Deriving Equation for Verilog Module

Deriving Equation for Verilog Module

By inspection of Table 1 we can make the following conclusions:

1. ALUControl[2] is equal to 1 when ALUOp[0] is equal to 1 in row 2 of Table 1

2. ALUControl[2] is equal to 1 when ALUOp[1] is equal to 1 and the values of the Funct bits (Funct[3:0]) satisfy some condition. The condition can found using K-Map

3. The Boolean Expression for ALUControl[2] can be set up using the reasoning from the 2 conclusions made previously into the following equation: Equation 1: ALUControl[2] = ALUOp[0] | ALUOp[1] & ( Exp1 ) Exp1 will be the Boolean expression found by K-Map. So the K-Map is set up as follows: Only look at Table 2 with the inputs and ALUControl[2] output column included

Funct Exp1

[3] [2] [1] [0]

0 0 0 0 0

0 0 1 0 1

0 1 0 0 0

0 1 0 1 0

1 0 1 0 1

The table above can serve as an incomplete truth table but a 4 variable system should have a 16-row truth table. Since the table above lists valid input combinations, minterms that are not shown can be considered as X (dont care) So the table is summarized as: Exp1 = m(2, 10) + X(1, 3, 6, 7, 8, 9, 11, 12, 13, 14, 15) Finally the K-Map for Exp1 will be:

00 01 11 10

00 0 X X 1

01 0 X X X

11 0 0 X X

10 X X X 1 Use the given K-Map to find Exp1 then substitute Exp1 into Equation 1.

By inspection of Table 1 we can make the following conclusions:

1. ALUControl[1] is equal to 1 in the first two rows of Table 1. When only ALUOp bits are considered for the first two rows of Table 1 possible combinations of values are ALUOp = 00, ALUOp = 01. Since ALUOp[1] has a value of 0 we can conclude that ALUControl[1] equals 1 when ALUOp[1] equals 0. Written as a Boolean expression we get: ALUControl[1] = ~ALUOp[1]

2. For the remainder of the Table 1, ALUControl[1] is equal to 1 when ALUOp[1] is equal to 1 and the values of the Funct bits (Funct[3:0]) satisfy some condition. The condition can found using K-Map

3. The Boolean Expression for ALUControl[1] can be set up using the reasoning from the 2 conclusions made previously into the following equation: Equation 2: ALUControl[1] = ~ALUOp[1] | ALUOp[1] & ( Exp2 ) Exp2 will be the Boolean expression found by K-Map. So the K-Map is set up as follows: Only look at Table 2 with the inputs and ALUControl[1] output column included

Funct Exp2

[3] [2] [1] [0]

0 0 0 0 1

0 0 1 0 1

0 1 0 0 0

0 1 0 1 0

1 0 1 0 1

The table above can serve as an incomplete truth table but a 4 variable system should have a 16-row truth table. Since the table above lists valid input combinations, minterms that are not shown can be considered as X (dont care) So the table is summarized as: Exp2 = m(0, 2, 10) + X(1, 3, 6, 7, 8, 9, 11, 12, 13, 14, 15) Use K-Map Minimization to find Exp2 then substitute Exp2 into Equation 2.

By inspection of Table 1 we can make the following conclusions: 1. ALUControl[0] is equal to 1 in the last two rows of Table 1. This means ALUOp[1] must equal 1 and the values of the Funct bits (Funct[3:0]) satisfy some condition. The condition can found using K-Map 2. The Boolean Expression for ALUControl[0] can be set up as follows: Equation 3: ALUControl[0] = ALUOp[1] & ( Exp3 ) Exp3 will be the Boolean expression found by K-Map. So the K-Map is set up as follows: Only look at Table 2 with the inputs and ALUControl[1] output column included

Funct Exp3

[3] [2] [1] [0]

0 0 0 0 0

0 0 1 0 0

0 1 0 0 0

0 1 0 1 1

1 0 1 0 1

The table above can serve as an incomplete truth table but a 4 variable system should have a 16-row truth table. Since the table above lists valid input combinations, minterms that are not shown can be considered as X (dont care) So the table is summarized as: Exp3 = m(5, 10) + X(1, 3, 6, 7, 8, 9, 11, 12, 13, 14, 15) Use K-Map Minimization to find Exp3 then substitute Exp3 into Equation 3