Quiz 2 in EE 2030 at Georgia Tech - Electrical and Computer Engineering, Exams of Computer Science

Quiz 2 for the ee 2030 course at the georgia institute of technology. The quiz covers topics such as binary number systems, logic gates, and state machines. Students are required to answer multiple-choice questions and fill in the missing parts in circuits and state diagrams.

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2012/2013

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GEORGIA INSTITUTE OF TECHNOLOGY
School of Electrical and Computer Engineering
EE 2030
QUIZ #2
Thursday, October 21, 1999
Name:
Last, First
Closed book, one page of handwritten notes allowed.
None of the problems require involved calculations. Reconsider your approach before
doing something tedious.
Clearly identify each answer.
Part pts Score
1 20
2 20
3 20
4 16
5 24
Total 100
pf3
pf4
pf5
pf8
pf9

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GEORGIA INSTITUTE OF TECHNOLOGY

School of Electrical and Computer Engineering

EE 2030

QUIZ

Thursday, October 21, 1999

Name:

Last, First

• Closed book, one page of handwritten notes allowed.

• None of the problems require involved calculations. Reconsider your approach before

doing something tedious.

• Clearly identify each answer.

Part pts Score

Total 100

Problem 1 (20 pts):

a. (8 pts) Convert the following numbers (all binary numbers are in 2’s complement form):

01101.101 binary =⇒ 13.625 decimal

-13 decimal =⇒ 1111 0011 binary (8 bit)

11010011 binary =⇒ -45 decimal

1010.0101 binary =⇒ -5.6875 decimal

01010101 binary =⇒ 85 decimal

17 decimal =⇒ 0001 0001 binary (8 bit)

-2 decimal =⇒ 1111 1110 binary (8 bit)

b. (4 pts) Add the following unsigned, 5 bit, binary numbers; indicate if there has been an overflow error:

1010

10000

overflow? N overflow? Y overflow? N overflow? N

c. (8 pts) Add the following 5 bit, 2’s complement binary numbers, indicate if there has been an overflow error:

1010

10000

overflow? Y overflow? N overflow? Y

overflow? N overflow? Y overflow? N

Problem 3 (20 pts):

a. (10 pts) Often, in digital communications, a code is sent to mark the beginning of a mes- sage. The state diagram shown below detects the sequence 11010 with overlapping codes not permitted. For this state machine there is one input, the digital signal x, and one output, the detect signal D. Fill in the state table below based on the state diagram.

000/

001/0 010/

011/

101/1 100/

0

0

1

0

1

1

0

1

0

1





Current State Input Next State Output S 2 S 1 S 0 x N S 2 N S 1 N S 0 D 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 1 0 1 0 0 0 1 0 0 0 1 1 0 0 1 0 1 0 1 0 0 0 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 1 0 0 0 1 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 0 0 1 1 0 1 1 0 0 0 1 1 1 0 0 x x x 0 1 1 0 1 x x x 0 1 1 1 0 x x x 0 1 1 1 1 x x x 0

The next states for the 110 and 111 are not specified so you could have them as “don’t cares” or you could have them transition into the 101 state.

b. (10 pts) Suppose we want to make an timer which counts down from 5 to 0. When the timer reaches 0 a bell rings and continues to ring until the OK button is pressed. When the OK button is pressed the timer starts over at 5 again. The OK button causes the OK input to go high. Draw a state diagram for a Moore type state machine that will implement the timer as described. Label the diagram clearly.

OK=

OK=

The 1 output causes the bell to ring.

Note, only state 000 has a 1 output.

Problem 5 (24 pts):

a. (8 pts) The bottom segment of a numeric LED display is only lit when the numbers 1, 4, 7, and 9 are displayed. Assume that the number to be displayed is determined by the binary number represented by the four input lines A 3 − A 0. Also assume that we do not care what is displayed if the input value is greater than 9. Fill in the truth table for the bottom LED driver (F 0 ).

A 3 A 2 A 1 A 0 F 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 1 1 0 0 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1 0 0 0 0 1 0 0 1 1 1 0 1 0 x 1 0 1 1 x 1 1 0 0 x 1 1 0 1 x 1 1 1 0 x 1 1 1 1 x

b. (4 pts) From your truth table, fill in the K-map for simplifying the expression for F 0. You do not need to simplify the expression, just fill in the table.

F 0 :

A 3 A 2

@^ A^1 A^0

@ @@

00

A 3

A 3

         

A 2

A 2

A 2

}

     }

A 1 A 1

︷ ︸︸ ︷︷ ︸︸ ︷

A 0 A 0 A 0

︸ ︷︷ ︸︸ ︷︷ ︸︸ ︷︷ ︸

0 1 x x

x x x x

c. (6 pts) The K-map corresponding to the top LED driver (F 6 ) is shown below. Write down the simplified expression for F 6.

F 6 :

A 3 A 2

@^ A^1 A^0

@ @@

00

A 3

A 3

         

A 2

A 2

A 2

}

     }

A 1 A 1

︷ ︸︸ ︷︷ ︸︸ ︷

A 0 A 0 A 0

︸ ︷︷ ︸︸ ︷︷ ︸︸ ︷︷ ︸

1 1 x x

x x x x

F 6 = A 0 A 2 + A 0 A 2 + A 1 + A 3

d. (6 pts) Implement F 6 using your choice of gates, multiplexers, or decoders.

A 0 A 1

A 2 A 3

F 6

F 6

A 3 A 2 A 1

A (^0) +V

+V

8-1 MUX

0 1 2 3 4 5 6 7

Here are two of many possible answers.