In digital electronics, a periodic signal (waveform) can be gated—i.e., enabled so it passes through, or disabled so it is blocked—by using either an AND gate with a suitably chosen enable input or an OR gate with the appropriate active-low/active-high control. Which statement about such waveform gating is accurate?

Introduction / Context:
In digital electronics, waveform gating is the practice of allowing or blocking a periodic signal using a logic control input. AND and OR gates are both used for this purpose, depending on whether the control is active-high or active-low.

Given Data / Assumptions:

• A periodic input signal (the waveform) must be either enabled (passed) or disabled (blocked).
• We have access to AND and OR gates and, if needed, the inverted form of the control signal.
• Active-high means control = 1 enables; active-low means control = 0 enables.

Concept / Approach:
Use an AND gate for active-high enabling (signal AND enable). Use an OR gate for active-low enabling by ORing the signal with the complement of the disable control, or equivalently by ORing the inverted waveform with an active-high control before reinverting. The choice depends on available polarities and required logic levels.

Step-by-Step Solution:
1) AND gating (active-high enable E): output = signal * E.2) OR gating (active-low enable En): output = signal OR 0 when En = 0; to block, force output to 1 or hold the line as needed with proper polarity planning.3) With appropriate inversion, both gates can implement enable or disable behavior.

Verification / Alternative check:
Truth tables for two-input AND/OR confirm these behaviors when one input is the control or its complement.

Why Other Options Are Wrong:
“Only AND…” and “Only OR…” ignore polarity transformations; “Incorrect” contradicts standard gating practice using either gate with proper inversion.

Common Pitfalls:
Forgetting control polarity; neglecting that an inverter may be needed; confusing blocking a waveform with forcing a constant level.

Final Answer:
Correct