- 2077 Chaitra · 2+6 marks
Spread spectrum disadvantages; CDMA implementation
Short answer
Spread spectrum is bandwidth inefficient for one user, has near-far problem, self-jamming, complex receivers, and strict synchronization. In DS-CDMA each user multiplies data by a unique high-rate code; the receiver correlates with the wanted code, recovering data while others remain noise. Example: orthogonal codes A=+1-1+1-1 and B=+1+1-1-1; A sends +1, B sends -1, channel sum 0 -2 +2 0; receiver A gives +1, receiver B gives -1.
What are the major disadvantages of spread spectrum system? Explain the implementation of CDMA system with appropriate example.
Disadvantages of spread spectrum systems
- Bandwidth inefficient for a single user – a signal of a few kHz is spread over MHz.
- Near-far problem – a nearby user's strong signal swamps far users; precise power control is needed.
- Self-jamming – codes are not perfectly orthogonal, so users interfere with each other.
- Complex receivers – code acquisition, tracking, RAKE receiver and high-speed chip processing.
- Strict synchronisation of PN codes is required.
Implementation of a CDMA system
In DS-CDMA every user transmits on the same carrier at the same time. Each user's data is multiplied by a unique high-rate PN (spreading) code. The receiver multiplies the received sum by the wanted user's code; only that user is despread back to narrowband, while all others stay as wideband noise.
Transmitter (user k)
data d_k(t) --(x)--(x)--> sum of all users
^ ^ in the channel
c_k(t) carrier
Receiver for user 1
r(t) --(x)--(x)--[Integrate over Tb]--> d1
^ ^
carrier c_1(t)
Steps:
- Data at bit rate R_b is multiplied by the code c_k(t), whose chip rate R_c >> R_b. Processing gain G_p = R_c / R_b.
- The spread signal modulates the carrier (BPSK/QPSK) and is sent.
- The receiver gets r(t) = Σ d_k(t)·c_k(t) + noise.
- Multiplying by c_1(t): since c_1·c_1 = 1, user 1's data is recovered; cross-terms d_k·c_k·c_1 remain spread because the codes have low cross-correlation.
- Integrating over one bit gives d_1; interference is reduced by G_p.
Example (chip level): Let code of A = +1 −1 +1 −1, code of B = +1 +1 −1 −1 (orthogonal).
- A sends bit +1 → +1 −1 +1 −1; B sends bit −1 → −1 −1 +1 +1.
- Channel sum = 0 −2 +2 0.
- Receiver A: (0)(+1)+(−2)(−1)+(+2)(+1)+(0)(−1) = 4 → 4/4 = +1 (A's bit).
- Receiver B: (0)(+1)+(−2)(+1)+(+2)(−1)+(0)(−1) = −4 → −1 (B's bit).
Real systems (IS-95) use 1.2288 Mcps, Walsh codes for channel separation, PN short codes for cell separation, power control and a RAKE receiver.
✓ Answer checked by Claude Opus 5.5 (high effort)
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