Shannon-Hartley theorem; energy/power of e^(−2t)u(t)

Shannon–Hartley channel capacity theorem

The theorem gives the maximum error-free data rate (channel capacity) of a band-limited channel disturbed by additive white Gaussian noise (AWGN):

$$C = B\log_2\left(1 + \frac{S}{N}\right) \ \text{bit/s}$$

where $B$ = channel bandwidth (Hz), $S$ = average signal power, $N = N_0B$ = noise power in the band.

Meaning:

• If the information rate $R \le C$, there exists a coding scheme that makes the error probability as small as desired.
• If $R > C$, error-free transmission is impossible whatever coding is used.
• Bandwidth and SNR can be traded: the same capacity can be obtained with less bandwidth and more power, or vice versa.
• As $B \to \infty$, capacity does not become infinite; it approaches $C_\infty = 1.44\,S/N_0$.

Example: a telephone channel with $B = 3.1$ kHz and SNR $= 30$ dB ($S/N = 1000$):

$$C = 3100\log_2(1 + 1000) = 3100 \times 9.967 \approx 30.9\ \text{kbit/s}$$

So no modem can reliably send more than about 30.9 kbit/s over such a line.

Is $x(t) = e^{-2t}u(t)$ an energy or power signal?

Energy:

$$\begin{aligned} E &= \int_{-\infty}^{\infty} |x(t)|^2\,dt = \int_{0}^{\infty} e^{-4t}\,dt \\ &= \left[\frac{e^{-4t}}{-4}\right]_0^{\infty} = 0 - \left(-\frac14\right) = \frac14\ \text{J} \end{aligned}$$

Power:

$$P = \lim_{T\to\infty}\frac{1}{T}\int_{0}^{T/2} e^{-4t}\,dt = \lim_{T\to\infty}\frac{1 - e^{-2T}}{4T} = 0$$

Answer: Energy $E = 0.25$ J (finite, non-zero) and power $P = 0$, so $x(t) = e^{-2t}u(t)$ is an energy signal.