EEE 120
Hardware Lab 1 Answer Sheet
Debugging a Half and Full Adder
Name:_ _
Task 1-1: Build and Test the SUM & CRY of the 1-Bit Half-Adder
Follow the testing procedures outlined in the laboratory manual on your 1-bit half adder circuit and
record your re
EEE120
Hardware Lab 3 Answer Sheet
Latches, Flip-Flops and Counters
Name:_
Task 3-1: Build an Active-High S-R Latch
Build in hardware an active-high S-R NOR latch. Test your circuit, following the testing protocol sequence
as outlined in the laboratory in
EEE 120
Simulation Lab 1 Answer Sheet
Half Adder, Increment & Twos Complement Circuit
Name:_ _
Date:_1/26/2016_
Task 1-1: Build and Test the 1-Bit Half-Adder
Include a picture of your Logisim circuit implementation of a 1-bit half adder here:
1
Follow the
CSE/EEE 120
Simulation Lab 3 Answer Sheet
Arithmetic and Logic Unit
Name:_
Date:_2-10-16_
Task 3-1: Build the NOT/NEG Circuit
Include a picture of your Logisim NOT/NEG circuit here:
Table 1 lists the functionality of the NOT/NEG circuit based on the input
CSE/EEE 120
Simulation Lab 2 Answer Sheet
4-Bit Full Adder, Multiplexer, Decoder & Buffer
Name:_
Date:_2/2/16_
Task 2-1: Design a Full Adder
Write down the canonical SOP expressions for the Cout and SUM function of a full adder. Be sure to check
the lab m
EEE 120
Simulation Lab 0 Answer Sheet
Simulator Tutorial: Using Logisim
Name:_
Date:_1/15/2016_
Task A: Creating a Subcircuit
Include a picture of your Logisim circuit SUB_TRY.circ here:
Include a picture of your Logisim circuit subdemo.circ here:
1
Creat
EEE120 Hardware Lab 2 Answer Sheet
TTL Characteristics, Open-Collector Buffers, Three-State Buffers
Name:_
Task 2-1: Gates with Common Outputs
Follow the testing procedure outlined in the laboratory instructions and record your results in Table 1.
Table 1
EEE 120
Hardware Lab 0 Answer Sheet
Using a Prototype Board and Checking Logic Circuits using a Voltmeter
Name:_ _
Task 0-1: Build the 3-Input AND Gate on a Breadboard and Test its Logic
Operation
Follow the testing procedures outlined in the Hardware Lab
15‘" AMC——>8 1999
1. (6 ? 3) + 4 — (2 — 1) = 5. To make this statement true, the question mark between the 6
and the 3 should be replaced by
(A) + (B) x (C) + (D) — (E) None ofthese.
2. What is the degree measure ofthe smaller angle
formed by the hands of
. Which one of the following bar graphs could represent the
9th AJHSME 1993 2
1. Which pair of numbers does NOT have a product equal to 36?
(A) {—4,—9} (13) {—3—12} (c) {3—72} ‘
3 6/ .
(13) {1,36} (E) {5,24}
49
. When the fraction —- is expressed in simpl
. mum—La ._._. x.$i_‘;h1.g;f‘n1_fvxlw ._-.H .4. r -._._.
110—9+8—7+6—5+4—3+2—1_
. Which of the following is not equal to - ?
8th AJHSME 1992 2
1~2+3—4+5—6+7—8+9
(A) —1 (3)1 (0)5 (D)9 (E) 10
5
4
(A);9 (BM: (0)113—2 (ml; (En-3g
. What is the largest differe
1.
3s
8.
3rd AJHSME 1987
.4 + .02 + .006 -
A) .012 B) .066 c) .12 D) .24 E) .426
.31. =
25
A) .008 B) .08 c) .a n) 1.25 E) 12.5
2(81+83+85+87+89+91+93+95+97+99)=
A) 1600 B) 1650 C) 1700 D) 1750 E) 1800
Martians measure angles in clerts. There are 500 cler
6th AJHSME 1990 2
1. What is the smallest sum of two 3-digit numbers that can be obtained
by placing each of the six digits 4,5,6,7,8,9 in one of the six boxes in this
addition problem? D I: D
+ E] El CI
2. Which digit of .12345, when changed to 9, gives
1.
