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010-111 - ACSM certified Personal Trainer - Dump Information
| Vendor | : | ACSM |
| Exam Code | : | 010-111 |
| Exam Name | : | ACSM certified Personal Trainer |
| Questions and Answers | : | 208 Q & A |
| Updated On | : | December 19, 2017 |
| PDF Download Mirror | : | 010-111 Brain Dump |
| Get Full Version | : | Pass4sure 010-111 Full Version |
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QUESTION: 196
Your client reports ankle edema. What would a certified Personal Trainer look for?
- Pain
- Atrophy
- Swelling
- Red coloration
Answer: C
QUESTION: 197
What is the role of High Density Lipoprotein in the blood?
- Transports cholesterol
- Increases anaerobic enzymes
- Increases triglycerides
- Decreases total cholesterol
Answer: A
QUESTION: 198
What is the definition of arteriosclerosis?
- Death of cardiac tissue
- Accumulation of plaque
- Hardening of the arteries
- Widening of the arteries
Answer: C
QUESTION: 199
What are three non-modifiable conditions that place someone at increased risk for the development of coronary artery disease?
- Advanced age, gender, family history
- Family history, obesity, diabetes mellitus
- Gender, family history, dyslipidemia
- Post-menopausal status, excessive alcohol consumption, advanced age
Answer: A
QUESTION: 200
What effect should a bronchodilator have on your asthmatic client?
- Increase airway resistance
- Decrease airway resistance
- Decrease blood pressure
- Increase blood pressure
Answer: B
QUESTION: 201
Which of the following blood lipids is influenced more by physical activity than by nutrition modification?
- LDL
- HDL
- VLDL
- Total cholesterol
Answer: B
QUESTION: 202
What are the acute affects of alcohol intake on exercise?
- Increases blood pressure and impairs exercise capacity
- Decreases metabolic rate and increases blood pressure
- Promotes dehydration and increases risk of heart arrhythmias
- Increases risk of heart arrhythmias and increases exercise capacity
Answer: C
QUESTION: 203
Which
of the following practices are NOT recommended for persons with asthma
who desire to participate in a strenuous aerobic exercise program?
- Exercise in an environment with warm, moist air.
- Self-administer prescribed medication as directed prior to or during the exercise session.
- Use a short, intense warm-up.
- Use a scarf or surgical mask in front of the mouth if exercising in cold weather.
Answer: C
QUESTION: 204
What are three non-modifiable conditions that place someone at increased risk for the development of coronary artery disease?
- Gender, family history, dyslipidemia
- Family history, obesity, diabetes mellitus
- Advanced age, gender, family history
- Post-menopausal status, excessive alcohol consumption, advanced age
Answer: C
QUESTION: 205
What is the leading cause of non-cardiovascular death in young athletes?
- Contact Sports Injuries
- Anemia
- Hypothermia
- Overheating
Answer: D
QUESTION: 206
A
client who reports a decreased Rating of Perceived Exertion at a given
treadmill running speed after taking albuterol, likely suffers from
which of the following conditions?
- Depression
- Asthma
- Bradycardia
- Claudication
Answer: B
QUESTION: 207
What is the exercise response to acute cigarette smoking?
- Respiration rate increases; blood pressure response to exercise decreases
- Likelihood of coronary artery spasm increases; blood pressure response to decreases
- Heart rate increases; likelihood of coronary artery spasm increases
- Likelihood of coronary artery spasm decreases; blood pressure response to exercise increases
Answer: C
QUESTION: 208
Which environmental trigger is NOT associated with exercise-induced asthma?
- dry air
- warm air
- dusty air
- pollutants
Answer: B
ACSM 010-111 Exam (ACSM certified Personal Trainer) Detailed Information
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C2020-001 - Fundamentals of making use of Maximo business Asset administration solutions V2No influence found, try new keyword!Programming in C#.70-483.1e.50q.examination 22 Kb Jan 01, 1970 ACSM certified very own trainer.010-111.1e.208q.exam 49 Kb Jan 01, 1970 CompTIA A Certification examination (220-802).220-802.5e.440q.exam 112 Kb Jan 01, 1970 enforcing Citrix NetScaler 10-5 for App and ...
