THE FOLLOWING DATA IS STORED IN THE MEMORY LOCATIONS STARTING FROM 3060H USING MEMORY RELATED INSTRUCTIONS (NOT MANUALLY). WRITE PROGRAM TO COUNT THE NUMBERS ONLY IF D6 AND D0 BITS ARE 1, ELSE REJECT THE NUMBER. STORE THE COUNTED NUMBERS STARTING FROM MEMORY LOCATION 3070H AND STORE THE COUNT IN 'D' REGISTER. DATA (H): 80, 71, 52, E7, 78, F2, 67, 35, 62, 08, 17.
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- Q1\ 1- Write a program in assembly language for the 8085 microprocessor to send 10 bytes of data located at the memory address (3000h) using SOD at a baud rate of 1200. Information: The 8085 processor operates at a frequency of 3.072 MHz. And two high pulses must be sent before each byte (start bits) and one low pulse after each byte (end bits). You can use this flowchart, but you should notice that this flowchart deals with one byte, and you are required to deal with 10 bytes Transmit No Set up Character Bit Counter Send Start Bit Wait Bit Time Get Character in Accumulator Output Bit Using Do Wait Bit Time Rotate Next Bit in Do Decrement Bit Counter Is It Last Bit? Yes Add Parity if Necessary • Send Two Stop Bits Return (a)Program Debugging: Debug the given program to move the data from external memory location 25h to P1 and then from P1 to R5 RETMON EQU OF000H ORG 100H А, РО А, 25H R5,A START MOV MOV MOV LCALL RETMON ENDProgramming Project: For Loops Write an Assembly Language Program that produces the Times Table for a specific integer value between 1 to 12. NOTE: You choose and hard code only one value (i.e. 7) and produce the times table for that value. Also note that since you are using registers and memory locations make sure that at each step you screen shot each result for each part of the Times Table. For example 7 x 12 = 84 but in Hexadecimal the value is 54. Sample code in C #include void main() ( int i, j; printf("Enter an integer: "); scanf("%d", &i); printf("Times %d Table\n", i). for(j = 0; j<= 12; j =j+1) { printf("%d x %d = %d\n", j, 1. j'i); } printf("\n");
- 1000 unsigned numbers located in memory starting at address F000H. Draw the flow diagram and write the program as a source code for suitable for 6802 microprocessor;store how many numbers greater than 65H are in address 30H.5- Create an algorithm in assembly that will compute the area of a triangle. Here is the state of the memory when starting the algorithm: Base is stored as an 8 bit unsigned integer in a memory location pointed to by the special register X. Height of the triangle is stored as an 8 bit integer in a memory location pointed to by the special register Y. Your computed area of the triangle should be stored in memory at a location pointed to by the special register Z. If multiple rows of memory are required, then Z indicates the starting address. Requirements Clearly list the assembly commands required for this algorithm. How many rows of program memory are required for this algorithm? How many clock cycles (according to the AVR ISA) are required for this algorithm? What addressing mode is used for each assembly instruction?Given the following data segment, write the statements to copy the value in num1 into num2 and store num1*2 in num3. Do not use the LA instruction. .data . word num2: .word num3: .word numl: 16 32 64
- STATEMENT- THE FOLLOWING DATA IS STORED IN THEMEMORY LOCATIONSSTARTING FROM 3060H USING MEMORY RELATED INSTRUCTIONS(NOT MANUALLY). WRITE PROGRAM TO COUNT THE NUMBERSONLY IF D6 AND D0 BITS ARE 1, ELSE REJECT THE NUMBER. STORETHE COUNTED NUMBERS STARTING FROM MEMORY LOCATION3070H AND STORE THE COUNT IN ‘D’ REGISTER.DATA (H): 80, 71, 52, E7, 78, F2, 67, 35, 62, 08, 17. Write program in assembly language on sim8085EXPLAIN ALL STEPS: Determine the output of the following Pep/9 machine language program if the input is tab. The left column is the memory address of the first byte on the line. 0000 D1FC150003 F1001F0006 D1FC150009 F10020000C D1FC15000F F100210012 D100200015 F1FC160018 D1001F001B F1FC16001E 005. Load the register (CL) from the memory location [050OH] then subtract the content of this register from the accumulator (AL). Correct the result as a (BCD) numbers. Let [0500H] 12H & AL 3FH %3D
- Complete the following sequence of MIPS instructions: li $t0, 42 la $s0, A addiu $s1, $s0, 100 loop: pos: bltu $s0, $s1, loop by dragging four items from the below list into the correct spaces. Assume that A is the name for the starting address of an array of words defined using "A: .space 100". Your code should store the integer 42 into each word of the array that contains a negative value. Iw $s2, 0($s0) sw $s2, 0($s0) bgez 0($s2), loop| sw $t0, -4($s0) bgez $s2, pos addiu $s0, $s0, 1 Iw $t0, 0($s0) move $s2, $s0 sw $t0, 0($s0) bgez $s2, loop bgez 0($s2), pos move $t0, $so addiu $s0, $s0, 4Topic: Assembly Language Write a program called bit_check.asm that jumps to a label if either bit 4, 5, or 6 is set in the BL register. If these bits are set the program should print "Bits 4, 5, or 6 are set." followed by a new line character. If they are not set, jump to a label to print "Bits 4, 5, and 6 are not set." followed by a new line character. There should be 4 tests, test with each bit position set and with none of them set. Use the print_string procedure to print your output. Remember, you can jump to the exit label from anywhere in your program.PART 2: REPETITION CONTROL STRUCTURE (WHILE, DO-WHILE) Instruction: A mathematician named Ulam proposed generating a sequence of numbers from any positive integer N greater than 1 using the following procedure: If N is 1, stop. If N is even, replace it with N/2. If N is odd, replace it with 3 * N + 1. Continue with this process until N reaches 1. Here are some examples of the Ulam sequence for the first few integers. 2, 1 3, 10, 5, 16, 8, 4, 2, 1 4, 2, 1 5, 16, 8, 4, 2, 1 6, 3, 10, 5, 16, 8, 4, 2, 1 Create a java program using while/do-while that accepts as input an integer value N (assume N> 1) and prints out the Ulam sequence that begins with the input value N. Sample Input/Output: Depicted below are sample outputs when the program is executed (the items in bold characters are input from the user, while the items in bold italic are calculated and printed by the program): Input N: 14 Ulam Sequence: 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 Input N: 5 Ulam…