A hollow rectangular waveguide is to be used to transmit signals at a carrier frequency of 6 GHz. Choose its dimensions so that the cutoff frequency of the dominant TE mode is lower than the carrier by 25% and that of the next mode is at least 25% higher than the carrier.

Dimensions; a = 3.33 cm and b = 2cm

Explanation:

The waveguide is hollow, thus, v_p = c

Now, for TE_mn modes;

The first mode is TE_10

The second mode is TE_01

So, we'll choose "b" such that;

a > b > a/2

Now, f_mn is given by the formula;

f_mn = v_po√((m/a) ² + (n/b)²)

So, f_10 = c/(2a) = = 0.75f_o

From the question, carrier frequency, f_o is given as 6GHz

Thus,

f_10 = 0.75 x 6 = 4.5 GHz = 4.5 x 10^(9) Hz

Also, a = c/(2f_10)

Where c is speed of light = 3 x 10^(8) m/s

Thus, a = 3 x 10^(8)/(2 x 4.5 x 10^(9))

a = 3.33 cm

Also, f_01 = c/(2b) = = 1.25f_o = 1.25 x 6GHz = 7.5 GHz = 7.5 x 10^(9) Hz

Thus, b = c/(2f_01)

b = 3 x 10^(8)/(2 x 7.5 x 10^(9)) = 2cm

answer:

the answer is d

explanation:

hes mad if he says,

explanation:

r!