Waveguide: Wavelength, Impedance and power calculation

Пікірлер: 7

  • @mohitrongare4033
    @mohitrongare40336 жыл бұрын

    Sir you are really gr8

  • @iqrachaudhary4360
    @iqrachaudhary43604 жыл бұрын

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

  • @aditipaul701
    @aditipaul7015 жыл бұрын

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

  • @pavantripathi6594

    @pavantripathi6594

    5 жыл бұрын

    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

    @SiddharthMishraLastOne

    Жыл бұрын

    @@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
    @aditipaul7015 жыл бұрын

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

  • @pavantripathi6594

    @pavantripathi6594

    5 жыл бұрын

    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.