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 = [tex]\frac{\frac{C}{ba}+ \frac{c}{a} }{2}[/tex]
The given frequency middle at the middle of operating frequency range
= 4.9 GHz = [tex]\frac{c + 2c }{ba}[/tex] = 3C / βa
α = [tex]\frac{3*3*10^{10} }{2*4.9*10^9}[/tex] = 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 = [tex]\sqrt[n]{1-(\frac{Fc}{f} )^{2} }[/tex]
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 = [tex]\frac{|2*10^6|^{2}* 9.18 * 4.59 }{4 * 285}[/tex] = 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 = [tex]\sqrt{\frac{u}{e} } = \sqrt{\frac{UoUr}{EoEr} }[/tex]
therefore for the given dielectric n = [tex]\sqrt{\frac{Uo}{Eo} } \sqrt{\frac{1}{2.5} } = \frac{377}{\sqrt{2.5} }[/tex] n = 238.43
ZTE = [tex]\sqrt[n]{1-(\frac{1.6}{2.45} )^{2} }[/tex]
therefore ZTE = 180.56 ohms
power carried 2 = [tex]\frac{|10^7|^2*9.18*4.59}{4*180.56} = 58*10^{11} N[/tex]
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