AFF02 Week 1 Lecture one 2No effect discovered, try new key phrase!Day Time Room Monday 9 – 10 a.m. 010-111 Wednesday 8 – 9a.m. 019 - 101 TIMETABLE- TUTORIALS ... TUTORIALS A+ A B+ B C+ C D E EX 85+ 78-eighty four seventy one-seventy seven sixty four-70 fifty seven-63 50-fifty six 40-49
Chapter1 - ITI 1100/section A winter 2011 DIGITAL equipment I...
Unformatted text
preview: ITI 1100/area A wintry weather 2011 DIGITAL system I Lectures:
Monday, eleven:30 – 13:00 room: STE G0103 Thursday, 13: 00 –14:30 room:
STE G0103 TUTORIAL: Thursday, 17:30 - 19:00 MNT 203 LAB LAB 1 Thursday,
19:00 - 22:00 CBY B302 LAB 2 Wednesday, 14:30 - 17:30 CBY B302 Professor
: Dr A. Karmouch, office CBY A508 Mid-term examination date: Saturday
March 5, 2011 (10:00-11:30) "bread 010-111board" Gervais Electronique
716 Ave Industrial cell: 738-3101
http://www.google.ca/search?hl=en&source=hp &q=bread
010-111board+graphic&meta=&aq=f&oq= 10% bargain for college
students path web page: http://www.web
page.uottawa.ca/~karmouch/educating/ Password: “elg1100b” route define
Digital Design 1. Binary methods. Digital methods. Binary Numbers.
quantity Base Conversions. Octal and Hexadecimal Numbers. enhances.
Signed Binary Numbers. Binary Codes. Binary Storage and Registers.
Binary arithmetic 2. Boolean Algebra and logic Gates. simple
Definitions. fundamental Theorems and houses of Boolean Algebra. Boolean
functions. Canonical and normal forms. other logic Operations. Digital
logic Gates. direction define Digital Design [2] 3. Gate level
Minimization The Map formulation. four Variable Map. manufactured from
Sums Simplification. Don’t Care conditions. NAND and NOR,
Implementation. different Two degree Implementations. exclusive OR
function. four. Combinational good judgment Combinational Circuits.
analysis procedure. Design method. Binary Adder Subtractor. Magnitude
Comparator. Decoders. Encoders. Multiplexers. route define Digital
Design [3] 5. Synchronous Sequential good judgment. Sequential Circuits.
Latches. Flip Flops. evaluation of Clocked Sequential Circuits. Design
system. 6. Registers advert Counters. Registers. Shift Registers. Ripple
Counters. Synchronous Counters. other Counters. Textbook purchasable on
the university of Ottawa book place. must buy! book Title: Digital
Design Authors: M Morris Mano & Michael D. Ciletti version: Fourth
edition ISBN: 0-13-198924-three writer: Pearson-Prentice hall route
format 7.5 Hours of scheduled guideline per week - 3 hours of Lecture -
1.5 hour of group Discussions. (beginning date: to be introduced in the
type) - 3 hours of Laboratory (starting date: to be introduced in the
classification) Laboratory every pupil will have a laboratory session
per week. There are six experiments to be performed, each requiring a
bunch coaching and completion file. The beginning date for Labs is the
week of Feb. 1, 2011, then every week Laboratory organizations will
consist of two college students handiest. students are required to
reside within the identical group and with the identical TA for the
whole semester. Laboratory every group performing the test is required
to record their information on paper and this should still be seen and
signed by using the TA. The records should be connected to the submitted
document. One lab file is anticipated from every community after each
lab. The lab document should be organized based on the guidelines
distinct in the lab guide. Grading Scheme Assignments 10% Laboratories
15% Mid time period exam 25% remaining exam 50% cheating and plagiarism
•cheating is any act that offers you unfair talents on the price of yet
another classmate. •Examples: –copying on assessments, homework
•Plagiarism see the following URL: http://www.uottawa.ca/plagiarism.pdf
•If we become aware of you are concerned in cheating or plagiarism you
might be turned over to the school, for investigation and sanctions •G R
