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# Signals and Systems: Energy and Power Signals

## 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) = 𝑖^{2}(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) = 𝑣^{2}(𝑡) = 𝑖^{2}(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(𝜔_{0}𝑡 + 𝜑).

### Solution

Given signal is,

𝑥(𝑡) = 𝐴 sin(𝜔_{0}𝑡 + 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

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

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

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

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

- Related Questions & Answers
- Signals and Systems: Even and Odd Signals
- Signals and Systems: Periodic and Aperiodic Signals
- Signals and Systems: Multiplication of Signals
- Signals and Systems – Rayleigh’s Energy Theorem
- Signals and Systems: Addition and Subtraction of Signals
- Signals and Systems: Real and Complex Exponential Signals
- Signals and Systems: Amplitude Scaling of Signals
- Signals and Systems – Classification of Signals
- Signals and Systems – Energy Spectral Density (ESD) and Autocorrelation Function
- Signals and Systems – Time Scaling of Signals
- Signals and Systems: Classification of Systems
- Signals and Systems – Parseval’s Power Theorem
- Signals and Systems – Properties of Even and Odd Signals
- Signals and Systems: Causal, Non-Causal, and Anti-Causal Signals
- Signals and Systems – Fourier Transform of Periodic Signals