Energy and Power Signals

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Energy Signal

A signal is said to be an energy signal if and only if its total energy E is finite, i.e., 0 < E < ∞. For an energy signal, the average power P = 0. The nonperiodic signals are the examples of energy signals.

Power Signal

A signal is said to be a power signal if its average power P is finite, i.e., 0 < P < ∞. For a power signal, the total energy E = ∞. The periodic signals are the examples of power signals.

Continuous Time Case

In electric circuits, the signals may represent current or voltage. Consider a voltage v(t) applied across a resistance R and i(t) is the current flowing through it as shown in the figure.

The instantaneous power in the resistance R is given by,

$$\mathrm{p(t) \:=\: v(t) \: \cdot \: i(t) \:\: \dotso \: (1)}$$

By Ohm's law,

$$\mathrm{p(t) \:=\: v(t)\frac{v(t)}{R} \:=\: \frac{v^{2}(t)}{R}\:\: \dotso \: (2)}$$

Also,

$$\mathrm{p(t) \:=\: i(t)R \: \cdot \: i(t) \:=\: i^2 (t)R \:\: \dotso \: (3)}$$

When the values of the resistance R = 1Ω, then the power dissipated in it is known as normalised power. Hence,

$$\mathrm{\text{Normalised power, }\: p(t) \:=\: v^2 (t) \:=\: i^2 (t) \:\: \dotso \: (4)}$$

If v(t) or i(t) is denoted by a continuous-time signal x(t), then the instantaneous power is equal to the square of the amplitude of the signal, i.e.,

$$\mathrm{p(t) \:=\: |x(t)|^2 \:\:\: \dotso \: (5)}$$

Therefore, the average power or normalised power of a continuous time signal x(t) is given by,

$$\mathrm{P\:=\:\lim_{T\rightarrow \infty}\frac{1}{T}\int_{-(T/2)}^{(T/2)}\left|x(t) \right|^{2}\: dt\:\:Watts\:\: \dotso \: (6)}$$

The total energy or normalised energy of a continuous time signal is defined as,

$$\mathrm{E\:=\:\lim_{T\rightarrow \infty }\int_{-(T/2)}^{(T/2)}\left|x(t)\right |^{2}\:dt\:\: Joules \:\:\dotso\: (7)}$$

Discrete Time Case

For the discrete time signal x(n), the integrals are replaced by summations. Hence, the total energy of the discrete time signal x(n) is defined as

$$\mathrm{E\:=\:\sum_{n\:=\:-\infty }^{\infty }\left | x(t) \right |^{2}}$$

The average power of a discrete time signal x(t) is defined as

$$\mathrm{P\:=\:\lim_{N\rightarrow \infty }\frac{1}{2N\:+\:1}\sum_{n\:=\:-N}^{N}\left | x(t) \right |^{2}}$$

Important Points

• Both energy and power signals are mutually exclusive, i.e., no signal can be both power signal and energy signal.
• A signal is neither energy nor power signal if both energy and power of the signal are equal to infinity.
• All practical signals have finite energy; thus they are energy signals.
• In practice, the physical generation of power signal is impossible since its requires infinite duration and infinite energy.
• All finite duration signals of finite amplitude are energy signals.
• Sum of an energy signal and power signal is a power signal.
• A signal whose amplitude is constant over infinite duration is a power signal.
• The energy of a signal is not affected by the time shifting and time inversion. It is only affected by the time scaling.

Numerical Example

Determine the power and energy of the signal x(t) = A sin(ω₀t + &varphi;).

Solution

Given signal is,

$$\mathrm{x(t) \:=\: A \sin(\omega_0 t \:+\: \varphi)}$$

Average Power of the Signal

$$\mathrm{P \:=\: \lim_{T\rightarrow \infty }\frac{1}{T}\int_{-(T/2)}^{(T/2)}\left | x(t) \right |^{2}\: dt}$$

$$\mathrm{\Rightarrow\: P\:=\:\lim_{T\rightarrow \infty }\frac{1}{T}\int_{-(T/2)}^{(T/2)}\left | A\: \sin (\omega _{0}t\:+\:\varphi ) \right |^{2}\: dt}$$

$$\mathrm{\Rightarrow\: P\:=\:\lim_{T\rightarrow \infty }\frac{A^{2}}{T}\int_{-(T/2)}^{(T/2)}\left |\frac{1\:-\:\cos (2\omega _{0}t\:+\:2\varphi )}{2} \right |\: dt}$$

$$\mathrm{\Rightarrow\: P\:=\:\lim_{T\rightarrow \infty }\frac{A^{2}}{2T}\int_{-(T/2)}^{(T/2)}\, dt \:-\: \frac{A^{2}}{2T}\int_{-(T/2)}^{(T/2)}\cos (2\omega _{0}t+2\varphi ) \: dt}$$

$$\mathrm{\Rightarrow \: P \:=\:\lim_{T\rightarrow \infty }\frac{A^{2}}{2T}\int_{-(T/2)}^{(T/2)}\: dt\:-\:0 \:=\: \lim_{T\rightarrow \infty }\frac{A^{2}}{2T}\left [ \frac{T}{2}\:+\:\frac{T}{2} \right ]\:=\:\frac{A^{2}}{2} }$$

Normalised Energy of the Signal

$$\mathrm{E\:=\:\int_{-\infty }^{\infty }\left | x(t)\right |^{2}\: dt\:=\:\int_{-\infty }^{\infty } \left | A\: \sin (\omega _{0}t\:+\:\varphi ) \right |^{2}\: dt}$$

$$\mathrm{\Rightarrow\: E\:=\:A^{2}\int_{-\infty }^{\infty }\left [ \frac{1\:-\:\cos (2\omega _{0}t\:+\:2\varphi )}{2} \right ]\: dt}$$

$$\mathrm{\Rightarrow\: E\:=\:\frac{A^{2}}{2}\int_{-\infty }^{\infty }dt\:-\:\frac{A^{2}}{2}\int_{-\infty }^{\infty }\cos (2\omega _{0}t\:+\:2\varphi )\: dt}$$

$$\mathrm{\Rightarrow\: E\:=\:\frac{A^{2}}{2}\left [ t \right ]_{-\infty }^{\infty }\:-\:0\:=\:\infty }$$