MCS-012 Solved Free Assignment 2024-25 Sem 2

Question 1:

(a) Please refer to Figure 4 of Unit 1 of Block 1 on page 11 of the Instruction execution example. Assuming a similar machine is to be used for the execution of the following three consecutive instructions: LOAD A ; Load the content of Memory location A into the Accumulator Register. ADD B ; Add the content of memory location B to Accumulator Register. STOR C ; Stores the content of the Accumulator register to memory location C. However, this machine is different from the example in Figure 4 in the following ways:

 Each memory word of this new machine is of 32 bits length.

 Each instruction is of length 32 bits with 12 bits for operation code (opcode) and 20 bits for specifying one direct operand. The size of operand is 32 bits.

 The Main Memory of the machine is of size 220 words.

 The three consecutive instructions are placed starting from memory location (11FFE)h ; operand A is at location (2FFFF)h and contains a value (111AB4C1)h, Operand B is at location (30000)h and contains a value (AAA1A1FE)h and operand C is at location (30001)h and contains a value (00000000)h.

 The AC, IR, MAR and MBR registers are of size 32 bits, whereas PC register is of size 20 bits. The initial content of the PC register is (11FFE)h.

(i) Draw a diagram showing the Initial State of the machine with the addresses and content of memory locations in hexadecimal. Show only those address locations of the memory that store the instruction and data. Also, show the content of all the stated registers.

(ii) Draw three more diagrams, each showing the state of the machine after execution of every instruction viz. LOAD, ADD and STOR. Show the changes in the values of Registers and memory locations, if any, due to the execution of the instruction. Show all the addresses and values in hexadecimal notations.

Ans:- Let's break down the problem and tackle each part systematically. Here's how the machine would look and how it progresses through the instruction execution.

Given Information:

1. **Memory Size**: 2^20 words (1 MB). Each word is 32 bits.

2. **Instruction Format**: 32 bits total.

- **Opcode**: 12 bits.

- **Operand Address**: 20 bits.

3. **Memory Locations**:

- **Instruction Start**: (11FFE)h

- **Operand A**: (2FFFF)h, value = (111AB4C1)h.

- **Operand B**: (30000)h, value = (AAA1A1FE)h.

- **Operand C**: (30001)h, value = (00000000)h.

4. **Registers**:

- **PC**: 20-bit (Initial content = (11FFE)h).

- **IR, AC, MAR, MBR**: 32-bit.

---

(i) Initial State of the Machine:

**Registers and Memory**:

| Register | Content | Description |

|----------|-----------------|------------------------------------------------|

| PC | 11FFE | Program Counter (Points to next instruction) |

| AC | 00000000 | Accumulator Register |

| IR | 00000000 | Instruction Register |

| MAR | 00000000 | Memory Address Register |

| MBR | 00000000 | Memory Buffer Register |

**Memory** (Instructions and Operands):

| Address | Content | Description |

|----------|-----------------|------------------------------------------------|

| 11FFE | Opcode: LOAD A | Instruction 1: LOAD A (OpCode + Address) |

| 11FFF | Opcode: ADD B | Instruction 2: ADD B (OpCode + Address) |

| 12000 | Opcode: STOR C | Instruction 3: STOR C (OpCode + Address) |

| 2FFFF | 111AB4C1 | Operand A |

| 30000 | AAA1A1FE | Operand B |

| 30001 | 00000000 | Operand C (initially 0) |

The exact instruction encodings depend on the opcode bit patterns, but we are only focusing on the hexadecimal representation of the addresses and values here.

---

(ii) Execution of Instructions:

We will now show the state of the machine after each instruction:

---

1. **After Executing `LOAD A`**:

- **Instruction**: LOAD A (11FFEh)

- **Action**: Load the content of memory location A (2FFFFh) into the Accumulator (AC).

**Changes**:

| Register | Content | Description |

|----------|-----------|--------------------------------------------------|

| PC | 11FFF | PC now points to the next instruction |

| AC | 111AB4C1 | Value from memory location (2FFFF)h loaded |

| IR | LOAD A | The instruction (LOAD A) is now in IR |

| MAR | 2FFFF | MAR holds the address of operand A |

| MBR | 111AB4C1 | MBR holds the value of operand A |

No changes in memory content.

---

2. **After Executing `ADD B`**:

- **Instruction**: ADD B (11FFFh)

- **Action**: Add the content of memory location B (30000h) to the Accumulator (AC).

**Changes**:

| Register | Content | Description |

|----------|------------|--------------------------------------------------|

| PC | 12000 | PC now points to the next instruction |

| AC | BBBC567F | (111AB4C1)h + (AAA1A1FE)h = (BBB C567F)h |

| IR | ADD B | The instruction (ADD B) is now in IR |

| MAR | 30000 | MAR holds the address of operand B |

| MBR | AAA1A1FE | MBR holds the value of operand B |

No changes in memory content.

---

3. **After Executing `STOR C`**:

- **Instruction**: STOR C (12000h)

- **Action**: Store the content of the Accumulator (AC) to memory location C (30001h).

**Changes**:

| Register | Content | Description |

|----------|------------|--------------------------------------------------|

| PC | 12001 | PC now points to the next instruction (if any) |

| AC | BBBC567F | No change |

| IR | STOR C | The instruction (STOR C) is now in IR |

| MAR | 30001 | MAR holds the address of operand C |

| MBR | BBBC567F | MBR holds the value to be stored in memory (AC) |

**Memory Changes**:

| Address | Content | Description |

|----------|-------------|--------------------------------------------------|

| 30001 | BBBC567F | Operand C now holds the result of ADD operation |

---

Summary:

- **Initial State**: Registers are empty except PC = 11FFEh, and the memory has three instructions starting at 11FFEh and operands at 2FFFFh, 30000h, and 30001h.

- **After LOAD A**: AC = 111AB4C1 (content from 2FFFFh).

- **After ADD B**: AC = BBBC567F (111AB4C1 + AAA1A1FE).

- **After STOR C**: Memory location 30001h contains BBBC567F (value of AC).

Let me know if you'd like further help with diagrams or explanations!

(b) Perform the following conversion of numbers:

i) Decimal (345789531)10 to binary and hexadecimal.

ii) Hexadecimal (ABC023DEF)h into Octal.

iii) String “MCS-12 Price in $” into UTF 8.

iv) Octal (12076543)O into Decimal.

Ans:- Let’s tackle each conversion step by step:

(i) **Decimal (345789531)₁₀ to Binary and Hexadecimal:**

- **Decimal to Binary**:

To convert decimal 345789531 to binary, repeatedly divide by 2 and note the remainders.

**Conversion**:

```plaintext

345789531 ÷ 2 = 172894765 remainder 1

172894765 ÷ 2 = 86447382 remainder 1

86447382 ÷ 2 = 43223691 remainder 0

43223691 ÷ 2 = 21611845 remainder 1

21611845 ÷ 2 = 10805922 remainder 1

10805922 ÷ 2 = 5402961 remainder 0

5402961 ÷ 2 = 2701480 remainder 1

2701480 ÷ 2 = 1350740 remainder 0

1350740 ÷ 2 = 675370 remainder 0

675370 ÷ 2 = 337685 remainder 0

337685 ÷ 2 = 168842 remainder 1

168842 ÷ 2 = 84421 remainder 0

84421 ÷ 2 = 42210 remainder 1

42210 ÷ 2 = 21105 remainder 0

21105 ÷ 2 = 10552 remainder 1

10552 ÷ 2 = 5276 remainder 0

5276 ÷ 2 = 2638 remainder 0

2638 ÷ 2 = 1319 remainder 0

1319 ÷ 2 = 659 remainder 1

659 ÷ 2 = 329 remainder 1

329 ÷ 2 = 164 remainder 1

164 ÷ 2 = 82 remainder 0

82 ÷ 2 = 41 remainder 0

41 ÷ 2 = 20 remainder 1

20 ÷ 2 = 10 remainder 0

10 ÷ 2 = 5 remainder 0

5 ÷ 2 = 2 remainder 1

2 ÷ 2 = 1 remainder 0

1 ÷ 2 = 0 remainder 1

```

So, **binary representation**:

```plaintext

(345789531)₁₀ = 10100101000111000101110110011₂

```

- **Decimal to Hexadecimal**:

To convert 345789531 to hexadecimal, divide by 16 and note the remainders.

**Conversion**:

```plaintext

345789531 ÷ 16 = 21611845 remainder 11 (B)

21611845 ÷ 16 = 1350740 remainder 5

1350740 ÷ 16 = 84321 remainder 4

84321 ÷ 16 = 5270 remainder 1

5270 ÷ 16 = 329 remainder 6

329 ÷ 16 = 20 remainder 9

20 ÷ 16 = 1 remainder 4

1 ÷ 16 = 0 remainder 1

```

So, **hexadecimal representation**:

```plaintext

(345789531)₁₀ = 149654B₁₆

```

---

(ii) **Hexadecimal (ABC023DEF)₁₆ to Octal**:

We first convert the hexadecimal to binary, and then from binary to octal.

- **Hexadecimal to Binary**:

| Hex | Binary |

|------|------------------|

| A | 1010 |

| B | 1011 |

| C | 1100 |

| 0 | 0000 |

| 2 | 0010 |

| 3 | 0011 |

| D | 1101 |

| E | 1110 |

| F | 1111 |

So, the binary form of `(ABC023DEF)₁₆` is:

```plaintext

1010 1011 1100 0000 0010 0011 1101 1110 1111₂

```

- **Binary to Octal**:

Group the binary digits into groups of three from right to left and convert each group into octal:

```plaintext

101 010 111 100 000 000 100 011 110 111 101 111₂

```

Convert to octal:

```plaintext

(527400437357)₈

```

---

(iii) **String “MCS-12 Price in $” to UTF-8**:

Each character in the string is encoded into its corresponding UTF-8 hexadecimal value:

- **M**: U+004D → 0x4D

- **C**: U+0043 → 0x43

- **S**: U+0053 → 0x53

- **-**: U+002D → 0x2D

- **1**: U+0031 → 0x31

- **2**: U+0032 → 0x32

- **(Space)**: U+0020 → 0x20

- **P**: U+0050 → 0x50

- **r**: U+0072 → 0x72

- **i**: U+0069 → 0x69

- **c**: U+0063 → 0x63

- **e**: U+0065 → 0x65

- **(Space)**: U+0020 → 0x20

- **i**: U+0069 → 0x69

- **n**: U+006E → 0x6E

- **(Space)**: U+0020 → 0x20

- **$**: U+0024 → 0x24

So, the UTF-8 encoding of the string “MCS-12 Price in $” is:

```plaintext

0x4D 0x43 0x53 0x2D 0x31 0x32 0x20 0x50 0x72 0x69 0x63 0x65 0x20 0x69 0x6E 0x20 0x24

```

---

(iv) **Octal (12076543)₈ to Decimal**:

To convert octal to decimal, multiply each digit by 8 raised to the power of its position (starting from 0 on the right):

Conversion**:

```plaintext

(12076543)₈ = 1×8⁷ + 2×8⁶ + 0×8⁵ + 7×8⁴ + 6×8³ + 5×8² + 4×8¹ + 3×8⁰

```

Calculate the powers of 8:

```plaintext

1×2097152 + 2×262144 + 0×32768 + 7×4096 + 6×512 + 5×64 + 4×8 + 3×1

= 2097152 + 524288 + 0 + 28672 + 3072 + 320 + 32 + 3

= 2626539

```

So, **decimal representation**:

```plaintext

(12076543)₈ = (2626539)₁₀

```

---

Let me know if you need any further clarifications!

(c) Simplify the following function using K-map: F(A, B, C, D) = Σ (0, 1, 2, 4, 8, 9, 10, 13) Draw the circuit for the resultant function using NAND gates.

Ans:- Let’s simplify the given function using a Karnaugh map (K-map) and then draw the circuit using NAND gates.

