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Main points of this exam paper are: Transform, Switch-Level Design, Computer Engineering, Implementation Using, CreateSwitch, Expression Below, Complements Are Available, TransistorsPossible, Boolean Expression, Expression Should Remain
Typology: Exams
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4 problems, 5 pages Exam One 6 February 2003
Instructions: This is a closed book, closed note exam. Calculators are not permitted. If you have a question, raise your hand and I will come to you. Please work the exam in pencil and do not separate the pages of the exam. For maximum credit, show your work. Good Luck!
Your Name ( please print ) ________________________________________________
1 2 3 4 total
4 problems, 5 pages Exam One 6 February 2003
Problem 1 (3 parts, 25 points) Switch-level Design
Part A (16 points) For each expression below, create a switch level implementation using N and P type switches. Assume both inputs and their complements are available. Your design should contain no shorts or floats. Use as few transistors as possible.
OUTx = ( A + ( B โ C ))โ ( D + E ) OUTy = ( A + B )( C + D + E )
Part B (9 points) Transform the following Boolean expression to a form where it can be implemented using switches (i.e., there should be no bars in the expression except for complements of the inputs A, B, C, etc.). The behavior of the expression should remain unchanged.
Out (^) X = (( A โ B )โ ( C โ ( D โ E )))+( E โ F โ G )=
4 problems, 5 pages Exam One 6 February 2003
Problem 3 (2 parts, 25 points) Karnaugh Maps
Part A (10 points) Given the following Karnaugh Map, circle and list all the prime implicants, indicating which are essential and write the simplified product-of-sums (POS) expression.
prime implicants essential?yes no
simplified POS expression Part B (15 points) Simplify the following POS expression using a Karnaugh Map. Circle and list all the prime implicants, indicating which are essential and write the simplified SOP expression.
Out =( A + B + C )( A + B + C )( A + B + C )
prime implicants essential?yes no
Simplified SOP expression
4 problems, 5 pages Exam One 6 February 2003
Problem 4 (2 parts, 20 points) Building Blocks
Part A (14 points) The truth table below describes the behavior of a 2-to-1 priority encoder. In 1 In 0 Out Valid 0 0 X 0 X 1 0 1 1 0 1 1 A.1 What is the priority of the inputs?
lowest priority highest priority
Implement this priority encoder by following these steps: A.2 Find the simplified Boolean expressions for Out and Valid using the Karnaugh maps below.
In 0
In 1
Out:
Out = (^) In 0
In 1
Valid:
Valid =
A.3 Use mixed logic to implement the priority encoder with OR gates and inverters only. Label all the inputs and outputs.
In 0 Out
In^1
Valid
Part B (6 points) Complete the truth table for the following unusual logic block.
1
What boolean function does this circuit implement?.
