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Power of an Energy Signal over Infinite Time
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.
Solution
Given signal is,
$$\mathrm{x(t)=rect(\frac{t}{\tau })}$$
The given signal x(t) is a rectangular function which is defined as
$$\mathrm{rect(\frac{t}{\tau })=\left\{\begin{matrix} 1\; \; for \left ( -\frac{\tau }{2} \right )<t<\left ( \frac{\tau}{2} \right )\ 0\; \;\;\; otherwise\ \end{matrix}\right.}$$
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.