**Step 1: Function F(A, B, C, D)**

We are given the function:

\[ F(A, B, C, D) = Σ (0, 1, 2, 4, 8, 9, 10, 13) \]

These are the minterms for which the function F is true.

**Step 2: K-map Setup**

The K-map for a 4-variable function has 16 cells. Let's label the rows and columns of the K-map:

\begin{array}{c|c c c c}

AB \backslash CD & 00 & 01 & 11 & 10 \\

\hline

00 & m_0 & m_1 & m_3 & m_2 \\

01 & m_4 & m_5 & m_7 & m_6 \\

11 & m_{12} & m_{13} & m_{15} & m_{14} \\

10 & m_8 & m_9 & m_{11} & m_{10} \\

\end{array}

We now place 1’s in the K-map cells that correspond to the minterms present in the function:

- $ m_0, m_1, m_2, m_4, m_8, m_9, m_{10}, m_{13} $

**Step 3: Filling the K-map**

\begin{array}{c|c c c c}

AB \backslash CD & 00 & 01 & 11 & 10 \\

\hline

00 & 1 & 1 & 0 & 1 \\

01 & 1 & 0 & 0 & 0 \\

11 & 0 & 1 & 0 & 0 \\

10 & 1 & 1 & 0 & 1 \\

\end{array}

**Step 4: Grouping the 1's**

We now group adjacent 1’s into the largest possible power of 2 groups to simplify the expression.

- **Group 1 (top-left 2×2 block)**: Covers minterms $ m_0, m_1, m_4, m_8 $.

- Common variables: $ A' $ and $ D' $

- Simplified term: $ A'D' $

- **Group 2 (bottom-left 2×2 block)**: Covers minterms $ m_8, m_9, m_{10}, m_{13} $.

- Common variables: $ B $ and $ C' $

- Simplified term: $ BC' $

- **Group 3 (minterms $ m_1, m_9 $)**: Horizontal pair of 1's in columns 01.

- Common variables: $ C'D $

- Simplified term: $ C'D $

**Step 5: Final Simplified Expression**

The final simplified Boolean function $ F(A, B, C, D) $ is the sum of the terms derived from the groups:

F(A, B, C, D) = A'D' + BC' + C'D

---

**Step 6: Circuit Using NAND Gates**

We can now draw the circuit for the simplified function using **NAND gates**. To implement the function using **only NAND gates**, we first need to express the function in a form suitable for NAND gates.

Expression:

F(A, B, C, D) = A'D' + BC' + C'D

**NAND Implementation** steps:

1. **Inverting inputs** (to get $ A' $, $ B' $, $ C' $, and $ D' $) using NAND gates.

2. **ANDing terms**:

- $ A'D' $: NAND gate to combine $ A' $ and $ D' $.

- $ BC' $: NAND gate to combine $ B $ and $ C' $.

- $ C'D $: NAND gate to combine $ C' $ and $ D $.

3. **ORing the terms** using a NAND-NAND combination (De Morgan’s Law).

The logic circuit will follow this structure:

1. **NAND gates for inverting**:

- $ A' = A \text{ NAND } A $

- $ D' = D \text{ NAND } D $

- $ B' = B \text{ NAND } B $

- $ C' = C \text{ NAND } C $

2. **NAND gates for AND**:

- $ A'D' = (A' \text{ NAND } D') $

- $ BC' = (B \text{ NAND } C') $

- $ C'D = (C' \text{ NAND } D) $

3. **NAND gates for OR** (Final Output):

The terms $ A'D' $, $ BC' $, and $ C'D $ can be ORed by using two NAND gates:

- $ (A'D') \text{ NAND } (BC') \text{ NAND } (C'D) $

This creates a circuit that implements the simplified function entirely using NAND gates.

---

Let me know if you'd like further clarification or an illustration of the circuit!

(d) Consider the Adder-Subtractor circuit as shown in Figure 3.15 page 76 of Block 1. What would be the values of various inputs and outputs; viz. Cin input to each full adder, A0, B0, A1, B1, A2, B2, A3, B3, S0, S1, S2, S3, Carry out bit, and overflow condition; if this circuit performs subtraction (A-B), when the value of A is 1010 and B is 1011.

Ans:- To answer this, let's break down the working of the Adder-Subtractor circuit. In such a circuit, subtraction is performed by adding the 2’s complement of the number $ B $ to $ A $. Here are the steps to determine the various inputs and outputs when performing $ A - B $.

**Step 1: Values of A and B**

- $ A = 1010 $ (in binary, which is $ 10 $ in decimal)

- $ B = 1011 $ (in binary, which is $ 11 $ in decimal)

We are performing $ A - B $, i.e., $ 1010 - 1011 $.

**Step 2: 2's Complement of $ B $**

To subtract $ B $, we first take the 2's complement of $ B $.

1. **Invert $ B $** (1's complement):

B = 1011 \quad \text{(invert)} \quad B' = 0100

2. **Add 1** to the 1's complement to get the 2's complement:

B' = 0100 + 1 = 0101

Thus, the 2’s complement of $ B $ is $ 0101 $. Now, we add this to $ A $.

**Step 3: Inputs to the Adder-Subtractor Circuit**

- **A** (inputs to the full adders):

A_3 = 1, A_2 = 0, A_1 = 1, A_0 = 0

- **B** (inputs to the full adders after applying XOR with the subtraction bit (1 for subtraction)):

- B's 2's complement is used in subtraction.

B_3 = 0, B_2 = 1, B_1 = 0, B_0 = 1 \quad \text{(this is the 2's complement of $ B $)}

- **Control bit (M)** for subtraction is set to **1**.

**Step 4: Full Adder Inputs and Outputs**

The circuit consists of four full adders, and we are performing addition between $ A $ and the 2's complement of $ B $. The following table summarizes the inputs and outputs for each full adder:

| Full Adder | $ A_i $ | $ B_i $ | $ Cin $ | Sum ($ S_i $) | Carry Out |

|------------|-----------|-----------|-----------|-----------------|------------|

| FA0 | 0 | 1 | 1 (from Cin for subtraction) | 0 | 1 |

| FA1 | 1 | 0 | 1 | 0 | 1 |

| FA2 | 0 | 1 | 1 | 0 | 1 |

| FA3 | 1 | 0 | 1 | 0 | 1 |

**Step 5: Overflow and Carry Out**

- **Carry Out**: The final carry-out from FA3 is **1**.

- **Overflow Condition**: Overflow occurs if the carry into the most significant bit (MSB) is different from the carry out of the MSB. In this case, since both are 1, **there is no overflow**.

**Step 6: Summary of Outputs**

- Inputs $ A_3 A_2 A_1 A_0 $ = 1010

- Inputs $ B_3 B_2 B_1 B_0 $ = 1011 (before complement), 0101 (2’s complement)

- **Sums** $ S_3 S_2 S_1 S_0 $ = 0000 (binary result of $ A - B = 1010 - 1011 = -1 $ in decimal, which is represented as 2's complement binary 1111).

- **Carry Out** = 1

- **Overflow** = 0

Thus, the binary result of $ 1010 - 1011 $ is $ 1111 $ (which represents $-1$ in 2’s complement), and there is no overflow. The carry-out bit is 1.

(e) Explain the functioning of a 2 × 4 decoder with the help of a logic diagram and an example input.

Ans:- **2 × 4 Decoder:**

A **2 × 4 decoder** is a combinational logic circuit that decodes a 2-bit binary input into one of four outputs. For each unique input combination, one and only one of the four outputs will be active (set to 1), while all others will be inactive (set to 0).

**Functionality of a 2 × 4 Decoder:**

- The decoder has **2 input lines** and **4 output lines**.

- Based on the binary input, one of the four outputs is selected and set to **1**, while the remaining three outputs are set to **0**.

- The input lines are typically labeled as $ A $ and $ B $.

- The output lines are typically labeled as $ Y_0, Y_1, Y_2, Y_3 $.

**Truth Table for the 2 × 4 Decoder:**

| Inputs | Output |

|--------|--------|

| A B | $ Y_0 $ | $ Y_1 $ | $ Y_2 $ | $ Y_3 $ |

|--------|--------|--------|--------|--------|

| 0 0 | 1 | 0 | 0 | 0 |

| 0 1 | 0 | 1 | 0 | 0 |

| 1 0 | 0 | 0 | 1 | 0 |

| 1 1 | 0 | 0 | 0 | 1 |

- When $ A = 0 $ and $ B = 0 $, the output $ Y_0 = 1 $, and all other outputs are $ 0 $.

- When $ A = 0 $ and $ B = 1 $, the output $ Y_1 = 1 $, and all other outputs are $ 0 $.

- When $ A = 1 $ and $ B = 0 $, the output $ Y_2 = 1 $, and all other outputs are $ 0 $.

- When $ A = 1 $ and $ B = 1 $, the output $ Y_3 = 1 $, and all other outputs are $ 0 $.

**Logic Diagram of a 2 × 4 Decoder:**

The logic diagram for the 2 × 4 decoder can be created using **AND gates** and **NOT gates**. Each output is a combination of the input lines $ A $ and $ B $ (or their complements), as follows:

- $ Y_0 = \overline{A} \cdot \overline{B} $ (Both inputs are 0)

- $ Y_1 = \overline{A} \cdot B $ (Input is 01)

- $ Y_2 = A \cdot \overline{B} $ (Input is 10)

- $ Y_3 = A \cdot B $ (Input is 11)

**Logic Diagram:**

```

A ------| |------|Y_2|

| AND | |

B ------| |------|

A ------| |------|Y_3|

| AND | |

B ------| |------|

A' -----| |------|Y_0|

| AND | |

B' -----| |------|

A' -----| |------|Y_1|

| AND | |

B -----| |------|

```

In the diagram:

- $ A' $ and $ B' $ represent the NOT (inverted) versions of the inputs $ A $ and $ B $, implemented using NOT gates.

- Each AND gate represents a condition where a specific combination of $ A $ and $ B $ determines which output is active.

**Working Example:**

Let’s consider an example with inputs $ A = 1 $ and $ B = 0 $:

1. The decoder inputs are $ A = 1 $ and $ B = 0 $.

2. Based on the logic:

- $ Y_2 = A \cdot \overline{B} = 1 \cdot 1 = 1 $ (since $ A = 1 $ and $ B = 0 $, $ Y_2 $ will be activated).

- All other outputs will be $ 0 $ because their input combinations are not satisfied.

Thus, for input $ A = 1 $ and $ B = 0 $, the output is $ Y_2 = 1 $, while $ Y_0 = 0 $, $ Y_1 = 0 $, and $ Y_3 = 0 $.

---

**Summary:**

- A 2 × 4 decoder takes 2 inputs and produces 4 outputs.

- For each combination of inputs, exactly one output is active (set to 1).

- The logic diagram uses AND gates and NOT gates to create the specific outputs based on input combinations.

- The truth table and the logic diagram illustrate the function and implementation of the decoder.

Let me know if you'd like further clarifications or additional details on the design!

(f) Assume that a source data value 1111 was received at a destination as 1011. Show how Hamming's Error-Correcting code bits will be appended to source data to identify and correct the error of one bit at the destination. You may assume that transmission error occurs only in the source data and not the source parity bits.

Ans:- **Hamming's Error-Correcting Code:**

Hamming's code is a method used to detect and correct a single-bit error in a transmitted data value. This is achieved by adding redundant parity bits to the data, which allows the receiver to detect and correct errors.

For a **4-bit data value**, the number of parity bits $ p $ required can be determined using the following relation:

2^p \geq p + d + 1

where $ d $ is the number of data bits (in this case, $ d = 4 $).

- For $ d = 4 $, we need at least $ p = 3 $ parity bits, as:

2^3 = 8 \geq 3 + 4 + 1 = 8

Thus, **3 parity bits** are sufficient for 4-bit data.

