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Main points of this past exam are: Minimised Expression, Boolean Algebra, Truth Table, Minimal Boolean Expression, K-Map, Minimised Logic Diagram, Truth Table, Input Multiplexer, Circuit, Operating System
Typology: Exams
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Semester 2 Examinations 2011
Module Code: COMH
School: Computing
Programme Title: Bachelor of Science (Honours) in WEB Development
Programme Code: KWEBD_8_Y
External Examiner(s): Mr. A .O’Leary Internal Examiner(s): Dr. John Creagh
Instructions: Answer Question 1 and TWO other questions.
Duration: 2 hours
Sitting: Semester 2 2011
Requirements for this examination: Answer Question 1 and TWO other questions.
Note to Candidates: Please check the Programme Title and the Module Title to ensure that you have received the correct examination. If in doubt please contact an Invigilator.
a) Using Boolean Algebra, find the minimised expression for the following truth table:
A B C Output 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 0
Draw the minimised logic diagram. (7 marks)
b) Using a K-map, derive the minimal Boolean expression for the following truth table:
A B C Output 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 0 1 1 1 1 (7 marks)
c) Draw the truth table and design a circuit for a 4 input multiplexer. (7 marks)
d) List the primary functions carried out by an Operating System. Briefly explain each item in your list. Note: Exclude user interfaces. ( 6 marks)
e) Describe the Von Neumann Architecture bottleneck. (7 marks)
f) Perform the following arithmetic, using 8 bits and using two’s complement representation: 5 – 29 (6 marks) [40 marks]
a) Convert the following Hexidecimal to Binary. A2D3 16 (3 marks)
b) What are the advantages and disadvantages of BCD code? (4 marks)
c) Convert the Binary 11012 into its corresponding Gray code. (3 marks)
d) Determine the Hamming code representation for the information bits (^101110111112) Also, determine if there is an error detected in the following Hamming code received (you must briefly explain you answer). 1011101100101112 (10 marks)
e) Assume the following 32-bit floating-point representation:
Using this format, code the following in floating point format: 39.3125 and -10.125 (10 marks) [30 marks]
sign bit of mantissa 8 bit exponent in two’s complement 23 bit mantissa