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DiKsa [7]
2 years ago
13

a transmitter is operating at 150 MHz with a power of 3 W into a one-quarter wavelength vertical antenna. The receiver, which is

32.2 km away, has an antenna with a gain of 8 dB. What is the received power?
Computers and Technology
1 answer:
Bogdan [553]2 years ago
4 0

The received power will be 1.243 nW

We're given:

frequency f = 150MHz

distance of the receiver d = 32.2 km=32200m

Power of transmitter P_{t} = 3W

Antenna gain = 8dB

To find :

Power received P_{r}

P_{r}= \frac{P_{t} *G_{t}*G_{r}* \lambda^2 }{16*\pi^2*d^2}

where G_{t is transmit gain and G_{r is receive gain as refrenced to isotropic source

⇒wavelength \lambda = \frac{c}{f} where c is the speed of the light

⇒  \lambda = \frac{3*10^8}{150*10^6} =2m

G_{t}= 1*1.64 ( value of dipole = 1.64)

Now,

Antenna gain=8dB ( in decibals)

⇒10log(x)=8

⇒x=10^0^.^8=6.3095

⇒ considering isotropic receiver

⇒G_{r}=6.3095*1.64=10.3477 (dipole =1.64)

Now , using the formula

P_{r}= \frac{3 *1.64*10.3477 *2^2 }{16*\pi^2*32200^2}=1.2437*10^-^9

Hence The received power will be 1.243 nW

Leaen more about communication devices here:

brainly.com/question/14530107

#SPJ10

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getchar();
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In higher languages like C# it becomes much simpler:

        static void Main(string[] args)
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            Console.ReadLine();
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        static void swaparrayends(int[] arr)
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