# frequency reuse ![](Public%20Extras/Doodles/frequency%20reuse-attachment.light.svg#invert) ![](Public%20Extras/Doodles/frequency%20reuse-attachment.png#invert) Neighbouring cells are assigned different groups of channels in order to minimise the interference. You will encounter at most 6 interfering frequencies in the first-tier interference. Replace a single high power transmitter with many low power transmitters. The same set of frequency channels is then reused at different geographical locations . Cells 1-7 form a cluster. ## Example Two 5MHz bands - 200kHz wide frequency channel. ### Scenario 1 Without frequency reuse, supports $\frac{5e6\times8}{200e3} = 200$ users ### Scenario 2 With cellular structure, if each frequency channel is reused L times: Support $\frac{5e6\times L\times8 }{200e3} = 200L$ users at once (if they space out!) Capacity increases almost linearly! #### But there is no free lunch! We haven't spoken about the performance of every user. If you can maintain the same performance, you do have the linear gain! we are missing out on the effects of interference! How do you ensure the per user performance after using frequency reuse. ## Mathematical Definitions Red uses the same frequency. ![](Public%20Extras/Doodles/frequency%20reuse-attachment-1.png#invert) ### Terms D is the centre to centre distance between reused cells in each cluster. $D_0$ is the centre to centre distance between cells within a cluster. Cluster Size: $N=\frac{D^2}{D_0^2}$ Frequency reuse factor $=\frac{1}{N}$ ### Frequency Reuse Pattern By exploiting the structure of hexagons (triangular) and the cosine rule: ![](Public%20Extras/Doodles/frequency%20reuse-attachment-2.png#invert) $D^2=(iD_0)^2+(jD_0)^2-2(iD_0)(jD_0)cos(120\degree)$ $N = i^2+ij+j^2$ where $i,j = 0,1,2,\ldots$ ### Typical values for the frequency reuse factor 1,3,4,7,9 #### Example: N=19 i = 3, j = 2 ![](Public%20Extras/Doodles/frequency%20reuse-attachment_0.light.svg#invert) ## Other values $R$ is the distance between BS and farthest MS (the longest distance in the cell). $\frac{D^2}{D_0^2}=N \Rightarrow D=\sqrt{N}D_0$ so $\Rightarrow \frac{D}{R}=\frac{\sqrt{N}D_0}{R}$ and $\frac{D_0}{2}=R\cos{30\degree}$ so $\frac{D}{R}=\sqrt{3N}$ ## How often can a frequency channel be reused? Depends on the performance - you can use it aggressively but the users see worse performance. Determined by [co-channel interference](co-channel%20interference.md) [^1] # References [^1]: