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This section includes 16 Mcqs, each offering curated multiple-choice questions to sharpen your Digital Signal Processing knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The condition with less number of samples L should be avoided. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 2. |
Which of the following is the disadvantage of Hanning window over rectangular window? |
| A. | More side lobes |
| B. | Less side lobes |
| C. | More width of main lobe |
| D. | None of the mentioned |
| Answer» D. None of the mentioned | |
| 3. |
The characteristic of windowing the signal called “Leakage” is the power that is leaked out into the entire frequency range. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 4. |
If x(n)=cosω0n and W(ω) is the Fourier transform of the rectangular signal w(n), then what is the Fourier transform of the signal x(n).w(n)? |
| A. | 1/2[W(ω-ω0)- W(ω+ω0)] |
| B. | 1/2[W(ω-ω0)+ W(ω+ω0)] |
| C. | [W(ω-ω0)+ W(ω+ω0)] |
| D. | [W(ω-ω0)- W(ω+ω0)] |
| Answer» C. [W(ω-ω0)+ W(ω+ω0)] | |
| 5. |
What is the Fourier transform of rectangular window of length L? |
| A. | \(\frac{sin(\frac{ωL}{2})}{sin(\frac{ω}{2})} e^{jω(L+1)/2}\) |
| B. | \(\frac{sin(\frac{ωL}{2})}{sin(\frac{ω}{2})} e^{jω(L-1)/2}\) |
| C. | \(\frac{sin(\frac{ωL}{2})}{sin(\frac{ω}{2})} e^{-jω(L-1)/2}\) |
| D. | None of the mentioned |
| Answer» D. None of the mentioned | |
| 6. |
If {x(n)} is the signal to be analyzed, limiting the duration of the sequence to L samples, in the interval 0≤ n≤ L-1, is equivalent to multiplying {x(n)} by? |
| A. | Kaiser window |
| B. | Hamming window |
| C. | Hanning window |
| D. | Rectangular window |
| Answer» E. | |
| 7. |
If the signal to be analyzed is an analog signal, we would pass it through an anti-aliasing filter with B as the bandwidth of the filtered signal and then the signal is sampled at a rate __________ |
| A. | Fs ≤ 2B |
| B. | Fs ≤ B |
| C. | Fs ≥ 2B |
| D. | Fs = 2B |
| Answer» D. Fs = 2B | |
| 8. |
THE_CONDITION_WITH_LESS_NUMBER_OF_SAMPLES_L_SHOULD_BE_AVOIDED.?$ |
| A. | True |
| B. | False |
| Answer» B. False | |
| 9. |
WHICH_OF_THE_FOLLOWING_IS_THE_DISADVANTAGE_OF_HANNING_WINDOW_OVER_RECTANGULAR_WINDOW??$ |
| A. | More side lobes |
| B. | Less side lobes |
| C. | More width of main lobe |
| D. | None of the mentioned |
| Answer» D. None of the mentioned | |
| 10. |
Which of the following is the advantage of Hanning window over rectangular window? |
| A. | More side lobes |
| B. | Less side lobes |
| C. | More width of main lobe |
| D. | None of the mentioned |
| Answer» C. More width of main lobe | |
| 11. |
The characteristic of windowing the signal called “Leakage” is the power that is leaked out into the entire frequency range.$ |
| A. | True |
| B. | False |
| Answer» B. False | |
| 12. |
If x(n)=cosω0n and W(ω) is the Fourier transform of the rectangular signal w(n), then what is the Fourier transform of the signal x(n).w(n)?$ |
| A. | 1/2[W(ω-ω<sub>0</sub>)- W(ω+ω<sub>0</sub>)]. |
| B. | 1/2[W(ω-ω<sub>0</sub>)+ W(ω+ω<sub>0</sub>)]. |
| C. | [W(ω-ω<sub>0</sub>)+ W(ω+ω<sub>0</sub>)]. |
| D. | [W(ω-ω<sub>0</sub>)- W(ω+ω<sub>0</sub>)]. |
| Answer» C. [W(‚âà√¨‚àö¬¢-‚âà√¨‚àö¬¢<sub>0</sub>)+ W(‚âà√¨‚àö¬¢+‚âà√¨‚àö¬¢<sub>0</sub>)]. | |
| 13. |
If {x(n)} is the signal to be analyzed, limiting the duration of the sequence to L samples, in the interval 0‚â§ n‚â§ L-1, is equivalent to multiplying {x(n)} by:$ |
| A. | Kaiser window |
| B. | Hamming window |
| C. | Hanning window |
| D. | Rectangular window |
| Answer» E. | |
| 14. |
The finite observation interval for the signal places a limit on the frequency resolution. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 15. |
What is the highest frequency that is contained in the sampled signal? |
| A. | 2Fs |
| B. | Fs/2 |
| C. | Fs |
| D. | None of the mentioned |
| Answer» C. Fs | |
| 16. |
If the signal to be analyzed is an analog signal, we would pass it through an anti-aliasing filter with B as the bandwidth of the filtered signal and then the signal is sampled at a rate: |
| A. | Fs ‚â§ 2B |
| B. | Fs ‚â§ B |
| C. | Fs ‚â• 2B |
| D. | Fs = 2B |
| Answer» D. Fs = 2B | |