Presentation is loading. Please wait.

Presentation is loading. Please wait.

Introduction to Microprocessors Chapter 2. Decimal or Base 10 Numbers  Have ten different digits (0-9)  It is a weighted number system. Each position.

Similar presentations


Presentation on theme: "Introduction to Microprocessors Chapter 2. Decimal or Base 10 Numbers  Have ten different digits (0-9)  It is a weighted number system. Each position."— Presentation transcript:

1 Introduction to Microprocessors Chapter 2

2 Decimal or Base 10 Numbers  Have ten different digits (0-9)  It is a weighted number system. Each position as you move to the left has a weight of the base (10 for decimal) raised to an integer power. The first position (from right to left) has power of zero, next position power of one…….  Ex: 123 = (1 x 10 2 ) + (2 x 10 1 ) + (3 x 10 0 )

3 Binary numbers  They have two digits (0 & 1)  To convert it to decimal:  ex1: 101 2 = (1 x 2 2 ) + (0 x 2 1 ) + (1 x 2 0 ) = 4 + 0 + 1 = 5 10  ex2: 1100 2 = (1 x 2 3 ) + (1 x 2 2 ) + (0 x 2 1 ) + (0 x 2 0 ) = 8+4+ 0 +0 =12 10  ex3: 11001 2 = (1 x 2 4 ) + (1 x 2 3 ) + (0 x 2 2 ) +  (0 x 2 1 ) + (1 x 2 0 ) = 16+ 8+0+ 0 +1 =25 10

4 Convert Decimal to Binary  Repeated division by 2  Ex: 18 10 = ? 2  18/2=9 and reminder is 0  9/2 = 4 and reminder is 1  4/2 =2 and reminder is 0  2/2 =1 and reminder is 0  ½ = 0 and reminder is 1  Therefore 18 10 = 10010 2

5 Hexadecimal (base 16)  Need 16 digits.  We use our ordinary decimal digits (0-9) and A-F

6 Hex to decimal ex1: 123 16 = (1 x 16 2 ) + (2 x 16 1 ) + (3 x 16 0 ) = 256 + 32 + 3 = 291 10 ex2: 1A 16 = (1 x 16 1 ) + (10 x 16 0 ) = 16 + 10 =26 10

7 Decimal to Hex Repeated division by 16 Ex1: 26 10 = ? 16 26/16=1 and reminder is 10 (A) 1/16 = 0 and reminder is 1 Therefore 26 10 = 1A 16 Ex2: 83 10 = 53 16

8 Determine next number in HEX  2E  9F  7CBF

9 Hex to Binary  Convert each digit to 4 bits binary equivalent  Example: AB35H 1010 1011 0011 0101

10 Binary to Hex  Break binary number into 4-bit sections from LSB to MSB  Convert the 4 bits binary to its hex equivalent  Example: 1010111110110010  Add zeros to the left if necessary A F B 2

11 Two’s complement signed number  Positive number ex: 00000101 2 = +5 10  Negative number is shown as 2’s complement of positive number  Write number as 8 bits  Complement each bit (0 ->1 and 1-> 0)  Add 1  Ex: -5 10 = ?  00000101 2 (1’s complement) => 11111010 +  1  11111011

12 Signed & unsigned numbers  8 bit unsigned can represent 0 to 255 10  8 bit signed can represent -128 to +127 10  2’s complement of a negative number is it’s corresponding positive number.  We do not have negative & positive zero!  If you have a negative number, How do you get its decimal equivalent??

13 2’s complement application  CPU do not need 2 separate hardware module to perform add & subtraction. Subtraction can be done as: a – b = a + (-b)

14 ASCII Code  American Standard Code for Information Interchange  7 bits needed. Ex: code for Q is 51(01010001)  One bit is added for Parity  Parity is even or odd  Even parity: Number of 1s are even ex: 11100001  Odd parity: Number of 1s are odd

15 Binary Coded Decimal (BCD)  Each decimal digit is represented by 4 bits  Example1: 53  Example 2: 9201 01010011 1001 001000000001

16 Application of BCD  Most common application of BCD is with seven-segment displays  BCD to seven segment decoder inputs one 4- bit BCD and outputs a segment pattern-code

17 Seven-segment  The display consists of seven individual bar LEDs  Displays decimal and hexadecimal values  The LED segments are labeled “a” through “g”  To illuminate a “0” on the display, segment “a” through “f” are illuminated

18 Overflow  Whenever two signed numbers are added or subtracted and the result is too large for the number of bits allocated

19 No overflow when adding a positive and a negative number No overflow when subtracting numbers with the same sign Overflow occurs when the result has “wrong” sign (verify!): Operation Operand A Operand B Result Indicating Overflow A + B  0  0  0 A + B  0  0  0 A – B  0  0  0 A – B  0  0  0 Detecting Overflow

20 Computer Programming  High level language  Assembly Language  Machine Language


Download ppt "Introduction to Microprocessors Chapter 2. Decimal or Base 10 Numbers  Have ten different digits (0-9)  It is a weighted number system. Each position."

Similar presentations


Ads by Google