Two Towers

The Two (cell) Tower Problem

These results are so counter intuitive, and so important that I wrote this page to convince others. I needed convincing too! As a reality check here is a simple case with specific numbers.

We have two cell towers and one cell phone. The phone is at distance 1 from tower 1 and distance 2 from tower 2. To simplify the math we assume that there is no multi-path. It works for multi-path and that is even more surprising. When tower j transmits at unit power the phone senses field strength 1/j.

We show how to minimize total transmitter power not only for cost saving, but even more because a signal for one phone, is noise to another—most likely a phone serviced by the same towers.

Assuming that phases are correctly chosen, what transmit power should each tower choose when transmitting to the phone to maximize reception at the phone, while spending unit total transmitter power? Noise is measured by power. Recall that the power is the square of the field strength. If tower j transmits with field strength S_j then the phone will see total field strength CS = S₁ + S₂/2 since transmission phase has been coordinated to be in phase at the phone. The total transmission power is TP = S₁² + S₂². To explore the set of power distributions between the two towers we let S₁ = cos θ and S₂ = sin θ. Note that TP = S₁² + S₂² = (cos θ)² + (sin θ)² = 1. Note that θ is not a phase but a parameter to choose transmission powers to maximize received power at phone while TP = 1. How do we choose θ to maximize CS?
CS = S₁ + S₂/2 = (cos θ) + (sin θ)/2.
θ CS
0 1
π/2 0.5
0.464 1.118
CS is maximized for θ = atan (½) = 0.464 . It saves power or minimizes total noise to turn on the farther tower to transmit at field strength 0.447 (= sin θ) and power 0.2 . The near tower’s field strength is 0.894 (= cos θ) and its power is 0.8 . The received power is 1.118² = 1.25 times the power if just tower 1 transmitted at unit power.

If the two towers are both one unit away then θ = atan (1) = π/4 and CS = √2 and the received power is twice that if just one tower transmitted, but again for the same total transmission power budget of 1.

Lest you think that we have invented a perpetual motion machine that cheats mother nature out of energy we can find a point where the two transmitter signals are exactly out of phase and and the signals entirely cancel each other. That point would be nearer tower 2 and probably not on the line between them. Exactly where depends on the ratio of carrier wave length and distance between towers. We have avoided that number above. Zero signal power is available there. If either transmitter were turned off, the signal would appear at that point.

In short we localize the delivery of power to the intended recipient. Yet more towers do much better.