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# DigitalCommons@University of Nebraska - Lincoln - page 64 / 69

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2 = 2 1 , 2 0 , , 2 ( + 1 )

, 2(+2)



• 2 , (this is a +1 or -1)

# Step 3:

Take the soft decision made in step 2 and multiply it with the first chip. By multiplying the first

chip by the bit that was sent (2), this gives the original value of 1. (See below). This is needed

to determine if the value 1 is positive or negative.

• 

2′ 1  =

2 2 1 , 2 0 , , 2 ( + 1 )

, 2(+2)

=

, 0 ,

, ( + 1 ) ( + 2 )

It should be noted that in Bit 2, 1 is the first chip. In proceeding Bits it will be different.

# Step 4:

Repeat steps 2 and 3, but use the proceeding Bits (3,4,5…) till Bit N+1.



3 = 3 2 , 3 1 , , 3 ( + 2 )

, 3(+3)

• 3 (this is +1 or -1)

• 

3′ 1  =

• 3

3 2 , 3 1 , , 3 + 2

, 3(+3)

= 2 ,

, , (+2)

, (+3)

• ..

Through Bit N+1

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