# Chapter 01 Lec 01

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DLF Digital Logic FundamentalsTheory (3 credit hr) Capt Mariam Kayani Practicle (1 credit hr) Ms Amin Akif

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Instructor Instructor: Capt Mariam Kayani mariam.signals@gmail.com 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

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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 June 11, 2011

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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 chessboard7

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

Decimal Number 7,392= 7x103 + 3x102 + 9x101 + 2x100 Thousands, hundreds, etcpower 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 (0,1,29) Decimal numbers are base 10.9

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 1s and 0s

Each coefficient aj is multiplied by 2j So 101 binary is 1 x 2 2 + 0 x 2 1 + 1 x 20 or 1x4 + 0x2 + 1x1=5

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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 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 (0,1,27) 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

Octal

Dec 0 1 2 3 4 5 6 7

Bin 000 001 010 011 100 101 110 111

Octal 0 1 2 3 4 5 6 7

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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

HexadecimalDec Bin 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 Hex 0 1 2 3 4 5 6 7 8 9 A B C D E F15

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

Number-base Conversion

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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 27 128 6 26 64 5 25 32 4 24 16 3 23 8 2 22 4 1 21 2 0 20 1

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ExampleWhat is 10011100 in decimal?27 26 25 24 23 22 21 20

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0

0

1

1

1

0

0

128+ 0 + 0 + 16 + 8 + 4 + 0 + 0 = 156 Since a 0 bit does not contribute anything in the sum, therefore a binary no. can be converted to decimal by adding only the 1 bits.18

Converting Decimal to Binary Example 41 41 divided by 2, giving quotient of 20 and reminder 1 20/2 , giving quotient of 10 and reminder 0 10/2 , giving quotient of 5 and reminder 0 5/2 , giving quotient of 2 and reminder 1 2/2 , giving quotient of 1 and reminder 0 , giving quotient of 0 and reminder 1 a0 a1 a2 a3 a4 a5

Therefore, the answer is (41)10= (a5a4a3a2a1a0)=(101001)219

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.20

Example (0.6875)10 integer fraction 0.6875 * 2 = 0.3750 * 2 = 0.7500 * 2 = 0.5000 * 2 = 1 0 1 1

coefficient

+ 0.3750 a-1 = 1 + 0.7500 a-2 =0 + 0.500 a-3 =1 + 0.000 a-4 =1

(0.1011)221

Octal Decimal Convert (231)8 to decimal

82 2

813

801

153

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 OctalInteger Fraction 4 + 0.104 0 + 0.832 6 + 0.656 5 + 0.248 1 + 0.984 7 + 0.872 (0.513)10= (0.406517)8 Coefficient a-1=4 a-2=0 a-3=6 a-4=5 a-5=1 a-6=7

0.513 X 8 = 0.104 X 8 = 0.832 X 8 = 0.656 X 8 = 0.248 X 8 = 0.984 X 8 =

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Dec

Hex 0 1 2 3 4 5 6 7 8 9 A B C D25

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

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

0x2AC 2 x 256 163 4096

+ 10 x 16 + 12 x 1 = 684 162 256 161 16 160 1

E F

Decimal Hex(684)10684 divided by 16, giving quotient of 42 , reminder 12 a0 42/16 , giving quotient of 2 and reminder 10 a1 10/16 , giving quotient of 0 and reminder 2

2AC26

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

2ac0010 1010 1100

0x2ac = 001010101100

No magic remember hex digit = 4 bits

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Binary to Hex Just convert groups of 4 bits

101001101111011 0101 0011 0111 10115 3 7 b

101001101111011 = 537B30

Arithmetic -- addition Binary similar to decimal arithmeticNo carries

0 1 1 0 0 + 1 0 0 0 1 1 1 1 0 1

1 0 1 1 0 0 Carries 1 0 1 1 0 + 1 0 1 1 1 1 0 1 1 0 1

1+1 is 2 (or 102), which results in a carry31

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Arithmetic -- subtraction

No borrows

1 0 1 1 0 - 1 0 0 1 0 0 0 1 0 0

0 0 1 1 0 Borrows 1 1 1 1 0 - 1 0 0 1 1 0 1 0 1 1 0 - 1 results in a borrowBorrow makes it (10)2= (2) 1033

Arithmetic -- multiplication1 0 1 1 X 1 0 1 1 0 1 1 0 0 0 0 1 0 1 1 1 1 0 1 1 134

Successive additions of multiplicand or zero, multiplied by 2 (102). Note that multiplication by 102 just shifts bits left.

Complements Simply Subtraction (Subtraction by addition) Rs Complement In Binary 2s complement In Decimal 10s complement

(R-1) Complement In Binary 1s complement In Decimal 9s 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 9s complement of 546700 is 999999-546700=453299 In simple words subtract each digit from 94 If n=4 then 2 = (16)10 = (10000)2 4 2 1 = (15)10 = (1111)2 Note: 1-0=1 and 1-1 =0 (Bit Changes) In simple words just change the bits

Binary: (2n -1 )-N

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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=01010037

Subtraction with r-Complement M-N Add the minuend, M to rs 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 rs complement of answer (Negative)38

Subtraction with rs Complement Using 10s complement subtract 72532-03250 Using 10s complement subtract 03250 -72532 Using 2s complement subtract 1010100 -1000011 Using 2s complement subtract 1000011- 1010100

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Subtraction with r-1 Com