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ECE 2030 Test 1 - Fall 2000, Exams of Computer Science

A closed book, closed notes test for ece 2030 students in the fall of 2000. The test includes five problems covering topics such as digital logic, switch level circuits, and karnaugh maps. Students are required to write clear answers, show their work, and adhere to the georgia tech academic honor code.

Typology: Exams

2012/2013

Uploaded on 04/08/2013

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ECE 2030 Test 1
Fall 2000
Dr. Heck
This is a closed book, closed notes test. No calculators are allowed.
You have 50 minutes to complete the test.
Please show all of your work.
Please abide by the Georgia Tech academic honor code. Violations will be handled in the
appropriate manner.
Solutions
pf3
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ECE 2030 Test 1

Fall 2000

Dr. Heck

  • This is a closed book, closed notes test. No calculators are allowed.
  • You have 50 minutes to complete the test.
  • Please show all of your work.
  • Please abide by the Georgia Tech academic honor code. Violations will be handled in the appropriate manner.

Solutions

Write concise, clear answers.

a) Why do we use both n-type and p-type transistors when building circuits to implement digital logic?

n-type and p-type transistor are not ideal switches. n-type is good at pulling low while p-type is good at pulling high. Using both compensates for the nonideal behavior.

b) What is the benefit of using Mixed Logic for gate design?

-easy to analyze and trouble shoot -easy to convert from the one set of gate implementation to another

Suppose F = ( A + B ) C + D ,

a) Create a switch level implementation for F using n-type and p-type transistors. Assume that both the inputs and their complements are available. Your design should contain no shorts and no floats.

F =ABC+D

b) Use mixed logic to create a gate level implementation for F using only NOR gates and inverters. Your design should include the complements of the inputs if they are needed. Determine the number of transistors needed for the design.

Number of transistors for part b) 3 × 4 + 3 × 2 = 18

+V

B

C

A

D

A B C

F D A B C F D

a) Draw the Karnaugh map for F = ( A + C ) B + AB

b) A truth table is given below. From the truth table, write the SOP and the POS expressions. Then, using a Karnaugh map, express the simplified SOP expression.

SOP = ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + AB CD

A B C D A B C D A B C D A B C D

POS A B C D A B C D A B C D A B C D A B C D

Simplified 

ABC

BCD

SOP AD ABD use either

A

BC

ABCD OUT

AB

CD