mobile telephones ‘OFF’ all through category No facet talking all over
class Write your questions and ask them once I invite you to accomplish
that Chapter 1 BINARY programs “Digital Age” The crucial tool of up to
date suggestions methods All computer systems have the equal fundamental
accessories. internal the computing device a glance inside a laptop
Block Diagram of a Digital computing device Arithmetic Operations What
happens interior the CPU in one computing device cycle executing the
operation 7 + 5 Digital programs Early computer systems had been
designed to operate numeric computations They used discrete points of
tips named digits (finite sets) DIGITAL systems: manipulate discrete
elements of counsel such because the 10 decimal digits or the 26 letters
of the alphabet we live in the “Digital Age”! Binary gadget and good
judgment Circuits •What sort of information do computers work with? –
Deep down internal, it’s all 1s and 0s •What can you do with 1s and 0s? –
Boolean algebra operations – These operations map at once to hardware
circuits (good judgment circuits) different Numbering techniques
•Decimal (Arabic): (0,1,2,3,four,5,6,7,8,9): example: (452968) •Octal:
(0,1,2,3,four,5,6,7): example (4073)
•Hexadecimal(0,1,2,three,four,5,6,7,8,9,A,B,C,D,E,F) instance: (2BF3)
•Binary: (0,1): illustration: 1001110001011 10 8 16 Base in Numbering
systems The decimal numbering system uses base 10. The values of the
positions are calculated via taking 10 to some energy. 1 one hundred 1
102 6 10 2 1 6 a hundred and one 2 100 . . three 1/10 three 10-1 7 1/a
hundred 7 10-2 5 Digits 1/a thousand Weights 5 10-three Digits Weights
Base 10 for decimal numbers? It uses 10 digits: The digits 0 via 9. Base
in Numbering techniques [2] The binary numbering system is called
binary since it uses base 2. The values of the positions are calculated
by means of taking 2 to a few power. Base 2 for binary numbers : It
makes use of 2 digits. The digits 0 and 1. illustration of Numbers There
are two feasible approaches of writing a host in a given device: 1-
Positional Notation 2- Polynomial representation Positional Notation N =
(an-1an-2 ... a1a0 . a-1a-2 ... a-m)r the place . = radix element r =
radix or base n = number of integer digits to the left of the radix
factor m = variety of fractional digits to the correct of the radix
factor an-1 = most big digit (MSD) a-m = least enormous digit (LSD)
Positional Notation The Decimal Numbering equipment The decimal
numbering device is a positional quantity equipment. example: (4 6 2
1)10 (1000 one hundred 10 1 ) 1 x 100 2 x one zero one 6 x 102 four x
103 Positional Notation Binary Numbering system The Binary Numbering
equipment is additionally a positional numbering equipment. as an
alternative of the use of ten digits, 0 - 9, the binary gadget makes use
of best two digits, the 0 and the 1. illustration of a binary number
& the values of the positions. 1 0 1 0 1 0 1 26 25 24 23 22 21 20 1
23 1 22 0 21 1 20 . 0 2-1 1 2-2 Binary digits, or bits Weights (in base
10) Polynomial Notation N = an-1 x rn-1 + an-2 x rn-2 + ... + a0 x r0 +
a-1 x r-1 ... + a-m x r-m = n −1 ∑ ai r i i= − m example: Positional (N)
Polynomial (N) N = (651.45)10 = 6 x 102 + 5 x a hundred and one + 1 x
100 + 4 x 10-1 + 5 x 10-2 crucial number programs There are three
critical quantity programs • Binary quantity equipment • Octal number
equipment • Hexadecimal quantity system Binary numbers Digits = 0, 1
Positional Polynomial (11010.11)2 = 1 x 24 + 1 x 23 + 0 x 22 + 1 x 21 + 0
x 20 + 1 x 2-1 + 1 x 2-2 –1 ok (kilo) = 210 = 1,024 –1M (mega) = 220 =
1,048,576 –1G (giga) = 230 = 1,073,741,824 changing Decimal to Binary N =
(an-1an-2 ... a1a0 . a-1a-2 ... a-m)r Integer Fractional Radix aspect
Integer half and Fractional half are converted otherwise changing the
Integer half • maintain dividing by using 2 until the quotient is 0.
