CDMA Receiver Simulator Show Theory & Math What’s Happening Inside the CDMA Receiver CDMA allows multiple users to share the same bandwidth by spreading each user’s data with a unique orthogonal code. s[n] = b[k] · c[n] r[n] = A₁s₁[n] + A₂s₂[n] x[n] = r[n] cos(2πfₙ/fₛ) z[n] = y[n] · c[n] b̂ = sign(Σ z[n]) Matched filtering maximizes SNR and allows bit recovery even in the presence of strong interferers (near–far problem). Codewards / Spread Spectrum c1 = [1, 1, -1, -1] c2 = [1, -1, 1, -1] User 1 Power: User 2 Power: User 1 Bits: User 2 Bits: Run Simulation Show Received Bits

DSSS-CDMA  This explanation builds intuition step by step for DSSS-CDMA , chip rate , processing gain , and how multiple users share the same analog carrier . What DSSS-CDMA Really Is All users transmit at the same time Over the same frequency band Using the same RF carrier Separated using codes FDMA → Different frequencies TDMA → Different time slots CDMA → Same time + same frequency, different codes Role of the Carrier Signal There is one common carrier frequency for all users. Carrier frequency: 900 MHz Bandwidth: 1.25 MHz (IS-95) The carrier is NOT assigned per user. s k (t) = b k (t) · c k (t) · cos(2πf c t) b k (t): user data (low rate) ...

  In a code-division multiple access (CDMA) system with N = 8 chips , the maximum number of users who can be assigned mutually orthogonal signature sequences is ______. Solution: Maximum Users in CDMA In a CDMA system, each user is assigned a unique signature sequence of length N chips . The maximum number of users that can be supported without mutual interference is equal to the number of mutually orthogonal sequences available. Key principle: The maximum number of mutually orthogonal sequences of length N is N . Given: N = 8 chips Maximum number of users = 8 Final Answer: 8 users

  The transmitted signal in a GSM system is of 200 kHz bandwidth , and 8 users share a common bandwidth using TDMA . If at a given time 12 users are talking in a cell , the total bandwidth of the signal received by the base station of the cell will be at least (in kHz). Solution: GSM Bandwidth with TDMA In a GSM system, one carrier channel has a bandwidth of 200 kHz and supports 8 users using TDMA. Bandwidth per user: Bandwidth per user = 200 kHz / 8 = 25 kHz Given that 12 users are talking simultaneously, the number of required carrier channels is: Number of carriers = 12 / 8 = 1.5 Since partial carriers are not possible, we need 2 full carriers . Total bandwidth received at the base station: Total Bandwidth = 2 × 200 kHz = 400 kHz Final Answer: 400 kHz

  Four messages band-limited to W, W, 2W, and 3W respectively are to be multiplexed using Time Division Multiplexing (TDM) . The minimum bandwidth required for transmission of this TDM signal is: Options: (a) W (b) 3W (c) 6W (d) 7W Solution: Minimum Bandwidth for TDM For a band-limited signal of bandwidth B , the minimum sampling rate required is 2B (Nyquist criterion). Sampling rates of the four messages: Bandwidth W → Sampling rate = 2W Bandwidth W → Sampling rate = 2W Bandwidth 2W → Sampling rate = 4W Bandwidth 3W → Sampling rate = 6W Total sampling rate: 2W + 2W + 4W + 6W = 14W In TDM, the minimum required transmission bandwidth is half the total sampling rate: Required Bandwidth = 14W / 2 = 7W Correct Answer: (d) 7W

  Quadrature Multiplexing Quadrature multiplexing (often called quadrature modulation / QAM/QPSK ) is a technique in which two independent signals are transmitted simultaneously over the same carrier frequency by placing them on orthogonal carriers . How it works Two carriers of the same frequency are used: In-phase carrier (I): cos(2πf c t) Quadrature carrier (Q): sin(2πf c t) These carriers are 90° out of phase (orthogonal). If: m 1 (t) → In-phase signal m 2 (t) → Quadrature signal The transmitted signal is: s(t) = m 1 (t) cos(2πf c t) + m 2 (t) sin(2πf c t) Why it works (key idea) ∫ cos(2πf c t) sin(2πf c t) dt = 0 So the two signals do not interfere with each other and can be separated at the receiver. Advantages Doubles data rate without increasing bandwidth Efficient use of spectrum Basis of QPSK, QAM, OFDM Where it is used Quadratur...

  A digital communication system uses a repetition code for channel encoding/decoding. During transmission, each bit is repeated three times (0 is transmitted as 000 and 1 is transmitted as 111). It is assumed that the source puts out symbols assumed independently and with equal probability. The decoder operates as follows: in a block of three received bits, if the number of zeros exceeds the number of ones, the decoder decides in favor of a 0; if the number of ones exceeds the number of zeros, the decoder decides in favor of a 1. Assuming a binary symmetric channel with crossover probability p = 0.1, the average probability of error is ________. 3-Repetition Code over a Binary Symmetric Channel Consider a Binary Symmetric Channel (BSC) with crossover probability p . Transmission rule: 0 → 000 1 → 111 The receiver applies majority decoding on the three received bits. Condition for Error A decoding error oc...

Instructions for Alamouti Scheme using BPSK Generate Message Generate NRZ Transmit from Antenna 1 Transmit from Antenna 2 Generate Message Generate NRZ Tx Antenna 1 Tx Antenna 2 Channel Coefficients Received Signal Channel Matrix h11= +j h21= +j Show Received Signal Perform Demodulation