By Dogan Ibrahim
На примере микроконтроллера PIC18F452/PIC18F4550/PIC18F258 и компилятора mikroC/CCSОсновные затрагиваемые темы:1) SD-карта
2) USB (на примере PIC18F4550)
3) CAN (на примере PIC18F258)
4) RTOS (общие представления, простейшая примитивная ОСь на CCS)Надо сказать, не очень "глубокая" книга. С другой стороны, не перегружена теорией и левым флудом. Вначале идет краткое описание модулей PICа. В середине и конце - практика, работа с IDE, схемы, собственно код...P.S. слово ZIGBEE, увиденное на обложке, в книге не ищите - это рекламный трюк издательства.Аннотация на английском языке:
The structure of the PIC 18FXXX sequence in addition to common oscillator, reset, reminiscence, and input-output circuits is totally distinct. After giving an advent to programming in C, the publication describes the undertaking improvement cycle in complete, giving info of the method of modifying, compilation, blunders dealing with, programming and using particular improvement instruments. the majority of the e-book offers complete info of attempted and confirmed hands-on initiatives, similar to the 12C BUS, USB BUS, CAN BUS, SPI BUS and real-time working systems.* a transparent creation to the PIC 18FXXX microcontroller's architecture
* 20 initiatives, together with constructing instant and sensor community functions, utilizing I2C BUS, USB BUS, CAN BUS and the SPI BUS, which offer the block and circuit diagram, application description in PDL, application directory and application description.
* various examples of utilizing developmental instruments: simulators, in-circuit debuggers (especially ICD2) and emulators
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Read Online or Download Advanced PIC microcontroller projects in C: from USB to ZIGBEE with the 18F series PDF
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Extra info for Advanced PIC microcontroller projects in C: from USB to ZIGBEE with the 18F series
Subtracting 127 from the mantissa, we obtain 252 – 127 ¼ 125. The EXOR of the signs of the numbers is 0. 375), which is the correct result. 4 Addition and Subtraction of Floating Point Numbers The exponents of floating point numbers must be the same before they can be added or subtracted. The steps to add or subtract floating point numbers are: Shift the smaller number to the right until the exponents of both numbers are the same. Increment the exponent of the smaller number after each shift.
6 First, divide the number into groups of four, then find the hexadecimal equivalent of each group: 10011111 = 1001 1111 9 F The hexadecimal number is 9F16. 7 Convert binary number 11101111000011102 into hexadecimal. 7 First, divide the number into groups of four, then find the hexadecimal equivalent of each group: 1110111100001110 = 1110 1111 0000 1110 E F 0 E The hexadecimal number is EF0E16. 8 Convert binary number 1111102 into hexadecimal. 8 Since the number cannot be divided exactly into groups of four, we have to insert, in this case, two zeros to the left of the number so the number of digits is divisible by four: 111110 = 0011 1110 3 E The hexadecimal number is 3E16.
18 Subtracting Binary Numbers To subtract one binary number from another, convert the number to be subtracted into negative and then add the two numbers. 26 Subtract binary number 0010 from 0110. 26 First, convert the number to be subtracted into negative: 0010 number to be subtracted 1101 complement 1 add 1 –––– 1110 Now add the two numbers: 0110 þ 1110 –––––– 0100 Since we are using only 4 bits, we cannot show the carry bit. 19 Multiplication of Binary Numbers Multiplication of two binary numbers is similar to the multiplication of two decimal numbers.