compile the remainders in reverse order. • example: (162) . 10 162 / 2
81 / 2 40 / 2 20 / 2 10 / 2 5/2 2/2 1/2 = eighty one = forty = 20 = 10
=5 =2 =1 =0 rem 0 rem 1 rem 0 rem 0 rem 0 rem 1 rem 0 rem 1 • Then
(162)10 = (10100010) 2 converting the Fraction half retain multiplying
the fractional part with the aid of 2 except it becomes 0. assemble the
integer elements (in ahead order). – youngsters this may additionally
now not terminate! –example: (0.375) 10 0.375 x 2 = 0.750 0.750 x 2 =
1.500 0.500 x 2 = 1.000 So, (.375) 10 = (.011)2 And (162.375) 10 =
(10100010.011)2 Why does this work? •This components will also be
utilized to transform from decimal to any base •are attempting changing
162.375 from decimal to decimal. 162 / 10 = 16 rem 2 16 / 10 = 1 rem 6 1
/ 10 = 0 rem 1 •each and every division “strips off” the rightmost
digit (the remainder). The quotient represents the final digits in the
quantity. Why does this work? [2] 0.375 x 10 = 3.750 0.750 x 10 = 7.500
0.500 x 10 = 5.000 • each and every multiplication “strips off” the
leftmost digit (the integer half). The fraction represents the last
digits. converting binary to decimal • to transform binary, or base 2,
numbers to decimal we first acquire the polynomial illustration of the
quantity, then sum the items. – example: (1101.01)2 1 23 1 22 0 21 1 20 .
0 2-1 1 2-2 Binary digits, or bits Weights (in base 10) (1 x 23) + (1 x
22) + (0 x 21) + (1 x 20) + (0 x 2-1) + (1 x 2-2) = eight + four + 0 + 1
+ 0 + 0.25 The decimal cost is: (13.25)10 Octal quantity device –Digits
= 0, 1, 2, 3, 4, 5, 6, 7 Positional = Polynomial (127.4)8= 1 x eighty
two + 2 x eighty one + 7 x eighty + four x 8-1 • Octal (base 8) digits
range from 0 to 7. when you consider that 8 = 23, one octal digit is
equivalent to 3 binary digits. changing Decimal to Octal Integer half:
preserve dividing by means of eight except the quotient is 0. bring
together the remainders in reverse order. Fractional half: keep
multiplying the fractional part by means of eight except it becomes 0.
assemble the integer ingredients (in forward order). equal components as
for the decimal to binary converting Octal to Decimal •to convert
Octal, or base eight, numbers to decimal we first attain the polynomial
representation of the number, then sum the products. instance
(127.four)8 = 1 x eighty two + 2 x 81 + 7 x 80 + 4 x eight-1 = (87.5)10
converting Binary to Octal •to transform from binary to octal, make
corporations of 3 bits, starting from the binary factor. Add 0s to the
ends of the quantity if necessary. Then convert each neighborhood of
bits to its corresponding octal digit. illustration (10110100.001011)2 =
=(010 110 a hundred . (2 6 4 . 001 011 )2 1 three )eight changing Octal
to Binary •to transform from octal to binary, exchange each and every
Octal digit with its equivalent 3-bit binary sequence. •instance
(261.35) eight = = (010 (2 110 6 001 1 . 011 . three one hundred and
one)2 5 )8 Hexadecimal numbers -Digits = 0, 1, 2, 3, 4, 5, 6, 7, eight,
9, A, B, C, D, E, F Positional Polynomial - (B65F)sixteen = eleven x 163
+ 6 x 162 + 5 x 161 + 15 x one hundred sixty • Hexadecimal (base
sixteen) digits are 0, 1, 2, three, 4, 5, 6, 7, eight, 9, A, B, C, D, E,
and F. given that 16 = 24, one hexa digit is reminiscent of four binary
digits. –It’s frequently less difficult to work with a host like B5
instead of 10110101. converting Decimal to Hexadecimal Integer part:
hold dividing by sixteen unless the quotient is 0. collect the
remainders in reverse order. Fractional part: maintain multiplying the
fractional part by means of sixteen unless it turns into 0. assemble the
integer constituents (in forward order). identical system as for the
decimal to binary conversion changing Hexadecimal to Decimal • to
transform Hexadecimal, or base 16, numbers to decimal, first acquire the
polynomial representation of the quantity, then sum the products.
