Interference is the major limiting factor in the performance of cellular radio systems.

Sources of Interference are:

• Another mobile in the same cell.
• A call in progress in the neighbouring cell.
• Other Base Station operating in the same frequency band
• Any other non cellular system which leaks energy into cellular frequency band. Effect of interference on voice channels results in cross talks. Effect on control channels results in dropped calls or missed calls. The two major types of system-generated cellular interference are co-channel interference and adjacent channel interference.

1) Co-channel Interference

Due to Frequency reuse, there are several cells in a given area that use the same set of frequencies. These are called as Co-channel cells and interference due to these cells is called as co-channel interference. Unlike thermal noise, which can be overcome by increasing the signal to-noise ratio (SNR), co-channel interference cannot be combated by simply increasing the carrier power of a transmitter. This is because an increase in carrier transmit power increases the interference to neighboring co-channel cells. S/I is related to cluster size, path loss exponent and can be derived mathematically. The S/I of the received signal should not be less than 18 dB. This rule was set by FCC.

Expression for S/I and Worst Case S/I Ratio

Consider the following scenario in which the desired signal strength is S and the cellphone is at a distance of $D_i$ from the $i^{th}$ interfering cell.

Let $i_0$ be the number of co channel interfering cells. Then the Signal to interference ratio for a mobile receiver which monitors a forward channel can be expressed as,

$$\frac{S}{I} = \frac{S}{\sum_{i=1}^{i_0} I_i}-----(1)$$

From Friis Free space equation, we know that the average received signal strength at any point is inversely proportional to the distance of separation between the transmitter and the receiver and is given by the following relation,

$$P_r = P_0 \times (\frac{d}{d_0})^{-n}-----(2)$$

Where $P_0$ is the reference power received at a reference distance $d_o$.

If R is the radius of the cell and D is the distance between cochannel cells ,then we can express S as,

$$S \alpha R^{-n}$$

$$And \ I \alpha D^{-n}$$

Hence Equation 1 can be written as,

$$\frac{S}{I} = \frac{R^{-n}}{\sum_{i=1}^{i_0} D^{-n}}-----(3)$$

Considering only first tier of interfering cells and assuming that all interfering cells are equidistant from the desired Base Station and is equal to distance D, then equation 3 can be written as,

$$\frac{S}{I} = \frac{D/R^n}{i_0} = \frac{\sqrt{(3 \times N)}^n}{i_0}-----(4)$$

Equation 4 relates S/I ratio to the cluster size which in turn determines the overall capacity of the system.

Worst Case S/I ratio

Approximate distances between the co-channel cells and the mobile at the edge is illustrated. As shown in above figure, worst case S/I will be experienced when the mobile station will be located at the corner of it’s own cell. It will receive minimum desired signal S and an increased amount of Interference signal from the cochannel cells.

From the approximate geometry shown, it can be seen that the mobile is at a distance of D+R from two farthest cells, D-R from the near most cells and at a distance of D from the other two cells. Assuming path loss exponent n=4, equation 3 can be rewritten as

$$\frac{S}{I} =\frac{R^{-4}}{2(D-R)^{-4} + 2(D + R)^{-4} + 2(D)^{-4}}$$

Taking $2R^{-4}$ common in the denominator

$$\frac{S}{I} = \frac{R^{-4}}{2R^{-4}[D/R -1)^{-4} + (D/R +1)^{-4}+ (D/R)^{-4}]}$$

$$\frac{S}{I} = \frac{R^{-4}}{2R^{-4}[(\sqrt{3N}-1)^{-4} + (\sqrt{3N}+1)^{-4} + (\sqrt{3N})^{-4}]}$$

Substituting value of N=7 in the above equation,

We calculate S/I= 17.27dB which is lesser than the threshold defined by FCC. Thus, we see that for a seven-cell cluster, the S/I ratio is slightly less than 18 dB for the worst case. To design the cellular system for proper performance in the worst case, it would be necessary to increase N to the next largest size, 12. Readers are encouraged to calculate and check the value of S/I for N =12.

Examples

1) In a N-cell reuse pattern (hexagonal geometry) with base stations at the centre of each cell with omni-directional antennas, What would be the D/R ratio required if a minimum value of C/I (S/I) = 18 dB must be ensured. Assume path loss exponent n = 3.1 and only tier 1 interferers.

Solution: We know that,$\frac{D}{R} = \sqrt{3 \times N}$

$\frac{S}{I} = \frac{D/R^n}{i_0} = \frac{\sqrt{(3 \times N)}^n}{i_0}$

$\frac{C}{I} = \frac{1}{i_0} (\frac{D}{R})^n$

$18 dB = 10 log (\frac{1}{i_0}(\frac{D}{R})^n)$

$10^{1.8} = \frac{1}{6} (\frac{D}{R})^{3.1}$

$\frac{D}{R} = 6.79$

2) Consider a N-cell reuse pattern (hexagonal geometry) with base stations at the centre of each cell with omni-directional antennas. What would be the value of N that will ensure that a minimum C/I = 18dB is maintained. Assume path loss exponent n = 2.73 and only tier 1 interferers are present.

$\frac{C}{I} = \frac{1}{i_0} (\frac{D}{R})^n$

$18 dB = 10 \ log (\frac{1}{i_0} (\sqrt{3N})^n)$

$10^{1.8} = 10 \ log (\frac{1}{6} (\sqrt{3N})^{2.73})$

N=25.8 ,Valid N that is greater than 25.8 is 27. Hence, Cluster size in this case is 27.

3) A cellular service provider implements a digital TDMA system which requires a minimum signal-to-interference ratio of 15 dB. Assume that only Tier-1 interferers need to be considered and the approximation that all Tier-1 interferers are equidistant from the mobile station. What is the optimal value of N considering omni-directional antennas ?

(Assume a path loss exponent of n = 4)

Minimum signal to interference ratio required =15 dB

Path Loss exponent(n)=4

Number of tier 1 interferers $i_o=6$

$\frac{1}{6} (\sqrt{3N})^4 \gt 15\ dB$

$\frac{1}{6} (\sqrt{3N})^4 \gt 10^{\frac{15}{10}}$

$N \gt 4.59$

Considering Valid cluster size, N = 7

2) Adjacent Channel Interference

Interference resulting from signals which are adjacent in frequencies to the desired signal is called Adjacent Channel Interference. It results in Near Far effect which is explained in the subsequent paragraph.

Consider a receiver and two transmitters, one close to the receiver, the other far away. If both transmitters transmit simultaneously and at equal powers, then receiver will receive more power from the nearer transmitter. This makes the farther transmitter more difficult, if not impossible, to understand. In short, the near–far problem is one of detecting or filtering out a weaker signal amongst stronger signals. To nullify Adjacent channel interference, Service Providers do not assign adjacent channels to the same cell. For e.g: For cluster size of 7, the allotment to Cell 1 will be Channels ( 1, 8, 15, 22 etc.)The allotment to cell 2 will be ( 2, 9, 16, 23 etc.)