--- title: "CCI vs ACI and co-channel reuse ratio proof" subject: "Wireless Communication" chapter: 2 key: "semester_7/wireless_communication:cci-aci-prove-reuse-ratio" source: https://www.bctengineer.com/q/semester_7/wireless_communication/cci-aci-prove-reuse-ratio site: BCT Engineering --- # CCI vs ACI and co-channel reuse ratio proof Wireless Communication · Chapter 2: Cellular Mobile Communication Concepts - Asked in: 2081 Bhadra, 3+3 marks - Video: [Frequency Reuse Factor | Co-Channel Interference & Signal To Interference Ratio](https://www.youtube.com/watch?v=JfIIiecbEg8) — ECIT Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/wireless_communication/cci-aci-prove-reuse-ratio - Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-2 ## Question Explain the difference between co-channel and adjacent channel interference. Prove that the co-channel reuse ratio is given by Q = √(3N), where N = i² + ij + j² is the cluster size. ## Short answer Co‑channel interference: same‑frequency cells; mitigated by large reuse distance (Q = D/R = √(3N)). Adjacent channel interference: neighboring frequencies leak due to imperfect filtering and near‑far effect, reduced by filtering and channel separation. The proof: moving i and j cells in hexagonal lattice, D² = 3R²(i²+ij+j²); from N = i²+ij+j², Q = √(3N). ## Answer #### Co-channel interference (CCI) - Interference between signals from **cells that use the same set of frequencies** (co-channel cells) under frequency reuse. - Cannot be removed by increasing transmitter power, because the interfering signal grows equally. It is reduced only by **physically separating co-channel cells** by a minimum distance D, i.e. a larger reuse ratio Q = D/R = √(3N), or by sectoring. - Measured by S/I = (√(3N))ⁿ / i₀ for the first tier of i₀ interferers. AMPS needs S/I ≥ 18 dB, which gives N = 7 with n = 4. #### Adjacent channel interference (ACI) - Interference from signals at **frequencies adjacent (next) to the wanted signal**, caused by imperfect receiver filters letting nearby frequencies leak into the passband. - Becomes serious through the **near–far effect**: a nearby transmitter on an adjacent channel overpowers a weak, distant wanted signal. - Reduced by careful **filtering** and **channel assignment** – not assigning adjacent channels in the same cell, keeping a large frequency separation between channels in a cell (e.g. channels 1, 22, 43, … in AMPS), and power control. | Point | Co-channel | Adjacent channel | |---|---|---| | Source | Same frequency in other cells | Neighbouring frequency | | Cause | Frequency reuse | Imperfect filtering, near–far effect | | Remedy | Larger D/R, sectoring | Better filters, channel separation, power control | | Depends on | Cluster size N | Filter selectivity, channel plan | #### Proof of Q = √(3N) **Assumptions:** hexagonal cells of radius R (centre to vertex); distance between centres of two adjacent cells = √3 R. **Locating a co-channel cell:** in a hexagonal layout, the nearest co-channel cell is found by moving *i* cells along a chain of hexagons, turning 60° anticlockwise, and moving *j* cells. ```text * Co-channel cell / / j cells / (after a 60° turn) A *-------------* i cells D = straight-line distance from A to the co-channel cell ``` The two paths have lengths i√3R and j√3R, and the angle between them (inside the triangle) is 120°. By the cosine rule: ```text D² = (i√3R)² + (j√3R)² − 2(i√3R)(j√3R) cos 120° = 3R²i² + 3R²j² + 3R²ij (cos 120° = −1/2) = 3R² (i² + ij + j²) ``` **Relating to cluster size:** the area of a hexagon is proportional to the square of the distance between centres of neighbouring hexagons. A cluster of N cells repeats with centre spacing D, while single cells repeat with spacing √3R, so ```text N = (D / √3R)² = i² + ij + j² ``` Substituting: ```text D² = 3R² N D = R √(3N) Q = D / R = √(3N) ``` | N | i, j | Q = √(3N) | |---|---|---| | 3 | 1, 1 | 3.00 | | 4 | 2, 0 | 3.46 | | 7 | 2, 1 | 4.58 | | 12 | 2, 2 | 6.00 | A larger Q means more separation between co-channel cells, so less co-channel interference but fewer channels per cell (less capacity). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Co-channel interference is caused by: A. Imperfect receiver filters passing neighbouring frequencies B. Cells that use the same set of frequencies under frequency reuse C. Multipath delay spread D. Using too many sectors in a cell Answer: B. CCI comes from co-channel cells, i.e. a consequence of frequency reuse. (Key point: Co-channel interference cause) 2. Why does increasing transmitter power not reduce co-channel interference? A. Higher power makes the receiver filters worse B. Higher power reduces D/R C. Co-channel cells then use different frequencies D. The interfering signal increases by the same amount, so S/I stays the same Answer: D. Raising power in all cells scales signal and interference equally; only larger D/R or sectoring helps. (Key point: Remedy for CCI) 3. Adjacent channel interference is mainly worsened by: A. The near-far effect, where a nearby adjacent-channel transmitter overpowers a weak wanted signal B. Using a large cluster size N C. Having too many first-tier co-channel cells D. Reusing the same frequency in distant cells Answer: A. ACI leaks through imperfect filters and is serious when an unwanted nearby transmitter is much stronger than the wanted one. (Key point: Adjacent channel interference) 4. In the proof of Q = sqrt(3N), the two path lengths i*sqrt(3)R and j*sqrt(3)R are joined with which angle in the cosine rule, and what result does it give? A. 60 degrees, giving D^2 = 3R^2(i^2 - ij + j^2) B. 90 degrees, giving D^2 = 3R^2(i^2 + j^2) C. 120 degrees, giving D^2 = 3R^2(i^2 + ij + j^2) D. 120 degrees, giving D^2 = R^2(i^2 + ij + j^2) Answer: C. With cos 120 = -1/2 the cross term becomes +3R^2 ij, so D^2 = 3R^2(i^2 + ij + j^2). (Key point: Cosine rule step) ## Related in Wireless Communication - [CCI/ACI, reuse ratio proof and channels per cell](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acci-aci-reuse-duplex-channels) — Wireless Communication, chapter 2 - [Co-channel and adjacent channel interference](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acochannel-adjacent-interference-terms) — Wireless Communication, chapter 2 - [Derivation of co-channel reuse ratio](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aderive-cochannel-reuse-ratio) — Wireless Communication, chapter 2 - [Frequency reuse and co-channel reuse ratio](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Afrequency-reuse-cochannel-reuse-ratio) — Wireless Communication, chapter 2 - [Prove Q = √3N; optimal N for 15 dB S/I (n = 4)](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Areuse-ratio-optimal-n-n4) — Wireless Communication, chapter 2 - [Numerical: co-channel distance for 7-cell reuse](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acochannel-distance-seven-cell) — Wireless Communication, chapter 2