I am trying to solve the following problem sin ce a week
Given: a satellite is flying in a Low Earth Orbit and the distance is 600 kilometers. The antenna on the satellite is a dish antenna with a diameter of 2 meter and an efficiency of 80%. The gain of the antenna array on the ground station is 12 dBi. The bandwidth of the communication signal is 100 kHz. Assume 2 dB losses in the
atmosphere. The system temperature of the receivers in the antenna array is 200 K.
If a fading margin of 10 dB is included in the link budget and a signal to noise ratio of 7dB is needed, calculate the needed power of the transmitter in the satellite and draw a link budget diagram. I can draw the link budget diagram and calulate the transmit power, but sincebandwith is given and not the frequency of operation I am a bit confused?
This is rather simple.
- The noise power is ktB. This gives 2.76e-16 or -125.6 dbm.
- The path loss to a receive antenna with area Ar is given by:
eff*Ar/(4*PI*dist*dist) . (the quotient of the area of the receive antenna by the surface area of a globe with radius equal to the distance.). Ar = PI/4*d*d. This gives a path loss of -122.55 dB.
- The necessary transmit power is now given by:
(receive power) - (path loss) - (Tx antenna gain) + (signal/noise) + (fade margin) .
Note that here the power is independent of the frequency, as the path loss goes up with the frequency, but also the gain of the 2 m dish antenna goes up. I have avoided the issue here, by calculating the loss of the 2 m antenna and the path loss together, and done this by dividing the area of the antenna, by the area of a globe around the earth station with the distance as diameter.
Best Regards,
Henri
Thanks but still you need the frequency to calculate the gain of the parabolic dish.
How can I do that without the frequency
Vikesh,
Yes and no.
The dependency of the gain of the parabolic dish is inverse to the dependency of the path loss to the frequency. Both effects cancel out, and make the situation not depending on the frequency.
Best Regards,
Henri.
Hi Vikesh
The problem has been deliberately posed so that the diameter of the transmitter and the gain of the receiver are given, whereas, if the frequency is not known, the calculation actually requires the gain of the transmitter and the effective aperture of the receiver!
However.
The gain of the transmitter is 4pi times the efficiency times the area / square wavelength and the effective area of the receiver is gain times square wavelength /4pi. When these are multiplied together the wavelength cancels. It doesn't appear anywhere else.
It is designed to test your understanding of the equations.
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