instance (B65F)sixteen = 11 x 163 + 6 x 162 + 5 x 161 + 15 x one hundred
sixty = (forty six,687)10 converting Hexadecimal to Binary • to
transform from hexadecimal to binary, substitute each hex digit with its
equal 4bit binary sequence. • illustration 261.3516 = = (2 6 1 . 3 5
)sixteen (0010 0110 0001 . 0011 0101)2 converting Binary to Hexadecimal
•to convert from binary to hex, make businesses of 4 bits, ranging from
the binary element. Add 0s to the ends of the number if necessary. Then
convert each community of bits to its corresponding hex digit. •example
(10110100.001011)2 = = (1011 (B 0100. 0010 1100)2 four . 2 C)sixteen D e
c im a l 0 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 three 1 four 1 5 B in a r y 0
0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 0 0 1
0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 O c ta l 0 1 2
three four 5 6 7 1 0 1 1 1 2 1 3 1 4 1 5 1 6 1 7 H e x 0 1 2 three 4 5 6
7 eight 9 A B C D E F ARITHMETIC OPERATIONS IN A BINARY equipment
Binary Addition 0 +0 -sum 0 ___ 0 + 1 -1 1 +0 -1 1 +1 -10 ( sum of 0 and
carryover of 1 ) Examples: 1001 +0110 1111 0001 +1001 1010 1100 + 0101 1
0001 carryover Binary Addition-Examples assess your work carry 11 1 10 1
101101 (forty five)10 +011101 Sum + 100 1 01 0 = seventy four (29)10 1 x
25 + 0 x 24 + 1 x 23 +1 x 22 + 0 x 21 + 1 x 20 = 32 + 0 + 8 + four + 1 =
45 Binary Addition- Examples Addition of three Binary Digits x 0 0 0 0 1
1 1 1 y CarryIn 0 0 0 1 1 0 1 1 0 0 0 1 1 0 1 1 Sum 0 1 1 0 1 0 0 1
CarryOut 0 0 0 1 0 1 1 1 Addition of giant Binary Numbers - illustration
showing greater numbers: 1010 0011 1011 0001 + 0111 0100 0001 1001 1
0001 0111 1100 1010 Binary Subtraction difference ___ 0 - 0 -0 0 - 1 -11
( Diff. of 1 and carryover of 1 ) 0 10100 101011 - 010101 0 0 10110 1
-0 -1 1 -1 -0 Binary Multiplication 0 x0 -Product 0 ___ A X B 0 x 1 -0 1
x0 -0 0 010000.010 0 001000.010 0000000000 0010000010 0000000000
0000000000 0000000000 0000000000 0010000010 1000110.000100 1 x1 -1
Binary Division 0 . . = 1 0 1 . . 1 1 enhances in Numbering techniques •
enhances are utilized in digital systems (computer systems) for
simplifying the Subtraction operation and for logical manipulation •
There are to type of enhances for every base ‘r’ system: 1- Radix
complement Ex. base 10base 2 r’s complement 10’s complement 2’s
complement 2- Diminished radix complement Ex. ( r-1) complement base 10
9’s complement Base 2 1’s complement Radix complement (r's complement)
[N]r = rn - (N)r where n is the variety of digits in (N)r. instance •2's
complement of (N)2 = (101001)2 [N]2 = 26 - (101001)2 = (a million)2 -
(101001)2 = (010111)2 • 10's complement of (N)10 = (72092)10 [N]10 =
(one hundred thousand)10 - (72092)10 = (27908)10. acquiring 2’s
complement • may also be acquired without delay from the given number
through 1-coping each and every little bit of the number starting at
least massive bit and continuing ahead probably the most large bit
except the first 1 has been copied. 2- After the primary 1 has been
copied substitute each and every of the last 0s and 1s with the aid of
1s and 0s respectively (a) [1010100]2 = 27 - (1010100)2 = (a million)2 -
(1010100)2 =(0101100)2 (b) [101001]2 = 26 - (101001)2 = (1000000)2 -
(101001)2= (010111)2 (a) 2 ’s 1010100 01 0 1 1 0 0 (b) 10 1 0 0 1 010
111 Diminished radix complement (r-1’s complement) [N]r-1 = (rn -1)r -
(N)r -9’s complement of [546700]9 = 999999 – 546700 = 453299 -1’s
complement of [1011000] = (10000000 – 1)2 – (1011000)2 = (0100111)2
obtaining 1’s complement • 1 ’s complement can be obtained without delay
from the given number by using replacing every of 0s and 1s by way of
1s and 0s of the quantity (i.e. complement every bit) [1011000] =
(10000000 – 1)2 – (1011000)2 = (0100111)2 1011000 1 ’s complement 01 0 0
1 1 1 Subtraction with 2’s Complement • 2’s complement are used to
convert subtraction to addition, which reduces hardware necessities
(most effective adders are needed). A - B = A + (-B) (add 2’s complement
of B to A) • 2’s Complement has the residences of the minus signal A +
(-A) = 0 A + 2’sA = 0 - (- A) = A A - B = A + (-B) 2’s 2’sA = A A - B = A
+ 2’sB Subtraction with 2’s Complement [2] Examples: A= 1010100 B=
1000011 • 2’s complement A - B = A + (-B)= A+ [B] = (1010100) +
(0111101) = (0010001) 1010100 + Discard conclusion elevate 0111101 1
0010001 Subtraction with 1’s Complement Examples: be aware: equal
proprieties minus signal as 2’s complement. A= 1010100 B= 1000011 •1’s
complement A - B = A+ [B]= (1010100) + (0111100) = (0010001) 1010100 +
Add end lift 0111100 1 0010000 + 1 0010001 Subtraction with
10s/9’sComplements (72)10 - (32)10 = (40)10 10’s Complement [32]= 102 –
(32)10 = (68)10 (seventy two)10 + (68)10 = 1 (40)10 9’s Complement [32]=
(102 -1) – (32)10 = (sixty seven)10 (72)10 + (67)10 = (1 + 39)10 =
(40)10 Signed Binary numbers • keep in mind that digital programs are
made with contraptions that tackle precisely two states : 0 and 1. •The
most effective states are “1” and “0”. There isn't any “-” state!
