Waveguide: Wavelength, Impedance and power calculation

[@iqrachaudhary4360]

Sir how ro design the waveguide antenna of frequency 10ghz and gain 10dbi... What are the larameters that are required?

[@mohitrongare4033]

Sir you are really gr8

[@aditipaul701]

for finding the impedence why do we take the ratio ix Ex/Hy not the total of (Ex+Ey)/(Hx+Hy)

[@pavantripathi6594]

The impedance is defined as ratio of transverse electric field mag(Et) and transverse magnetic field mag(Ht). Transverse means the components which are perpendicular to the direction of propagation. In case of waveguide, Et is the vector sum of Ex and Ey and Ht is the vector sum of Hx and Hy. Follow the lecture from 17:50 min. Ask if there is doubt.

[@SiddharthMishraLastOne]

@@pavantripathi6594 So (Ex+Ey)/(Hx+Hy) is correct, right? It should be total Transverse components. Then why did Sir only take Ex/Hy or -Ey/Hx to calculate the impedance?

[@aditipaul701]

why do we take f10 as fundamental freq not f01??

[@pavantripathi6594]

Your question is good. Refer to the Rectangular Waveguide: TE mode lecture, the professor has gone through the problem of Waveguide WR-75 which is widely used in practice. We calculated for TE10: f10=7.87 Ghz ; for TE01: f01=15.74 Ghz. We see that the cut off frequency of TE10 mode is less than TE01 mode and hence TE10 is considered as dominant mode. Also if we look at the formula of cut off frequency f_cmn=1/(2√μϵ) √((m/a)^2+(n/b)^2 ) -----f_cmn=1/(2a√μϵ) √((m)^2+(a/b)^2 n^2 ) and we know a>b therefore a/b >1 hence frequency at m=1 and n=0 will be smaller than frequency at m=0 and n=1. Thus we consider f10 as first mode or dominant mode. Hope this clears your query. Ask if its not clear.