**Step 1: Assign Parity Bits to the Data**

Let’s take the source data **1111** (before transmission) and append the parity bits. The positions of the parity bits are powers of 2: positions 1, 2, and 4 in the codeword. The data bits are placed in the remaining positions (3, 5, 6, and 7).

So, the 7-bit codeword structure is as follows (with P representing parity bits and D representing data bits):

P_1 \, P_2 \, D_3 \, P_4 \, D_5 \, D_6 \, D_7

Where:

- $ D_3 = 1 $, $ D_5 = 1 $, $ D_6 = 1 $, and $ D_7 = 1 $.

Now, we need to calculate the values of the parity bits.

### **Step 2: Calculate Parity Bits**

- $ P_1 $: Covers bits 1, 3, 5, 7.

P_1 = D_3 \oplus D_5 \oplus D_7 = 1 \oplus 1 \oplus 1 = 1

- $ P_2 $: Covers bits 2, 3, 6, 7.

P_2 = D_3 \oplus D_6 \oplus D_7 = 1 \oplus 1 \oplus 1 = 1

- $ P_4 $: Covers bits 4, 5, 6, 7.

P_4 = D_5 \oplus D_6 \oplus D_7 = 1 \oplus 1 \oplus 1 = 1

Thus, the complete Hamming codeword to be transmitted is:

P_1 P_2 D_3 P_4 D_5 D_6 D_7 = 1 1 1 1 1 1 1

The source data with Hamming code is $ 1111111 $.

**Step 3: Error Detection and Correction at the Receiver**

Now, assume the received codeword at the destination is **1011111**. We know there has been a 1-bit error.

The received codeword is:

1 0 1 1 1 1 1

We will check the parity bits to detect the error.

- **Check $ P_1 $**: Covers bits 1, 3, 5, 7.

1 \oplus 1 \oplus 1 \oplus 1 = 0 \quad (\text{no error in this group})

- **Check $ P_2 $**: Covers bits 2, 3, 6, 7.

0 \oplus 1 \oplus 1 \oplus 1 = 1 \quad (\text{error in this group})

- **Check $ P_4 $**: Covers bits 4, 5, 6, 7.

1 \oplus 1 \oplus 1 \oplus 1 = 0 \quad (\text{no error in this group})

**Step 4: Determine the Error Position**

The results from checking the parity bits indicate where the error occurred. We represent the results of the checks as binary digits, with $ P_4 $ contributing the most significant bit (MSB), followed by $ P_2 $ and $ P_1 $.

- $ P_4 = 0 $

- $ P_2 = 1 $

- $ P_1 = 0 $

Thus, the binary error position is $ 010_2 = 2 $. This means there is an error in **bit position 2** of the received codeword.

**Step 5: Correct the Error**

Now that we know the error is in bit 2, we can correct the codeword by flipping bit 2. The received codeword was **1011111**, so we change bit 2:

1011111 \quad \xrightarrow{\text{flip bit 2}} \quad 1111111

The corrected codeword is **1111111**, which matches the original transmitted codeword.

**Step 6: Extract the Original Data**

Once the error is corrected, we can remove the parity bits to retrieve the original data. The data bits are at positions 3, 5, 6, and 7:

D_3 D_5 D_6 D_7 = 1111

Thus, the original data is **1111**, which has been successfully retrieved after error correction.

---

**Summary of Steps:**

1. **Append parity bits** to the 4-bit data value using Hamming's code.

2. **Transmit** the 7-bit codeword.

3. **Check the parity bits** at the receiver to detect any error.

4. **Determine the error position** using the results of the parity bit checks.

5. **Correct the error** by flipping the bit at the detected position.

6. **Extract the original data** from the corrected codeword.

This method allows for the detection and correction of a single-bit error during transmission.

(g) Explain the functioning of the RS flip flop with the help of a logic diagram and characteristic table. Also, explain the excitation table of this flip-flop.

Ans:- **RS Flip-Flop (Reset-Set Flip-Flop)**

The **RS flip-flop** (also called **SR flip-flop**) is a simple bistable memory element used in digital logic circuits. It has two stable states, allowing it to store a single bit of information (either 0 or 1). The flip-flop has **two inputs**, typically labeled as:

- **S (Set)**: Used to set the flip-flop (i.e., output becomes 1).

- **R (Reset)**: Used to reset the flip-flop (i.e., output becomes 0).

**Logic Diagram of RS Flip-Flop:**

There are two common types of RS flip-flops:

1. **RS Flip-Flop using NOR gates**.

2. **RS Flip-Flop using NAND gates**.

1. **RS Flip-Flop using NOR Gates:**

A basic RS flip-flop can be constructed using two **NOR gates**, with the inputs connected in a cross-coupled fashion. The output of each NOR gate feeds into one input of the other NOR gate.

Here’s the logic diagram for an RS flip-flop using **NOR gates**:

```

S ------| |------ Q

| NOR |

Q' <----| |----|

| |

R ------| | |

| NOR | |

Q <-----| |<---|

```

- **S (Set)**: Input that sets the output to 1.

- **R (Reset)**: Input that resets the output to 0.

- **Q**: Output (normal output).

- **Q'**: Complementary output (inverse of $ Q $).

**Characteristic Table of RS Flip-Flop:**

The **characteristic table** describes how the outputs $ Q $ and $ Q' $ behave for different combinations of the inputs $ S $ and $ R $.

| S | R | Q (Next State) | Description |

|---|---|----------------|------------------------|

| 0 | 0 | Q (No change) | No change in the output |

| 0 | 1 | 0 | Reset state (Q = 0) |

| 1 | 0 | 1 | Set state (Q = 1) |

| 1 | 1 | Invalid | Invalid condition |

- **S = 0, R = 0**: The output **remains unchanged** (holds its previous state).

- **S = 0, R = 1**: The flip-flop is **reset**, so the output $ Q $ becomes 0.

- **S = 1, R = 0**: The flip-flop is **set**, so the output $ Q $ becomes 1.

- **S = 1, R = 1**: This is an **invalid condition** for the RS flip-flop using NOR gates because both inputs cannot be high simultaneously. This leads to an undefined state, as both $ Q $ and $ Q' $ would be forced to 0, violating the complementary relationship $ Q = \overline{Q'} $.

**Functioning of RS Flip-Flop:**

- **Set State (S = 1, R = 0)**:

- When the **S input is 1** and the **R input is 0**, the output $ Q $ is set to 1 (i.e., the flip-flop stores 1).

- The flip-flop "remembers" this state even if the inputs return to 0.

- **Reset State (S = 0, R = 1)**:

- When the **S input is 0** and the **R input is 1**, the output $ Q $ is reset to 0 (i.e., the flip-flop stores 0).

- **Hold State (S = 0, R = 0)**:

- If both **S and R are 0**, the output remains in its **previous state**. No change occurs.

- **Invalid State (S = 1, R = 1)**:

- When both inputs are 1, the output becomes undefined (both $ Q $ and $ Q' $ could be forced to 0, which contradicts the rule that $ Q = \overline{Q'} $).

**Excitation Table of RS Flip-Flop:**

The **excitation table** describes what inputs $ S $ and $ R $ are required to achieve a desired transition from the current state $ Q $ to the next state $ Q^+ $. It is the inverse of the characteristic table.

| Current State (Q) | Next State (Q⁺) | S | R |

|-------------------|-----------------|---|---|

| 0 | 0 | 0 | 0 |

| 0 | 1 | 1 | 0 |

| 1 | 0 | 0 | 1 |

| 1 | 1 | 0 | 0 |

- To transition from **0 to 0**, the inputs must be **S = 0, R = 0** (no change).

- To transition from **0 to 1**, the inputs must be **S = 1, R = 0** (set operation).

- To transition from **1 to 0**, the inputs must be **S = 0, R = 1** (reset operation).

- To stay at **1 to 1**, the inputs must be **S = 0, R = 0** (no change).

**Summary:**

- The **RS flip-flop** stores a single bit of data and can be set, reset, or left unchanged depending on the input values of $ S $ and $ R $.

- The **characteristic table** shows how the flip-flop's output responds to different inputs.

- The **excitation table** shows what inputs are required to move from one state to another.

This flip-flop can be used in memory storage elements and sequential circuits. However, the **S = 1, R = 1** case is invalid, so the design must ensure that this combination never occurs.

(h) Explain the functioning of the master-slave flip-flop with the help of a diagram.

Ans:- **Master-Slave Flip-Flop:**

A **Master-Slave flip-flop** is a sequential circuit that eliminates the problem of the input changes affecting the output in real-time, as can happen with simpler flip-flops like the SR or JK flip-flops. This design ensures that the output is updated only at specific times, typically synchronized with a clock signal. It consists of two flip-flops:

1. **Master flip-flop**: Receives and stores the input data when the clock signal is high (1).

2. **Slave flip-flop**: Transfers the stored data from the master to the output when the clock signal is low (0).

The **Master-Slave flip-flop** uses two **gated flip-flops** arranged in series and controlled by the same clock signal, but with opposite triggering (positive and negative edge). The output of the slave is the final output of the entire system.

**Basic Functioning:**

- When the clock is **high** (1), the **master flip-flop** is **enabled** and captures the input data, but the slave flip-flop is disabled.

- When the clock is **low** (0), the **slave flip-flop** is **enabled**, and it receives the data from the master flip-flop, which is now disabled.

This sequential operation ensures that input changes are not immediately reflected at the output, allowing for more stable data storage.

**Master-Slave Flip-Flop Diagram:**

Here’s a basic conceptual diagram of a **Master-Slave flip-flop** using two **SR flip-flops**. The circuit operates with a clock signal $ CLK $ that controls the transfer of data.

```

+--------------------+

| |

| Master |

CLK ----->| Flip-Flop (SR) |-----> Qm (Master Output)

S ------>| |

R ------>| |

+--------------------+

| |

| Slave |

CLK'----->| Flip-Flop (SR) |-----> Q (Final Output)

| (Inverted CLK) |

| |

+--------------------+

```

- The **master flip-flop** is triggered on the **rising edge** of the clock signal, while the **slave flip-flop** is triggered on the **falling edge** of the clock signal (or its inverted version).

- **S** and **R** are the set and reset inputs.

- **Qm** is the output of the master flip-flop.

- **Q** is the final output of the slave flip-flop.

**Working of Master-Slave Flip-Flop:**

1. **Clock High (1):**

- The **master flip-flop** is active and captures the inputs $ S $ and $ R $.

- The **slave flip-flop** is inactive during this phase and does not change its state.

- The master flip-flop updates its state based on the input, but this state is not immediately passed to the output.

2. **Clock Low (0):**

- The **slave flip-flop** becomes active and takes the state of the master flip-flop as its input.

- The **master flip-flop** is now inactive and holds its state until the clock goes high again.

- The state of the master flip-flop is passed to the output of the slave flip-flop, which becomes the final output $ Q $.

This arrangement ensures that the input is "sampled" and then held in the master during the clock's high phase, while the output only changes when the clock goes low, preventing any immediate feedback loops or output glitches.

**Advantages of Master-Slave Flip-Flop:**

- **Edge-triggered**: It solves the issue of glitches that can arise in level-triggered flip-flops by ensuring that inputs affect the output only on specific clock edges.

- **Stable output**: The output does not change as soon as the input changes, ensuring that the output is stable between clock pulses.

- **Synchronous operation**: The output only changes at specific clock times, allowing it to be easily integrated into clocked digital circuits.

**Summary:**

- The **Master-Slave flip-flop** uses two flip-flops (master and slave) to ensure stable data storage and synchronized outputs.

- The **master** captures the input when the clock is high, and the **slave** transfers the data to the output when the clock is low.

- The design prevents unwanted output changes during the clock's active phase and helps eliminate the problem of race conditions.

This type of flip-flop is commonly used in **edge-triggered** systems and in clocked sequential circuits like registers and counters.

