transmitted power in cooperative communication

Hi , I have a system of source node,destination node and a N relays between them. if I have a total transmitted power Pt ,then the transmitted power of each node (source and relays) is Pt/(N+1) ;is it ???

Answer 1

[Walter Roberson]

That does not sound like a MATLAB question.

The transmitted power of each node would only be Pt/(N+1) if the nodes are transmitting at equal power. Unless your nodes are equally spaced taking into account attenuation then that will either fail or be a waste of power.

What I mean by "taking into account attenuation" is that you might have three nodes, A, B, C, where the physical distance between A and B might be the same as B to C, but the pathway between A to B might be "free space" where the attenuation is just through signal spreading, but the pathway between B and C might (for example) pass through a plant (such as a tree) that does not block the signal but partly absorbs it. If the power sent by A to B is "just enough" for the path to be reliable taking into account noise, then that same power transmitted by B would fail in crossing the lower-quality path of equal length between B and C. If you increase the signal levels to the point where the power sent by B to C is "just enough" to be reliable taking into account the noise and signal absorption of the obstacle, then if that same power were used to send from A to B, that would be a waste -- A to B does not need that much power. Insisting on equal power is therefore either a failure or a waste in this kind of situation.

Now, it could theoretically be the case that the physical distance between B and C is less than the physical distance between A and B, such that after taking into account absorption by the tree, the same power used from A to B is just right for the lower-quality but shorter pathway between B and C. Possible, but unlikely, especially as much more typical is for absorption to vary with air temperature and air humidity.

To avoid path loss and to avoid transmission waste, it is a better practice for each node to dynamically adjust power as it figures out how "effectively far" the target node is.

There are various ways for the nodes to dynamically adjust power.

For example, if the equivalent of TCP is used, where the target node has to respond to specifically say that it received the data, then the controlling node (A) can monitor the power with which it received the acknowledgement from the recipient (B); if the acknowledging node includes information about the power level it transmitted the signal at, then the controlling node can figure out how much the signal was attenuated, and can adjust its own transmitting power accordingly.

If the equivalent of UDP is used instead, where the target node is not required to respond to specifically say that it received the data, then a system of occasional "beacons" can be used. For example, at periodic intervals, each node could transmit at full strength a "I am here!" signal; each node that can hear the beacon would then take note of the power with which it received the beacon, and with the assumption that the sender used full power can figure out the attenuation. (Requiring that full power be used for the beacon is not necessarily the best idea: as the battery runs down, those full power bursts could end up consuming a substantial fraction of the remaining power. There are approaches to deal with that...)

[lam lam]

ok ,thanks a lot for the detailed response, in fact, its about a sensors network which is illustrated in the attached figure(we have L relays). the distance dsr is different from dsd .the channels source-relays ,source destination (direct link) and relays-destination is assumed to be a rayleigh channels with awgn noise. in this conditions we can use the same transmitted power (Pt/L) ???

[lam lam]

[Walter Roberson]

Yes, you can use the same transmitted power, but that transmitted power would have to be the maximum of the powers necessary to for any of the individual steps, taking into account the distances and local causes of attenuation.

Imagine, for example, a vine partly blocking the path between the second relay from the top and the destination: that would attenuate the signal, forcing the system to use a higher transmit power for that link even if the distance is the same as the other distances. If you require that all of the nodes use the same power, then you are requiring that the source and all of the other relays use that higher power even though there is no obstacle for them.

I would also point out that geometrically, in order for the relays to all be the same distance from the destination, the relays would have to be on a sphere around the destination; and in order for the relays to all be the same distance from the source, the relays would have to be on a sphere around the source. If you draw out two circles of different radius (since dsr and drd can be different distances), you will find that in 2 dimensions there are only exactly two points that can meet that requirement. This suggests that it is much much more likely that the distances between the source and relays are not all the same, and that the distances between the relays and the destination are not all the same. You could use equal power for the situation, but you do not need equal power.

It is not uncommon for cooperative radio systems to be set up to figure out the lowest energy route, including passing data between relays. For example such a system might notice the extra energy required because of an obstacle (such as the vine mentioned above) and figure out that it is better to avoid that relay.