Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

ECE 251 Homework solutions, Exercises of Digital Electronics

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

Typology: Exercises

2018/2019

Uploaded on 09/26/2019

dresre
dresre 🇺🇸

3 documents

1 / 5

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
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 2’s complement representations of +66 and –66 using 8-bits are:
,
1.()bLong division:
38 0
19 1
9 1
4 0
2 0
1 1
0
PAGE 1
pf3
pf4
pf5

Partial preview of the text

Download ECE 251 Homework solutions and more Exercises Digital Electronics in PDF only on Docsity!

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 2’s complement representations of +66 and –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 2’s complement representations of +72 and –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 –17, we complement it and verify that it represents +17:

The 2’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 –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: