Combined Bank
Post: Officer(it)
Exam Date: 04.10.2024, Exam Taker: BSC
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.

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}
\[ X = \{1, 3, 5, 8\},\] \[ \quad Y = \{1, 2, 5, 6, 9, 10\} \]
Below is the table showing the conversion steps:
| Sr. no. | Exprsn | Stack | Postfix |
|---|---|---|---|
| 0 | ( | ||
| 1 | P | ( | P |
| 2 | = | ( | P = |
| 3 | 12 | ( | P = 12 |
| 4 | / | (/ | P = 12 |
| 5 | ( | (/ ( | P = 12 |
| 6 | 7 | (/ ( | P = 12 7 |
| 7 | – | (/ (- | P = 12 7 |
| 8 | 3 | (/ (- | P = 12 7 3 |
| 9 | ) | (/ | P = 12 7 3 – |
| 10 | + | (+ | P = 12 7 3 – / |
| 11 | 2 | (+ | 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.
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>
✅ 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:
\[ \text{Probability} = \frac{6}{36} = \frac{1}{6} \]
