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PC Bus and Interrupt System
The PC Bus utilized a bus controller, address latches, and data transceivers (bidirectional data buffers).
1) Bus controller :( Intel 8288 Bus Controller) coordinates activities on bus. It converts clock signal and CPU status into bus control signals. These control signals direct operations of data transceivers, latches and the I/O bus
2) Address latches: these are buffers for the address lines. They consider 2 reasons, fill the speed gap between the CPU and other devices; and permit the CPU pins to be utilized for other purposes.
3) Data transceivers: it is bidirectional data buffers
Interrupt processing: interrupt processing follows the below steps:
Once the external device recognizes the acknowledge, then it places the interrupt vector number on the data bus (through interrupt controller, in the case of IBM PC)
After the CPU receives the interrupt vector, it start the standard interrupt-initiation sequence: forming the interrupt vector address; then it is starting execution of the interrupt handler routine.
Compute the Fibonacci sequence - assembly program: Problem: Fibonacci In this problem you will write a program that will compute the first 20 numbers in the Fibonacci sequ
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init_lcd ;(this initialises a 2 row lcd) bcf TRISA,0 ;PORTA bit 0 as an output (lcd RS pin) bcf TRISA,1 ;PORTA bit 1
Port Mapped I/O or I/O Mapped I/O I/O devices are mapped into a separate address space. This is generally accomplished by having a different set of signal lines to denote a mem
SEG : Segment of a Label:- The SEG operator is which is used to decide the segment address of the, variable, label or procedure and substitutes the segment base address in plac
to find the matrix addition
External System Bus Architecture : This is a 16 bit processor with 40 pins. It has twenty address pins and out of which sixteen are utilized as data pins. This concept of by us
The Alpha : The development of the Alpha chip start in the year 1988 The new chip used 64 bit technology, let users to pack more complexity into their programs than exis
a- Trace the following program fragment and find out the content of ax after the the execution of the program. X db 5,7 -3,-9,4,-7,9 Mov
Opcode : The opcode generally appear in the first byte.but in a few instructions, a register objective is in the first byte and few other instructions may have their 3-bits of
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