Energy of a Power Signal over Infinite Time

Electronics & Electrical Electron Digital Electronics

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.

Manish Kumar Saini

Published on 12-Nov-2021 11:31:02

Previous Page Print Page