because of hardware obstacles computers symbolize negative numbers by
using the leftmost bit for the signal bit. -- “0” shows a positive
quantity, -- while a “1” suggests a bad quantity Signed Magnitude •The
leftmost bit shows the signal of the number. The remaining bits give the
magnitude of the quantity the usage of eight bits to signify binary
quantity the value in the instance is: signal magnitude 1 0 0 0 0 0 1 1
-3 = 10000011 = 1/ (signal bit) 0000011 signal Magnitude illustration is
good for having the potential for a human to study 010-111 and
understand what quantity is represented Signed Complement (a) Signed
Magnitude illustration 1 0 0 0 0 0 1 1 -three = 10000011 = 1/ (signal
bit) 0000011 (b) Signed 1’s illustration 1 1 1 1 1 1 0 0 -3 = 10000011 =
1/ (signal bit) 1111100 (c) Signed 2’s representation 1 1 1 1 1 1 0 1
-three = 10000011 = 1/ (signal bit) 1111101 fixed-length Registers •All
functional digital contraptions have fastened-length registers •This
capability that numbers in a computer are represented by way of a hard
and fast number of bits –The earliest microprocessors had been 4-bit
gadgets –Intel 8080 and the 6502 (Apple II) chips had been 8bit –Intel
8088 (IBM notebook) and Motorola 68000 (Mac) are sixteen-bit gadgets
–Pentium chips and PowerPC chips are 32-bit latitude of a bunch Overflow
all the way through addition •a set-size register can handiest hold a
number numbers For a 4-bit device, the latitude of high-quality integers
is 0 - 15 For an 8-bit gadget the range of fine integers is 0 – 255
When adding wonderful integers, Overflow happens when the sum falls
outside the latitude of the register Overflow in Signed enhances •when
numbers are handled as signed complement, a “raise” of 1 from the
addition of essentially the most tremendous bits doesn't point out an
overflow, 3 00011 + (-three)+11101 = 00000, with a raise of “1”
(2’complement) : We be aware of that addion operation in 2’s complement
the conclusion-raise is discarded ! •For signed complement, overflow
happens when: The addition of two advantageous numbers outcomes in a bad
number OR The addition of two poor numbers outcomes in a favorable
number Overflow Examples •In a 6-bit register with signed 2’s complement
+ 17 = 010001 + 16 = +010000 =100001 100001 = - (11111) = -(31)10
instead of + (33)10 •equal with a 7-bit register + 17 = 0 010001 +
sixteen = +0 010000 = 0100001 0100001 = + 33 No Overflow Binary codes:
BCD (1) • To symbolize guidance as strings of alphanumeric characters. •
Binary Coded Decimal (BCD) – Used to represent the decimal digits 0 -
9. – four bits are used. – each bit place has a weight associated with
it (weighted code). – Weights are: 8, four, 2, and 1 from MSB to LSB
(known as 84-2-1 code). Binary codes: BCD (2) – BCD Codes: 0 three 6 9
