Which of the following statement/s are CORRECT ? (A) NRZ is a bipolar scheme…

Answer: The correct statements are the ones that assert that NRZ-L and NRZ-I have an average signal rate of N/2; that Manchester combines ideas of RZ and NRZ-L; and that NRZ-L and NRZ-I can have DC-component problems.

• Statement about NRZ being a bipolar scheme with positive voltage representing bit 0: Incorrect. NRZ is a level-based code and the polarity convention (which voltage means 0 or 1) is arbitrary. Bipolar AMI is a specific bipolar variant, but NRZ in general is not inherently bipolar.

• Statement that NRZ-L and NRZ-I both have an average signal rate of N/2: Correct. For random data, NRZ-L changes level when consecutive bits differ, which occurs about half the time on average, giving roughly N/2 transitions per second. NRZ-I produces a transition on each '1'; with random bits the probability of a '1' is 0.5, so transitions average N/2 as well.

• Statement that Manchester combines the idea of RZ and NRZ-L: Correct. Manchester encoding inserts a guaranteed mid-bit transition (providing timing similar to RZ) while conveying bit value via level/transition convention, effectively combining level information with a return-to-zero–style timing reference.

• Statement that NRZ-L and NRZ-I both have DC-component problems: Correct. Long runs of identical bits produce little low-frequency variation in both schemes, creating a DC component that can cause baseline wander and make clock recovery difficult.

• Statement that Manchester and differential Manchester require three times the minimum bandwidth of NRZ: Incorrect. Manchester-style encodings generally have higher bandwidth than NRZ (typically about twice the NRZ bandwidth) because they enforce transitions within each bit; they do not require three times the NRZ bandwidth.

Therefore the correct set is the one that contains the statements about average signal rate being N/2, the combination of RZ and NRZ-L in Manchester, and the DC-component problems of NRZ-L/NRZ-I.