Power of an Energy Signal over Infinite Time

Electronics & ElectricalElectronDigital Electronics

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.,

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

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

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

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

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

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.


Given signal is,

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

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

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

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.

Published on 12-Nov-2021 11:05:51