0000 0011 0110 1001 1 0001 4 0100 7 0111 2 5 8 0010 0101 one thousand –
Used to encode numbers for output to numerical displays –example:
(9750)10 = (1001011101010000)BCD Binary codes: ASCII [2] • ASCII
(American common Code for suggestions Interchange) (see desk 1.7 of
textbook) – Most general character code. – instance: ASCII code
representation of the be aware ‘Digital’ persona D i g i t a l Binary
Code 1000100 1101001 1100111 1101001 1110100 1100001 1101100 Hexadecimal
Code forty four 69 67 sixty nine seventy four 61 6C apply complications
Solved in the category Examples: Signed complements 2’s signal-bit
(9)10 +( 6)10 0 0 0 1001 0110 1111 (9)10 0 1001 - ( 6)10 1 0 1010 0011
(6)10 0 0110 - ( 9)10 1 0111 1 1101 Examples: Signed complements 1’s
signal-bit (9)10 0 0 0 +( 6)10 1001 0110 1111 (9)10 0 - ( 6)10 1 0 1
1001 1001 0010 = (0010) + (0001) = (0011) (6)10 0 0110 - ( 9)10 1 0110 1
1 a hundred complications question (a) Convert le following binary
quantity into (i) Octal, (ii) Decimal, (iii) hexadecimal 10101101.10110
(b) Convert A= 16.25 and B = eight.25 into binary, use 7 bits to
represent the integer half and three bits to signify the fractional
half, then operate right here operations I) ii) iii) vi) C= A + B D =A–B
E= A x B F=A B note: Compute C and D (a) using non-signed binary
numbers and devoid of complements (b) the use of signed 2’s complement
(c) Convert le following number into (i)Decimal, (ii) Octal, (iii)
binary (FD8.C2B)sixteen issues 10101101.10110 solutions: i) Octal one
hundred and one (2 iii) Hexa (010 5 one zero one.101 100) 5 . 5 4)8 (
1010 1101.1011 0000 )2 (A D . B 0)sixteen ii) Decimal (10101101.10110)2 =
1 x 27 + 0 x 26 + 1 x 25 +0 x 24 + 1 x 23 + 1 x 22 +0 x 21 + 1 x 20 + 1
x 2-1 + 0 x 2-2 1 x 2-3 + 1 x 2-four + 0 x 2-5 = (173.6875)10
complications Integer half • Conversion de (sixteen) . 10 sixteen / 2
eight/2 four/2 2/2 1/2 =8 =four =2 =1 =0 the rest 0 the rest 0 the rest 0
remainder 0 the rest 1 • Then (16)10 = (ten thousand) 2 complications
Fractional part – converting: (0.25) 10 •then, (.25) 10 = (.01)2 0.25 x 2
= 0.50 0.50 x 2 = 1.00 and (16.25) 10 = (ten thousand.01)2 identical as
for A B = 8.25: (8.25) 10 = (1000.01)2 representation the usage of 7
bits and 3 bits A= (16.25) 10 = (0010000.010)2 B= (eight.25) 10 =
(0001000.010)2 issues A +B Non Signed Binary 0010000.010 0001000.010 0
011000.a hundred A 0010000.010 - B 0001000.010 0001000.000 complications
Signed 2’ complement A 0 010000.010 +B 0 001000.010 0 011000.a hundred A
0 010000.010 + (-B) 1 110111.110 0 001000.000 issues A X B 0 010000.010
0 001000.010 0000000000 0010000010 0000000000 0000000000 0000000000
0000000000 0010000010 1000110.000100 complications .B A. Dividend ten
thousand.010 1000010 01000 0000 1000010 Divider a thousand .010 1.1..
