Provide the solutions with explanations, in English

Here are the solutions with explanations for the given 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 pattern into sets of four bits from right to left. 1000 \ 1100 \ 0100 Step 2: Convert each 4-bit group into its hexadecimal equivalent. • 1000_2 = 8_16 • 1100_2 = C_16 (since 12_10 = C_16) • 0100_2 = 4_16 Step 3: Combine the hexadecimal digits. The hexadecimal representation is 8C4_16. 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 fourth bit from the right, as there are four bits after the binary point. 1000 \ 1100 . 0100_2 Step 2: Convert the integer part (1000 \ 1100_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 × 0.5 + 1 × 0.25 + 0 × 0.125 + 0 × 0.0625 = 0.25_10 Step 4: Add the integer and fractional parts. 140 + 0.25 = 140.25_10 140.25_10 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. Step 1: Separate the 12-bit binary pattern into an 8-bit mantissa and a 4-bit exponent. Mantissa: 1000 \ 1100_2 Exponent: 0100_2 Step 2: Convert the 8-bit mantissa (1000 \ 1100_2) from two's complement to denary. Since the most significant bit (MSB) is 1, the mantissa is negative. The value of an 8-bit two's complement number is calculated as: -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 = -116_10 For a normalized floating-point number with an 8-bit mantissa, the denary value of the mantissa is typically interpreted as a fraction by dividing by 2^n-1 where n is the number of bits (or 2^7 for 8 bits). Mantissa value M = (-116)/(2^7) = (-116)/(128) = -0.90625_10. Step 3: Convert the 4-bit exponent (0100_2) from two's complement to denary. Since the MSB is 0, the exponent is positive. 0 × 2^3 + 1 × 2^2 + 0 × 2^1 + 0 × 2^0 = 0 + 4 + 0 + 0 = 4_10 Exponent value E = 4_10. Step 4: Calculate the final denary value using the formula M × 2^E. Value = -0.90625 × 2^4 Value = -0.90625 × 16 Value = -14.5_10 -14.5_10 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): This register holds the memory address of the next* instruction to be fetched from main memory. It is automatically incremented after each instruction fetch. CIR (Current Instruction Register): This register stores the instruction that has just been fetched from memory and is currently being decoded and executed by the CPU. MBR (Memory Buffer Register) / MDR (Memory Data Register): This register temporarily holds data or an instruction that has been read from memory or is about to be 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. Here are the descriptions for the missing steps: Step 1: Fetch Address Register Transfer Notation: MAR ← [PC] Description: The address of the next instruction, currently stored 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 main memory and placed into the Memory Buffer Register (MBR). This occurs concurrently with the PC being incremented (Step 2a). Step 4: Decode Instruction Register Transfer Notation: Decode [CIR] Description: The instruction held in the Current Instruction Register (CIR) is decoded by the Control Unit to determine the operation to be performed and identify any operands required. c) State two examples when the status register might have a bit set. The status register (or flag register) contains individual bits that indicate the outcome of an arithmetic or logical operation. 1. Zero Flag (Z): This bit is set to 1 if the result of an arithmetic or logical operation is zero. For example, after a subtraction operation where the two operands were equal. 2. Carry Flag (C): This bit is set to 1 if an arithmetic operation (like addition) produces a carry-out from the most significant bit, indicating that the result exceeded the capacity of the register for unsigned numbers. It can also be set for a borrow in subtraction. That's the last one for today! Send me the next one 📸