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Page 1: Chapter 01 Lec 01

DLF DLF

Digital Logic Fundamentals Digital Logic Fundamentals

Theory (3 credit hr)– Capt Mariam KayaniTheory (3 credit hr)– Capt Mariam Kayani

Practicle (1 credit hr) – Ms Amin AkifPracticle (1 credit hr) – Ms Amin Akif

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Instructor• Instructor: Capt Mariam Kayani

[email protected]

• Telephone :119 ext 33246

• Office Hours:

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Course Resources

• Digital Design (3rd/4th Edition)

By Morris Mano

• Topics from different sources will be indicated during the semester

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RULES

• Submit your assignments to class senior before

start of class on due date

• No Late submissions. (No exceptions!)

• Assignments with too much similarity will be

penalized appropriately

• If u miss a quiz/Exam College SOP will be

followed 4

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Objective of Course

• Introduction to concepts of digital logic, gates, and the digital circuits

• Design and analysis of combinational circuits• Design and analysis of sequential circuits

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Lecture - 01

April 9, 2023April 9, 2023

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Digital Systems• Digital System play a prominent role in this digital age

– Communication, medical treatment, internet, DVD, CD, etc…

• Digital Computer follow a sequence of instructions, called programs, that operate on given data– User can specify and change program or data according to needs

• Like Digital Computers, most digital devices are programmable

• Digital Systems have the ability to Manipulate discrete elements of information.– Any set that is restricted to a finite number of elements contains

discrete information• 10 Decimal digits• 26 Alphabet letters• 52 Playing cards• 64 squares of a chessboard

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Digital Systems• Digital Systems can do hundreds of millions of

operations per second• Extreme reliability due to error-correcting codes• A Digital System is interconnection of digital

modules• To understand Digital module, we need to know

about digital circuits and their logical functions• Hardware Description Language (HDL) is a

programming language that is suitable for describing digital circuit in a textual form– Simulate a digital system to verify operation

before it is built 8

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Decimal Number

• 7,392= 7x103 + 3x102 + 9x101 + 2x100

– Thousands, hundreds, etc…power of 10 implied by position of coefficient

• Generally a decimal number is represented by a series of coefficients– a6 a5 a4 a3 a2 a1 a0 (.) a-1 a-2 a-3 a-4

• aj cofficient are any of the 10 digit (0,1,2…9)

• Decimal numbers are base 10.9

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Binary Number

• Digital Systems manipulate discrete quantities of information in binary form

• Strings of binary digits (“bits”)

• Two possible values 0 and 1

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Binary Numbers• Each digit represents a power of 2• Coefficient have two possible values 0 and 1• Strings of binary digits (“bits”)

–n bits can store numbers from 0 to 2n -1–n bits can store 2n distinct combinations of 1’s

and 0’s

• Each coefficient aj is multiplied by 2j

• So 101 binary is 1 x 22 + 0 x 21 + 1 x 20

or 1 x 4 + 0 x 2 + 1 x 1 = 5 11

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BITs & Bytes• A bit (short for binary digit) is the smallest unit

of data in a computer. – A bit can hold only one of two values: 0 or 1,

corresponding to the electrical values of off or on, respectively.

– Because bits are so small, you rarely work with information one bit at a time

– A byte is a unit of measure for digital information.

– A single byte contains eight consecutive bits

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Special Powers of 2

210 (1024) is Kilo, denoted "K"

220 (1,048,576) is Mega, denoted "M"

230 (1,073, 741,824)is Giga, denoted "G"

240 (1,099,511,627,776 ) is Tera, denoted “T"

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Octal• Octal is base 8• A number is represented by a series of

coefficients

– a6 a5 a4 a3 a2 a1 a0. a-1 a-2 a-3 a-4

• aj cofficient are any of 8 digit (0,1,2…7)

• Need 3 bits for representation• Example:

(127.4)8= 1 X 82 +2 X 81 +7 X 80 +

4 X 8-1

64+16+7+.5= (87.5)10

Dec Bin Octal

0 000 0

1 001 1

2 010 2

3 011 3

4 100 4

5 101 5

6 110 6

7 111 7

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Hexadecimal• Hexadecimal is base 16• A number is represented by a series

of coefficients– a6 a5 a4 a3 a2 a1 a0. a-1 a-2 a-3 a-4

• aj cofficient are any of 16 digit (0,1,2,3,4,5,

6,7,8, 9,A,B,C,D,E,F)• Need 4 bits for representation• (B65F)16

11 X 163 +6 X 162 + 5 X 161 +15 X 160

= 11x4096 + 6x256 +5x16 +15= 45056 + 1536 + 80 +15 = 46,687

Dec Bin Hex

0 0000 0

1 0001 1

2 0010 2

3 0011 3

4 0100 4

5 0101 5

6 0110 6

7 0111 7

8 1000 8

9 1001 9

10 1010 A

11 1011 B

12 1100 C

13 1101 D

14 1110 E

15 1111 F

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Number-base Conversion

• A number in one base (10,2,8, 16) can be converted into its equivalent in another base.

