?A 68000 system receives parallel data and transmits it serially in a cycle of 1
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?A 68000 system receives parallel data and transmits it serially in a cycle of 10 clock pulses (interrupts from clock). At pulse 1, data is read from Parallel in as a whole byte. At pulse 2 through 9, this byte is transmitted out serially via bit b1 of the data bus to Serial Out starting with the second least significant bit (b1). During serial transmission, other bits of the data bus are all zero At pulse 10, there is no activity. The process is repeated endlessly. Any time the Reset button is pressed, the conversion starts fresh with a new byte and a 10-pulse cycle. Use the Table below to figure out Interrupt Level, Vector Address, and ILP Register bits.
?A 68000 system receives parallel data and transmits it serially in a cycle of 10 clock pulses (interrupts from clock). At pulse 1, data is read from Parallel in as a whole byte. At pulse 2 through 9, this byte is transmitted out serially via bit b1 of the data bus to Serial Out starting with the second least significant bit (b1). During serial transmission, other bits of the data bus are all zero At pulse 10, there is no activity. The process is repeated endlessly. Any time the Reset button is pressed, the conversion starts fresh with a new byte and a 10-pulse cycle. Use the Table below to figure out Interrupt Level, Vector Address, and ILP Register bits. Write a complete (start-end) 68000 Assembly language program with comments for this system.Explanation / Answer
;lets assume ;D0=FFFFFFFF D1=01234567 MOVE.B D1,D0 ;copies a Byte from source ;D0=FFFFFF67 D1=01234567 .... ;lets assume ;D0=FFFFFFFF D1=01234567 MOVE.W D1,D0 ;copies a Word from source ;D0=FFFF4567 D1=01234567 ;lets assume ;D0=FFFFFFFF A1=01234567 MOVE.W A1,D0 ;copies a Word from source ;D0=FFFF4567 A1=01234567 .... ;lets assume ;D0=01234567 A1=FFFFFFFF MOVE.W D0,A1 ;copies a Word from source ;D0=01234567 A1=00004567 ; sign to A1, changed .... ;lets assume ;D0=0000FFFF A1=00000000 MOVE.W D0,A1 ;copies a Word from source ;D0=0000FFFF A1=FFFFFFFF ; sign to A1, changed ;lets assume ;D0=FFFFFFFF A1=00001000 ;mem.addr. $1000=01234567 MOVE.L (A1)+,D0 ;copies the Long word starting at address stored in A1, to D0 ;then increment A1 by 4 ;D0=01234567 A1=00001004 ;mem.addr. $1000=01234567 ;lets assume ;D0=FFFFFFFF A1=000010A8 ;mem.addr. $10A4=01234567 MOVE.L -(A1),D0 ;first decrements A1 by 4, A1=000010A4 ;then copies the Long word starting at address stored in A1, to D0 ;D0=01234567 ;mem.addr. $10A4=01234567
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