CS302 Assignment Solution & Comment Semester: Fall 2016 Digital Logic Design

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 numbersystem.

(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).