ece 251 homework 3 solutions, Exercises of Digital Electronics

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NEW JERSEY INSTITUTE OF TECHNOLOGY DEPT. OF ELECTRICAL & COMPUTER ENGINEERING

ACADEMIC YEAR 2011- SEMESTER 1

ECE251 Digital DESIGN

Solution HW

( a)

(b)

(c)

( d)

(a)

( b)

(c)

(a) Given. The truth table:

n xyz xy+z y+xz F 0 000 0 0 0 1 001 1 0 0 2 010 0 1 0 3 011 1 1 1 4 100 0 0 0 5 101 1 1 1 6 110 1 1 1 7 111 1 1 1

(b) Given. Truth table:

n xyz xy+z y+xz F 0 000 0 0 0 1 001 1 0 0 2 010 0 1 0 3 011 1 1 1 4 100 0 0 0 5 101 1 1 1 6 110 1 1 1 7 111 1 1 1

n wxyz F 0 0000 0 0 0 0 0 1 0001 1 0 0 1 1 2 0010 0 0 0 0 0 3 0011 0 0 0 1 1 4 0100 0 0 0 0 0 5 0101 1 0 0 0 1 6 0110 0 0 0 0 0 7 0111 0 0 0 0 0 8 1000 0 0 0 0 0 9 1001 1 0 0 0 1 10 1010 0 0 0 0 0 11 1011 0 0 0 0 0 12 1100 0 1 1 0 1 13 1101 1 1 0 0 1 14 1110 0 0 1 0 1 15 1111 0 0 0 0 0

4. We have:

The implementation:

Figure 1: Minimum implementation

(a) (b)

6.

(a) (b) (c)

7. We use the identity:. (a) (b) (c) (d) 8. (a) The Boolean function is computed from inputs to output:

Figure 2: Computation of the raw Boolean function (b) We simplify the Boolean expression from (a):

(c) The implementation:

Figure 3: Minimum SOP implementation

9. We have:

Figure 4: Computation of Boolean function

(a) (b)