Question

2. Given the last NINE digits. Write out minterms with these numbers as subscripts of mi. You may remove the duplicated terms.

Given the NINE numbers are 5, 1, 1, 4, 6, 0, 0, 4, and 2. By removing a duplicated number ‘1’, '4', '0', the minterms are m0 and m4.
Then, answer the following SIX questions.
(a) Suppose there are FOUR input variables a,b,c, and d, and one output F1. OR the above
minterms together to obtain a canonical SOP. Write down the canonical SOP of F1.
(b) ADD 4 to each subscript of the minterms in (a) to get a new canonical SOP F2. Write
down the canonical SOP of F2.
(c) Convert the canonical SOP of F2 obtained in (b) to its equivalent canonical POS.
(d) Construct the truth table of the Boolean function of F1 and F2 obtained in (a) and (b).
(e) Write out the corresponding K-maps of the Boolean function of F1 and F2.
(f) Try to simplify the Boolean function of F1 and F2 by K-map obtained in (e).
3. Considering the canonical SOP F1 obtained in Q2, answer the following FOUR questions.
(a) MINUS 2 to each subscript of the minterms of F1 to get a new canonical SOP F3 that has
only THREE input variables a,b, and c. If the corresponding result is less than 0, set it to 0.
Simplify F3 by K-map.
(b) Draw out the logic diagram of F3 by three basic logic gates.
(c) Draw out the logic diagram of F3 by a 3-8 decoder.
(d) Draw out the logic diagram of F3 by a 8-to-1 multiplexer.

Answer 1

(a) The minterms are m0 = b'c'd' + a'c'd' + a'b'd' + a'b'c' and m4 = b'c'd + a'b'd + a'bc'd + a'bc' + abcd. ORing these together gives the canonical SOP of F1: F1 = m0 + m4 = b'c'd' + a'c'd' + a'b'd' + a'b'c' + b'c'd + a'b'd + a'bc'd + a'bc' + abcd

(b) Adding 4 to each subscript gives: F2 = m4,4 + m8,8 = b'c'd' + a'b'c'd + a'bc'd + abcd + b'c'd + a'b'c'd + a'bc' + abcd = b'c'd' + a'b'c'd + a'bc'd + 2abcd + a'bc'

(c) To obtain the POS of F2, apply DeMorgan's law to each term: F2 = (b+c+d)(a+c+d)(a'+b'+d')(a'+b'+c')' + (b+c+d)(a'+b+c+d')(a+b'+c+d')(a+b+c'+d')'(a'+b+c') + (b'+c+d')(a+b'+c+d')(a'+b+c+d')(a+b+c+d) = Π(0,2,5,6,9,11,14)'

(d) The truth table for F1 is:

a | b | c | d | F1 --+---+---+---+--- 0 | 0 | 0 | 0 | 1 0 | 0 | 0 | 1 | 1 0 | 0 | 1 | 0 | 1 0 | 0 | 1 | 1 | 1 0 | 1 | 0 | 0 | 1 0 | 1 | 0 | 1 | 1 0 | 1 | 1 | 0 | 1 0 | 1 | 1 | 1 | 1 1 | 0 | 0 | 0 | 1 1 | 0 | 0 | 1 | 0 1 | 0 | 1 | 0 |

Step-by-step explanation: