Exam Solutions for ECE 2030 Computer Engineering Fall 2002 - Problem 1 to 4, Exams of Computer Science

The solutions to exam two for the computer engineering course ece 2030 in fall 2002. It includes the conversion of decimal and hexadecimal values to binary and octal notation, arithmetic operations using two's complement and unsigned representations, completion of a truth table for a circuit, implementation of a transparent latch, and analysis of state machine sequencer energy consumption.

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ECE 2030 10:00pm Computer Engineering Fall 2002
4 problems, 4 pages Exam Two Solutions 30 October 2002
Problem 1 (3 parts, 30 points) Numbers and Arithmetic
Part A (9 points) Convert some decimal values (and powers of two) into binary notation:
binary notation decimal notation
10101010 2 + 8 + 32 + 128 = 170
11111.1111 (32-1).(1-.0625)=31.9375
249 512 trillion (512T)
Part B (9 points) Convert the following hexadecimal values into octal notation:
hexadecimal notation octal notation
4321 0100001100100001 = 41441
14.9 00010100.1001 = 24.44
55.AA 01010101.10101010 = 125.524
Part C (12 points) For each problem below, (a) compute the operations using the rules of
arithmetic, (b) indicate whether an error occurs assuming all numbers are expressed using a four
bit two’s complement representation, and (c) indicate whether an error occurs assuming all
numbers are expressed using a four bit unsigned representation.
1 1
+ 1 0 1 1
1 1 0 0
+ 1 1 0 0
1 1 1 1
- 0 0 0 1
1
- 1 1 1 1
result 1 1 1 0 1 0 0 0 1 1 1 0 0 0 1 0
signed
error? NO NO NO NO
unsigned
error? NO YES NO YES
1
pf3
pf4

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4 problems, 4 pages Exam Two Solutions 30 October 2002

Problem 1 (3 parts, 30 points) Numbers and Arithmetic

Part A (9 points) Convert some decimal values (and powers of two) into binary notation:

binary notation decimal notation

10101010 2 + 8 + 32 + 128 = 170

11111.1111 (32-1).(1-.0625)=31.

249 512 trillion (512T)

Part B (9 points) Convert the following hexadecimal values into octal notation:

hexadecimal notation octal notation

4321 0100001100100001 = 41441

14.9 00010100.1001 = 24.

55.AA 01010101.10101010 = 125.

Part C (12 points) For each problem below, (a) compute the operations using the rules of arithmetic, (b) indicate whether an error occurs assuming all numbers are expressed using a four bit two’s complement representation, and (c) indicate whether an error occurs assuming all numbers are expressed using a four bit unsigned representation.

result

signed error?

NO NO NO NO

unsigned error?

NO YES NO YES

4 problems, 4 pages Exam Two Solutions 30 October 2002

Problem 2 (3 parts, 27 points) Storage

Part A (9 points) For the circuit below, complete the truth table describing its behavior. Then label the status of each row from the following choices: SET, RESET, HOLD, META-STABLE.

A B Out status

0 0 Qo HOLD 1 0 0 RESET 0 1 1 SET

A

B

Out

1 1 0 METASTABLE

Part B (9 points) Now implement a transparent latch using this circuit plus additional basic gates (AND, OR, NAND, NOR, and NOT). Label inputs IN and EN. Label output OUT. Do not attempt to employ mixed logic notation.

IN OUT

EN

Part C (9 points) Assume the following signals are applied to a register with write enable. Draw the output signal Out. Draw a vertical line where In is sampled. Assume Out starts at zero.

WE

In

Out

4 problems, 4 pages Exam Two Solutions 30 October 2002

Problem 4 (3 parts, 22 points) Pass Gate Fun

Part A (6 points) For the pass gate symbol shown on the left below, draw its implementation on the right. Be sure that the named terminals (A, B, C, and D) on the right correspond to the icon terminals on the left.

A

B

C

D

A

B

C

D

Part B (8 points) Implement a 1 to 2 demultiplexer using only pass gates and inverters. Label all inputs (IN, S) and outputs (O 0 , O 1 ).

Out

In

Select

Out

Part C (8 points) Using the 1 to 2 demux implemented in part B (in icon form), implement a 1 to 4 demux. Label all inputs (IN, S 0 , S 1 ) and outputs (O 0 , O 1 , O 2 , O 3 ).

IN

OUT 0

S

1 to 2 OUT 1

IN

OUT 0

S

1 to 2 OUT 1

IN

OUT 0

S

1 to 2 OUT 1

IN

OUT 0

OUT 1

OUT 2

OUT 3

S 1 S 0