# Exchange diameter for power in parabolic antenna

If I want to reduce the diameter of a satellite dish, how much power should I add to the transmitter? how is this formula? As I have already been informed a little in this other question I asked: X-Band range and data I would need a 3 meter diameter parabolic antenna and 20W of power roughly, how can I add power, for example 200W and reduce the diameter of the parabolic antenna to 0.5 meters?

We assume a transmitter with 20 W on one side and a receiving antenna with 3 m diameter on the other side.

When we increase the power from 20 W to 200 W, we have 10 times more power. We may reduce the area of the antenna to 1/10 and its diameter to 1/sqrt(10). Instead of 3 m we may use now 0.95 m.

We need 2 KW instead of 20 W to reduce the diameter by 1/sqrt(100) = 1/10, from 3 m to only 0.3 m.

The distance between transmitter and receiver is the same for 20 W, 200 W and 2 kW. The diameter of the transmitting antenna is not changed.

But what if both the transmitting and the receiving antennas diameter is decreased? The beam width of the transmitter would increase and the received signal would be weaker. For small beam widths a reduction of the antenna diameter by a factor of 0.5 would double the beam width.

If we use 1.5 m diameter instead of 3 m on both sides, we need to increase the power by a factor of 4 (the square of 2) for both sides. So need 16 times the power to compensate, 320 W instead of 20 W.

For a reduction by the factor of 0.1, from 3 m down to 0.3 m, we need 10000 times more power, 200 kW instead of 20 W.

So exchange diameter for power is impossible for satellites.

• Wait, if I understood, I am talking about an antenna that will be transmitting and receiving, my satellite receives orders from the earth by UHF, and sends them by this X-Band antenna to another satellite, then this antenna is receiving and transmitting, given that it also receives information from the other satellite. This X-Band antenna has to be able to reach Titan, where the other satellite will be. The satellite has 650W available, I plan to make a 1 meter deployable parabolic antenna, what power would I need? I honestly do not understand the formula that you put. Sorry Mar 17, 2020 at 18:57
• @ValentinoZaffrani Another example: Two satellites with 3 m antennas each and 20 W transmitters. We reduce the diameters on both sides from 3 to 1 m, that is 1/3. The area is reduced by 1/9. The effect of larger beam width is 1/9 too. We multiply 1/9 with 1/9 and get 1/81. To compensate we need to increase the power by the factor of 81 and get 1620 W instead of 20 W. But such a network will not work over the distance from Earth to Titan anyway, the maximum distance is much smaller. There is a reason for the large DSN antennas used for that distance.
– Uwe
Mar 17, 2020 at 19:24
• Cassini antenna: 20W of power and 4m diameter parabolic antenna. the problem is that in the earth, they have a very big parabolic antenna, me no. so how i can compensate it? Mar 17, 2020 at 22:44
• @ValentinoZaffrani There is no compensation for a 34 m antenna with a 20 kW transceiver that would fit into a satellite. A 16 m antenna with a 100 kW transmitter would not help.
– Uwe
Mar 17, 2020 at 23:19