE kk j k j xy jk j ωμ ω ωω μ ⎛⎞ + = ∂∂ + ⎜⎟ ⎝⎠ −= += + =− ε ε εε ε ε E 3. Let the dimensions equate to a = 6 cm and b = 15 cm, it does not change the final results if

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450fall09hw12_Sol - ECE 450 Fall 2009 Homework 12 Solutions...

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