Question

Design a rectangular metallic waveguide to be used for transmission of electromagnetic power at 2.45 GHz. This frequency should be at the middle of the operating frequency band. The design should also allow maximum power transfer without sacrificing the operating bandwidth. In each case, you should use a safety factor of 10 and neglect ohmic loss in the conductors.
a. Design an air-filled guide to meet the given specifications. Find the maximum power the waveguide can transmit. The breakdown electric field in air is assumed to be 2 x10^6 V/m
b. Now assume that a electric material is used to fill the waveguide. The material is characterized by ε = 2.5 εo, μ = μ o and σ = 0. The breakdown field in the dielectric is 10^7 V/m. How many times more power can be transmitted by this waveguide?

Answer 1

Answer:

A) 1.4 *10^11 watts

B)  41.42 ≈ 41 TIMES

Explanation:

Designing a rectangular metallic wave guide using the given data

Electromagnetic power = 2.46 GHz also at the middle of operating frequency

A) Design an air-filled guide to meet the given specifications.

operating frequency range =  C / αa < f < C / a

2.45 GHz = $\frac{\frac{C}{ba}+ \frac{c}{a} }{2}$  

The given frequency middle at the middle of operating frequency range

= 4.9 GHz = $\frac{c + 2c }{ba}$  = 3C / βa

α = $\frac{3*3*10^{10} }{2*4.9*10^9}$  = 45/4.9 = 9.18 cm

note: to operate in dominant mode aspect ratio should be  b = α/2

therefore b = 4.59 cm

Also Maximum power can be carried by wave guide only in dominant mode

i.e TE10 mode

power carried = I E I^2ab / 4Zte   using this formula

ZTE = impedance when operated in TE mode = $\sqrt[n]{1-(\frac{Fc}{f} )^{2} }$

Fc = cutoff frequency = (3*10^16) / (2*9.18) = 1.6GHz

F = operating frequency = 2.45 GHz

n = freespace impedance = 377 ohms

input all the given values back to ZTE  equation

ZTE = 285 ohms

power carried = $\frac{|2*10^6|^{2}* 9.18 * 4.59 }{4 * 285}$  =  4*10^12 * 0.036

THEREFORE power carried 1 = 1.4 *10^11 watts

B) The dielectric materials given data/parameters

∈ = 2.5 ∈o   ∪ = ∪o

breakdown field = 10^7

free space impedance  n = $\sqrt{\frac{u}{e} } = \sqrt{\frac{UoUr}{EoEr} }$

therefore for the given dielectric n = $\sqrt{\frac{Uo}{Eo} } \sqrt{\frac{1}{2.5} } = \frac{377}{\sqrt{2.5} }$     n = 238.43

ZTE = $\sqrt[n]{1-(\frac{1.6}{2.45} )^{2} }$    

therefore ZTE = 180.56 ohms

power carried 2 = $\frac{|10^7|^2*9.18*4.59}{4*180.56} = 58*10^{11} N$

To calculate the number of time power can be transmitted by the waveguide = power carried 2 / power carried 1

=  58*10^11 / 1.4*10^11  = 41.42 ≈ 41