(i) Represent (129. 5)10 and (-1.125)10 in IEEE 754 single-precision and double-precision formats.

Ans:-**IEEE 754 Floating Point Representation:**

The IEEE 754 standard defines the representation of floating-point numbers in two main formats:

1. **Single Precision (32-bit)**

2. **Double Precision (64-bit)**

Each format consists of three parts:

- **Sign bit (S)**: Indicates the sign of the number (0 for positive, 1 for negative).

- **Exponent (E)**: A biased exponent, which allows both positive and negative exponents.

- **Mantissa (M)** (also called Fraction): Represents the significant digits of the number in normalized form.

**Steps for Converting Decimal to IEEE 754:**

1. **Convert the number to binary**.

2. **Normalize the binary number** (move the decimal point so there’s only one digit before the decimal point).

3. **Determine the sign bit** (0 for positive numbers, 1 for negative numbers).

4. **Calculate the biased exponent**:

- The exponent is biased by adding a fixed value (127 for single-precision, 1023 for double-precision).

5. **Determine the mantissa**: Drop the leading 1 (since it’s implied in normalized numbers) and take the remaining binary digits for the mantissa.

6. **Fill in the bits for single or double precision**.

---

**Example 1: Represent $ (129.5)_{10} $ in IEEE 754 Formats**

**Step 1: Convert 129.5 to Binary**

- Convert the integer part $ 129_{10} $:

129_{10} = 10000001_2

- Convert the fractional part $ 0.5_{10} $:

0.5_{10} = 0.1_2

Thus, $ 129.5_{10} = 10000001.1_2 $.

**Step 2: Normalize the Binary Number**

Normalize $ 10000001.1_2 $ by moving the decimal point to just after the first digit:

1.00000011_2 \times 2^7

The exponent is $ 7_{10} $.

**Step 3: Determine the Sign Bit**

- Since $ 129.5 $ is positive, the sign bit $ S = 0 $.

**Step 4: Calculate the Biased Exponent**

For **single-precision** (32-bit):

- Bias is $ 127 $, so the biased exponent is:

E = 127 + 7 = 134_{10} = 10000110_2

For **double-precision** (64-bit):

- Bias is $ 1023 $, so the biased exponent is:

E = 1023 + 7 = 1030_{10} = 10000000110_2

**Step 5: Determine the Mantissa**

The normalized number is $ 1.00000011_2 $, so the mantissa is $ 00000011000000000000000 $ (for single-precision) or $ 0000001100000000000000000000000000000000000000000000 $ (for double-precision), as we drop the leading 1.

---

**IEEE 754 Single-Precision (32-bit) Representation of $ 129.5_{10} $:**

- **Sign bit (S)**: $ 0 $

- **Exponent (E)**: $ 10000110_2 $ (which is 134 in decimal)

- **Mantissa (M)**: $ 00000011000000000000000 $

Final representation:

0 \, 10000110 \, 00000011000000000000000

This is $ 0x43006000 $ in hexadecimal.

---

**IEEE 754 Double-Precision (64-bit) Representation of $ 129.5_{10} $:**

- **Sign bit (S)**: $ 0 $

- **Exponent (E)**: $ 10000000110_2 $ (which is 1030 in decimal)

- **Mantissa (M)**: $ 0000001100000000000000000000000000000000000000000000 $

Final representation:

0 \, 10000000110 \, 0000001100000000000000000000000000000000000000000000

This is $ 0x405800C000000000 $ in hexadecimal.

---

**Example 2: Represent $ (-1.125)_{10} $ in IEEE 754 Formats**

#### **Step 1: Convert -1.125 to Binary**

- Convert the integer part $ 1_{10} $:

1_{10} = 1_2

- Convert the fractional part $ 0.125_{10} $:

0.125_{10} = 0.001_2

Thus, $ 1.125_{10} = 1.001_2 $.

#### **Step 2: Normalize the Binary Number**

The number $ 1.001_2 $ is already normalized:

1.001_2 \times 2^0

The exponent is $ 0_{10} $.

**Step 3: Determine the Sign Bit**

- Since $ -1.125 $ is negative, the sign bit $ S = 1 $.

#### **Step 4: Calculate the Biased Exponent**

For **single-precision** (32-bit):

- Bias is $ 127 $, so the biased exponent is:

E = 127 + 0 = 127_{10} = 01111111_2

For **double-precision** (64-bit):

- Bias is $ 1023 $, so the biased exponent is:

E = 1023 + 0 = 1023_{10} = 01111111111_2

**Step 5: Determine the Mantissa**

The normalized number is $ 1.001_2 $, so the mantissa is $ 00100000000000000000000 $ (for single-precision) or $ 0010000000000000000000000000000000000000000000000000 $ (for double-precision), as we drop the leading 1.

---

**IEEE 754 Single-Precision (32-bit) Representation of $ -1.125_{10} $:**

- **Sign bit (S)**: $ 1 $

- **Exponent (E)**: $ 01111111_2 $ (which is 127 in decimal)

- **Mantissa (M)**: $ 00100000000000000000000 $

Final representation:

1 \, 01111111 \, 00100000000000000000000

This is $ 0xBF900000 $ in hexadecimal.

---

**IEEE 754 Double-Precision (64-bit) Representation of $ -1.125_{10} $:**

- **Sign bit (S)**: $ 1 $

- **Exponent (E)**: $ 01111111111_2 $ (which is 1023 in decimal)

- **Mantissa (M)**: $ 0010000000000000000000000000000000000000000000000000 $

Final representation:

1 \, 01111111111 \, 0010000000000000000000000000000000000000000000000000

This is $ 0xBFF2000000000000 $ in hexadecimal.

---

**Summary:**

1. **Single-precision representation of $ 129.5_{10} $**: $ 0x43006000 $

2. **Double-precision representation of $ 129.5_{10} $**: $ 0x405800C000000000 $

3. **Single-precision representation of $ -1.125_{10} $**: $ 0xBF900000 $

4. **Double-precision representation of $ -1.125_{10} $**: $ 0xBFF2000000000000 $

Question 2:

(a) Refer to the Figure 2(b) on page 8 in Unit 1 of Block 2. Draw the Internal organisation of a 16×2 RAM. Explain all the Input and Output of this organisation. Also, answer the following:

(i) How many data input and data output lines does this RAM need? Explain your answer.

(ii)How many address lines are needed for this RAM? Give reasons in support of your answer

Ans:- **Internal Organization of a 16×2 RAM**

A **16×2 RAM** (Random Access Memory) configuration means that it can store **16** words, each containing **2 bits** of data. The internal organization typically consists of:

- **16 memory cells** arranged in a grid, where each cell holds 2 bits.

- **Address lines** to select a specific word in memory.

- **Data lines** for input and output of the 2 bits of data.

Here's a simplified diagram of a 16×2 RAM:

```

+----------------------+

| 16 x 2 RAM |

+----------------------+

| Address Lines |

| (A0, A1, A2, A3) |

+----------------------+

| Memory Cells |

| +----+ +----+ |

| | 00 | | 00 | |

| +----+ +----+ |

| | 01 | | 01 | |

| +----+ +----+ |

| | 10 | | 10 | |

| +----+ +----+ |

| | 11 | | 11 | |

| +----+ +----+ |

+----------------------+

| Data Input Lines (D0, D1) |

+-----------------------------+

| Data Output Lines (D0, D1) |

+-----------------------------+

```

**Explanation of Inputs and Outputs:**

- **Address Lines (A0, A1, A2, A3)**:

- Used to select one of the 16 memory locations. The number of address lines needed is determined by the formula $ n = \log_2(N) $, where $ N $ is the number of locations. For 16 locations, $ \log_2(16) = 4 $, so **4 address lines** are required.

- **Data Input Lines (D0, D1)**:

- Used to input data into the RAM. Since each word in the RAM is 2 bits wide, **2 data input lines** are needed.

- **Data Output Lines (D0, D1)**:

- Used to output data from the RAM. Like the data input lines, **2 data output lines** are also needed.

---

**Answering the Questions:**

(i) How many data input and data output lines does this RAM need? Explain your answer.

**Answer**:

- The **16×2 RAM** needs **2 data input lines** and **2 data output lines**. This is because each memory location can store 2 bits of data. Therefore, to read from and write to these locations, 2 lines are necessary for both input and output.

(ii) How many address lines are needed for this RAM? Give reasons in support of your answer.

**Answer**:

- **4 address lines** are needed for the **16×2 RAM**. This requirement arises from the number of unique memory locations that need to be addressed. Since the RAM has **16** locations, the number of address lines can be calculated as follows:

n = \log_2(N) = \log_2(16) = 4

Thus, **4 address lines** can uniquely select each of the 16 locations, ranging from **0000 (0)** to **1111 (15)** in binary.

---

**Summary:**

- **Internal Organization**: The RAM consists of 16 memory cells, with each cell capable of storing 2 bits of data. It requires 4 address lines to select a cell and 2 input/output lines for data operations.

- **Data Input Lines**: 2 lines.

- **Data Output Lines**: 2 lines.

- **Address Lines**: 4 lines, to access the 16 memory locations.

(b) A computer has 4 K Word RAM with each memory word of 8 bits. It has cache memory, having 16 blocks, having a size of 16 bits (2 memory words). Show how the main memory address (3AC)h will be mapped to the cache address, if

(i) Direct cache mapping is used

(ii) Associative cache mapping is used

(iii)Two-way set associative cache mapping is used.

You should show the size of the tag, index, main memory block address and offset in your answer.

Ans:- To map the main memory address $(3AC)_{16}$ to cache addresses in a system with a **4 K Word RAM** and a **cache memory** of **16 blocks**, we need to analyze the organization of both the main memory and cache memory.

**Main Memory Organization**

- **Total main memory size**:

4 \text{ K words} = 4 \times 1024 = 4096 \text{ words}

- **Word size**: 8 bits (1 byte).

Thus, the total memory size in bytes is:

4096 \text{ words} \times 1 \text{ byte/word} = 4096 \text{ bytes} = 0x1000 \text{ bytes}.

**Cache Memory Organization**

- **Cache size**: 16 blocks.

- **Block size**: 16 bits = 2 bytes = 2 words.

Thus, the total cache size in bytes is:

16 \text{ blocks} \times 2 \text{ bytes/block} = 32 \text{ bytes}.

**Address Breakdown**

1. **Main Memory Address Size**:

- Since the main memory has **4096 words**, we need:

\text{Address bits} = \log_2(4096) = 12 \text{ bits}.

2. **Cache Block Size**:

- Each block holds 2 words (16 bits):

\text{Block offset} = \log_2(2) = 1 \text{ bit} \text{ (to select 1 word out of 2)}.

3. **Number of Cache Blocks**:

- There are 16 blocks:

\text{Index bits} = \log_2(16) = 4 \text{ bits} \text{ (for selecting a cache block)}.

4. **Tag Size**:

- The tag size can be determined as follows:

\text{Tag bits} = \text{Total address bits} - (\text{Index bits} + \text{Block offset}) = 12 - (4 + 1) = 7 \text{ bits}.

**Address Breakdown Summary:**

- **Total Address Bits**: 12 bits.

- **Tag Bits**: 7 bits.

- **Index Bits**: 4 bits.

- **Block Offset**: 1 bit.

**Address Conversion for $ (3AC)_{16} $**

Converting $ (3AC)_{16} $ to binary:

- $ (3AC)_{16} = 0011\ 1010\ 1100 $

- In binary, $ (3AC)_{16} = 0011\ 1010\ 1100 $.

**Mapping for Different Cache Organizations**

(i) **Direct Cache Mapping**

In direct cache mapping, each block from main memory maps to exactly one block in cache.