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Converting Binary to Decimal

• The conversion of a num in base r to decimal is done by expanding the num in a power series and adding all the terms

• Multiply digit by power of 2.

7 6 5 4 3 2 1 0

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 1

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Example

1 0 0 1 1 1 0 0

128+ 0 + 0 + 16 + 8 + 4 + 0 + 0 = 156128+ 0 + 0 + 16 + 8 + 4 + 0 + 0 = 156

•Since a 0 bit does not contribute anything in the Since a 0 bit does not contribute anything in the sum, therefore a binary no. can be converted to sum, therefore a binary no. can be converted to decimal by adding only the 1 bits. decimal by adding only the 1 bits.

What is 10011100 in decimal?What is 10011100 in decimal?

27 26 25 24 23 22 21 20

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Converting Decimal to Binary

• Example 41• 41 divided by 2, giving quotient of 20 and reminder 1 a0• 20/2 , giving quotient of 10 and reminder 0 a1• 10/2 , giving quotient of 5 and reminder 0 a2• 5/2 , giving quotient of 2 and reminder 1 a3• 2/2 , giving quotient of 1 and reminder 0 a4• ½ , giving quotient of 0 and reminder 1 a5

Therefore, the answer is (41)10= (a5a4a3a2a1a0)=(101001)2

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Number With Radix Point

• If the number includes a radix point, it is necessary to separate the number into an integer part and a fraction part.

• The conversion of a fractional part in base r is done by multiplying the number and all successive integers are accumulated instead of reminders.

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Example

• (0.6875)10

integer fractioncoefficient

• 0.6875 * 2 = 1 + 0.3750 a-1 = 1• 0.3750 * 2 = 0 + 0.7500 a-2=0• 0.7500 * 2 = 1 + 0.500 a-3=1• 0.5000 * 2 = 1 + 0.000 a-4=1

(0.1011)2

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Octal Decimal

• Convert (231)8 to decimal

82 81 80

153

2 3 1

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Decimal Octal

• Convert decimal 153 to Octal

• 153 divided by 8, giving quotient of 19 , reminder 1 a0• 19/8 , giving quotient of 2 and reminder 3 a1• 10/8 , giving quotient of 0 and reminder 2 a2

153 = (231)8

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Decimal Octal (Fraction)• Convert decimal 0.513 to Octal

Integer Fraction Coefficient

0.513 X 8 = 4 + 0.104 a-1=4

0.104 X 8 = 0 + 0.832 a-2=0

0.832 X 8 = 6 + 0.656 a-3=6

0.656 X 8 = 5 + 0.248 a-4=5

0.248 X 8 = 1 + 0.984 a-5=1

0.984 X 8 = 7 + 0.872 a-6=7

(0.513)10= (0.406517)8

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Hex Decimal

• Just multiply each hex digit by decimal value, and add the results.

4096163

116256160161162

0x2AC0x2AC

2 2 x 256x 256 + 10 + 10 x 16x 16 + 12 + 12 x 1x 1 = 684= 684

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14

13

12

11

10

9

8

7

6

5

4

3

2

1

0

Dec

F

E

D

C

B

A

9

8

7

6

5

4

3

2

1

0

Hex

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Decimal Hex

(684)10

684 divided by 16, giving quotient of 42 , reminder 12 a042/16 , giving quotient of 2 and reminder 10 a110/16 , giving quotient of 0 and reminder 2

