A ground receiver station is receiving a sign | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) $1$. Calculate the maximum line-of-sight distance for space wave communication: $d_m = \sqrt{2Rh} = \sqrt{2 	imes 6.4 	imes 10^6 	imes 300} = \

Vedclass · Accepted Answer

Sky wave propagation

Vedclass · Answer

(B) $1$. Calculate the maximum line-of-sight distance for space wave communication: $d_m = \sqrt{2Rh} = \sqrt{2 	imes 6.4 	imes 10^6 	imes 300} = \sqrt{3840 	imes 10^6} \approx 61.97\, km \approx 62\, km$.$2$. Since the distance between the transmitter and receiver is $100\, km$,which is greater than $d_m$ $(100\, km > 62\, km)$,space wave propagation is not possible.$3$. Calculate the critical frequency of the ionosphere: $f_c = 9(N_{\max})^{1/2} = 9 	imes (10^{12})^{1/2} = 9 	imes 10^6\, Hz = 9\, MHz$.$4$. Since the signal frequency $(5\, MHz)$ is less than the critical frequency $(9\, MHz)$,the signal can be reflected by the ionosphere.$5$. Therefore,the signal is received via sky wave propagation.

Similar Questions

$A$ signal of $5\ kHz$ frequency is amplitude modulated on a carrier wave of frequency $2\ MHz$. The frequencies of the resultant signal is/are

What type of modulation is employed in India for radio transmission?

$A$ transmitting antenna at the top of a tower has height $32\, m$ and the height of the receiving antenna is $50\, m$. What is the maximum distance between them for satisfactory communication in line of sight $(LOS)$ mode? ........$km$

The amplitude of $15 \sin (1000 \pi t)$ is modulated by $10 \sin (4 \pi t)$ signal. The amplitude modulated signal contains frequency(ies) of
$A) 500 \text{ Hz}$
$B) 2 \text{ Hz}$
$C) 250 \text{ Hz}$
$D) 498 \text{ Hz}$
$E) 502 \text{ Hz}$
Choose the correct answer from the options given below:

Vedclass Test Series

Exam Paper Generator

Online Exam Module