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Assignment No. 01
Semester: Fall  2016
Digital Logic Design – CS302

Topics Covered: Number systems to

Boolean Algebra & Logic Simplification

 

Total Marks: 20

 

Due Date: 14 Nov, 2016

 

Objectives:

To understand different Number Systems with its conversion from one to another and Boolean Algebra with its implementation with Logic Gates.

Instructions:

Please read the following instructions carefully before submitting assignment:

It should be clear that your assignment will not get any credit if:

 

  • The assignment is submitted after due date.
  • The assignment is submitted via email.
  • The assignment is copied from Internet or from any other student.
  • The submitted assignment does not open or file is corrupt.
  • It is in some format other than .doc/docx.

 

Note: All types of plagiarism are strictly prohibited.

For any query about the assignment, contact at CS302@vu.edu.pk

 

 

 

Important!

You have to provide all the steps of processing in all questions otherwise, marks will be deducted.

Question No. 01                                                                                                                          5 Marks

In the binary number system, we represent numeric values using two different symbols which is typically 0 and 1. Suppose we have a tertiary number system, which consists of three different symbols i.e. 0, 1 and 2. You have to perform the following operation defined in a given expression where we have to subtract a tertiary number from a decimal number and have to express the output in a binary number system.

(576)10 – (110002)3 = (_?_)2

 

Question No. 02 (a)                                                                                                                     7 Marks

Suppose we have a digital circuit expressed through the following truth table. You have to write simplified Boolean expression using Boolean Algebra.

Note: You have to write the Boolean Algebra Rule name for each simplification step.

 

A

B

C

D

Output=Y

0

0

0

0

1

0

0

0

1

1

0

0

1

0

0

0

0

1

1

0

0

1

0

0

0

0

1

0

1

1

0

1

1

0

0

0

1

1

1

1

1

0

0

0

1

1

0

0

1

1

1

0

1

0

0

1

0

1

1

0

1

1

0

0

0

1

1

0

1

0

1

1

1

0

0

1

1

1

1

0

 

Question No. 02 (b)                                                                                                                    8 Marks

Draw the circuit diagram for the simplified Boolean expression obtained from Question No. 02 (a).

 

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Replies to This Discussion

Please Discuss here about this assignment.Thanks

Our main purpose here discussion not just Solution

We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.

Solution of Question 1 is in attachment

Correct

Attachments:

Rizwan Ahmed  thanks for sharing 

wrong sir

mry first ka answer (11111010)2 hai or second ka A.B+C+D
first wale mn mn ne phly tertiary ko decimal mn convert kia
phir dono decimal values ko minus kia or answer ko binary mn convert kr dia

These are Ternary Number System not  tertiary.

conversion of ternary number is totally same as binary but we convert decimal to binary then we divide decimal value into 2 now we have ternary number i this condition you'll divide decimal to 3 exactly same method as binary.

Answer 01:

Method 01:

ð (576)10 = (210100)3

ð  (210100)3 - (110002)3 = (_?_)2

ð  210100 – 110002 = (11111010)2

Method 02:

ð (110002)3 = (326)10

ð  (576)10 - (326)10 = (250)10

ð  (250)10 (11111010)2

Answer: 

3

576

3

192 – 0

3

  64 – 0

3

  21 – 1

            3

   7 –  0

 

   2 -  1

 

 

ð  (210100)3 - (110002)3 = (_?_)2

210100 – 110002 = (11111010)2

question number 2nd b solve kr dn.. plzz

Dear Students Don’t wait for solution post your problems here and discuss ... after discussion a perfect solution will come in a result. So, Start it now, replies here give your comments according to your knowledge and understandings....

mn ne 2nd question kr lia hai..pr confrm ni hai

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