Problem 3. Consider the direct-sequence CDMA system as de- scribed in Lecture 15. At the receiver suppose that instead of using the local code qi(t) we instead use q₁(t+A) where A is some ± fraction of a bit time, i.e., the local code may be shifted one direction or the other. Compute the degradation (in dB) to E₁/No due to a nonzero A at the output of the corre- lator for BPSK signaling. You may assume for the local code that adjacent chips are equally likely to match or differ. Receiver CDMA Operations (cont.) Now, d,(t), i = 1,2,..., N is a constant (±1) over T, seconds. Note that q(t) = 1. So, for User 1 we get (ideally) Ть r(t)q₁(t)dt = d₁ 0 Where d₁ just denotes the constant value of d₁(t) at any time in the bit interval which is either +1 or -1 in the given 7,-second interval. Each user repeats the above operations using their particular code. CDMA Receiver Block Diagram From satellite Address Code Sync Correlator Phase-coherent carrier Tb Jo Bit Decision 0 CDMA Receiver

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Problem 3.
Consider the direct-sequence CDMA system as de-
scribed in Lecture 15. At the receiver suppose that instead of using the local
code qi(t) we instead use q₁(t+A) where A is some ± fraction of a bit time,
i.e., the local code may be shifted one direction or the other. Compute the
degradation (in dB) to E₁/No due to a nonzero A at the output of the corre-
lator for BPSK signaling. You may assume for the local code that adjacent
chips are equally likely to match or differ.
Transcribed Image Text:Problem 3. Consider the direct-sequence CDMA system as de- scribed in Lecture 15. At the receiver suppose that instead of using the local code qi(t) we instead use q₁(t+A) where A is some ± fraction of a bit time, i.e., the local code may be shifted one direction or the other. Compute the degradation (in dB) to E₁/No due to a nonzero A at the output of the corre- lator for BPSK signaling. You may assume for the local code that adjacent chips are equally likely to match or differ.
Receiver CDMA Operations (cont.)
Now, d,(t), i = 1,2,..., N is a constant (±1) over T, seconds.
Note that q(t) = 1.
So, for User 1 we get (ideally)
Ть
r(t)q₁(t)dt = d₁
0
Where d₁ just denotes the constant value of d₁(t) at any time in the bit interval
which is either +1 or -1 in the given 7,-second interval.
Each user repeats the above operations using their particular code.
CDMA Receiver Block Diagram
From satellite
Address
Code Sync
Correlator
Phase-coherent carrier
Tb
Jo
Bit
Decision
0
CDMA Receiver
Transcribed Image Text:Receiver CDMA Operations (cont.) Now, d,(t), i = 1,2,..., N is a constant (±1) over T, seconds. Note that q(t) = 1. So, for User 1 we get (ideally) Ть r(t)q₁(t)dt = d₁ 0 Where d₁ just denotes the constant value of d₁(t) at any time in the bit interval which is either +1 or -1 in the given 7,-second interval. Each user repeats the above operations using their particular code. CDMA Receiver Block Diagram From satellite Address Code Sync Correlator Phase-coherent carrier Tb Jo Bit Decision 0 CDMA Receiver
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