CS302 - Digital Logic Design Assignment No. 1 Solution Fall 2016 Due Date: Nov 14, 2016

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).

 

BEST OF LUCK

 

Tags: -, 1, 14, 2016, Assignment, CS302, Date:, Design, Digital, Due, More…Fall, Logic, No., Nov, Solution

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Please help to solve assignment

Solution of Question 1 is in attachment...

Correct

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Nh mere bhai Base 2 me answer dyna he na k base 3 me

yes brother, its true... but 75% correct.. cuz ab is answer ko binary mein convert karna baki... overall sahi hai

g

(576)10 =  (1001000000)2

(110002)3 = (101010000)2

(1001000000)2 – (1001010000)2 = (11111010)2

Question No. 02 (a) ka koi idea dyn plz

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