When it is executed on a microprocessor of clock

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When it is executed on a microprocessor of clock frequency 5 Mhz, takes 415 or 0.8 microseconds to complete its execution. The same instruction when run on a microprocessor with 4Mhz clock frequency takes, 414, i.e. 1 microseconds. NOP is used to produce the delay, without effecting the actual running of the program. Consider the following examples, in which you are trying a produce a time delay of 1 ms, over a microprocessor, with a clock frequency of 5 MHz. Time taken for each clock cycle is 1/(5 MHz), which is equal to 0.2 micro seconds. To get a delay of 1 ms, we need to execute 100010.2 or 5000 processor clock cycles. The following program can be used to produce the delay, with the counter value correctly initialized. clock cycles MOV CX,N ; 4 N will vary depending on the ; amount of delay required DELAY: NOP ; 3 NOP ; 3 LOOP DELAY ; 17 or 5 LOOP instruction takes 17 clock cycles when the condition is true and 5 clock cycles otherwise. The condition will be true, 'N' number of times and false only once, when the control comes out of the loop.
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To calculate 'N': Total clock cycles (CT) = MOVE clock cycles + N(2*NOP clock- Awmbly Language Programming (Part-I) cycles + 17) - 12 (when CX = 0) 5000 = 4 + N(6 + 17) - 12 N = (5000 + 8)/23 = 218 = ODAh Therefore the counter, CX, should be initialized by ODAh, in order to get the required delay. Example 2 - Array Application This program adds a 5-byte number in one array, called NUMl to a Ebyte number in another array called NUM2 The sum is stored in an array called SUM, with the state of the carry flag stored in byte 6 of SUM. The first value in each array is the LSB of that number. ALGORITHM : Make count = LEN Clear the carry flag Load address of NUMl REPEAT Put byte from N U M l in accumulator Add byte rrom NUM2 to accumula~or + carry Decrement counl Increment to next address UNTIL count = O Rotate carry into LSB of accumulator Mask all but LSB of accumulator Store carry result, address pointer in correct position. ;PORTS : None uscd ;PROCEDURES : None used ;REGISTERS : Uses CS, DS, AX, CX, BX, DX DATA SEGMENT NUMl DB OFFh, 10h , Olh , l l h , 20h NUM2 DB lOh, 2Oh, 30h, 40h , OFFh SUM DB 6DUP(O) DATA ENDS
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\lic.roprocessor and A\sernhly 1.11npunp Proprnmnling LEN EUU 05h ; constant for length of thc array CODE SEGMENT ASSUME CS:CODE, DS:DATA START: MOV MOV MOV MOV MOV CLC AGAIN: MOV ADC MOV I NC LOOP RCL AND MOV FINISH: NOP NOP CODE ENDS END AX, DATA ; 1nitiali;rcdata segment DS, AX ; register BX, 00 ; Load dispIaccment of 1st number CX, 0000 ; Clear counter CL, LEN ; Set up count ; Clear carry. Ready for addition AL, NUMl[BX] ; Get a byte from NUMl AL, NUM2[BX) ; Add to byte from NUM2 SUM[BX], AL ; Store in SUM array BX AGAIN ; Continue until no more bytes AL, Olh ; Move carry into bit 0 of AL AL, Olh ; Mask all but bit 0 of AL SUM[BX],AL ; Put carry into 6th byte START Example 3 - Code Conversion Consider the following program which converts a +digit BCD numbcr into its binary equivalent, The BCD number is srored as a word in memory location cal BCD. The result is stored in location HEX.
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  • Spring '10
  • Ramon
  • Assembly Language, X86, Binary numeral system, ........., Binary-coded decimal, ax

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