Exam text content

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ELT-47426 Transmission lines and waveguides

Small Exam II, Feb 28th 2014. Answer to all guestions.

Jari Kangas
(a) A rectangular waveguide with dimensions a = 1.5 cm, b = 0.8 cm is filled with a dielectric

I
with permeability uo and relative permittivity 4. The magnetic field inside the guide is
given as
Hz =2sin (=) cos (27) sin(m10'4 — 82) A/m.
a

Find: (4 p. in total)
i. The mode that is propagating inside the guide.
ii. The cutoff freguency. €, < 2.95 j0' Hz 2 ls
iii. The propagation constant. P> 17 22 + s E
iv. The wave impedance. 16=?21/4 / =
(b) Consider rectangular airfilled waveguide. Determine criteria for having maximum band-
width for such a structure such that it also offers maximum power withstand for dominant hd
a s£

w

mode. Determine the corresponding cut-off wavenumber. (4 p.)
2. Look at the pictures of a simulation case on the other paper. Explain what kind of settings
(parameters, conditions) are inherent in the simulation case. Consider them from EM field [

k

analysis point of view. (4P.) —«
3. (a) Consider transmission line and waveguide analysis, your task is to explain key differences 2
between them. (3 p.)
(b) Explain how to check orthogonality of different modes in a spherical cavity. Reason also
. l

why orthogonality of different modes is of interest and where it can be used. (3 p.)

Miscellaneous information
e Resonance freguency of a rectangular cavity resonator can be expressed as

1 12 2
fo 7 aVt+(7)

where /19 = 4 x 10 7H/m and € = 8.854 x 10 F/m.

 

For rectangular waveguides average power is given and field components are related as:
2 0353 GWH
(6 Eh, == and (Pu) < yi8 Pii
8? M (n?a? + m?0?) 8? (n?a? + m20?)

1 : E
jAV:E, — juu2 x ViH,).

a == I
1 = =P (i8V:H> + jwe2 x V:E)), Ei= P=R (

Constants in free space and some formulas:

€ speed of light in vacuum c = 2.99792458 « 108m /s = 3 x 108m/s

e intrinsic impedance 10 = as = 1207 O

e ec*=l1+oif|x| small

 

Ei
Eigenfreguency=9.554751e8 Surface: Electric field norm (V/m) Arrow Surface: Electric displacement field

 

 

JI T T = T T T T T T A 1698.6
”i
0,025 1600
0.02 1400
ST 1200
) 1000
1 800
-0.005 600
008 e
1.02 200
-0.025 +; L L 1 1

L 1 L L L L

-0.08 -0.06 -0.04 -0.02 o 0.02 0.04 0.06 0.08 0.1 0.12 W 5.4238x10%

 

Eigenfreguency=9.554751e8 Surface: Magnetic field, z component (A/m)

 

0,045 = , - : , ä , <a. 7
L A -3.8878x10
o
0.025 -0.5
0.02 4
0,015 15
0.01 a
0.005 - 25
0,005 - 3
-0.005 3.5
-0.01 4
-0.015 45
-0.02 55
-0.025 L , i : ; : i i ;

 

-0.08 -0.06 -0.04 -0.02 o 0.02 0.04 0.06 0.08 0.1 0.12 V -5.404

Eigenfreguency=9.554751e8 Surface: Magnetic flux density, z component (T)

 

0,045 r i - - r - ' - i i : A 4.8856x10
0.04 etä
0,035 x107
S
oli 2
0.01 z
0.005 -30
88 50
0. -5
-0,015
-0.02 -60
-0.025 6, ; i ; i

L L L fi 1

-0.08 -0.06 -0.04 -0.02 o 0.02 0.04 0.06 0.08 0.1 0.12 V -6.7909x10*