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Karteikarten

Midterm Study

Lucy House Microprocessors Course Englisch 43 Karten

Midterm Study

Karten in diesem Stapel (43)

  1. 1Microprocessor vs MicrocontrollerMicroprocessor vs Microcontroller
  2. 2Microprocessor is really just the ________.CPU (Central Processing Unit)
  3. 3The CPU size is specified by _________.The number of bits with which it can work at a time.
  4. 4RAM stands forRandom Access Memory
  5. 5ROM stands forRead only memory
  6. 6ALU Schematic "V"ALU Schematic "V"
  7. 7Harvard ArchitectureHarvard Architecture
  8. 8Princeton ArchitecturePrinceton Architecture
  9. 9Bit0 or 1
  10. 10ByteCalled Char in our programs
    Made of 8 bits
  11. 11ShortMade of 16 bits
    2 addresses
  12. 12Integer4 addresses
  13. 13Range for Unsigned Data Types0 to 255
    0 to 65535
    0 to 4,294,967,295
  14. 14Range for Signed Data Types-128 to 127
    -32768 to 32767
    -2,147,483,648 to 2,147,483,647
  15. 15Declaring variablesunsigned char x
  16. 16To mask a bit,and it with a reg that has a 1 in the spot in question.
    Ex, to see if char x is set in its 3rd digit, perform x&0x04 or x&0b00000100
  17. 17Reading hex0x13 = 1*16^1 + 3*16^0
  18. 18Setting a bitUse "OR"
    If reg is 11101000 and we want to set bit 1 (second from left), 11101000|00000010, which will only effect that bit.
  19. 19Clearing a bitClear by making zero in bit in question and all else 1's
  20. 20delay_ms declarationvoid delay_ms(unsigned short delay_time)
  21. 21delay_ms subroutine script{
    // Declare variables
    unsigned short i;
    while (1) {
    for (i=0; i<7600; i++) {
    // do nothing
    }
    // decrement delay_time
    delay_time--;
    // see if it is time to leave
    if (delay_time == 0) {
    // break out of this infinite loop
    break;
    }
    }
    }
  22. 22the contents of a memory location*(0x400E100)
  23. 23*(char *)0x400E1000=0x0F;*(The memory 8 bits long *) at address 0x400E1000 is set = 0x0F;
  24. 24*(volatile int *) 0x400E1074*(The memory 32 bits long *) at address 0x400E1074.
    Even if you haven't changed the contents you can still read it (thats what volatile does)
  25. 25The Atmel board we are using this quarter is _________________ architecture.Harvard
  26. 26Little EndianLSB first
  27. 27Big EndianMSB first
  28. 28The Atmel board is a ______ Endian device,Little
  29. 29For a timing diagram, the number is read fromright to left. The start bit is 0, the stop bit is 1. Before and after the transmission of the signal, idle, which is high.
  30. 30baud rate is in unitsbits per second
  31. 31bit time is1/(baud rate)
  32. 32On the board, a _____-bit address stores ______-bits of data32, 8
  33. 33x=0xAB; char_ptr=&x; //What's this doing?char_ptr is equal to the address of x
  34. 34for PORTB_PIO_CODR, you clear a bit bymaking it 1
  35. 35UART CharacteristicsUniversal Asynchronous Reciever Transmitter
  36. 36RS-232 CharacteristicsRecommended Specification (RS)
    Baud Rate can be specified: MCK/(16*CD)
    RHR: recieving hold register
    THR: transmitting hold register
    Slower
  37. 37Cursor Config[1c
  38. 38Clear screen, home cursor[j
  39. 39move to row a, column b, where a and b are numbers[a;bH
  40. 40how to make a subroutine that returns a valueunsigned char SubRoutine()
  41. 41how to make a subroutine that takes in a value (and does not output anything)void SubRoutine(unsigned char A)
  42. 42how to make a subroutine that takes in and puts out valuesunsigned char SubRoutine(unsigned char A)
  43. 43—

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Course Microprocessors Karteikarten am Midterm Study, erstellt von Lucy House am 12/04/2016.
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Lucy House
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