Power of an Energy Signal over Infinite Time

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What is an Energy Signal?

A signal is said to be an energy signal if and only if its total energy (E) is finite. That means 0 < E < ∞. The average power of an energy signal is zero over infinite time (i.e., P = 0). The non-periodic signals are examples of energy signals.

Power of an Energy Signal

Consider a continuous-time energy signal x(t). The energy of the signal x(t) is finite, i.e.,

$$\mathrm{E \:=\: \int_{-\infty }^{\infty }x^{2}(t)dt \:=\: finite\: \: \dotso \:(1)}$$

Hence, the power of the signal x(t) is,

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

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

$$\mathrm{\mathrm{\Rightarrow P \:=\: \lim_{T \:\rightarrow\: \infty} \frac{1}{2T}\int_{-\infty }^{\infty }x^{2}(t)dt}\: \: \dotso \:(2)}$$

From equations (1) and (2), we get,

$$\mathrm{P \:=\: \lim_{T \:\rightarrow\: \infty}\frac{1}{2T}[E]}$$

$$\mathrm{\because \: \lim_{T\:\rightarrow\: \infty}\frac{1}{2T} \:=\: \frac{1}{\infty} \:=\:0}$$

$$\mathrm{\therefore\: P\:=\:\lim_{T\:\rightarrow\: \infty}\frac{1}{2T}[E] \:=\: 0 \:\times\: E\:=\:0}$$

Therefore, the power of an energy signal is zero over infinite time.

Numerical Example

Determine whether the signal $\mathrm{x(t) \:=\:rect(\frac{t}{\tau })}$ is an energy signal or not. If it is, then calculate the energy and the power of the energy signal.

Solution

Given signal is,

$$\mathrm{x(t) \:=\: rect(\frac{t}{\tau })}$$

The given signal x(t) is a rectangular function which is defined as

$$\mathrm{rect(\frac{t}{\tau }) \:=\: \left\{\begin{matrix} 1\: \: for \left ( -\frac{\tau }{2} \right )\lt t \lt \left(\frac{\tau}{2} \right)\: 0\: \:\:\: \text{otherwise}\ \end{matrix}\right.}$$

The figure shows the graphical representation of the signal x(t). It is a nonperiodic signal, so it can be an energy signal.

$$\mathrm{E \:=\: \int_{-\infty }^{\infty }x^{2}(t)dt \:=\: \int_{(-\tau /2)}^{(\tau /2)}(1)^{2}dt}$$

$$\mathrm{\Rightarrow \: E \:=\: \frac{\tau}{2} \:+\: \frac{\tau}{2} \:=\: \tau}$$

Thus, the energy of the signal is finite and is E = τ joules.

Power of the signal −

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

$$\mathrm{\Rightarrow\: P\:=\:\lim_{T\:\rightarrow\: \infty }\frac{1}{2T}\left [ \tau \right ] \:=\: 0}$$

The power of the given signal is zero over infinite time. Hence, it is an energy signal.