Explain PROM, EEPROM, RAM, and address mapping in computer organization.

1.i.a) Explanation and practical application of PROM, EEPROM, RAM:

• PROM (Programmable Read-Only Memory): A type of ROM that can be programmed once by the user or manufacturer. Once programmed, its contents cannot be erased or modified.
  • Application: Firmware in older embedded systems, such as early video game cartridges or simple microcontrollers.
• EEPROM (Electrically Erasable Programmable Read-Only Memory): A type of ROM that can be erased and reprogrammed electrically, byte by byte. It retains data without power.
  • Application: Storing configuration data in devices like BIOS settings in computers, smart cards, or calibration data in sensors.
• RAM (Random Access Memory): Volatile memory used for temporary storage of data and program instructions that the CPU is currently using. Data is lost when power is removed.
  • Application: Main memory in computers, where the operating system, running applications, and active data are stored for quick access by the CPU.

1.i.b) Address mapping in computer organization:
Address mapping is the process of translating logical addresses generated by the CPU into physical addresses in main memory. This technique is crucial for implementing virtual memory, allowing programs to operate within a larger address space than the physical memory available, and for memory protection.

1.i.c) What is cache memory?
Cache memory is a small, high-speed memory located between the CPU and main memory. It stores frequently accessed data and instructions, enabling faster retrieval by the CPU and significantly reducing the average memory access time.

1.ii) Logic circuit and truth table for membership requirements:

The membership requirements are:

1. Over the age of 18 (A=1) AND has been recommended (B=1). This is represented by (A AND B).
2. Either working full-time (C=1) OR retired (D=1), but not both. This is represented by (C XOR D).

Both conditions must be met for membership approval (X=1).
Therefore, the overall logic expression is: $X = (A \text{ AND } B) \text{ AND } (C \text{ XOR } D)$.

a) Logic circuit to represent the membership requirements:

      A ---|      |
           | AND  |---
      B ---|      |   |
                       | AND --- X
      C ---|      |   |
           | XOR  |---
      D ---|      |

(Note: This is a textual representation. In a diagram, the AND and XOR gates would be drawn with their standard symbols.)

b) Truth table for the logic expression: $X = (A \text{ XOR } B) \text{ AND NOT } C$

2.i.a) What is 2's complement? Why do modern computers use 2's complement?

• 2's complement: A mathematical operation on binary numbers that allows for the representation of signed integers (positive and negative numbers) in a way that simplifies arithmetic operations. For a positive number, its 2's complement is its standard binary representation. For a negative number, it is found by inverting all bits of its positive counterpart (1's complement) and then adding 1.
• Why modern computers use it: Modern computers use 2's complement because it simplifies the design of arithmetic logic units (ALUs). It allows subtraction to be performed using addition (by adding the 2's complement of the subtrahend), eliminating the need for separate subtraction hardware. It also provides a unique representation for zero and handles both positive and negative numbers uniformly in arithmetic operations.

2.i.b) Represent the decimal number 127 and -128 in 8-bit 2's complement.

• Decimal 127 in 8-bit 2's complement:

  1. 127 is a positive number.
  2. Convert 127 to binary: $127 = 64 + 32 + 16 + 8 + 4 + 2 + 1 = 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^1 + 2^0$.
  3. In 8 bits, this is: $\boxed{\text{01111111}}$. (The most significant bit is 0, indicating a positive number).

• Decimal -128 in 8-bit 2's complement:

  1. The range for 8-bit 2's complement is -128 to 127.
  2. To represent -128, we can use the following method:
     • Start with the binary representation of its absolute value, 128: $10000000_2$.
     • Invert all bits (1's complement): $01111111_2$.
     • Add 1 to the result: $01111111_2 + 1_2 = 10000000_2$.
  3. Therefore, -128 in 8-bit 2's complement is: $\boxed{\text{10000000}}$. (The most significant bit is 1, indicating a negative number).