Provide answers to all these questions with explanations.(In English)

here are the solutions to your questions: 2. (i) The binary pattern 1000 1100 0100 could be interpreted in different ways. a) Determine its hexadecimal representation. Step 1: Group the binary digits into sets of four from right to left. 1000 1100 0100 Step 2: Convert each group of four binary digits to its hexadecimal equivalent. • 1000_2 = 8_16 • 1100_2 = C_16 • 0100_2 = 4_16 The hexadecimal representation is 8C4_16. b) Determine its value in denary if it represents an unsigned fixed-point number with four bits after the binary point. Step 1: Place the binary point after the first eight bits, as there are four bits after the binary point. 10001100.0100_2 Step 2: Convert the integer part (10001100_2) to denary. 1 · 2^7 + 0 · 2^6 + 0 · 2^5 + 0 · 2^4 + 1 · 2^3 + 1 · 2^2 + 0 · 2^1 + 0 · 2^0 = 128 + 0 + 0 + 0 + 8 + 4 + 0 + 0 = 140_10 Step 3: Convert the fractional part (0100_2) to denary. 0 · 2^-1 + 1 · 2^-2 + 0 · 2^-3 + 0 · 2^-4 = 0 + 0.25 + 0 + 0 = 0.25_10 Step 4: Add the integer and fractional parts. 140 + 0.25 = 140.25 The value in denary is 140.25. c) Determine its value in denary if it represents a normalized floating-point number with an 8-bit mantissa and a 4-bit exponent, both in two's complement. The binary pattern is 100011000100_2. • Mantissa (8 bits): 10001100_2 • Exponent (4 bits): 0100_2 Step 1: Determine the value of the 8-bit mantissa in two's complement, interpreted as a fraction. The most significant bit (MSB) is the sign bit. For a two's complement fraction, the value is calculated as -(b_7 · 2^0) + b_6 · 2^-1 + b_5 · 2^-2 + b_4 · 2^-3 + b_3 · 2^-4 + b_2 · 2^-5 + b_1 · 2^-6 + b_0 · 2^-7. For 10001100_2: = -(1 · 2^0) + (0 · 2^-1) + (0 · 2^-2) + (0 · 2^-3) + (1 · 2^-4) + (1 · 2^-5) + (0 · 2^-6) + (0 · 2^-7) = -1 + 0 + 0 + 0 + 0.0625 + 0.03125 + 0 + 0 = -1 + 0.09375 = -0.90625 So, the mantissa value is -0.90625. Step 2: Determine the value of the 4-bit exponent in two's complement. For 0100_2: The MSB is 0, so it's a positive number. 0 · 2^3 + 1 · 2^2 + 0 · 2^1 + 0 · 2^0 = 0 + 4 + 0 + 0 = 4 So, the exponent value is 4. Step 3: Calculate the final denary value using the formula: Value = Mantissa × 2^Exponent. Value = -0.90625 × 2^4 Value = -0.90625 × 16 Value = -14.5 The value in denary is -14.5. 2. (ii) The figure below shows various steps in the fetch-execute cycle. PC, CIR and MBR are processor registers. a) State the function of each of the registers PC, CIR and MBR. • PC (Program Counter): Stores the memory address of the next instruction to be fetched from memory. • CIR (Current Instruction Register): Holds the instruction that is currently being decoded and executed by the CPU. • MBR (Memory Buffer Register): Temporarily stores data or instructions that are being read from or written to memory. b) Describe the missing steps 1, 2b and 4 using either register transfer notation or a written description. Steps 2a and 2b occur at the same time. Step 1: Fetch Address • Register transfer notation: MAR <- [PC] • Description: The address of the next instruction, held in the Program Counter (PC), is transferred to the Memory Address Register (MAR). Step 2b: Fetch Instruction • Register transfer notation: MBR <- Memory[MAR] • Description: The instruction located at the memory address specified by the MAR is fetched from memory and placed into the Memory Buffer Register (MBR). This occurs concurrently with Step 2a (incrementing the PC). Step 4: Fetch Operand (if required) • Register transfer notation: MAR <- [Address part of CIR], then MBR <- Memory[MAR] • Description: If the instruction in the CIR requires an operand from memory, the address of that operand is extracted from the instruction and placed into the MAR. Then, the operand is fetched from memory and placed into the MBR. c) State two examples when the status register might have a bit set. Two examples when the status register (or flag register) might have a bit set are: • Zero Flag: This bit is set to '1' if the result of an arithmetic or logical operation is zero. For example, after a subtraction where the two operands were equal. • Carry Flag: This bit is set to '1' if an arithmetic operation (like addition) produces a carry-out from the most significant bit, or if a borrow is required during a subtraction. You're doing great! What's next?