Quotient Binary common sense • Binary good judgment offers with 1 -
Variables that can tackle two discrete values Values can also be
referred to as proper, False, yes, no, and so forth. 2 - Operations that
anticipate LOGICAL which means Binary common sense is akin to Boolean
algebra Boolean Algebra •primary arithmetic required for the description
of digital circuits • used to describe the different interconnections
of digital circuits • the variable used in the Boolean algebra are
referred to as Boolean variables we will examine two-valued Boolean
algebra and services with simplifications the usage of fundamental
Boolean Identities Two-valued Boolean Algebra • It incorporates 1-
Boolean Variables - precise by letters of the alphabet equivalent to A,
B, C, x, y, z and so forth. - each and every variable can have two and
most effective two distinct values: 1 and nil (genuine, False) - can
also be a characteristic of every other Boolean variables (F=ABC) 2-
Boolean Operations -There are three basic logical operations: AND, OR,
and not primary Boolean Operations- AND operation • Represented through a
dot or by way of the absence of an operator example: examine: x.y = or
xy=z x AND y is equal to z Interpretation: Z = 1 if and handiest if x= 1
AND y= 1 otherwise z = 0 actuality desk: Don’t confuse this with binary
multiplication operation x y xy 0 0 0 0 1 0 1 0 0 1 1 1 reality desk
offers the price of z for all feasible values of x and y fundamental
Boolean Operations- OR operation • Represented through a plus sign (+)
example: read 010-111: x+y=z x OR y is equal to z Interpretation: z = 1
if x= 1 or if y= 1 or if each x =1 and y =1. z = 0 if x = 0 and y=1
reality desk: x Don’t confuse this with binary addition operation y x+y 0
0 0 0 1 1 1 0 1 1 1 1 reality desk offers the cost of z for all
feasible values for x and y basic Boolean Operations- now not operation •
Represented by using a chief or an overbar (also called complement)
instance: study 010-111: x’ = z (or x = z) no longer x is equal to z
Interpretation: z = “what x is not” x= 1 then z=0; x= 0 then z=1 truth
table: x x’ 0 1 1 0 fact table gives the cost of z for all possible
values for x Binary logic and Binary signals • For simplicity, we
regularly still write digits instead: – 1 is right – 0 is fake • we can
use this interpretation together with particular operations to design
services and common sense circuits for doing arbitrary computations.
logic Gates • good judgment gates are digital circuits that operate on
one or more input signal to supply an output sign •fundamental
operations can be carried out in hardware using a basic good judgment
gate. –Symbols for each of the common sense gates are proven beneath.
–These gates output the product, sum or complement of their inputs
common sense Operation: illustration: good judgment gate: AND (product)
of two inputs x.y, or xy OR (sum) of two inputs x+y no longer
(complement) With one input x’ Gates with distinctive Inputs • AND and
OR Gates may additionally have greater than 2 enter signals Binary
indicators •computers use voltages to characterize suggestions. •Two
voltage levels are used to characterize a binary cost “1” and “0” • Some
digital methods as an example may define that: - Binary ‘0” is the same
as 0 Volt - Binary “1” is the same as four Volt It’s effortless for us
to translate these voltages into values 1 and 0. Volts four 1 0 0 Binary
logic and Binary signals •It’s also feasible to believe of voltages as
representing two logical values, real and false. These logical values
are known as Boolean values Volts 4 proper False 0 common sense Gates -
alerts example two input alerts 1 0 0 one output signal good judgment
Gates - signals example 2 input alerts 1 1 1 1 output signal Timing
Diagram –input and output alerts basic and different common sense gates
•fundamental common sense gate •AND •OR •not These are called “primary
common sense gates” as all other gates and digital Circuits can be made
out of these gates. • different common sense gates •NAND •NOR •XOR •XNOR
These are known as “normal good judgment gates” as any digital circuit
can be designed via simply using these gates The NAND & NOR Gates •
we are able to use a NAND and NOR gates to put in force all three of the
fundamental operations (AND,OR,no longer). they are mentioned to be
functionally finished both NAND and NOR gates are very useful as any
design can be realized the use of either one. •it is easier to build
digital circuits the use of all NAND or NOR gates than to mix AND,OR,
and not gates. •NAND/NOR gates are typically sooner and cheaper to
produce. The NAND Gate •The NAND gate is a mix of an AND gate followed
by an inverter (now not gate). •we will use a NAND gate to implement all
three of the simple operations (AND,OR,no longer). •one of these gate
is declared to be functionally comprehensive. A B F= A.B = A+B A 0 0 1 1
B 0 1 0 1 F 1 1 1 0 The XOR Gate (exclusive-OR) • here's a XOR gate. •
XOR gates assert their output when precisely some of the inputs is
asserted, hence the name. • The operator image for this operation is ⊕ 1
⊕ 1 = 0 and 1 ⊕ 0 = 1. A B F= A⊕ B = AB + AB A 0 0 1 1 B 0 1 0 1 F 0 1 1
0 The XNOR Gate • This services as an unique-NOR gate, or comfortably
the complement of the XOR gate. • The image for this operation is 1 1 = 1
and 1 0 = 0. A B F A B F Z= F= A ⊕ B = (AB) + (A ⋅ B) = AB + A' B' 0 0 1
1 0 1 0 1 1 0 0 1 ...View Full document
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