Philadelphia University

Faculty of Engineering

Marking Scheme

Exam Paper

BSc CE

Logic Circuits (630211)

First Exam First semester Date: 19/11/2012

Section 1

Weighting 20% of the module total

Lecturer: Dr. Qadri Hamarsheh

Coordinator: Dr. Qadri Hamarsheh

Internal Examiner: Eng. Anis Nazer

Marking Scheme

Logic Circuits (630211)

The presented exam questions are organized to overcome course material through 4 questions.

The all questions are compulsory requested to be answered.

Marking Assignments

Question 1 This question is attributed with 7 marks if answered properly; the answers are as following:

1.  Convert 201(3) to base ten:

a)  / 5 / b)  / 19
c)  / 8 / d)  / 98

2.  The decimal equivalent of the hexadecimal number (F.4)16 is ------.

a)  / 14.25 / b)  / 15.25
c)  / 14.75 / d)  / 15.75

3.  The BCD code of the decimal number 937.25 is ------.

a)  / 100100110111.00100101 / b)  / 100011111.010101
c)  / 1110101001. 11001 / d)  / None of the above

4.  The 2's complement representation of 34(10) is:

a)  / 11011110 / b)  / 01100010
c)  / 00100010 / d)  / 11100010

5.  If you apply DeMorgan’s theorem, the complement of the expression A+B+CD, you get:

a)  / AB(C+D) / b)  / A+B+C+D
c)  / A+B+C+D / d)  / A+B(CD)

6.  The equivalent canonical (standard) form for the following logical expression F=AB+C is

a)  / F=ABC+ABC+ABC+ABC
b)  / F= ABC+ABC+ABC+ABC
c)  / F=ABC+ABC+ABC+ABC+ABC
d)  / None of the above

7.  Which of the following is equal to FA,B=m0, m3

a)  / (A+B)A+B / b)  / A+BA+B
c)  / A.B+A.B / d)  / AB+AB

Question 2 This question is attributed with 3 marks if answered properly; the answers are as following:

Solution
1.  Digital systems are generally easier to design.
2.  Ease of manipulation.
3.  Digital representation is very well suited for numerical and non-numerical information processing.
4.  Accuracy and precision are greater.
5.  Operations can be programmed.
6.  Digital circuits are less affected by noise.
7.  Low cost.
8.  Easy to duplicate similar circuits.
9.  Stable (reliable, repeatable).
10.  Easily controllable by computer.

Question 3 This question is attributed with 7 marks if answered properly; the answers are as following:

Question 3.a (2 marks)

Solution:
AB+AC+BC
= AB+BCA+A+AC
=AB+ABC+ABC+AC
=ABC+1+ACB+1
=AB+AC

Question 3.b (3 marks)

Solution:

Question 3.c (2 marks)

Solution:
AB + AB + AB = AB+B+ AB
= A•1 + AB
= A + AB
= A + AB + 0
= A + AB + AA
= A + BA + A
= A + B•1
= A + B

Question 4 This question is attributed with 3 marks if answered properly; the answers are as following:

Solution:
a)  g= d+ec+cde (2 marks)
b)  g=cd+ce+cde (1 mark)