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# CS302 Assignment no:1 Fall 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:

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

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### Replies to This Discussion

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

how you have solved Qno 1 .. .plz shere your idea  to convert decimal number into tertiary.

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.

i had convert ternary number into decimal number by sum of weight mathod just like we do for binary then by converting this into 10 base number we ll have both  decimal numbers by subtracting these number i had convert this answer into binary number by remainder mathod and

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

 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

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