1. **Calculate the Block Address in Main Memory**:

- Since each block is 2 words (or 2 bytes), the block address is given by:

\text{Block Address} = \text{Main Memory Address} \div \text{Block Size} = 3AC_{16} \div 2 = (3AC_{10} / 2 = 3B6)_{16}.

2. **Extract Index and Tag**:

- Block address $ 3B6 $ in binary:

0011\ 1011\ 0110

- **Split the address**:

- **Tag**: The first 7 bits: $ 0011101 $

- **Index**: The next 4 bits: $ 0110 $ (which is $ 6_{10} $).

- **Block Offset**: The last 1 bit: $ 0 $.

**Direct Mapping Summary**:

- **Tag**: $ 0011101 $ (7 bits)

- **Index**: $ 0110 $ (4 bits)

- **Offset**: $ 0 $ (1 bit)

---

(ii) **Associative Cache Mapping**

In associative mapping, any block from main memory can be stored in any cache block.

1. **The entire 12-bit address is used for tag**:

- No index bits are required since any block can go into any cache line.

**Associative Mapping Summary**:

- **Tag**: $ 001110101100 $ (12 bits)

- **Index**: Not applicable (not used).

- **Offset**: $ 0 $ (1 bit).

---

(iii) **Two-Way Set Associative Cache Mapping**

In two-way set associative mapping, the cache is divided into 2 sets, each containing 8 blocks.

1. **Calculate the number of sets**:

- Number of sets = 16 blocks / 2 ways = 8 sets.

- **Index bits for sets**:

\text{Index bits} = \log_2(8) = 3 \text{ bits}.

2. **Tag Size**:

- Total address bits = 12.

- Tag bits = 12 - (3 index bits + 1 offset bit) = 8 bits.

### **Mapping Summary for Two-Way Set Associative**:

- **Tag**: $ 00111010 $ (8 bits)

- **Index**: $ 110 $ (3 bits)

- **Offset**: $ 0 $ (1 bit)

---

**Final Summary of Mappings**:

|-----------------------------|------------------|------------|-------------|

This provides a comprehensive view of how the main memory address $ (3AC)_{16} $ is mapped to cache addresses using different mapping techniques.

(c) What are the different kinds of interrupts? Explain the process of handling an interrupt with the help of a diagram.

Ans:- Interrupts are signals that inform the CPU that an event needs immediate attention. They allow the CPU to respond to various events asynchronously, ensuring efficient processing. Below are the different types of interrupts and a detailed explanation of the interrupt handling process, illustrated with a diagram.

**Types of Interrupts**

1. **Hardware Interrupts**:

- Generated by hardware devices (e.g., keyboard, mouse, disk drives) to signal that they require processing.

- Examples include:

- I/O interrupts (e.g., keypress, mouse movement)

- Timer interrupts (e.g., system clock tick)

- Power failure interrupts

2. **Software Interrupts**:

- Generated by programs when they require system calls (e.g., requesting services from the operating system).

- Examples include:

- System calls (e.g., file operations, process control)

- Exceptions or traps (e.g., division by zero, invalid memory access)

3. **Timer Interrupts**:

- A specific type of hardware interrupt that is generated by the system timer at regular intervals.

- Used for multitasking and time-slicing among processes.

4. **External Interrupts**:

- Triggered by external events, such as hardware malfunctions or signals from other systems.

5. **Internal Interrupts** (Traps):

- Generated by the CPU when it detects an error condition during instruction execution (e.g., arithmetic overflow).

**Interrupt Handling Process**

The process of handling an interrupt involves several steps, as illustrated in the following diagram:

```plaintext

+-----------------+

| CPU Running |

| a Program |

+-----------------+

| (1) Interrupt Occurs

+-----------------+

| Save CPU State |

| (Registers, PC)|

+-----------------+

| (2) Determine Interrupt Type

+-----------------+

| Identify the |

| Interrupt Source|

+-----------------+

| (3) Execute Interrupt Service Routine (ISR)

+-----------------+

| Process the |

| Interrupt |

+-----------------+

| (4) Restore CPU State

| (Registers, PC)

+-----------------+

| Resume |

| Previous Task |

+-----------------+

```

**Steps in Handling an Interrupt**

1. **Interrupt Occurs**:

- The CPU is executing a program when an interrupt signal is received. This can be from hardware or software.

2. **Save CPU State**:

- The current state of the CPU, including the program counter (PC) and registers, is saved onto the stack. This allows the CPU to return to the exact state it was in before the interrupt occurred.

3. **Determine Interrupt Type**:

- The CPU identifies the source of the interrupt (hardware or software) and the specific interrupt type. This is typically done using an interrupt vector table, which maps interrupt requests to their respective service routines.

4. **Execute Interrupt Service Routine (ISR)**:

- The CPU executes the corresponding ISR, which is a special function designed to handle the specific interrupt. This routine may involve reading data from an I/O device, processing information, or performing any necessary actions.

5. **Restore CPU State**:

- After the ISR has completed, the saved state of the CPU (from the stack) is restored. This includes the registers and the program counter.

6. **Resume Previous Task**:

- The CPU resumes execution of the interrupted program as if the interrupt had never occurred.

**Summary**

Interrupts are vital for responsive and efficient computing, allowing the CPU to deal with various events as they occur. The handling process ensures that the system can pause its current operations, respond to the interrupt, and then resume its previous task seamlessly. This mechanism supports multitasking and enhances overall system performance.

(d) What is a DMA? What are the advantages of using DMA? Explain the functions of a DMA interface with the help of a block diagram.

Ans:- **Direct Memory Access (DMA)**

**Direct Memory Access (DMA)** is a feature that allows hardware devices to transfer data directly to and from system memory without continuous intervention from the CPU. This capability enhances the efficiency of data transfers, particularly for high-speed devices like disk drives, graphics cards, and network interfaces.

**Advantages of Using DMA**

1. **Increased Efficiency**:

- DMA allows peripherals to access memory directly, freeing up the CPU from being involved in every data transfer. This results in better utilization of CPU resources.

2. **Higher Data Transfer Rates**:

- With DMA, data can be transferred in blocks rather than byte by byte, which increases the speed of data transfers, especially for large amounts of data.

3. **Reduced CPU Overhead**:

- By offloading data transfer tasks to the DMA controller, the CPU can focus on executing other processes, leading to improved system performance.

4. **Concurrent Operations**:

- The CPU and peripherals can operate concurrently, as the CPU can execute instructions while the DMA controller handles data transfers.

5. **Lower Latency**:

- DMA reduces the latency in data transfers, as it minimizes the number of interrupts sent to the CPU, making the system more responsive.

**Functions of a DMA Interface**

A DMA interface facilitates the communication between peripherals and memory. It includes several key functions, as represented in the following block diagram:

**Block Diagram of DMA Interface**

```plaintext

+-------------------------+

| CPU |

| |

| +------------------+ |

| | | |

| | Memory | |

| | | |

| +--------+---------+ |

| | |

| v |

| +------------------+ |

| | | |

| | DMA | |

| | Controller | |

| | | |

| +--------+---------+ |

| | |

| v |

| +------------------+ |

| | | |

| | Peripheral | |

| | Device | |

| | | |

| +------------------+ |

+-------------------------+

```

**Components of DMA Interface**

1. **DMA Controller**:

- The DMA controller manages the data transfer between memory and peripherals. It handles the necessary control signals for initiating and terminating data transfers.

2. **Memory**:

- This is where data is temporarily stored. The DMA controller accesses memory directly for read and write operations.

3. **Peripheral Device**:

- This is the hardware component (e.g., disk drives, network cards) that requires data transfer to/from memory. The DMA controller interfaces with this device to manage data transfers.

### **Functions of DMA Interface**

1. **Address Generation**:

- The DMA controller generates the appropriate memory addresses for data transfer. It keeps track of the source and destination addresses during transfers.

2. **Control Signals**:

- The DMA controller generates control signals to manage the timing of the data transfer. It informs both the CPU and the peripheral device when to read or write data.

3. **Data Transfer**:

- The actual data transfer occurs between the peripheral and memory without involving the CPU. The DMA controller moves data blocks efficiently.

4. **Interrupt Handling**:

- Once the data transfer is complete, the DMA controller may send an interrupt to the CPU to indicate that the operation has finished. This allows the CPU to perform any necessary post-transfer processing.

5. **Channel Management**:

- In systems with multiple DMA channels, the controller manages which peripheral has access to the memory at any given time, ensuring smooth operation among all devices.

(e) Assume that a disk has 128 tracks, with each track having 64 sectors and each sector is of size 1 M Bytes. The cluster size in this system can be assumed to be 2 sectors. A file having the name assignmentmcs012.txt is of size 16 MB. Assume that it is a new disk, and the first 16 clusters are occupied by the Operating System. Rest all the clusters are free. How can this file be allotted space on this disk? Also, show the content of FAT after the space allocation to this file. You may make suitable assumptions.

Ans:- To allocate space for the file **assignmentmcs012.txt** on a disk with the specified characteristics, let’s break down the allocation process step by step.

Disk Specifications Recap:

- **Number of Tracks**: 128

- **Sectors per Track**: 64

- **Sector Size**: 1 MB

- **Cluster Size**: 2 sectors (thus, 2 MB per cluster)

- **File Size**: 16 MB

- **Occupied Clusters by OS**: 16 clusters (the first 16 clusters)

### 1. **Calculate Clusters Required for the File**

- Since each cluster is **2 MB**, to find the number of clusters required for a **16 MB** file:

\text{Number of clusters required} = \frac{\text{File Size}}{\text{Cluster Size}} = \frac{16 \text{ MB}}{2 \text{ MB/cluster}} = 8 \text{ clusters}

2. **Identify Available Clusters**

- The first 16 clusters are occupied by the Operating System, so the first available cluster for file allocation starts at **Cluster 17**.

- The following clusters can be allocated:

- **Cluster 17**

- **Cluster 18**

- **Cluster 19**

- **Cluster 20**

- **Cluster 21**

- **Cluster 22**

- **Cluster 23**

- **Cluster 24**

3. **Cluster Allocation**

The clusters allocated for the file **assignmentmcs012.txt** will be **Clusters 17 to 24**. Each of these clusters can hold 2 MB of data, totaling 16 MB.

4. **Content of the FAT (File Allocation Table)**

In a FAT system, each cluster entry indicates the next cluster in the file or a special end-of-file marker.

**FAT Content Before Allocation**

Assuming an empty FAT (where all entries are marked as 0 for free):

- **Clusters 0 to 15**: Occupied by OS (let’s represent this as -1 or a specific marker indicating they are not available).

- **Clusters 16**: Free (represented by 0).

- **Clusters 17-24**: Free (represented by 0).

**FAT Content After Allocation**

After allocating the clusters for **assignmentmcs012.txt**, the FAT entries will look like this:

| **Cluster Number** | **FAT Entry (Next Cluster)** |

|---------------------|-------------------------------|

| 0 | -1 (OS Occupied) |

| 1 | -1 (OS Occupied) |

| 2 | -1 (OS Occupied) |

| 3 | -1 (OS Occupied) |

| 4 | -1 (OS Occupied) |

| 5 | -1 (OS Occupied) |

| 6 | -1 (OS Occupied) |

| 7 | -1 (OS Occupied) |

| 8 | -1 (OS Occupied) |

| 9 | -1 (OS Occupied) |

| 10 | -1 (OS Occupied) |

| 11 | -1 (OS Occupied) |

| 12 | -1 (OS Occupied) |

| 13 | -1 (OS Occupied) |

| 14 | -1 (OS Occupied) |

| 15 | -1 (OS Occupied) |

| 16 | 0 |

| 17 | 18 |

| 18 | 19 |

| 19 | 20 |

| 20 | 21 |

| 21 | 22 |

| 22 | 23 |

| 23 | 24 |

| 24 | -1 (End of File) |

| 25 | 0 |

| ... | 0 |

| 127 | 0 |

**Summary**

- The file **assignmentmcs012.txt** of size 16 MB requires 8 clusters for allocation (Clusters 17 to 24).

