 
			 
			MCQOPTIONS
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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 | |