What is the energy and power of signal x(t) = sint

What is the energy and power of signal x(t) = sint

The correct answer and explanation is:

To determine the energy and power of the signal x(t)=sin⁡(t)x(t) = \sin(t), we must use standard definitions from signal analysis.

1. Energy of the Signal

The energy EE of a signal x(t)x(t) is given by: E=∫−∞∞∣x(t)∣2dtE = \int_{-\infty}^{\infty} |x(t)|^2 dt

For x(t)=sin⁡(t)x(t) = \sin(t): E=∫−∞∞sin⁡2(t) dtE = \int_{-\infty}^{\infty} \sin^2(t) \, dt

But this integral diverges because sin⁡2(t)\sin^2(t) oscillates and does not decay over time. Since: sin⁡2(t)=1−cos⁡(2t)2\sin^2(t) = \frac{1 – \cos(2t)}{2}

Integrating this over all time: ∫−∞∞1−cos⁡(2t)2dt=∞\int_{-\infty}^{\infty} \frac{1 – \cos(2t)}{2} dt = \infty

So, the energy of sin⁡(t)\sin(t) is infinite, meaning it is not an energy signal.

2. Power of the Signal

The average power PP of a signal is: P=lim⁡T→∞12T∫−TT∣x(t)∣2dtP = \lim_{T \to \infty} \frac{1}{2T} \int_{-T}^{T} |x(t)|^2 dt

For x(t)=sin⁡(t)x(t) = \sin(t): P=lim⁡T→∞12T∫−TTsin⁡2(t) dtP = \lim_{T \to \infty} \frac{1}{2T} \int_{-T}^{T} \sin^2(t) \, dt

Using the identity sin⁡2(t)=1−cos⁡(2t)2\sin^2(t) = \frac{1 – \cos(2t)}{2}, we get: P=lim⁡T→∞12T∫−TT1−cos⁡(2t)2dt=lim⁡T→∞12T⋅[12⋅∫−TT(1−cos⁡(2t))dt]P = \lim_{T \to \infty} \frac{1}{2T} \int_{-T}^{T} \frac{1 – \cos(2t)}{2} dt = \lim_{T \to \infty} \frac{1}{2T} \cdot \left[ \frac{1}{2} \cdot \int_{-T}^{T} (1 – \cos(2t)) dt \right] =12T⋅[12⋅(2T−0)]=12= \frac{1}{2T} \cdot \left[ \frac{1}{2} \cdot (2T – 0) \right] = \frac{1}{2}

Thus, the average power is 0.5\boxed{0.5}.

Conclusion

• The signal x(t)=sin⁡(t)x(t) = \sin(t) has infinite energy, so it is not an energy signal.
• It has finite power, with an average power of 0.5\boxed{0.5}, so it is a power signal.

This classification is common for periodic signals like sine and cosine waves. Energy signals are usually non-periodic and decay over time (e.g., pulses), whereas periodic signals like sin⁡(t)\sin(t) repeat indefinitely and carry constant average power.