2AC

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Binary to Octal

• Partition Binary number into group of three digits each

• The corresponding octal digit is then assigned to each group

• (10 110 001 101 011 . 111 100 000 110)2

• (10 110 001 101 011 . 111 100 000 100)2 = (26153.7406)8

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Octal to Binary

• Each Octal digit is converted to its three digit binary equivalent

• (26153.7406)8 = (010 110 001 101 011 . 111 100 000 110)2

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Hex to Binary

• Convention – write 0x before number• Hex to Binary – just convert digits

2ac2ac

00100010 10101010 11001100

0x2ac = 0010101011000x2ac = 001010101100

No magic – remember hex digit = 4 bitsNo magic – remember hex digit = 4 bits29

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Binary to Hex

• Just convert groups of 4 bits

101001101111011101001101111011

10111011

55 33 77 bb

101001101111011 = 537B101001101111011 = 537B

0101 0101 0111 0111 0011 0011

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Arithmetic -- addition

• Binary similar to decimal arithmetic

+ 10001

00110

10111

No carries 001101

+ 11101

01101

1 10110

Carries

1+1 is 2 (or 102), which results in a carry

3131

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Arithmetic -- subtraction

- 01001

01101

00100

No borrows 01100

- 11001

01111

11010

Borrows

0 - 1 results in a borrowBorrow makes it (10)2 =(2) 10

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Arithmetic -- multiplication

1101

1 110

0000

1 11101

Successive additions of multiplicand or zero,multiplied by 2 (102).

Note that multiplication by 102 just shifts bits left.

1101

101 X

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Complements

• Simply Subtraction

(Subtraction by addition)– R’s Complement

• In Binary 2’s complement• In Decimal 10’s complement

– (R-1) Complement• In Binary 1’s complement• In Decimal 9’s Complement

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Diminished Radix Complement

• Given a number N in base r having n digits, it is (rn -1 )-N• Decimal: (10n -1 )-N

– If n=6 then 106-1=1000000-1=999999– 9’s complement of 546700 is 999999-546700=453299– In simple words subtract each digit from 9

• Binary: (2n -1 )-N– If n=4 then 24= (16)10= (10000)2

– 24 – 1 = (15)10= (1111)2

– Note: 1-0=1 and 1-1 =0 (Bit Changes)– In simple words just change the bits

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Radix Complement• Given a number N in base r having n

digits, it is (rn –N)• Simply add one to the radix-1 complement

(rn –N) = [(rn -1 )-N] +1

• Decimal: – 2389 7610+1=7611

• Binary: (2n -1 )-N– 101100010011+1=010100

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Subtraction with r-Complement• M-N• Add the minuend, M to r’s complement

of Subtrahend, N– M+ (rn -N)= M-N+ rn

• If M GTE N then sum will produce end carry rn. Ignore it

• If M LT N (No Carry) then take r’s complement of answer (Negative)

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Subtraction with r’s Complement

• Using 10’s complement subtract 72532-03250

• Using 10’s complement subtract 03250 -72532

• Using 2’s complement subtract 1010100 -1000011

• Using 2’s complement subtract 1000011- 1010100

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Subtraction with r-1 Complement

• Similar to r’s complement• But since r-1 complement is 1 less than r complement,

Carry is added back to get the result• If no carry, result is negative1’s complement to get the

answer– 1010100-1000011– 1000000-1010100

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Signed Binary Numbers• Need notation for negative values• Everything must be represented by binary

digits• Signed magnitude convention

– Left most bit can be used • 0 Positive• 1 Negative• 01001 is +9 and 11001 is -9 (Not 25. Convention known in

advance)

• Signed Complement (Store negative as comps)– Signed 1’s complement (8 bits)11110110 – Signed 2’s complement (8 bits)11110111– Signed Magnitude (8 bits) 10001001

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Complements

• Simply Subtraction (Subtraction by addition)– R’s Complement

• In Binary 2’s complement• In Decimal 10’s complement

– (R-1) Complement• In Binary 1’s complement• In Decimal 9’s Complement

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Diminished Radix Complement• Given a number N in base r having n

digits, it is (rn -1 )-N• Decimal: (10n -1 )-N

– If n=6 then 106-1=1000000-1=999999– 9’s complement of 546700 is 999999-

546700=453299– In simple words subtract each digit from 9

• Binary: (2n -1 )-N– If n=4 then 24= (16)10= (10000)2

– 24 – 1 = (15)10= (1111)2

– Note: 1-0=1 and 1-1 =0 (Bit Changes)– In simple words just change the bits

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Radix Complement

• Given a number N in base r having n digits, it is (rn –N)

• Simply add one to the radix-1 complement(rn –N) = [(rn -1 )-N] +1

• Decimal: – 2389 7610+1=7611

• Binary: (2n -1 )-N– 101100010011+1=010100

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Subtraction with r-Complement• M-N• Add the minuend, M to r’s complement

of Subtrahend, N– M+ (rn -N)= M-N+ rn

• If M GTE N then sum will produce end carry rn. Ignore it

• If M LT N (No Carry) then take r’s complement of answer (Negative)

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Subtraction with r’s Complement

