--- title: "Reverse CDMA (IS-95) channel" subject: "Wireless Communication" chapter: 8 key: "semester_7/wireless_communication:reverse-cdma-is95-channel" source: https://www.bctengineer.com/q/semester_7/wireless_communication/reverse-cdma-is95-channel site: BCT Engineering --- # Reverse CDMA (IS-95) channel Wireless Communication · Chapter 8: Wireless Systems and Standards - Asked in: 2081 Baisakh, 4 marks - Video: [W 1.26 IS 95 Reverse Access and Traffic channels and 64ary modulation (2G CDMA)](https://www.youtube.com/watch?v=yJ8eHyYLj5M) — Ramanikumar animations -Telecom and Wireless - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/wireless_communication/reverse-cdma-is95-channel - Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-8 ## Question Draw reverse CDMA (IS-95) channel. ## Short answer IS-95 reverse channel uses access and traffic channels. Data is convolutionally coded (r=1/3, K=9), interleaved, 64-ary Walsh modulated, randomized, spread by user-specific long code and quadrature short PN codes with OQPSK at 1.2288 Mcps; no pilot, non-coherent detection. ## Answer The reverse (uplink) channel of IS-95 carries signals from the mobile to the base station on a 1.25 MHz carrier at 1.2288 Mcps. It has two kinds of channels: **access channels** (4800 bps, for call origination and replies to pages) and **reverse traffic channels** (user data). Users are separated by their **long PN code** (user-specific mask); since there is no pilot, the base station uses **non-coherent 64-ary orthogonal (Walsh) modulation**. ```text data 9600/4800/2400/1200 bps | [Conv. encoder r=1/3, K=9] -> 28.8 ksps | [Symbol repetition] -> 28.8 ksps | [Block interleaver, 20 ms] | [64-ary orthogonal modulator] 6 code symbols -> 1 Walsh word (64 chips) 4.8 ksym/s -> 307.2 kcps | [Data burst randomizer] (gating for low rates) | (XOR)<--[Long code generator] (user mask) | 1.2288 Mcps +--(x) I short PN --[filter]--> cos wct --+ | (+)-> out +--(x) Q short PN-[1/2 chip delay]-[filter] --> sin wct ---------+ (OQPSK) ``` **Steps** 1. Variable-rate vocoder data in 20 ms frames. 2. Rate 1/3, K = 9 convolutional coding → 28.8 ksps (lower rates repeated). 3. Block interleaving over 20 ms. 4. **64-ary orthogonal modulation**: each group of 6 symbols selects one of 64 Walsh codes → 307.2 kcps. 5. **Data burst randomizer**: at lower rates, only the needed power-control groups are sent (pseudo-random gating), saving power and reducing interference. 6. **Long code spreading** (42-bit, user mask from ESN) → 1.2288 Mcps; identifies the user. 7. **Quadrature spreading** by I and Q short PN codes, Q branch delayed by half a chip → **OQPSK**, which reduces envelope variations and suits the mobile's power amplifier. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does the IS-95 reverse link use non-coherent 64-ary orthogonal modulation? A. Because Walsh codes separate users on the reverse link B. Because the base station has no PN code C. To increase the chip rate to 1.2288 Mcps D. There is no pilot on the reverse link, so coherent detection is not possible Answer: D. Without a reverse pilot, the base station uses non-coherent detection of 64-ary Walsh words. (Key point: Reverse modulation) 2. What separates different users on the reverse channel? A. The user-specific long PN code (mask) B. Different Walsh codes assigned by the BS C. Different carrier frequencies D. Time slots Answer: A. Long code spreading with a user mask identifies each mobile. (Key point: User separation) 3. In the 64-ary orthogonal modulator, what happens to the code symbols? A. Each group of 6 symbols selects one of 64 Walsh words of 64 chips B. Each symbol selects one of 6 Walsh words C. Each 64 symbols give 6 chips D. Symbols are XORed with a pilot Answer: A. 28.8 ksps /6 = 4.8 ksym/s, each becoming 64 chips = 307.2 kcps. (Key point: 64-ary mapping) 4. Why is the Q branch delayed by half a chip in the reverse link? A. To synchronise with the pilot B. To form OQPSK, reducing envelope variation for the mobile's power amplifier C. To separate users D. To add error correction Answer: B. OQPSK avoids 180° phase transitions, easing amplifier requirements. (Key point: OQPSK) ## Related in Wireless Communication - [Forward CDMA (IS-95) channel](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aforward-cdma-is95-channel) — Wireless Communication, chapter 8, asked 4× - [Forward/reverse channel and 5G versus 4G](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aforward-reverse-5g-vs-4g) — Wireless Communication, chapter 1, asked 2× - [Forward and reverse channel](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aforward-reverse-channel) — Wireless Communication, chapter 1 - [Forward/reverse channel and features of 5G](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aforward-reverse-5g-features) — Wireless Communication, chapter 1 - [Channelization code and IS-95 forward channels](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Achannelization-code-is95-forward) — Wireless Communication, chapter 8 ## Related in other subjects - [Slope change 1 in 95 to 1 in 1420 with jump](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aslope-change-95-to-1420-jump-height-or-spillway) — Hydraulics, chapter 9