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Combined Bank

Post: Officer(it)
Exam Date: 04.10.2024, Exam Taker: BSC
1. You are required to convert a 32 bit digital number to an analogue voltage over the voltage range of 0 to 3.3V with a Digital to Analogue converter (DAC). What is the resolution of the analogue output ?

The resolution of a Digital-to-Analog Converter (DAC) is the smallest possible change in output voltage per step of the digital input. It can be calculated using the formula:

\[ \text{Resolution} = \frac{\text{Voltage Range}}{\text{Number of Steps}} \]

For a 12-bit DAC, the number of steps is given by:

\[ \text{Number of Steps} = 2^{12} = 4096 \]

Given a voltage range of 0 to 3.3V, the resolution is:

\[ \text{Resolution} = \frac{3.3V}{4096} \approx 0.00080586V \]

Since 1V = 1000mV, converting this to millivolts:

\[ 0.00080586V \times 1000 = 0.80586 \text{ mV} \]

Thus, the resolution of the analogue output is approximately 0.80586 mV per step.

2. Find the sets X and Y If XUY = {1,2,3,5,6,8,9,10}, X∩Y={1,5},Y−X= {2,6,9,10}

X U Y = {1,2,3,5,6,8,9,10}
X ∩ Y = {1,5}
Y - X = {2,6,9,10}

Y  = (X ∩ Y) U (Y - X)
   = {1,5} U {2,6,9,10}
   = {1, 2, 5, 6, 9, 10} (ans.)

X = (X U Y) \ (Y - X)
  = {1,2,3,5,6,8,9,10} \ {2,6,9,10}
  = {1, 3, 5, 8}

\[ X = \{1, 3, 5, 8\},\] \[ \quad Y = \{1, 2, 5, 6, 9, 10\} \]

3. Convert the infix expression P = 12/(7-3)+2 to postfix expression and evaluate it.

Below is the table showing the conversion steps:

Sr. no.ExprsnStackPostfix
0(
1P(P
2=(P =
312(P = 12
4/(/P = 12
5((/ (P = 12
67(/ (P = 12 7
7(/ (-P = 12 7
83(/ (-P = 12 7 3
9)(/P = 12 7 3 –
10+(+P = 12 7 3 – /
112(+P = 12 7 3 – / 2
12)P = 12 7 3 – / 2 +

Evaluation Steps 😛 = 12 7 3 – / 2 +

✔ 12 → Push onto stack → Stack: \([12]\)

✔ 7 → Push onto stack → Stack: \([12, 7]\)

✔ 3 → Push onto stack → Stack: \([12, 7, 3]\)

✔ – → Pop 7 and 3, perform \( 7 – 3 = 4 \) → Push 4 → Stack: \([12, 4]\)

✔ / → Pop 12 and 4, perform \( 12 / 4 = 3 \) → Push 3 → Stack: \([3]\)

✔ 2 → Push onto stack → Stack: \([3, 2]\)

✔ + → Pop 3 and 2, perform \( 3 + 2 = 5 \) → Push 5 → Stack: \([5]\)

The evaluated value of the expression is: P = 5


🎥 Video Solution:
Infix to Postfix Without Stack


🎥 Video Solution:
Infix to Postfix algorithm.

4. Explain how encapsulation and inheritance are advantageous in Object Oriented Programming?

Advantages of Encapsulation:

Data Security: It hides internal details, preventing unauthorized access and misuse.
Easy Maintenance: Since data access is controlled, changes in the code don’t affect other parts.
Better Flexibility: Internal workings of a class can be modified without affecting how others interact with it.
Less Complexity: Users only need to know how to use the class, not its internal details.
Improved Debugging: Restricting access to data makes it easier to find and fix bugs.

Advantages of Inheritance:

Code Reusability: A new class can use the properties and methods of an existing class, reducing duplicate code.
Clear Structure: It helps in organizing code by creating a hierarchy of classes.
Easy Expansion: New functionality can be added to a class without modifying existing code.
Simplifies Management: Subclasses inherit properties from the parent class, making code more organized and efficient.

🎥 Video Solution: How encapsulation and inheritance are advantageous in OOP>

5. If you throw two unbiased dice (each with six sides) together, what is the proabability that sum of two upward faces will be 7? Explain your answer.

✅ Total Possible Outcomes

When two unbiased six-sided dice are thrown, the number of possible outcomes is calculated as:

\[ \text{Total outcomes} = 6 \times 6 = 36 \]

Each die can show one of six numbers (1 to 6), and there are two dice, so we multiply 6 by 6.

✅ Desired Outcomes

We want to find the probability that the sum of the numbers on the two upward faces is 7. The possible outcomes that result in a sum of 7 are:

\[ (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) \]

These are the only pairs that sum to 7. There are 6 such pairs.

✅ Probability Calculation

The probability of an event is calculated as the ratio of the number of favorable outcomes to the total number of outcomes:

\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \]

Substituting the values:

\[ \text{Probability} = \frac{6}{36} = \frac{1}{6} \]

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