• Using 10’s complement subtract 72532-03250

• Using 10’s complement subtract 03250 -72532

• Using 2’s complement subtract 1010100 -1000011

• Using 2’s complement subtract 1000011- 1010100

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Signed Binary Numbers• Need notation for negative values• Everything must be represented by binary digits• “Signed magnitude convention” is where;

– Left most bit can be used • 0 Positive• 1 Negative• 01001 is +9 and 11001 is -9 (Not 25. Convention

known in advance)

• Signed Complement sys

(Store negative no is rep by its complement)– Signed 1’s complement (8 bits)11110110 – Signed 2’s complement (8 bits)11110111– Signed Magnitude (8 bits) 10001001

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Arithmetic -- addition

• Binary similar to decimal arithmetic

+ 10001

00110

10111

No carries 001101

+ 11101

01101

1 10110

Carries

1+1 is 2 (or 102), which results in a carry

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Arithmetic -- subtraction

- 01001

01101

00100

No borrows 01100

- 11001

01111

11010

Borrows

0 - 1 results in a borrowBorrow makes it (10)2 =(2) 10

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Arithmetic -- multiplication

1101

1 110

0000

1 11101

Successive additions of multiplicand or zero,multiplied by 2 (102).

Note that multiplication by 102 just shifts bits left.

1101

101 X

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BCD (Binary Coded Decimal)

• Representing the decimal digits by mean of code that contains 1 , 0 i.e. Decimal digits stored in binary

– Four bits/digit (Use 10 instead of 16)– Num with k decimal digit will require 4k bits – BCD is base as its equivalent binary number 0 – 9– Binary combination 1010-111 are not used in BCD– Example 931 is coded as 1001 0011 0001

– Decimal 15 is BCD 0001 0101 in Binary it was 1111

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BCD Addition• Since each digit is max 9 Sum will always

be less than 19= 9+9+1(carry)• Two BCD digits are added as binary

numbers– When binary sum is more than binary (1001)2,

result is invalid (unlike Hex last 6 were ignored)

• Addition of 6=(0110)2 make a correct BCD and produces a carry– Binary Sum carry and Decimal Carry differ by

16-10=6

• 4+5, 4+8, 8+9• 184+576

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Decimal Arithmetic• Representation of signed decimal number in BCD is

similar to the rep of signed number in binary– Sign and mag system– Sign complement system

• Designate a plus with 0 and minus with 9• For Addition, add all digits including the sign digit and

discard the carry. This assumes that all –ive no are in 10’s complement form– (+375) + (-240) = +135

• Method for Subtraction is same as binary signed number

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Binary Codes for Numbers• Binary codes for decimal digits require 4 bits per digit• Many codes are use 4 bits in 10 distinct possible combinations (out of 16)• 2421 and Excess 3 are self complementing (1 and 0 9’s Comp of decimal)• Contents can be interpreted differently. Contents can be interpreted differently. • What decimal value does 1100001111001001 represent in different What decimal value does 1100001111001001 represent in different

binary codes?binary codes?

Dec Binary BCD Excess-3 2421 84-2-1 0 0 0000 0011 0000 0000 1 1 0001 0100 0001 0111 2 10 0010 0101 0010 0110 3 11 0011 0110 0011 0101 4 100 0100 0111 0100 0100 5 101 0101 1000 1011 1011 • 110 0110 1001 1100 1010 • 111 0111 1010 1101 1001 • 1000 1000 1011 1110 1000 • 1001 1001 1100 1111 1111

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Other Codes Exist

Gray Code – to represent te digital data when it is converted from analog data– Only one bit changes at a time– 0000,0001,0011,0010,0110,0111,0101,0100,1

100,1101,1111,1110,1010,1011,1001,1000– Why is this useful?

• 01111000 All Four bits need to be changed, may cause intermediate erroneous number

– Application of Gray code is when analog data are rep by a continuous change of shaft position

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Character Codes• ASCII

– Many applications require handling of not only numbers but letters and special characters

– Stands for American Standard Code for Information Interchange

– 7 Bits to store 128 characters – In ASCII, every letter, number, and punctuation

symbol has a corresponding number, or ASCII code– This encoding system not only lets a computer store

a document as a series of numbers, but also lets it share such documents with other computers that use the ASCII system.

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Error Detection

• 8th bit is sometimes added to the ASCII character to represent its parity

• A parity bit is an extra bit included with a message to make the total num of 1’s either even or odd.

