Energy of a Power Signal over Infinite Time

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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,

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

Therefore, the energy of the signal is given by,

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

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

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

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,

$$\mathrm{x(t) \:=\: \sin^2 \omega t}$$

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.