Homework #7 KEY

Section 2.6 page 180 in textbook

1. #30 part a Find the one’s complement representations, using bit strings of length six, of 22₁₀

010110₂

2. #30 part c Find the one’s complement representations, using bit strings of length six, of -7₁₀

+7 = 000111₂ -7 = 111000₂

3. #36 part a Find the two’s complement representations, using bit strings of length six, of 22₁₀

010110₂

remember: positive numbers are the same in all three representations

4. #36 Part c Find the two’s complement representations, using bit strings of length six, of -7₁₀

-7₁₀ = 111001₂

The next 6 questions relate to the Boolean function described below by a Karnaugh map:

	cd
ab	\	00	01	11	10
	00	1			1
	01	1	x		1
	11	x		1
	10	1		1

5. find its simplest Product of Sums expression

	cd
ab	\	00	01	11	10
	00	0	1	1	0
	01	0	x	1	0
	11	x	1	0	1
	10	0	1	0	1

F’(a,b,c,d) = a’d + c’d + acd’

F’’(a,b,c,d) = F(a,b,c,d) = (a’d + c’d + acd’)’ = (a’’+d’)(c’’+d’)(a’+c’+d’’) = (a+d’)(c+d’)(a’+c’+d)

6. find its equivalent Sum of Products expression

	cd
ab	\	00	01	11	10
	00	1			1
	01	1	x		1
	11	x		1
	10	1		1

F(a,b,c,d) = c’d’ + a’d’ + acd

7. Using truth tables, show that your answers to 5 and 6 produce the same function

a	b	c	d	a'	c'	d'	c'd'	a'd'	acd	SOP	a+d'	c+d'	a'+c'+d	POS
0	0	0	0	1	1	1	1	1	0	1	1	1	1	1
0	0	0	1	1	1	0	0	0	0	0	0	0	1	0
0	0	1	0	1	0	1	0	1	0	1	1	1	1	1
0	0	1	1	1	0	0	0	0	0	0	0	1	1	0
0	1	0	0	1	1	1	1	1	0	1	1	1	1	1
0	1	0	1	1	1	0	0	0	0	0	0	0	1	0
0	1	1	0	1	0	1	0	1	0	1	1	1	1	1
0	1	1	1	1	0	0	0	0	0	0	0	1	1	0
1	0	0	0	0	1	1	1	0	0	1	1	1	1	1
1	0	0	1	0	1	0	0	0	0	0	1	0	1	0
1	0	1	0	0	0	1	0	0	0	0	1	1	0	0
1	0	1	1	0	0	0	0	0	1	1	1	1	1	1
1	1	0	0	0	1	1	1	0	0	1	1	1	1	1
1	1	0	1	0	1	0	0	0	0	0	1	0	1	0
1	1	1	0	0	0	1	0	0	0	0	1	1	0	0
1	1	1	1	0	0	0	0	0	1	1	1	1	1	1

8. Write the answer to #56 in disjunctive normal form –

It is actually already in disjunctive normal form: c’d’ + a’d’ + acd

If you expand it so that each minterm contains each variable,

this is what you get (minus the green minterms which are duplicates of the minterms below them)

a’d’ = a’bd’ + a’b’d’ = a’bcd’ + a’bc’d’ + a’b’cd’ + a’b’c’d’ +

c’d’ = ac’d’ + a’c’d’ = abc’d’ + ab’c’d’ + a’bc’d’ + a’b’c’d’ +

acd = abcd + ab’cd

9. Write the answer to #65 in conjunctive normal form

It is actually already in conjunctive normal form: (a+d’)(c+d’)(a’+c’+d)

If you expand it so that each maxterm contains each variable,

this is what you get:

(a+b+c+d’)(a+b+c’+d’)(a+b’+c+d’)(a+b’+c’+d’)(a’+b’+c+d’)(a’+b+c+d’)(a+b’+c+d’)(a+b+c+d’)(a’+b’+c’+d)(a’+b+c’+d)

Note that the underlined maxterms are duplicates as are the outlined maxterms

10. Using MMLogic (or any other circuit design program) - construct the circuit for either #5 or #6 (inputs should be switches, output should be a lamp). Bring a disk with your circuit on it to class (as well as a printout).