Channel Assignment Strategies for Cellular Phone Systems

Nowadays people benefit a lot from mobile communication systems. One of recent major breakthroughs in solving the problem of spectral congestion is the cellular concept, which means replacing a large transmitter with many little transmitters, each providing coverage to a small piece of the service area. Neighboring transmitters are assigned different groups of channels so that the interference between transmitters is minimized. Each cellular transmitter is assigned a radio channel to be used within a small geographic area called a cell. By limiting the coverage area to within the boundaries of a cell, the same channel may be used to cover different cells that are separated from one another by distances large enough to keep interference levels within tolerable limits. The design process of selecting and allocating channel groups for all of the cellular transmitters in a system is called frequency reuse or frequency planning. The hexagonal cell shape shown in the picture of the cover is conceptual and is a simplistic model of radio coverage for each transmitter, but it has been universally adopted since the hexagon permits easy and manageable analysis of a cellular system. Our team has made a comprehensive study on a practical problem of designing mobile communication systems. We know that there are several constraints on frequency assignment. First, any two transmitters within a distance four times as much as the length of the hexagon side cannot be assigned the same channel. Second, due to spectral spreading, transmitters within distance twice as much as the length of the hexagon side must be assigned channels which differ by at least k. We have developed quite an efficient strategy dependent on k to assign channels as few as possible to a given service area according to these rules. If we put our strategy into practice under simple situations, we have 7 channels to be the fewest when k equals 1, 9 channels when k equals 2, and 12 channels when k equals 3. When the situation becomes complicated as k > 3, we find the smallest number of channels needed is a linear function 2k + 7, which is a pretty simple solution. Thus we have the entire results as follows: span =    2k + 5, k = 1, 2 2k + 6, k = 3 2k + 7, k ≥ 4. 2 Patterns of our assignments are as follows: 2k+5 2k+7 3 k+5 k+4 2 1 k+3 2k+6