Tiffany D. answered • 03/12/21
B.S. Electrical and Computer Engineering
⦁ List the minterms of the following expression.
X = ~A~BC~D + ~ABC~D + A~B~C~D + A~B~CD + A~BC~D + A~BCD + AB~CD
X = ∑ m(0010 + 0110 + 1000 + 1001 + 1010 + 1011 + 1101)
X = ∑ m(2 + 6 + 8 + 9 + 10 + 11 + 13)
minterms: m2, m6, m8, m9, m10, m11, m13
⦁ Create a table of first level minterms of the expression.
Groups are based on the number of 1s in each binary representation.
For ex: minterms m1 (0001) and m8 (1000) both have one 1, so they belong in the same group.
- None of the minterms have zero 1s, so there isn't a group for that one.
- m2 (0010) and m8 (1000) have one 1, so they belong in a group.
- m6 (0110), m9 (1001), and m10 (1010) have two 1s, so they belong in a group.
- m11 (1011) and m13 (1101) have three 1s, so they belong in a group.
Group | Minterm | ABCD
---------------------------------------
0 | m2 | 0010
| m8 | 1000
---------------------------------------
1 | m6 | 0110
| m9 | 1001
| m10 | 1010
--------------------------------------
2 | m11 | 1011
| m13 | 1101
⦁ Create a table of second level minterms of the expression.
Compare the minterms in successive groups, meaning if there is a change in only 1-bit position, group those minterms and place a '_' symbol in that position, leaving the other bits the same.
For ex: m0 (0000) and m1 (0001) only differ in their least significant bit. Therefore, they would be a matched pair and the representation would be 000_.
Groups are separated by which groups from the previous table we are comparing in consecutive order.
For ex: Group 0 consists of the comparisons between Group 0 and Group 1 of the previous table. Group 1 consists of the comparisons between Group 1 and Group 2.
m2-m6 only differ in the 3rd most least significant bit. In m2, there is a 0 in this place, and in m6, there is a 1 in this place. Therefore, in this table, there will be a '_' in the 3rd bit's place.
0010
0110
0_10
Group | Matched Pair | ABCD
-----------------------------------------------------
0 | m2m6 | 0_10
| m8m9 | 100_
| m8m10 | 10_0
-----------------------------------------------------
1 | m9m11 | 10_1
| m9m13 | 1_01
| m10m11 | 101_
This step will be repeated until there are no more matched pairs.
Compare pairs in group 0 with each pair in group 1.
m8m10 and m9m11 share the same bits except for one so they move on.
Group | Matched Pair | ABCD
-------------------------------------------
0 | m8m10,m9m11 | 10_ _
There are no more matching pairs, so we are left with the pairs left standing.
(m2,m6) = 0_10 = ~AC~D
(m9,m13) = 1_01 = A~CD
(m8m10,m9m11) = 10_ _ = A~B
⦁ List the prime implicants of the expression.
Prime implicants: ~AC~D, A~CD, A~B
⦁ Check the coverage of each prime implicant and determine the final reduced expression for the expression.
~AC~D: 2 = 1, 6 = 1
A~CD: 9 = 1, 13 = 1
A~B: 8 = 1, 9 = 1, 11 = 1, 10 = 1
X = ~AC~D + A~CD + A~B
