The 8-bit registers R1, R2, R3, and R4 initially have the following values: R1 - 1111 0010, R2- 1 1 1 1 1 1 1 1 , R3- 1011 1001 , R4 1 1 101010 Determine the 8-bit values in each register after the execution of the following sequence of microoperations. R1 <-R1 + R2 R3<- R3 ^ R4, R2 <- R2 + 1 R1 <- R1 – R3
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3 The 8-bit registers R1, R2, R3, and R4 initially have the following values:
R1 - 1111 0010, R2- 1 1 1 1 1 1 1 1 , R3- 1011 1001 , R4 1 1 101010
Determine the 8-bit values in each register after the execution of the following sequence of
microoperations.
R1 <-R1 + R2
R3<- R3 ^ R4, R2 <- R2 + 1
R1 <- R1 – R3
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- Q2- Write a program in assembly language for the 8085 microprocessor to receive one byte of data via the SID and store it at the memory address (3000H to 3009H) using a baud rate of 1200. Information: The 8085 processor operates at a frequency of 3.072 MHz . When receive the required bytes, you must adhere to the following: The bits of two high bits will be received at the beginning of the reception(start bits 1 1 ), after that the data bits will be received, after that the low bit of the stop bit will be received (stop bit 0 ). The following flowchart will help you. The solution must be integrated and include the calculation of the baudrate delay timeQ1- Write a program in assembly language for the 8085 microprocessor to receive 10 bytes of data via the SID and store it at the memory address (3000H to 3009H) using a baud rate of 1200. Information: The 8085 processor operates at a frequency of 3.072 MHz. When you receive each byte of the required bytes, you must adhere to the following: The bits of two high bits will be received at the beginning of the reception (start bits), after that the data bits will be received, after that the low bit of the stop bit will be received (stop bit). The following flowchart will help you, but you should notice that this flowchart deals with one byte, and you are required to deal with 10 bytes The solution must be integrated and include the calculation of the baudrate delay time Of+CD!HID+[00 Yes SIDATA Read SID Start Bit? Wait for Half-Bit Time Set up Bit Counter Wait Bit Time Read SID Save Bit Decrement Bit Counter All Bits Received? Add Bit to Previous Bits Go Back to Get Next Bit Return IMUNIThe 8-bit register AR, BR, CR, and DR initially have the following values: [5]AR = 11010010; BR = 11111111; CR = 10101001; DR = 10101010Determine the 8-bit values in each register after the execution of the following sequence ofmicrooperations.AR AR + BR Add BR + ARCR CR AND DR, BR BR + 1 AND DR to CR, Increment BRAR AR - CR Subtract CR from AR
- Design a binary multiplier that multiplies two 8-bit binary number by following design rules that shown in class. The Q and B are the two separate 8-bit binary inputs, C is the 3-bit sequence counter and R is the 16-bit result. (Note: Explain the registers that you will use to establish given process.)(Hint: Morris Mano, Computer System Architecture, Chapter 10.6.) QxB B Clock Start -Busy/ReadyAnswer by True or False and correct the False: 1- The instruction : MOV [Dx+SI], Ax is allowed T 2- The instruction : MOV ES:[SI], Ax is not allowedY 3- Translate (XLAT) converts the data in AL into a number stored at the memory location addressed by Bx plus AL. 4- DWORD PTR is used to address a 16-bit data of a number stored in memory F 5- REP is an instruction used to repeat any instruction as much as the value in Cx register. FThe 8-bit register AR, BR, CR, and DR initially have the following values:AR = 11010010; BR = 11111111; CR = 10101001; DR = 10101010Determine the 8-bit values in each register after the execution of the following sequence ofmicrooperations.AR <- AR + BR Add BR + ARCR <- CR AND DR, BR <- BR + 1 AND DR to CR, Increment BRAR <- AR - CR Subtract CR from AR
- Q1:/ Show the contents in hexadecimal of registers PC, AR, DR, AC, IR and SC of the basic computer when an instruction at address 021 in the basic computer has I = 1, an operation code of the ADD instruction, and an address part equal to 051. The memory word at address 051 contains 0083. The memory word at address 083 contains B8F2. The memory word at address 038 contains A837 and the content of AC is A937. Give the answer in a table with six columns, one for each register and a row for each timing signal. (All numbers are in hexadecimal) uipors - eaPlease complete the binary encoding for the following BL instruction at address Ox2000 using the given information. Please write the answer in a group of four bits as follows: 0010 0010 0010 1100. Address Instructions 0x2000 bl sub 0x2004 add r3, r2, r1 0x2008 sub r6, r5, r3 sub: stmfd r13!, {r0-r12, 0X200C r14] 0x2010 add r7, r2, r5 Ox2014 mov r3, r2Given the following instruction format specification 19 15 14 13 11 10 8 7 M Dest Op1 Opcode Op2 11101100110100110001 0 what binary value will determine the operation performed by the following machine code instruction? Please provide answer in binary, include only the bits which determine the operation performed.
- a. Find the address accessed by each of the following instructions. If DS = 0100H, BX= 0120H, DATA = 0140H, and SI = 0050H and real mode operation:1. MOV DATA[SI], ECX2. MOV BL, [ BX+SI]b. Descriptor contains a base address of 00260000H, a limit of 00110H, and G = 1,determine starting and ending locations are addressed by the descriptor for aCore2.Part 3: Putting it all together Consider the situation where the EAX and EBX registers are storing the values listed below. EAX: 01100001 10101110 00101111 11111111 EBX: 11101010 00000000 00000010 10101101 We want to add the values of EAX and EBX together. When we perform this addition, the result is stored EBX, overwriting the value that was already in there. What is the result of the addition stored in EBX? Give your result in binary. What value is stored in BL? Give your result in hexadecimal. What flags are set after the operation is complete?The 8-bit register AR, BR, CR, and DR initially have the following values: [5]AR = 11010110; BR = 11100111; CR = 10110001; DR = 10111010Determine the 8-bit values in each register after the execution of the following sequence of microoperations.AR AR + BR Add BR + ARCR CR AND DR, BR BR + 1 AND DR to CR, Increment BRAR AR - CR Subtract CR from AR