Why is the average power of an energy signal zero?

P = lim E/2T, and finite E divided by 2T → ∞ gives 0

Because the signal is zero everywhere

Why is the energy of a power signal infinite?

E = lim P·2T, and a non-zero P times 2T → ∞ diverges

Which statement is correct?

Every energy signal is a power signal

Every periodic signal is an energy signal

Power signals have finite energy

2074 Bhadra · 2+5 marks

Power of energy signal is zero

Short answer

An energy signal has finite total energy E (0<E<∞) and zero average power; example: e^(-at)u(t) with a>0 gives E=1/(2a), P=0. A power signal has finite nonzero power P and infinite energy; example: A cos(ω₀t) gives P=A²/2, E→∞. Thus power of an energy signal is zero and energy of a power signal is infinite.

Differentiate between energy signal and power signal. Show by giving suitable example that power of the energy signal is zero and energy of the power signal is infinite.

Point	Energy signal	Power signal
Energy	Finite, $0<E<\infty$	Infinite
Average power	Zero	Finite, $0<P<\infty$
Duration	Usually finite or decaying	Infinite duration
Typical type	Aperiodic, transient	Periodic or random
Examples	Pulse, $e^{-at}u(t)$	$A\cos\omega t$ , $u(t)$
Spectrum	Energy spectral density	Power spectral density

E = \lim_{T\to\infty}\int_{-T}^{T}|x(t)|^2dt, \qquad P = \lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}|x(t)|^2dt

Power of an energy signal is zero

Take $x(t) = e^{-at}u(t)$ , $a > 0$ .

E = \int_0^{\infty}e^{-2at}dt = \frac{1}{2a} \quad\text{(finite)}

P = \lim_{T\to\infty}\frac{1}{2T}\int_0^{T}e^{-2at}dt = \lim_{T\to\infty}\frac{1-e^{-2aT}}{4aT} = 0

In general, $P = \lim_{T\to\infty} E/2T$ ; a finite $E$ divided by $2T \to \infty$ gives zero.

Energy of a power signal is infinite

Take $x(t) = A\cos\omega_0 t$ .

P = \frac{1}{T_0}\int_0^{T_0}A^2\cos^2\omega_0 t\,dt = \frac{A^2}{2} \quad\text{(finite)}

E = \lim_{T\to\infty}\int_{-T}^{T}A^2\cos^2\omega_0 t\,dt \approx \lim_{T\to\infty}\frac{A^2}{2}(2T) = \infty

In general, $E = \lim_{T\to\infty} P\cdot 2T$ ; a non-zero $P$ multiplied by $2T \to \infty$ gives infinity. Hence a signal cannot be both an energy and a power signal.

✓ Answer checked by Claude Opus 5.5 (high effort)

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