2.1.4.2 Realize Logical Expression using Linked LUTs

Below is an example on linking LUT0 to LUT1 and how to fill out the truth tables for both LUTs.

Imagine the following logical expression: (A•B⁢⊕ C)  + (D•E−).

The truth table for LUT0 may be created. LUT0 will take care of the first part of the logical expression (A•B⊕C).

This will give the following truth table:
Table 2-10. LUT0 Truth Table
CBAOUT
0000
0010
0100
0110
1000
1011
1101
1110

From the above truth table, 0x60 may go into LUT0 TRUTH register to realize the first part of the logical expression.

Now, the truth table for LUT1 must be created. Before this can be done, it must be decided what input to use on LUT1. All inputs can be used, and in this example LUT0 out is linked to LUT1 input 1. This will be equal to the second column of the truth table. If input 0 was used, it would have been the first column that may be used and the third if input 2 was used.

To make the development of the truth table for LUT1 easier, the expression could be simplified since LUT0 already has handled the first part. The expression can be viewed like this when creating the truth table for LUT1: (X⁢+D•E−), where X = (A•B⊕C)
Table 2-11. LUT1 Truth Table
EX DOUT
0000
0011
0101
0111
1000
1010
1101
1111

From the above truth table, 0xCE may go into LUT1 TRUTH register to realize the second part of the logical expression.