2. The product
31+2+
'ﬁ 217 30"
4. The ﬁgure consists of alternating light and
4th AJHSME 1988 2
The diagram shows part of a scale of a
measuring device. The arrow indicates an
approximate reading of
A) 10.05 B) 10.15 C) 10.25 10 11
D) 10.3 E) 10.6
8x
5th AJHSME 1989 2
1. (1+11+21+31+41) +(9+19+29+39+49)=
A) 150 B) 199 C) 200 D) 249 E) 250
2 2 4 6 _ I) _
' T6 + TM + 175M ‘ f -
A) .012 B) .0246 C) .12 D) .246 E) 246
3. Which of the following numbers is the largest?
A) .99 B) .9099 C) .9 D) .909 E) .90
IV\
1.
2nd AJI-ISME 1986
In July 1861, 366 inches of rain fell in Cherrapunji, India.
What was the average rainfall in inches per hour during that
month?
366 366X31 366X24
A) 31x24 B) 24 C) 31
31x24
D) 366 E) 366X31X24
Which of the following numbers has
1 mx}
1st AJHSME 1985 2
,_.
wu
xx
uno
><><
\lv—I
II
A
A) l B) 0 C) 49 D) 49
E) 50
90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 =
A) 845 G B) 945 C) 1005 D) 1025 E) 1045
107 =
5x104
A) .002 B) .2 C) 20 D) 200 E) 2000
The area of polygon ABCDEF,
in s
CSE230/EEE230 Computer Organization & Assembly Language Midterm Examination, October 11th, 2006 Arizona State University Rules
1. The exam is open book and open notes; Laptop usage is not allowed during the exam. 2. Work on Problems 1 to 3. They add up to
7.3 The key features of solutions to these problems: Low temporal locality for data means accessing variables only once. High temporal locality for data means accessing variables over and over again. Low spatial locality for data means no marching through
CSE230/EEE230 Homework Assignment #6
Due 1:40pm Monday, December 4, 2006, in the classroom. Notes: 1. Your submission will be accepted until Wednesday, December 6 without penalty. (After Dec. 4, you can stop by the instructor's or the TA's office to drop
6.1
a. Shortening the ALU operation will not affect the speedup obtained from pipelining. It would not affect the clock cycle. b. If the ALU operation takes 25% more time, it becomes the bottleneck in the pipeline. The clock cycle needs to be 250 ps. The
CSE230/EEE230 Homework Assignment #5
Due 1:40pm Wednesday, November 22, 2006, in the classroom. Note: You are encouraged to turn in your solutions on or before November 22. But your submission will be accepted until Monday, November 27 without penalty.
So
4.14 The total execution time for the machines are as follows: Computer A = 2 + 20 + 200 seconds = 222 seconds Computer B = 5 + 20 + 50 seconds = 75 seconds Computer C = 10 + 20 + 15 seconds = 45 seconds Thus computer C is faster. It is 75/45 = 5/3 times
CSE230/EEE230 Homework Assignment #4
Due 1:40pm Wednesday, November 1, 2006, in the classroom. Reading assignment: Read Section 5.6 Solve the following problems from Chapters 4 & 5 in the textbook. Total: 70 points 4.14 (5 points) 4.15 (5 points) 5.1 (10
3.1 0000 0000 0000 0000 0001 0000 0000 0000two 3.2 1111 1111 1111 1111 1111 1000 0000 0001two 3.9 The problem is that A_lower will be sign-extended and then added to $t0. The solution is to adjust A_upper by adding 1 to it if the most significant bit of A
CSE230/EEE230 Homework Assignment #3
Due 1:40pm Monday, October 9, 2006, in the classroom.
Solve the following problems from Chapters 3 & 4 in the textbook. Total: 70 points 3.1 (5 points) 3.2 (5 points) 3.9 (10 points) 3.30 (10 points) 3.42 (10 points) 3
CSE230/EEE230 Homework Assignment #2
Due 11:59pm Wednesday, September 20, 2006, via the Digital Drop Box. * Please submit your solutions to HW Assignment 2 via the Digital Drop Box on the Blackboard. * All of your source code and any supplemental document
The full score will be 60 point. 1.46 Magnetic disk: Time for 1/2 revolution = 1/2 rev*1/7200 minutes/rev*60 seconds/minutes = 4.17 ms Time for 1/2 revolution = 1/2 rev*1/10,000 minutes/rev*60 seconds/minutes = 3 ms 1.47 (DVD): Bytes on center circle = 1.
CSE230/EEE230 Homework Assignment #1
Due 1:40pm Wednesday, September 6, 2006, in the classroom.
Solve the following problems from Chap. 1 of the textbook. 1.46, 1.47, 1.48, 1.50, 1.52, 1.53, 1.54