For x(t) = 2e^(-5t)u(t-1), what is the energy integral used?

integral from 1 to infinity of 4e^(-10t) dt

integral from 0 to infinity of 2e^(-5t) dt

integral from 1 to infinity of 2e^(-10t) dt

integral from -infinity to infinity of 4e^(-5t) dt

What is the average power and why?

0, since finite energy is divided by an infinite time 2T

Equal to the energy, since E is small

Infinite, since the signal never ends

0.4, the coefficient of the integral

How is the signal classified?

Neither energy nor power signal

2082 Bhadra (new course) · 4 marks

Energy and power of 2e^(−5t)u(t−1)

Short answer

Energy is computed by integrating |x(t)|^2 = 4e^(-10t) from t=1 to ∞: E = 0.4e^(-10) ≈1.816×10^(-5) J. Power is the limit of average power over infinite interval, P = 0. Finite energy and zero power make it an energy signal.

Calculate the energy and power of the signal x(t) = 2e^(−5t)u(t−1).

The signal $x(t) = 2e^{-5t}u(t-1)$ is zero for $t < 1$ and decays exponentially for $t \ge 1$ .

Energy:

\begin{aligned} E &= \int_{-\infty}^{\infty}|x(t)|^2\,dt = \int_{1}^{\infty}4e^{-10t}\,dt \\ &= 4\left[\frac{e^{-10t}}{-10}\right]_1^{\infty} = 4\left(0 + \frac{e^{-10}}{10}\right) \\ &= 0.4\,e^{-10} = 1.816\times 10^{-5}\ \text{J} \end{aligned}

Power:

P = \lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}|x(t)|^2dt = \lim_{T\to\infty}\frac{1}{2T}\cdot 0.4\left(e^{-10} - e^{-10T}\right) = 0

Answer: $E = 0.4e^{-10} \approx 1.816\times10^{-5}$ J and $P = 0$ . The energy is finite, so $x(t)$ is an energy signal.

✓ Answer checked by Claude Opus 5.5 (high effort)

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