The signal x(t) = e^(−2t) · u(t) (a decaying exponential for t ≥ 0, zero before). How should it be classified?
APower signal — it has a well-defined nonzero average value over its active region
BEnergy signal — it decays to zero, so the integral of |x(t)|² converges to a finite number
CNeither — signals only defined for t ≥ 0 cannot be classified
DBoth — it has finite energy and also non-zero instantaneous power
A decaying exponential goes to zero as t → ∞, so the integral of |x(t)|² from −∞ to ∞ converges to a finite value — finite energy. Since energy is finite, average power P = E / ∞ = 0. This makes it an energy signal. Options A and D represent common confusions: a signal that decays to zero cannot sustain a nonzero long-run average power, and the two classes are mutually exclusive — finite E forces P = 0, so a signal cannot be both.
Question 2 Multiple Choice
A student computes that a signal has infinite total energy (E = ∞). They conclude it must therefore be a power signal. Is this reasoning correct?
AYes — any signal with infinite energy must have finite nonzero average power
BNo — a signal can have infinite energy and also infinite average power, belonging to neither class
CYes — infinite energy means the signal persists, which always produces a stable average power
DNo — signals with infinite energy are always periodic and should be treated as power signals
The student's reasoning is incomplete. While it is true that power signals have infinite energy, not all signals with infinite energy are power signals. A signal like x(t) = t (linearly growing) has both infinite total energy and infinite average power — it belongs to neither class. To confirm a power signal, you must explicitly compute the average power P = lim(T→∞) (1/2T) ∫|x(t)|² dt and verify it converges to a finite, nonzero value.
Question 3 True / False
A signal that is bounded — meaning |x(t)| ≤ M for most t and some finite M — is expected to have finite total energy.
TTrue
FFalse
Answer: False
This is a critical trap. A sine wave sin(2πft) is bounded (never exceeds 1), but its total energy is infinite because it persists over all time. Energy requires both bounded amplitude AND that the signal decays fast enough for the integral of |x(t)|² to converge. Boundedness controls amplitude; it says nothing about duration. A bounded signal that never decays to zero will have infinite energy and is a power signal, not an energy signal.
Question 4 True / False
An energy signal and a power signal are mutually exclusive classifications — no signal can belong to both categories simultaneously.
TTrue
FFalse
Answer: True
By definition, an energy signal has finite total energy E, which forces P = lim(1/2T)∫|x|² dt → 0 as T → ∞ (energy spread over ever-growing interval averages to zero). A power signal requires finite nonzero P, which forces E = ∞. These conditions are contradictory: a signal cannot simultaneously have finite E and finite nonzero P. (A third class exists — signals where both E and P are infinite — but no signal can satisfy both the energy-signal condition and the power-signal condition.)
Question 5 Short Answer
Why does classifying a signal as an energy signal or a power signal matter for choosing analysis tools?
Think about your answer, then reveal below.
Model answer: The classification determines which mathematical framework applies. Energy signals fit naturally into the Fourier transform framework — the transform exists in a classical sense when energy is finite (the signal's spectrum is well-defined via direct integration). Power signals require the power spectral density framework using autocorrelation and limiting arguments, because their Fourier transforms don't converge in the classical sense. Using the wrong tool — trying to take the direct Fourier transform of a sine wave — produces distributional objects (Dirac deltas) that require careful handling.
This is why classification is not just a definitional exercise — it is a prerequisite for every downstream technique. Filtering, correlation, and spectral estimation all depend on knowing which class of signal you're working with. Getting the classification wrong leads to using a tool on a signal for which it was not designed, producing either mathematical errors or results that must be interpreted with significant care.