MCQOPTIONS
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MCQOPTIONS
This section includes 7 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 anti-symmetric condition with M even is not used in the design of which of the following linear-phase FIR filter? |
| A. | Low pass |
| B. | High pass |
| C. | Band pass |
| D. | Bans stop |
| Answer» B. High pass | |
| 2. |
Which of the following is not suitable either as low pass or a high pass filter? |
| A. | h(n) symmetric and M odd |
| B. | h(n) symmetric and M even |
| C. | h(n) anti-symmetric and M odd |
| D. | h(n) anti-symmetric and M even |
| Answer» D. h(n) anti-symmetric and M even | |
| 3. |
What is the number of filter coefficients that specify the frequency response for h(n) anti-symmetric? |
| A. | (M-1)/2 when M is even and M/2 when M is odd |
| B. | (M-1)/2 when M is odd and M/2 when M is even |
| C. | (M+1)/2 when M is even and M/2 when M is odd |
| D. | (M+1)/2 when M is odd and M/2 when M is even |
| Answer» C. (M+1)/2 when M is even and M/2 when M is odd | |
| 4. |
What is the number of filter coefficients that specify the frequency response for h(n) symmetric? |
| A. | (M-1)/2 when M is odd and M/2 when M is even |
| B. | (M-1)/2 when M is even and M/2 when M is odd |
| C. | (M+1)/2 when M is even and M/2 when M is odd |
| D. | (M+1)/2 when M is odd and M/2 when M is even |
| Answer» E. | |
| 5. |
The roots of the equation H(z) must occur in ________________ |
| A. | Identical |
| B. | Zero |
| C. | Reciprocal pairs |
| D. | Conjugate pairs |
| Answer» D. Conjugate pairs | |
| 6. |
The roots of the polynomial H(z) are identical to the roots of the polynomial H(z-1). |
| A. | True |
| B. | False |
| Answer» B. False | |
| 7. |
If H(z) is the z-transform of the impulse response of an FIR filter, then which of the following relation is true? |
| A. | z<sup>M+1</sup>.H(z<sup>-1</sup>)=&pm;H(z) |
| B. | z<sup>-(M+1)</sup>.H(z<sup>-1</sup>)=&pm;H(z) |
| C. | z<sup>(M-1)</sup>.H(z<sup>-1</sup>)=&pm;H(z) |
| D. | z<sup>-(M-1)</sup>.H(z<sup>-1</sup>)=&pm;H(z) |
| Answer» E. | |