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# spread the transmitted bit it is multiplied by the spreading sequence. So the transmitted signal

through the communication channel becomes  ∗  = (1, 1, 1, 1). Assuming the spreading

sequence at the receiver is the same as the transmitter, the dot product of the transmitted data and

the spreading sequence can be taken 1, 1, 1, 1 ∙  = 1 1 + 1 ∗ −1 + 1 1 + 1

# 1 = −4. . The next step is to make a hard decision based on the dot product of the received

signal and spread sequence. Since the dot product equals -4 and -4 is less than zero, the despread

bit is -1, which matches the original bit we transmitted. The iterative detector model makes

decisions in a similar matter, however, in addition the N components that are used in the dot

product, it incorporates N more components from the SESS signal (as explained in section 2.4-

2.6).

# The difference between the channel performance of Rayleigh fading and worst-case

channel when an iterative detector is present, can most likely be attributed to the difference in the

channel that lead to the hard decision that is made. In the jamming channel for low ρ values, a

single chip value may be jammed with a very large amount. This single chip is weighted on the

hard bit decision so much that it dominates the decision. For instance, in the example in the

previous paragraph, if one chip of the data at the receiver is heavily jammed the dot product can

be represented as (where the 20 represents the jammed chip, as it changed from -1 to 20):

1, 1, 1, 20 ∙  = 1 1 + 1 ∗ −1 + 1 1 + 20 1 = 17

# This shows how the worst-case jamming can effect on a spreading system. Adding an iterative

detector is certainly going to help by adding more chips to the hard decision, and it does by 6 db,

but it does not come close the 15 db in a Rayleigh fading channel. It would appear that a greater

the N, the greater the chance that the jamming will not affect the system. However, it has been

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