Signal classification: If g(t) = A (a non-zero constant) for all t, is g(t) an energy signal, a power signal, or neither?
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AEnergy signal
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BPower signal
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CNeither energy nor power signal
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DInsufficient data
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EEnergy signal only if A = 0
Answer
Correct Answer: Power signal
Explanation
Introduction / Context:Signals are commonly classified as energy or power signals based on the finiteness of their total energy and average power. Constant signals are a classic test case that clarifies the definitions.
Given Data / Assumptions:
- g(t) = A, with A ≠ 0 constant for all t.
- Energy: E = ∫{−∞}^{∞} |g(t)|^2 dt.
- Average power: P = lim{T→∞} (1/(2T)) ∫{−T}^{T} |g(t)|^2 dt.
Concept / Approach:
An energy signal has finite E and zero power. A power signal has finite, non-zero P and infinite energy (over infinite duration). A nonzero constant over infinite time accumulates infinite energy but has finite average power equal to A^2.
Step-by-Step Solution:
Compute energy: E = ∫ |A|^2 dt from −∞ to ∞ = ∞ (diverges).Compute average power: P = lim{T→∞} (1/(2T)) ∫{−T}^{T} |A|^2 dt = lim{T→∞} (1/(2T)) (2T |A|^2) = |A|^2.Hence, g(t) is a power signal (finite, non-zero average power; infinite total energy).Verification / Alternative check:
For periodic or constant signals, power is finite and equal to the time average of |g(t)|^2. Here that equals A^2, consistent with power-signal classification.
Why Other Options Are Wrong:
- Energy signal requires finite energy; here energy diverges.
- “Neither” is incorrect because average power is finite and well-defined.
- “Insufficient data” is incorrect; information given is adequate.
Common Pitfalls:
- Assuming finite amplitude implies finite energy even over infinite duration.
- Confusing definitions of energy and power criteria.
Final Answer:
Power signal