- The FAT is updated to reflect the chain of clusters allocated to this file, with the last cluster (Cluster 24) indicating the end of the file with a special marker (typically -1 or a specific end-of-file marker).

(f) Explain the following, giving their uses and advantages/disadvantages, if needed. (Word limit for the answer of each part is 50 words ONLY)

(i) Access time of disks

(ii) CD-ROM

(iii) Classification of Printers

(iv) Scanner

(v) Refresh rates of monitors

(vi) Devices for data backup

Ans:- (i) **Access Time of Disks**

Access time is the time taken by a disk to locate and retrieve data. It consists of seek time (time to position the read/write head) and rotational latency (time for the sector to rotate under the head). **Advantage**: Faster data retrieval. **Disadvantage**: Affected by fragmentation.

(ii) **CD-ROM**

A **Compact Disc-Read Only Memory (CD-ROM)** is an optical disc used to store data, typically up to 700 MB. **Advantages**: Low cost, portability. **Disadvantages**: Read-only, slower access compared to hard drives.

(iii) **Classification of Printers**

Printers are classified as **impact printers** (e.g., dot matrix) and **non-impact printers** (e.g., inkjet, laser). **Impact printers** create images by striking the paper, while **non-impact** printers use methods like spraying ink. Non-impact printers are quieter and faster.

(iv) **Scanner**

A scanner is a device that converts physical documents into digital images. **Advantages**: Digitizes documents for editing or sharing. **Disadvantages**: Quality varies by resolution, slower for large-scale scanning.

(v) **Refresh Rates of Monitors**

Refresh rate is the number of times a monitor updates its image per second, measured in Hertz (Hz). **Advantages**: Higher refresh rates provide smoother visuals. **Disadvantages**: Higher rates consume more power.

(vi) **Devices for Data Backup**

Common data backup devices include **external hard drives**, **cloud storage**, and **tape drives**. **Advantages**: Provides data security in case of failure. **Disadvantages**: Physical devices can fail, cloud storage requires internet access.

Question 3:

(a) A single-core uniprocessor system has 16 General purpose registers. The machine has RAM of size 1 M memory words. The size of every general-purpose register and memory word is 32 bits. The computer uses fixed-length instructions of size 32 bits each. An instruction of the machine can have two operands. One of these operands is a direct memory operand and the other is a register operand. An instruction of a machine consists of bits for operation code, bits for memory operand and bits of register operand. The machine has about 64 different operation codes. The machine also has special purpose registers, which are other than general purpose registers. These special purpose registers are – Program Counter (PC), Memory Address Register (MAR), Data Register (DR) and Flag registers (FR). The first register among the general-purpose registers can be used as Accumulator Register. The size of Integer operands on the machine may be assumed to be equal to the size of the accumulator register. To execute instructions, the machine has another special purpose register called Instruction Register (IR) of size 32 bits, as each instruction is of this size. Perform the following tasks for the machine. (Make and state suitable assumptions, if any.)

(i) Design suitable instruction formats for the machine. Specify the size of different fields that are needed in the instruction format. Also, indicate how many bits of the instructions are unused for this machine. Explain your design of the instruction format. What would be the size of each register?

(ii) Illustrate two valid instructions of the machine by drawing a diagram that shows instructions and related data in registers and memory.

(iii) Assuming that an instruction is first fetched to the Instruction Register (IR), its memory operand is brought to the DR register and the result of an operation is stored in the Accumulator register, write and explain the sequence of micro-operations to fetch and execute an addition instruction that adds the contents of a memory operand with the contents of a register operand. The result is stored in the accumulator register. Make and state suitable assumptions, if any.

Ans:- (i) **Design of Instruction Format**

The machine has 64 different operation codes (opcodes), 16 general-purpose registers, and 1M (2^20) memory words. The instruction is 32 bits long and consists of the following fields:

**Fields in the Instruction Format:**

1. **Opcode (6 bits)**: Since there are 64 opcodes, we need 6 bits to represent them (as 2^6 = 64).

2. **Register Operand (4 bits)**: There are 16 general-purpose registers, so 4 bits are required to address them (as 2^4 = 16).

3. **Memory Operand (20 bits)**: The RAM has 1M memory words, so 20 bits are required to address them (as 2^20 = 1M).

**Instruction Format Layout:**

|-----------------|---------------------------|---------------------------|-----------------|

| 6 | 4 | 20 | 2 |

- **Opcode**: 6 bits to specify the operation.

- **Register Operand**: 4 bits to specify which general-purpose register is involved.

- **Memory Operand**: 20 bits to specify the direct memory operand.

- **Unused bits**: 2 bits remain unused since 32 bits are available, but we only need 30 bits for the fields.

**Size of Registers:**

Each general-purpose register, memory word, and the accumulator are all **32 bits** in size, as per the machine specification.

(ii) **Two Example Instructions**

1. **ADD R2, [0x003A1F]**

- **Opcode**: `ADD` (Assume opcode = 000001)

- **Register Operand**: `R2` (register number 2 = 0010)

- **Memory Operand**: `0x003A1F` (20-bit memory address)

Instruction in binary (32 bits):

```

000001 0010 00000011101000011111

```

2. **LOAD R1, [0x001F3C]**

- **Opcode**: `LOAD` (Assume opcode = 000010)

- **Register Operand**: `R1` (register number 1 = 0001)

- **Memory Operand**: `0x001F3C` (20-bit memory address)

Instruction in binary (32 bits):

```

000010 0001 00000001111100111100

```

(iii) **Micro-Operations for Fetch and Execute Cycle**

Assume that the machine fetches the instruction, decodes it, fetches the memory operand into the **Data Register (DR)**, performs the addition, and stores the result in the **Accumulator**. Below is the sequence of micro-operations for an addition instruction:

**Instruction**: `ADD R2, [0x003A1F]`

This instruction adds the value from memory location `0x003A1F` to the value in register `R2`, and stores the result in the accumulator register.

**Step 1: Fetch Instruction**

The program counter (PC) points to the memory address of the next instruction. The machine fetches the instruction into the **Instruction Register (IR)**.

1. `MAR ← PC`

Load the **Memory Address Register (MAR)** with the content of the **PC**.

2. `MBR ← Memory[MAR]`

Load the **Memory Buffer Register (MBR)** with the instruction at the memory location pointed to by **MAR**.

3. `IR ← MBR`

Load the instruction from **MBR** to the **Instruction Register (IR)**.

4. `PC ← PC + 1`

Increment the program counter to point to the next instruction.

**Step 2: Decode Instruction**

The instruction in the **IR** is decoded to identify the opcode, register operand, and memory operand.

1. `Decode IR`

Identify the **opcode**, **register** (R2), and **memory address** (0x003A1F).

**Step 3: Fetch Memory Operand**

The machine fetches the memory operand from the specified memory address into the **Data Register (DR)**.

1. `MAR ← 0x003A1F`

Load the **Memory Address Register** with the memory address `0x003A1F`.

2. `MBR ← Memory[MAR]`

Fetch the memory operand from address `0x003A1F` into the **Memory Buffer Register**.

3. `DR ← MBR`

Load the **Data Register (DR)** with the value from the memory.

**Step 4: Execute Addition**

The value in **Register R2** is added to the memory operand in the **DR**, and the result is stored in the accumulator register.

1. `Accumulator ← R2 + DR`

Add the content of **R2** to **DR** and store the result in the **Accumulator**.

**Summary of Micro-Operations:**

- **Fetch**:

- Load instruction from memory into **IR**.

- Increment **PC**.

- **Decode**:

- Identify **opcode** and operands from **IR**.

- **Fetch Memory Operand**:

- Fetch memory operand into **DR**.

- **Execute**:

- Add **R2** and **DR**.

- Store result in **Accumulator**.

This process ensures the instruction is fetched, decoded, and executed efficiently.

(b) Assume that you have a machine, as shown in section 3.2.2 of Block 3 having the set of microoperations as given in Figure 10 on page 62 of Block 3. Consider that R1 and R2 both are 8-bit registers and contain 01111110 and 11010101 respectively. What will be the values of select inputs, carry-in input, and the result of the operation (including carry-out bit) if the following micro-operations are performed on these registers? (For each micro-operation you may assume the initial value of R1 and R2 as given above)

(i) Increment R2

(ii) Subtract R2 from R1

(iii) AND of R1 with R2

(iv) Shift left R1

Ans:- To perform these micro-operations on registers **R1** (01111110) and **R2** (11010101), we refer to the typical ALU control inputs and results as described in Figure 10 (Section 3.2.2 of Block 3).

(i) **Increment R2**

This operation increments the value of **R2** by 1.

- **Initial Value of R2**: `11010101`

- **Operation**: `R2 ← R2 + 1`

- **Select Inputs**: The ALU select inputs for the addition operation (typically `000` or `001` depending on the system's specification).

- **Carry-In**: `0` (No carry-in is needed for increment)

- **Result**:

```

11010101 + 1 = 11010110

```

- **Final Value of R2**: `11010110`

- **Carry-Out**: `0` (no overflow as the result fits in 8 bits)

(ii) **Subtract R2 from R1**

This operation performs `R1 - R2`, which is equivalent to adding the two's complement of **R2** to **R1**.

- **Initial Values**:

- **R1**: `01111110`

- **R2**: `11010101`

- **Operation**: `R1 ← R1 - R2`

- **Select Inputs**: ALU select inputs for subtraction (typically `010` for subtraction, but could depend on system).

- **Carry-In**: `1` (for two's complement subtraction)

- **Result**:

```

01111110 - 11010101 = 10101001

```

- **Final Value of R1**: `10101001`

- **Carry-Out**: `0` (no overflow in the subtraction)

(iii) **AND of R1 with R2**

This operation performs a bitwise AND between **R1** and **R2**.

- **Initial Values**:

- **R1**: `01111110`

- **R2**: `11010101`

- **Operation**: `R1 ← R1 AND R2`

- **Select Inputs**: ALU select inputs for AND operation (typically `100` or `010` for AND).

- **Carry-In**: `0` (not needed for AND operation)

- **Result**:

```

01111110 AND 11010101 = 01010100

```

- **Final Value of R1**: `01010100`

- **Carry-Out**: `0` (AND operation does not involve carry)

(iv) **Shift Left R1**

This operation shifts the bits of **R1** left by one position.

- **Initial Value of R1**: `01111110`

- **Operation**: `R1 ← R1 << 1`

- **Select Inputs**: ALU select inputs for shift-left (typically `110` for logical shift left).

- **Carry-In**: `0` (not needed for shift operation)

- **Result**:

```

01111110 << 1 = 11111100

```

- **Final Value of R1**: `11111100`

- **Carry-Out**: `0` (the leftmost bit of the original value, 0, is shifted out)

Summary of Micro-Operations:

|---------------------|------------|------------|-----------|-------------|-----------|

| Increment R2 | - | 11010101 | R2 + 1 | 11010110 | 0 |

| Subtract R2 from R1 | 01111110 | 11010101 | R1 - R2 | 10101001 | 0 |

| AND R1 with R2 | 01111110 | 11010101 | R1 AND R2 | 01010100 | 0 |

| Shift Left R1 | 01111110 | - | R1 << 1 | 11111100 | 0 |

Let me know if you'd like to explore more about any specific micro-operation!

(c) Consider that an instruction pipeline has four stages namely instruction fetch (INFE), Instruction decode and Operand Fetch (IDOF), Instruction Execute (INEX) and store results (STRE). Draw an instruction pipeline diagram showing the execution of five sequential instructions using this pipeline. Explain, what problem may occur, if the 2nd instruction is a conditional jump instruction?

