ECE 251 Homework solutions, Exercises of Digital Electronics

homework solutions of some assignments from jacob savir's ece 251 course

Typology: Exercises

2018/2019

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NEW JERSEY INSTITUTE OF TECHNOLOGY
DEPT. OF ELECTRICAL & COMPUTER ENGINEERING
ACADEMIC YEAR 2011-2012
SEMESTER 1
ECE251 Digital DESIGN
SOLUTION HW2
1. We have:
1.()a Long division:
66 0
33 1
16 0
8 0
4 0
2 0
1 1
0
Therefore, the 2โ€™s complement representations of +66 and โ€“66 using 8-bits are:
,
1.()bLong division:
38 0
19 1
9 1
4 0
2 0
1 1
0
PAGE 1
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NEW JERSEY INSTITUTE OF TECHNOLOGY

DEPT. OF ELECTRICAL & COMPUTER ENGINEERING

ACADEMIC YEAR 2011-

SEMESTER 1

ECE251 Digital DESIGN

SOLUTION HW

  1. We have: 1.()a Long division:

66 0 33 1 16 0 8 0 4 0 2 0 1 1 0

Therefore, the 2โ€™s complement representations of +66 and โ€“66 using 8-bits are:

,

1.()b Long division:

Therefore,.

1.()c Long division:

72 0 36 0 18 0 9 1 4 0 2 0 1 1 0

Therefore, the 2โ€™s complement representations of +72 and โ€“72 using 8-bits are:

,.

1.()d Long division:

55 1 27 1 13 1 6 0 3 1 1 1 0

Therefore,.

  1. From Pb. 1 we have:

, ,

The addition:

To check if the result represents โ€“17, we complement it and verify that it represents +17:

The 2โ€™s complement of 11101111 is 00010001, which has the value +17.

Expected

Carries

Discard overflow bit

The result is 10000001. To verify that it represents โ€“127, we complement it and verify that we get +127. The complemented pattern is 01111111 which, in fact, has the value of 127.

  1. We show that the truth tables of the left hand side (LHS) of the equation matches the truth table of the right hand side (RHS) of the equation.

LHS 000 0 0 0 0 001 0 0 0 0 010 1 0 0 1 011 1 0 0 1 100 0 1 0 1 101 0 1 0 1 110 0 0 0 0 111 0 0 1 1

RHS

Since LHS=RHS, the proposition is proved.

  1. (^) We have:
  2. The two logic implementations are: