# Signals and Systems: Energy and Power Signals

Electronics & ElectricalElectronDigital Electronics

## 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,

𝑝(𝑡) = 𝑣(𝑡) ∙ 𝑖(𝑡) … (1)

By Ohm’s law,

$$\mathrm{p(t)=v(t)\frac{v(t)}{R}=\frac{v^{2}(t)}{R}\, \, \, \, \cdot \cdot \cdot (2)}$$

Also,

𝑝(𝑡) = 𝑖(𝑡)𝑅 ∙ 𝑖(𝑡) = 𝑖2(𝑡)𝑅     … (3)

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

Normalised power, 𝑝(𝑡) = 𝑣2(𝑡) = 𝑖2(𝑡)     … (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.,

𝑝(𝑡) = |𝑥(𝑡)|2     … (5)

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

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

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

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

## 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 + 𝜑).

### Solution

Given signal is,

𝑥(𝑡) = 𝐴 sin(𝜔0𝑡 + 𝜑)

### 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

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

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

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

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

Published on 13-Nov-2021 10:24:47