Note that parity detects only simple errors– One, three, etc. bits

• More complex methods exist• Some that enable recovery of original info

– Cost is more redundant bits58

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Even Parity

• ExamplesA (01000001) with even parity is 01000001

C (01000011) with even parity is 11000011

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Odd Parity

• Similar except make the number of 1’s odd

• ExamplesA (01000001) with odd parity is 11000001

C (01000011) with odd parity is 01000011

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Binary Storage and Registers• Physical existence in information storage medium

for storing individual bits• A binary cell is a device that posses stable stages

and is capable of storing one bit of information• A Register is a group of binary cells.

– Can store any discrete quantity of information that contains n bits.

– 1100001111001001 is a 16 bit register– 2n possible states to store 0 to 2n -1

number– Contents can be interpreted differently 61

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Register Transfers

• Basic Operation in digital systems• When Key is pressed 8 bit

alphanumeric character code in to Input Register

• Contents of Input Register are transferred to eight least significant cells of a Processor Register

• After every transfer input register is cleared for new keystroke

• Each eight bit character transfer to the processor register is preceded by shift of previous character to next eight cells on its left

• When Processor Register is full, its contents are transferred to the Memory Register 62

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• Adding two 10 bit binary numbers

• Memory Unit• Processor Unit

Manipulation of binary variable

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Binary Logic• Variables have two possible distinct values, 0

and 1

• Three Logic operations– AND “ . ” If and only if all variable are 1– OR “ + ” If any one or more of the variable is 1– NOT “ ’ ” Complement (Reverse)

• Truth Table of Logical Operations

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Representation of Binary variables• Different Digital Systems represent 0 and 1

differently– Logical 0 as 0 volts. Logical 1 as 4 volts– Range

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Logic Gates are electronic circuits that operate on one or more input signals to produce an output signal

Logic Gates

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Gates with Multiple Inputs

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Binary Codes for Numbers• Binary codes for decimal digits require 4 bits per digitDec Binary BCD Excess-3 2421 84-2-1

0 0000 0000 0011 0000 0000

1 0001 0001 0100 0001 0111

2 0010 0010 0101 0010 0110

3 0011 0011 0110 0011 0101

4 0100 0100 0111 0100 0100

5 0101 0101 1000 1011 1011

0110 0110 1001 1100 1010

0111 0111 1010 1101 1001

1000 1000 1011 1110 1000

1001 1001 1100 1111 1111

Example : 395

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Binary Storage and Registers• Physical existence in information storage medium

for storing individual bits• A binary cell is a device that posses stable stages

and is capable of storing one bit of information• A Register is a group of binary cells.

– Can store any discrete quantity of information that contains n bits.

– 1100001111001001 is a 16 bit register– 2n possible states to store 0 to 2n -1

number– Contents can be interpreted differently 69

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Register Transfers• Basic Operation in digital systems• When Key is pressed 8 bit

alphanumeric character code in to Input Register

• Contents of Input Register are transferred to eight least significant cells of a Processor Register

• After every transfer input register is cleared for new keystroke

• Each eight bit character transfer to the processor register is preceded by shift of previous character to next eight cells on its left

• When Processor Register is full, its contents are transferred to the Memory Register

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• Adding two 10 bit binary no• Memory Unit• Processor Unit

Manipulation of binary variable

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Binary Logic

• Variables have two possible distinct values, 0 and 1

• Three Logic operations– AND “ . ” If and only if all variable are 1– OR “ + ” If any one or more of the variable is 1– NOT “ ’ ” Complement (Reverse)

• Truth Table of Logical Operations

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Representation of Binary variables

• Different Digital Systems represent 0 and 1 differently– Logical 0 as 0 volts. Logical 1 as 4 volts– Range

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Logic Gates are electronic circuits that operate on one or more input signals to produce an output signal

Logic Gates

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Gates with Multiple Inputs

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General Queries

• How to determine the base of number?– Example

• 14/2=5• 24+17= 40

• 54/4 = ?

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General Queries• How to fine the base Given the solution of the quadratic

Equation?– Example

• X2 -11x +22 = 0 is x=3, x=6 what is the base of the number

Substitute both the values of x in the equation

(3)^2 – 11(3) +22 = 0

9 – (33) + (22) = 0

9 – (3xr^1 + 3xr^0) + (2xr^1 + 2xr^0) =0

9 – 3r -3 + 2r + 2 =0

r = 8 77

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General Queries

• How to determine the r’s/(r-1)’complement of number with fraction ?– Example

• 2’s Complement (0.0110)2

r^n – N = 2^0-(0.0110)2 = (1)2– (0.0110)2 = 0.1010• 1’s Complement (0.0110)2

1- r^n –N = (1- 2^-4)10 – (0.0110)2 = (0.935)10- (0.0110)2 (0.1111)2- (0.0110)2 = (0.1001)2

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