MCQOPTIONS
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MCQOPTIONS
This section includes 13 Mcqs, each offering curated multiple-choice questions to sharpen your Network Theory knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Find the function f (t) in terms of unit step function in the graph shown below. |
| A. | 4t [u (t) – u (t + 5)] |
| B. | 4t [u (t) + u (t + 5)] |
| C. | 4t [u (t) – u (t – 5)] |
| D. | 4t [u (t) + u (t – 5)] |
| Answer» D. 4t [u (t) + u (t – 5)] | |
| 2. |
In the graph shown below, find the expression f (t). |
| A. | 2t |
| B. | 3t |
| C. | 4t |
| D. | 5t |
| Answer» D. 5t | |
| 3. |
Find the expression of f (t) in the graph shown below. |
| A. | 10t [u (t) – u (t – 1)] – (-10t+20) [u (t-1) – u (t-3)] + (20t – 40) [u (t-3) – u (t-4)] |
| B. | 10t [u (t) – u (t – 1)] – (-10t+20) [u (t-1) – u (t-3)] – (20t – 40) [u (t-3) – u (t-4)] |
| C. | 10t [u (t) – u (t – 1)] + (-10t+20) [u (t-1) – u (t-3)] + (20t – 40) [u (t-3) – u (t-4)] |
| D. | 10t [u (t) – u (t – 1)] + (-10t+20) [u (t-1) – u (t-3)] – (20t – 40) [u (t-3) – u (t-4)] |
| Answer» D. 10t [u (t) – u (t – 1)] + (-10t+20) [u (t-1) – u (t-3)] – (20t – 40) [u (t-3) – u (t-4)] | |
| 4. |
Find the function f3 (t) from the time t = 3 to 4 sec. |
| A. | (20t – 40) [u (t-3) – u (t-4)] |
| B. | (20t + 40) [u (t-3) – u (t-4)] |
| C. | (20t + 40) [u (t-3) + u (t-4)] |
| D. | (20t – 40) [u (t-3) + u (t-4)] |
| Answer» B. (20t + 40) [u (t-3) – u (t-4)] | |
| 5. |
Find the function f2 (t) from the time t = 1 to 3 sec. |
| A. | (-10t+20) [u (t-1) +u (t-3)] |
| B. | (-10t+20) [u (t-1) – u (t-3)] |
| C. | (-10t-20) [u (t-1) + u (t-3)] |
| D. | (-10t-20) [u (t-1) – u (t-3)] |
| Answer» C. (-10t-20) [u (t-1) + u (t-3)] | |
| 6. |
The total period of the function shown in the figure is 4 sec and the amplitude is 10. Find the function f1 (t) from t = 0 to 1 in terms of unit step function. |
| A. | 10t [u (t) – u (t + 1)] |
| B. | 10t [u (t) + u (t – 1)] |
| C. | 10t [u (t) + u (t + 1)] |
| D. | 10t [u (t) – u (t – 1)] |
| Answer» E. | |
| 7. |
The Laplace transform of a function f (t) is? |
| A. | \(\int_0^{\infty}\) f(t) e-st |
| B. | \(\int_{-\infty}^0\) f(t) e-st |
| C. | \(\int_0^{\infty}\) f(t) est |
| D. | \(\int_{-\infty}^0\) f(t) est |
| Answer» B. \(\int_{-\infty}^0\) f(t) e-st | |
| 8. |
FIND_THE_FUNCTION_F_(T)_IN_TERMS_OF_UNIT_STEP_FUNCTION_IN_THE_GRAPH_SHOWN_IN_QUESTION_9.?$ |
| A. | 4t [u (t) – u (t + 5)]. |
| B. | 4t [u (t) + u (t + 5)]. |
| C. | 4t [u (t) – u (t – 5)]. |
| D. | 4t [u (t) + u (t – 5)]. |
| Answer» D. 4t [u (t) + u (t ‚Äö√Ñ√∂‚àö√ë‚àö¬® 5)]. | |
| 9. |
Find the expression of f (t) in the graph shown in question 5? |
| A. | 10t [u (t) – u (t – 1)] – (-10t+20) [u (t-1) – u (t-3)] + (20t – 40) [u (t-3) – u (t-4)]. |
| B. | 10t [u (t) – u (t – 1)] – (-10t+20) [u (t-1) – u (t-3)] – (20t – 40) [u (t-3) – u (t-4)]. |
| C. | 10t [u (t) – u (t – 1)] + (-10t+20) [u (t-1) – u (t-3)] + (20t – 40) [u (t-3) – u (t-4)]. |
| D. | 10t [u (t) – u (t – 1)] + (-10t+20) [u (t-1) – u (t-3)] – (20t – 40) [u (t-3) – u (t-4)]. |
| Answer» D. 10t [u (t) ‚Äö√Ñ√∂‚àö√ë‚àö¬® u (t ‚Äö√Ñ√∂‚àö√ë‚àö¬® 1)] + (-10t+20) [u (t-1) ‚Äö√Ñ√∂‚àö√ë‚àö¬® u (t-3)] ‚Äö√Ñ√∂‚àö√ë‚àö¬® (20t ‚Äö√Ñ√∂‚àö√ë‚àö¬® 40) [u (t-3) ‚Äö√Ñ√∂‚àö√ë‚àö¬® u (t-4)]. | |
| 10. |
Find the function f2 (t) from the time t = 1 to 3 sec. |
| A. | (-10t+20) [u (t-1) +u (t-3)]. |
| B. | (-10t+20) [u (t-1) – u (t-3)]. |
| C. | (-10t-20) [u (t-1) + u (t-3)]. |
| D. | (-10t-20) [u (t-1) – u (t-3)]. |
| Answer» C. (-10t-20) [u (t-1) + u (t-3)]. | |
| 11. |
The unit step is not defined at t =? |
| A. | 0 |
| B. | 1 |
| C. | 2 |
| D. | 3 |
| Answer» B. 1 | |
| 12. |
In the bilateral Laplace transform, the lower limit is? |
| A. | 0 |
| B. | 1 |
| C. | ‚àû |
| D. | – ∞ |
| Answer» E. | |
| 13. |
Laplace transform changes the ____ domain function to the _____ domain function. |
| A. | time, time |
| B. | time, frequency |
| C. | frequency, time |
| D. | frequency, frequency |
| Answer» C. frequency, time | |