Ans:- Instruction Pipeline Diagram for 5 Sequential Instructions

In an instruction pipeline, each stage of the pipeline can process a different instruction concurrently. The pipeline consists of four stages:

1. **INFE** (Instruction Fetch)

2. **IDOF** (Instruction Decode and Operand Fetch)

3. **INEX** (Instruction Execute)

4. **STRE** (Store Results)

Below is a diagram showing the execution of five sequential instructions using the 4-stage pipeline:

|-------|---------------|---------------|---------------|---------------|---------------|

| 1 | INFE | | | | |

| 8 | | | | | STRE |

- **Cycle 1**: Instruction 1 is fetched.

- **Cycle 2**: Instruction 1 moves to decode and fetch operands, while Instruction 2 is fetched.

- **Cycle 3**: Instruction 1 moves to execution, Instruction 2 to decode, and Instruction 3 is fetched.

- This continues with each stage processing a different instruction concurrently.

Problem with Conditional Jump Instruction

If the second instruction is a **conditional jump**, it introduces a **control hazard** (or branch hazard). This is a problem where the pipeline doesn't know if the branch will be taken until the instruction is executed (during the **INEX** stage). If the jump is taken, the instructions that have already been fetched and decoded (like Instructions 3, 4, and 5 in the example) are no longer valid since the program execution flow would now change.

Here’s what could happen:

1. **Incorrect Fetching**: Instructions 3, 4, and 5 will have been fetched based on the assumption that the instructions after the jump are sequential, but if the jump is taken, these instructions will be invalid.

2. **Stalling the Pipeline**: The pipeline might need to be stalled (i.e., paused) until the conditional jump is resolved. This leads to **pipeline bubbles** or empty stages in the pipeline while waiting for the result of the branch instruction.

3. **Flushing**: If the branch is taken, the instructions that were fetched but shouldn't be executed will need to be **flushed** from the pipeline, which introduces delay.

Handling the Problem:

- **Branch Prediction**: Some systems use branch prediction techniques to guess whether the branch will be taken or not, allowing the pipeline to continue executing without stalling as much.

- **Delay Slot Filling**: Some architectures fill the slots after a branch instruction with instructions that will always be executed, regardless of whether the branch is taken or not, to minimize wasted cycles.

Effect in the Diagram:

In this case, if instruction 2 is a conditional jump and the pipeline predicts the wrong outcome, instructions 3, 4, and 5 would need to be flushed, and new instructions fetched after instruction 2 completes execution.

(d) Explain the structure and operation of the micro-programmed control unit with the help of a diagram.

Ans:- Structure and Operation of the Micro-Programmed Control Unit

A **micro-programmed control unit** is responsible for generating the control signals that direct the operations of the computer's hardware. Instead of hardwiring the control logic, the control signals are generated by reading **microinstructions** from a **control memory**. The micro-programmed control unit is flexible and easier to modify compared to a hardwired control unit.

Structure of the Micro-Programmed Control Unit

The key components of a micro-programmed control unit include:

1. **Control Memory (CM)**: This is the memory that stores the microprogram. A microprogram is a sequence of **microinstructions** that specify the control signals required to execute a machine instruction.

2. **Microinstruction Register (MIR)**: This register holds the current microinstruction that is being executed.

3. **Control Address Register (CAR)**: This register holds the address of the next microinstruction to be fetched from the control memory.

4. **Control Data Register (CDR)**: This register temporarily holds the microinstruction read from the control memory.

5. **Sequencing Logic**: This determines the next microinstruction address and loads it into the Control Address Register (CAR). The logic can perform branching based on the outcome of conditions or it can simply fetch the next sequential microinstruction.

6. **Control Signals**: The microinstructions generate control signals that manage the operations of the CPU components such as registers, ALU, memory, and buses.

Diagram of a Micro-Programmed Control Unit

```

+-------------------------------------------------+

| |

| +----------------------------------+ |

| | Control Memory (CM) | |

| +----------------------------------+ |

| | |

| +--------------| |

| | v |

| +-------------+ +--------+ |

| | Sequencing | | CDR | |

| | Logic | +--------+ |

| +-------------+ | |

| | v |

| v +------------+ |

| +--------+ | MIR | |

| | CAR |<----+------------+ |

| +--------+ | |

| | |

| +-------------------+----------------+ |

| | | |

| v v |

| Control Signals Conditional |

| to CPU (ALU, Signals |

| Registers, Memory) |

+-------------------------------------------------+

```

Operation of the Micro-Programmed Control Unit

1. **Fetching a Microinstruction**:

- The address of the next microinstruction to be executed is stored in the **Control Address Register (CAR)**.

- This address is used to fetch a microinstruction from the **Control Memory (CM)**.

- The fetched microinstruction is loaded into the **Control Data Register (CDR)**.

2. **Decoding the Microinstruction**:

- The microinstruction from the CDR is transferred to the **Microinstruction Register (MIR)**.

- The microinstruction in the MIR is decoded to generate the appropriate control signals for the CPU components (ALU, registers, memory, etc.).

3. **Executing the Microinstruction**:

- The control signals generated by the microinstruction direct the CPU hardware to perform a specific operation, such as moving data between registers, performing an ALU operation, or accessing memory.

4. **Sequencing the Next Microinstruction**:

- The **Sequencing Logic** determines the address of the next microinstruction to be executed.

- This address is loaded into the **Control Address Register (CAR)**.

- The sequence of microinstructions can proceed either sequentially (next address) or conditionally (branching based on flags or status).

5. **Generating Control Signals**:

- Control signals are generated by decoding the microinstruction, and these signals control the various parts of the CPU (e.g., enabling registers, setting the ALU mode, triggering memory access, etc.).

Advantages of Micro-Programmed Control Unit

1. **Flexibility**: The control unit can be easily modified by changing the microprogram stored in the control memory.

2. **Simplified Design**: Complex instructions can be broken down into smaller microoperations, making the control unit simpler to design.

3. **Ease of Troubleshooting**: Since the control signals are generated by a program (the microprogram), it’s easier to diagnose and fix errors compared to hardwired control units.

Disadvantages of Micro-Programmed Control Unit

1. **Slower than Hardwired Control**: The process of fetching and decoding microinstructions adds overhead, making it slower than a hardwired control unit.

2. **Memory Requirement**: It requires additional memory to store the microinstructions, leading to more hardware resources.

(e) Explain the use of large register file in RISC. Also, explain the optimisation of RISC pipelining.

Ans:- Use of Large Register File in RISC

A **Reduced Instruction Set Computer (RISC)** architecture emphasizes efficiency and speed by simplifying the instruction set and relying on a large register file. The **large register file** plays a crucial role in enhancing the performance of RISC processors. Here’s why:

1. **Minimizing Memory Accesses**

- In RISC architecture, operations are performed primarily between registers rather than between registers and memory. A large number of registers reduces the need to frequently load data from and store data to memory, which is relatively slow. This optimizes performance by minimizing the latency involved in memory accesses.

2. **Efficient Function Calls and Context Switching**

- A large register file allows the RISC processor to allocate **register windows** for handling subroutine or function calls. Each subroutine is given a separate window of registers, reducing the overhead of saving and restoring register values during function calls. This helps in faster context switching and makes recursive function calls more efficient.

3. **Support for Register-to-Register Operations**

- RISC architectures typically prefer register-to-register operations rather than memory-to-register or memory-to-memory operations, which are slower. Having more registers allows for greater flexibility in storing temporary data and intermediate results within the registers, improving instruction throughput.

4. **Avoiding Register Spills**

- In smaller register files, when there are too many values to store, the values must be "spilled" into memory, leading to additional memory accesses. A large register file reduces the chance of register spilling, thereby improving overall performance.

5. **Parallelism and Pipelining**

- With more registers available, it becomes easier to keep multiple instruction pipelines busy, as different sets of registers can be used for different stages of instruction execution. This contributes to the overall efficiency and speed of RISC pipelines.

Optimisation of RISC Pipelining

Pipelining in RISC architecture is essential for improving the instruction throughput by allowing multiple instructions to be processed simultaneously in different stages of execution. RISC pipelines are optimized for speed and efficiency through several key techniques:

1. **Uniform Instruction Length**

- RISC processors typically use **fixed-length instructions**. This uniformity simplifies instruction fetching and decoding, allowing for easy partitioning of instructions into pipeline stages. The processor knows exactly how long each instruction is, avoiding variability that could introduce delays in the pipeline.

2. **Simplified Instruction Set**

- The simplicity of the RISC instruction set (few instructions with uniform execution times) ensures that most instructions can be executed within a single cycle or a small number of cycles. This predictability reduces pipeline stalls and ensures that all pipeline stages are kept busy.

3. **Load/Store Architecture**

- RISC processors typically separate **memory access instructions (load/store)** from **arithmetic and logical instructions**. This means that data transfer between memory and registers only occurs through specific load and store instructions, while all other instructions operate exclusively on data within registers. This separation avoids complex memory operations in the pipeline stages, allowing for more efficient pipelining.

4. **Branch Prediction and Delayed Branching**

- **Branch instructions** (e.g., conditional jumps) can disrupt the pipeline by introducing **control hazards**. To optimize pipelining in the presence of branches, RISC processors employ techniques like:

- **Branch Prediction**: Predicting whether a branch will be taken or not allows the processor to speculatively execute instructions ahead of time, reducing the impact of control hazards.

- **Delayed Branching**: In this technique, the processor always executes the instruction immediately following a branch, regardless of whether the branch is taken or not. This ensures that the pipeline is not wasted while waiting for the branch decision to be made.

5. **Register Renaming**

- **Register renaming** helps to avoid **data hazards** (conflicts where one instruction depends on the result of a previous instruction) by allowing multiple instances of the same register name to be used simultaneously in different instructions. This ensures that instructions can be issued and executed out of order while keeping the pipeline busy.

6. **Superscalar Execution**

- Modern RISC processors often employ **superscalar execution**, where multiple instruction pipelines operate in parallel. Each pipeline can handle different instructions concurrently, thus increasing the instruction throughput.

7. **Hazard Detection and Forwarding**

- RISC processors are equipped with **hazard detection units** that monitor potential **data hazards** (where one instruction requires data from a previous instruction that has not yet completed). They use techniques like **forwarding** (or **bypassing**) to pass results from one pipeline stage to another without having to wait for the data to be written to a register, reducing the number of pipeline stalls.

8. **Pipelined ALU and Memory Access**

- Both the Arithmetic Logic Unit (ALU) and memory access stages are highly optimized in RISC processors to avoid bottlenecks. For example, **pipelined ALUs** allow for multiple arithmetic operations to be processed in different stages of completion, keeping the pipeline active.

Question 4:

(a) Write a program using 8086 assembly Language (with proper comments) that accepts two different digits as input from the keyboard. Each digit is converted to its binary equivalent value. These converted digits are stored in registers BL and CL. The program then stores the smaller of these two values in AL register. The program also checks if the present AL value is larger than all the values contained in a byte array of size 6, which is stored in the memory. If so, then a value 1 is moved to DL register, else a value 0 is moved to DL register. You may assume the byte array has the values 02h, 03h, 05h, 01h, 02h, 03h. Make suitable assumptions, if any.

Ans:- Here is the 8086 assembly language program to achieve the described functionality:

Program Explanation:

1. **Input Two Digits**: The program accepts two digits from the keyboard.

2. **Conversion**: The digits are converted to their binary equivalent and stored in `BL` and `CL`.

3. **Comparison**: The smaller value between `BL` and `CL` is stored in the `AL` register.

4. **Array Check**: The program compares the value in `AL` with a byte array `[02h, 03h, 05h, 01h, 02h, 03h]`. If `AL` is greater than all elements in the array, `DL` is set to 1; otherwise, `DL` is set to 0.

Assumptions:

- Input digits are assumed to be in ASCII format.

- The byte array is predefined and stored in memory.

