Antenna Exam Question (orbiting satellite)

[samSmith]

Homework Statement

A radiometer system on a satellite orbiting the Earth at an altitude of 1000km has an antenna of gain 46 dBi pointing at the earth’s surface. The antenna is matched to a receiver with an input noise temperature of 50K and a bandwidth of 10MHz. The sky noise temperature is 10K. The earth’s surface temperature is 300K. The surface field reflection coefficient is 0.7. Estimate the approximate area of the earth’s surface that is “seen” by the system. Estimate the total noise power at the receiver input and the fractional change in this noise power caused by a 1K change in the earth’s surface temperature

Homework Equations

Pd = (Gtx * Pin)/(4*pi*r^2), Pn = KTB , other basic antenna equations

The Attempt at a Solution

My attempt is probably wrong but here it goes! So I believe the first step is to calculate the Noise Power of the transmitting antenna, this is done by,

P = KTB
P = 1.38*10^-23 * 50 *10*10^6
= 6.9 * 10^-15 WNow this is the bit I am uncertain about but the formulae fits the given variables.

Pd = Gtx Pin/ (4piR^2)

Substituting the numbers gives a power density of 2.18 x 10^-17 W. However I am unsure of what to do next, I've tried a few things but suspect them of being wrong!

Any help given will be appreciated! Thanks!

 

[mark726]

To estimate the approximate area of the earth's surface that is "seen" by the system, you can use the formula:

A = (4*pi*r^2) * (1 - reflection coefficient)

Where A is the area, r is the distance from the satellite to the earth's surface (in this case, 1000km), and the reflection coefficient is given as 0.7.

Plugging in the numbers, we get:

A = (4*pi*1000^2) * (1 - 0.7)
= 1.24 * 10^10 square meters

This is the approximate area of the earth's surface that is "seen" by the system.

To estimate the total noise power at the receiver input, we can use the formula:

Pn = KTB

Where Pn is the total noise power, K is the Boltzmann constant (1.38 * 10^-23), T is the temperature (in Kelvin), and B is the bandwidth (10MHz).

Plugging in the numbers, we get:

Pn = (1.38 * 10^-23) * (10 * 10^6) * (50)
= 6.9 * 10^-15 W

This is the total noise power at the receiver input.

To estimate the fractional change in this noise power caused by a 1K change in the earth's surface temperature, we can use the formula:

Fractional change = (Pn2 - Pn1) / Pn1

Where Pn2 is the total noise power at a 1K higher temperature, and Pn1 is the original total noise power.

Plugging in the numbers, we get:

Fractional change = (6.9 * 10^-15 - 6.9 * 10^-15) / 6.9 * 10^-15
= 0

This means that a 1K change in the earth's surface temperature would not cause any change in the total noise power at the receiver input.