Question

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?

Answer 1

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

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