8086 Assembly Code:

```asm

.MODEL SMALL

.STACK 100H

.DATA

byteArray DB 02h, 03h, 05h, 01h, 02h, 03h ; Array of 6 bytes

prompt1 DB 'Enter first digit: $'

prompt2 DB 'Enter second digit: $'

.CODE

MAIN PROC

MOV AX, @DATA ; Initialize data segment

MOV DS, AX

; Input first digit

MOV AH, 09H ; Display prompt

LEA DX, prompt1

INT 21H

MOV AH, 01H ; Read character from keyboard

INT 21H

SUB AL, 30H ; Convert ASCII to binary (AL = first digit)

MOV BL, AL ; Store first digit in BL

; Input second digit

MOV AH, 09H ; Display prompt

LEA DX, prompt2

INT 21H

MOV AH, 01H ; Read character from keyboard

INT 21H

SUB AL, 30H ; Convert ASCII to binary (AL = second digit)

MOV CL, AL ; Store second digit in CL

; Compare BL and CL, store smaller in AL

CMP BL, CL ; Compare BL and CL

JBE BL_IS_SMALLER ; If BL <= CL, jump to BL_IS_SMALLER

MOV AL, CL ; If CL is smaller, move CL to AL

JMP COMPARE_ARRAY ; Jump to array comparison

BL_IS_SMALLER:

MOV AL, BL ; If BL is smaller or equal, move BL to AL

COMPARE_ARRAY:

LEA SI, byteArray ; Load address of the byte array

MOV CX, 06 ; Set counter to 6 (array size)

MOV DL, 01H ; Assume AL is larger than all elements

COMPARE_LOOP:

CMP AL, [SI] ; Compare AL with array element

JBE SET_ZERO ; If AL <= array element, set DL to 0

INC SI ; Move to the next element

LOOP COMPARE_LOOP ; Repeat until all elements are checked

JMP END_PROGRAM

SET_ZERO:

MOV DL, 00H ; AL is not larger than all elements

END_PROGRAM:

; Exit the program

MOV AH, 4CH

INT 21H

MAIN ENDP

END MAIN

```

Explanation of Key Parts:

1. **Input Handling**:

- The program uses `INT 21H` interrupt with `AH = 01H` to read the digit inputs from the keyboard. The ASCII values are converted to binary by subtracting `30H`.

2. **Comparison**:

- After the input, the digits in `BL` and `CL` are compared using the `CMP` instruction, and the smaller value is moved to the `AL` register.

3. **Array Comparison**:

- The program compares the value in `AL` with the six elements of the predefined byte array. If `AL` is greater than all the elements, `DL` is set to `1`; otherwise, it's set to `0`.

4. **Looping**:

- The comparison with the byte array is done using a `LOOP` structure, which iterates over the array and compares each element.

Output:

- After the program executes, `DL` will hold the value `1` if `AL` is greater than all the array elements, or `0` otherwise.

Let me know if you'd like further details or explanations!

(b) Differentiate between the FAR and NEAR procedure calls in 8086 micro-processor. Assuming that a stack is used for implementing procedure calls, explain how call and return statements of 8086 microprocessor would use stack for NEAR and FAR procedure calls and return from a call. Also, assuming that two parameters are to be passed to a procedure using stack, explain how they will be passed to the procedure and accessed in the procedure. You need not write the assembly code but draw necessary diagrams to illustrate the concept.

Ans:- Differentiation between FAR and NEAR Procedure Calls in 8086 Microprocessor:

In the **8086 microprocessor**, procedure calls can be categorized into **NEAR** and **FAR** procedure calls based on how control is transferred and how the address is handled.

1. **NEAR Procedure Call**:

- **Definition**: A **NEAR call** is a **procedure call** within the same code segment.

- **Addressing**: It only requires the **offset** of the procedure because both the calling procedure and the called procedure are in the same code segment.

- **Stack**: During a NEAR call, only the **Instruction Pointer (IP)** is pushed onto the stack.

- **Return**: For returning from a NEAR call, only the **IP** is popped from the stack.

2. **FAR Procedure Call**:

- **Definition**: A **FAR call** is a **procedure call** to a different code segment.

- **Addressing**: It requires both the **segment** and the **offset** of the procedure since the procedure is in a different code segment.

- **Stack**: During a FAR call, both the **Code Segment (CS)** and **IP** are pushed onto the stack.

- **Return**: For returning from a FAR call, both the **CS** and **IP** are popped from the stack.

---

Stack Usage in 8086 for NEAR and FAR Procedure Calls

1. **NEAR Procedure Call**:

- When a NEAR procedure is called, the **IP** (Instruction Pointer) is pushed onto the stack.

- The processor jumps to the procedure using only the new **offset** (IP) within the same code segment.

- On returning from the NEAR call, the processor pops the **IP** from the stack to resume execution.

**Diagram for NEAR Call:**

| **Before Call** | **Stack after NEAR CALL** |

|------------------|----------------------------|

| - | IP of return address |

**Diagram for NEAR Return:**

| **Before Return**| **Stack after NEAR RET** |

|------------------|----------------------------|

| IP of return addr| - |

2. **FAR Procedure Call**:

- When a FAR procedure is called, both the **CS** (Code Segment) and **IP** (Instruction Pointer) are pushed onto the stack.

- The processor loads the new **CS** and **IP** to transfer control to a procedure in a different segment.

- On returning from the FAR call, both the **CS** and **IP** are popped from the stack to resume execution.

**Diagram for FAR Call:**

| **Before Call** | **Stack after FAR CALL** |

|------------------|----------------------------|

| - | CS of return address |

| | IP of return address |

**Diagram for FAR Return:**

| **Before Return**| **Stack after FAR RET** |

|------------------|----------------------------|

| IP of return addr| - |

| CS of return addr| |

---

Call and Return for NEAR and FAR Procedure Calls:

- **NEAR CALL**:

1. **CALL**: Pushes **IP** of the return address onto the stack.

2. **RET**: Pops the **IP** from the stack to resume execution after the procedure.

- **FAR CALL**:

1. **CALL**: Pushes **CS** and **IP** of the return address onto the stack.

2. **RET**: Pops the **CS** and **IP** from the stack to resume execution after the procedure.

---

Passing Parameters Using the Stack

To pass parameters to a procedure via the stack, the calling program pushes the parameters before the `CALL` instruction. The called procedure can then access these parameters by referencing the stack.

1. **Calling the Procedure**:

- The caller pushes the parameters onto the stack in reverse order (last parameter first).

- Then the `CALL` instruction is executed, which pushes the return address onto the stack.

**Stack before CALL** (with 2 parameters: `param1`, `param2`):

| **Stack** |

|-------------------|

| param1 |

| param2 |

| Return address |

2. **Accessing Parameters in the Procedure**:

- Inside the called procedure, the stack pointer (SP) will point to the return address. The parameters can be accessed relative to the stack pointer (e.g., `param1` can be accessed at `[SP+2]`, `param2` at `[SP+4]`).

**Diagram of Stack in the Procedure**:

| **Stack** | **Contents** |

|--------------------|-------------------|

| param2 | `[SP+4]` |

| param1 | `[SP+2]` |

| Return address | `[SP]` |

---

Summary:

- **NEAR Calls** only push the **IP** onto the stack, while **FAR Calls** push both **CS** and **IP**.

- The **stack** is used for parameter passing and return address storage. Parameters are pushed onto the stack before calling the procedure, and they can be accessed within the procedure using offsets from the stack pointer.

(c) Explain the following in the context of 8086 Microprocessor with the help of an example or a diagram:

(i) Explain the use of Segment Registers in 8086 microprocessor

(ii) Explain the use of the flags - CF, ZF, OF, DF

(iii) Explain the Instructions – XLAT, MUL, SAR, RCL

Ans:- Explanation of 8086 Microprocessor Concepts

(i) Use of Segment Registers in 8086 Microprocessor

The **8086 microprocessor** uses segment registers to divide memory into different segments, allowing efficient access and organization of code and data. This segmentation helps manage the limited address space of 16 bits.

**Segment Registers**:

1. **CS (Code Segment)**: Holds the starting address of the segment containing the executable code.

2. **DS (Data Segment)**: Points to the segment containing data.

3. **SS (Stack Segment)**: Contains the starting address of the stack.

4. **ES (Extra Segment)**: Used for additional data storage, often for string operations.

**Example**:

Assume the following memory layout:

- Code Segment starts at `0x1000`

- Data Segment starts at `0x2000`

- Stack Segment starts at `0x3000`

```assembly

MOV AX, [DS:1000h] ; Load data from Data Segment into AX

CALL 1000h ; Call a procedure in Code Segment

```

**Diagram**:

```

Memory Layout:

+----------------+

| Code Segment |

| (CS) |

| 0x1000 |

+----------------+

| Data Segment |

| (DS) |

| 0x2000 |

+----------------+

| Stack Segment |

| (SS) |

| 0x3000 |

+----------------+

```

(ii) Explanation of Flags - CF, ZF, OF, DF

Flags in the 8086 microprocessor are special bits in the **Flags Register** that indicate the status of the processor after arithmetic and logical operations.

1. **CF (Carry Flag)**:

- Indicates an overflow in unsigned arithmetic operations.

- Set to `1` if there is a carry out from the most significant bit.

**Example**:

```assembly

ADD AL, 0FFh ; If AL + 0FFh results in 100h, CF is set to 1.

```

2. **ZF (Zero Flag)**:

- Set to `1` if the result of an operation is zero.

- Useful for conditional branching.

**Example**:

```assembly

SUB AX, AX ; ZF is set to 1 because the result is zero.

```

3. **OF (Overflow Flag)**:

- Indicates an overflow in signed arithmetic operations.

- Set to `1` if the signed result is too large for the destination operand.

**Example**:

```assembly

ADD AL, 7Fh ; If AL = 7Fh and we add 01h, AL becomes 80h. OF is set.

```

4. **DF (Direction Flag)**:

- Determines the direction of string operations.

- If `DF = 0`, the string operations process from low memory to high. If `DF = 1`, they process from high to low.

**Example**:

```assembly

CLD ; Clear Direction Flag to process strings from low to high.

MOVSI ; Move string with DF = 0.

```

(iii) Explanation of Instructions – XLAT, MUL, SAR, RCL

1. **XLAT (Translate)**:

- Used to convert a byte from a source to a byte from a destination using a lookup table.

- The address for translation is in `AL`, and the table is pointed to by `BX`.

**Example**:

```assembly

; Assuming a table in memory pointed by DS:BX

XLAT ; Translates AL using the table at DS:BX

```

**Diagram**:

```

Memory:

+---+---+---+---+

| 0 | 1 | 2 | 3 |

+---+---+---+---+

AL (index) --> Use value at DS:BX+AL

```

2. **MUL (Multiply)**:

- Performs unsigned multiplication of the accumulator (`AL` or `AX`) with a specified operand.

- The result is stored in the `AX` register (for byte multiplication) or `DX:AX` (for word multiplication).

**Example**:

```assembly

MOV AL, 5

MUL BL ; AL * BL -> AX

```

3. **SAR (Shift Arithmetic Right)**:

- Shifts the bits of the operand to the right and fills the leftmost bits with the sign bit (preserves the sign for signed numbers).

- Used for signed division by powers of two.

**Example**:

```assembly

SAR AX, 1 ; Arithmetic shift right AX by 1

```

4. **RCL (Rotate through Carry Left)**:

- Rotates the bits of a register left through the carry flag.

- The leftmost bit is moved into the carry flag, and the carry flag is moved into the rightmost bit.

**Example**:

```assembly

RCL AX, 1 ; Rotate AX left through CF

```

**Diagram for RCL**:

```

Before RCL: AX = 1010 1101

After RCL: CF = 0, AX = 0101 1011

```

This explanation provides a concise overview of segment registers, flags, and key instructions within the context of the 8086 microprocessor, supplemented with examples and diagrams for clarity.

MCS-012 Solved Free Assignment 2024-25 Sem 2

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