Solved Problems on Introduction to Computer Engineering - Exam 3 | ECE 2030, Exams of Electrical and Electronics Engineering

Material Type: Exam; Professor: Lee; Class: Intro to Computer Engr; Subject: Electrical & Computer Engr; University: Georgia Institute of Technology-Main Campus; Term: Fall 2008;

Typology: Exams

Pre 2010

Uploaded on 08/04/2009

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ECE2030A Fall 2008 Prof. H.-H. S. Lee
Georgia Tech Page 1 of 8
ECE2030A Fall 2008
Introduction to Computer Engineering
Exam #3
50 min in-class exam.
Close books, close notes.
NO calculator.
This exam is given under the Georgia Tech Honor Code System. You must
observe and sign the Honor Pledge: “I have neither given nor received aid on
this exam.” Your print name and signature below signifies your compliance
with this honor code.
Name (Print): __________SOLUTION____________________________
Signature: _________________________________________________
1. _____________ (30 pts)
2. _____________ (30 pts)
3. _____________ (40 pts)
Total (100 pts) ____________
pf3
pf4
pf5
pf8

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Georgia Tech Page 1 of 8

ECE2030A Fall 2008

Introduction to Computer Engineering

Exam

50 min in-class exam.

Close books, close notes.

NO calculator.

This exam is given under the Georgia Tech Honor Code System. You must

observe and sign the Honor Pledge: “I have neither given nor received aid on

this exam.” Your print name and signature below signifies your compliance

with this honor code.

Name (Print): __________ SOLUTION ____________________________

Signature: _________________________________________________

1. _____________ (30 pts)

2. _____________ (30 pts)

3. _____________ (40 pts)

Total (100 pts) ____________

Georgia Tech Page 2 of 8

A^ Y

B (^) X

O3 O2 O1 O

CLR 4 bit Ripple Counter (^) CLK

  1. (30%) General Questions. (Circle one of the answers) 1.1. How many output bits will be generated when performing 5-bit by 5-bit integer multiplication? (a) 5 bits (b) 10 bits ANSWER (c) 25 bits (d) 32 bits

1.2. Given the latch shown below, which of the following could cause an unstable state? (a) A = B = 0 ANSWER (X=Y=1) (b) A = 1, B = 0 (c) A = 0, B = 1 (d) A = B = 1

1.3. Which of the following memory contains the least number of transistors? (a) One bit DRAM ANSWER (b) One bit RS Latch (2 nd^ most) (c) One bit Register (most) (d) One bit SRAM (2 nd^ least)

1.4. What is the capacity of 2 k^ x N where k= 21, N=32? (a) 8 M Bytes ANSWER (2 21 x 2^5 = 2^26 = 64 M Bits = 8 M Bytes) (b) 64 M Bytes (c) 2 M Bytes (d) 32 M bits

1.5. Which type of the following memory requires periodic refreshing to not lose data? (a) D Flip-Flop (b) DRAM ANSWER (has a capacitor that needs to be refreshed) (c) SRAM (d) On-chip L1 Cache

1.6. Given the Mod-N counter shown below, what is N? (a) 6 (b) 7 ANSWER (c) 8 (d) 9

O3:O1 = 0110, therefore resets after 6. Counts from 0 to 6, therefore is a modulo-7 counter.

Georgia Tech Page 4 of 8

2.2. (10%) Given the following input waveforms CLK and X, please draw the corresponding timing diagram (or waveform) of Y and Z for each of the following circuits. Assume the initial states of all latches or flip-flops are unknown.

D Q

C

D Flip-

Flop

D Q

C

D Flip-

Flop

X

CLK

Z

CLK

X

Y

Y

Z

Q

Q

Georgia Tech Page 5 of 8

2.3. (10%) Given the following input waveforms CLK and X, please draw the corresponding timing diagram (or waveform) of Y and Z for the following circuits using Toggle Flip Flops. Assume the initial states of all latches or flip-flops are Zero.

TE Q

C

Toggle

Flip-Flop

TE Q

C

X

CLK

Z

CLK

X

Y

Y

Z

Toggle

Flip-Flop

Georgia Tech Page 7 of 8

3.2. (15%) Derive the State Table and simplify the equations with Karnaugh maps.

Present State Input (X) Next State Output P1 P0 U'/D N1 N0 C1 C 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 1 0 1 1 0 0 0 1 1 0 0 1 1 1 0 1 0 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 0 1 1

C1 = P1. C0 = P0.

Alternate:

Present State Input (X) Next State Output P1 P0 U'/D N1 N0 C1 C 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 1 0 1 0 1 0 0 1 1 0 0 0 0 1 0 0 1 1 1 1 1 0 1 0 1 0 1 1 1 0 1 1 1 1 1 1 1 1 0 1 0

C1 = N1. C0 = N0.

Georgia Tech Page 8 of 8

The Next State logic for either method is the same.

For N1: P0 X (^00 01 11 ) P 0 0 0 0 1 1 1 0 1 1

(P0 P1) X P1 P

N1 P0 X P1 X P1P

For N0: P0 X (^00 01 11 ) P 0 1 0 0 0 1 1 1 0 1

(P0 P1)X P1P

N0 P0X P1X P1P

3.3. (15%) Draw the sequential logic of this counter of your design. You do not need to draw the internals of the D Flip-Flop, simply draw each D Flip-Flop as a block.

The diagram below shows the answer for when C1 = N1, and C0 = N0.

The diagram for the other solution is the same, except C1 is pulled from P1, and C0 from P0.

D F/F

D

C

Q

Q

D F/F

D

C

Q

Q

U/D

CLK C C

N

N1 P

P