Explain the functions of each of the following Intel 8085 microprocessor signals: (i) RESET-IN; (ii) IO/m; (iii) RD.

Here are the solutions to the questions from the image: SECTION A: MICROPROCESSOR SYSTEMS (a) Explain the functions of each of the following Intel 8085 microprocessor signals: (i) RESET-IN: This is an active-low input signal. When the RESET-IN pin is held low, the microprocessor resets its Program Counter (PC) to 0000H, clears all internal registers and flags, and restarts program execution from memory location 0000H. (ii) IO/M: This is a status output signal. When this pin is high, it indicates that the CPU is performing an Input/Output (I/O) operation. When it is low, it indicates that the CPU is performing a memory operation. This signal helps external devices distinguish between memory and I/O access. (iii) RD: This is an active-low output signal. When the RD pin is asserted (pulled low), it indicates that the CPU is performing a read operation, meaning it is fetching data from either memory or an I/O device. (b) Complete the table by stating the addressing mode and size in bytes for each instruction. | Instruction | Addressing mode | Size in bytes | | :---------- | :------------------ | :------------ | | LXI H, 20FFH | Immediate | 3 | | ADD M | Register Indirect | 1 | | STAX B | Register Indirect | 1 | | DAA | Implied | 1 | (c) Write an assembly language program to perform each of the following operation: (i) logically AND (10101101)_2 with (01110010)_2; First, convert the binary numbers to hexadecimal: (10101101)_2 = AD_16 (01110010)_2 = 72_16 `assembly MVI A, ADH ; Load the first operand (ADh) into the Accumulator MVI B, 72H ; Load the second operand (72h) into Register B ANA B ; Perform logical AND between Accumulator and Register B. Result is stored in Accumulator. HLT ; Halt the program ` (ii) multiply F2C_16 by 8. Multiplication by 8 is equivalent to three left shifts. In the 8085, a 16-bit left shift can be achieved by adding the 16-bit number to itself using the DAD H instruction. `assembly LXI H, 0F2CH ; Load the 16-bit number F2C (0F2CH) into the H-L register pair DAD H ; H-L = H-L + H-L (multiplies by 2) DAD H ; H-L = H-L + H-L (multiplies by 4) DAD H ; H-L = H-L + H-L (multiplies by 8) HLT ; Halt the program. The result (F2C * 8) is in the H-L pair. ` (d) State three Intel 8085 microprocessor software interrupts. 1. RST 0 2. RST 1 3. RST 2 (Other valid answers include RST 3, RST 4, RST 5, RST 6, RST 7) (e) Describe each of the interrupt instructions: (i) SIM (Set Interrupt Mask): This instruction is used to set the interrupt mask for the RST 7.5, RST 6.5, and RST 5.5 hardware interrupts. It also controls the serial output data (SOD) pin. The content of the Accumulator determines the mask settings and the state of the SOD pin. (ii) RIM (Read Interrupt Mask): This instruction reads the current status of the interrupt masks and pending interrupts into the Accumulator. It also reads the serial input data (SID) pin. The Accumulator will contain information about the interrupt masks, pending interrupts, and the state of the SID pin. (f) With the aid of a flowchart, describe the polling technique of interrupt service. The polling technique for interrupt service involves the CPU periodically checking the status of various I/O devices to determine if any require service. Instead of being interrupted by a device, the CPU actively queries each device in a predetermined sequence. If a device's status flag indicates it needs attention, the CPU executes the corresponding service routine. If not, it moves on to check the next device. This method is simple but can be inefficient if many devices are polled frequently without needing service. [node distance=1.5cm, auto, >=stealth'] startstop = [rectangle, rounded corners, minimum width=3cm, minimum height=1cm, text centered, draw=black, fill=blue!20] process = [rectangle, minimum width=3cm, minimum height=1cm, text centered, draw=black, fill=orange!20] decision = [diamond, minimum width=3cm, minimum height=1cm, text centered, draw=black, fill=green!20] arrow = [thick,->,>=stealth] (start) [startstop] Start; (main_program) [process, below of=start] Execute Main Program; (check_dev1) [decision, below of=main_program, yshift=-0.5cm] Device 1 needs service?; (service_dev1) [process, right of=check_dev1, xshift=2.5cm] Service Device 1; (check_dev2) [decision, below of=check_dev1] Device 2 needs service?; (service_dev2) [process, right of=check_dev2, xshift=2.5cm] Service Device 2; (check_devN) [decision, below of=check_dev2, yshift=-0.5cm] Device N needs service?; (service_devN) [process, right of=check_devN, xshift=2.5cm] Service Device N; (loop_back) [process, below of=check_devN, yshift=-0.5cm] Continue Main Program / Loop; [arrow] (start) -- (main_program); [arrow] (main_program) -- (check_dev1); [arrow] (check_dev1) -- node Yes (service_dev1); [arrow] (service_dev1) |- (loop_back); [arrow] (check_dev1) -- node No (check_dev2); [arrow] (check_dev2) -- node Yes (service_dev2); [arrow] (service_dev2) |- (loop_back); [arrow] (check_dev2) -- node No (check_devN); [arrow] (check_devN) -- node Yes (service_devN); [arrow] (service_devN) |- (loop_back); [arrow] (check_devN) -- node No (loop_back); [arrow] (loop_back) -- (main_program); % Loop back to main program or re-check devices