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Energy of a Power Signal over Infinite Time
What is a Power Signal?
A signal is said to be a power signal if its average power (P) is finite, i.e., 0 < 𝑃 < ∞. The total energy of a power signal is infinity over infinite time, i.e., 𝐸 = ∞. The periodic signals are the examples of power signals.
Energy of a Power Signal
Consider a continuous-time power signal x(t). The power of the signal x(t) is finite and is given by,
$$\mathrm{P=\lim_{T\rightarrow \infty }\frac{1}{2T}\int_{-T }^{T }x^{2}(t)dt\; \; ...(1)}$$
Therefore, the energy of the signal is given by,
$$\mathrm{E=\lim_{T\rightarrow \infty }\int_{-T }^{T }x^{2}(t)dt}$$ $$\mathrm{\Rightarrow E=\lim_{T\rightarrow \infty }\left [2T\cdot \frac{1}{2T}\int_{-T }^{T }x^{2}(t)dt \right ]}$$ $$\mathrm{\Rightarrow E=\lim_{T\rightarrow \infty }2T\left [\lim_{T\rightarrow \infty } \frac{1}{2T}\int_{-T }^{T }x^{2}(t)dt \right ]\; \; ...(2)}$$
Using equations (1) and (2), we get,
$$\mathrm{\Rightarrow E=\lim_{T\rightarrow \infty }2T\cdot P=\infty}$$
Therefore, the energy of the power signal is infinite over infinite time.
Numerical Example
Determine whether the signal 𝑥(𝑡) = sin2 𝜔𝑡 is a power signal or not. If it is, then calculate the power and energy of the signal.
Solution
Given signal is,
𝑥(𝑡) = sin2 𝜔𝑡
As the given signal x(t) is a squared sine wave, it is a periodic signal, so it can be a power signal.
The average power of the signal −
$$\mathrm{P=\lim_{T\rightarrow \infty }\frac{1}{2T}\int_{-T }^{T }x^{2}(t)dt}$$ $$\mathrm{\Rightarrow P=\lim_{T\rightarrow \infty }\frac{1}{2T}\int_{-T }^{T }\left [\sin ^{2}\omega t \right ]^{2}dt=\lim_{T\rightarrow \infty }\frac{1}{2T}\int_{-T }^{T }\sin ^{4}\omega t\; dt}$$
By the definition of standard trigonometric relation, we get,
$$\mathrm{\sin ^{4}\omega t=\frac{1}{8}(3-4\cos 2\omega t+\cos 4\omega t)}$$ $$\mathrm{\therefore P=\lim_{T\rightarrow \infty }\frac{1}{2T}\int_{-T}^{T}\frac{1}{8}(3-4\cos 2\omega t+\cos 4\omega t)dt}$$ $$\mathrm{\Rightarrow P=\lim_{T\rightarrow \infty }\frac{1}{2T}\int_{-T}^{T}\frac{3}{8}\: dt-\lim_{T\rightarrow \infty }\frac{1}{2T}\int_{-T}^{T}\frac{4}{8}cos 2\omega t\: dt+\lim_{T\rightarrow \infty }\frac{1}{2T}\int_{-T}^{T}\frac{1}{8}\cos 4\omega t\: dt}$$ $$\mathrm{\Rightarrow P=\lim_{T\rightarrow \infty }\frac{1}{2T}\left ( \frac{3}{8} \right )\left [ t \right ]_{-T}^{T}-0+0}$$ $$\mathrm{\Rightarrow P=\lim_{T\rightarrow \infty }\frac{1}{2T}\left ( \frac{3}{8} \right )\left [ T+T \right ]=\frac{3}{8}}$$
Therefore, the power of the given signal is finite and is equal to 𝑃 = 3⁄8 watts.
Now, the energy of the signal −
$$\mathrm{E=\lim_{T\rightarrow \infty }\int_{-T }^{T }x^{2}(t)dt=\lim_{T\rightarrow \infty }\int_{-T }^{T }\left [\sin ^{2}\omega t \right ]^{2}dt}$$ $$\mathrm{\Rightarrow E=\lim_{T\rightarrow \infty }\int_{-T}^{T}\frac{1}{8}(3-4\cos 2\omega t+\cos 4\omega t)dt=\lim_{T\rightarrow \infty }\left ( \frac{3}{8} \right )\left [ t \right ]_{-T}^{T}}$$ $$\mathrm{\Rightarrow E=\lim_{T\rightarrow \infty }\left ( \frac{3}{8} \right )\left [2T \right ]=\infty }$$
Thus, the energy of the given power signal is infinity over infinite time.
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