---
title: "Shannon-Hartley theorem; energy/power of e^(−2t)u(t)"
subject: "Communication Systems"
chapter: 1
key: "semester_6/communication_systems:shannon-hartley-energy-e2t"
source: https://www.bctengineer.com/q/semester_6/communication_systems/shannon-hartley-energy-e2t
site: BCT Engineering
---

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

Communication Systems · Chapter 1: Introduction

- Asked in: 2080 Chaitra, 3+5 marks
- Video: [Channel Capacity by Shannon-Hartley | Basics, Proof & Maximum Bandwidth Condition](https://www.youtube.com/watch?v=3ekWsXeZ8TM) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/shannon-hartley-energy-e2t
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-1

## Question

With a suitable example explain Shannon-Hartley channel capacity theorem. Determine whether the given signal x(t) = e^(−2t)u(t) is a power or an energy signal.

## Answer

#### 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**.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The Shannon-Hartley capacity of an AWGN channel of bandwidth B and signal-to-noise ratio S/N is:
   A. C = B (1 + S/N)
   B. C = B log2(S/N)
   C. C = B log2(1 + S/N) bit/s
   D. C = log2(1 + B S/N) bit/s
   Answer: C. Capacity grows linearly with bandwidth but only logarithmically with SNR. (Key point: Shannon-Hartley formula)

2. What does the theorem say about an information rate R > C?
   A. Error-free transmission is impossible whatever coding is used
   B. Error-free transmission is possible with enough bandwidth only
   C. Error-free transmission is possible with a good enough modulator
   D. The error probability is exactly 0.5
   Answer: A. For R <= C there exists a code with arbitrarily small error probability; for R > C none exists. (Key point: Meaning of capacity)

3. For B = 3.1 kHz and SNR = 30 dB (S/N = 1000), the capacity is approximately:
   A. 3.1 kbit/s
   B. 9.97 kbit/s
   C. 93 kbit/s
   D. 30.9 kbit/s
   Answer: D. C = 3100 x log2(1001), about 3100 x 9.967, which is about 30.9 kbit/s. (Key point: Telephone channel example)

4. Why is x(t) = e^(-2t) u(t) an energy signal?
   A. Its energy is 1/4 J (finite, non-zero) and its average power is 0
   B. Its energy is infinite and its power is 1/4 W
   C. Its energy is 1/2 J and its power is non-zero
   D. Its energy and power are both infinite
   Answer: A. E = integral of e^(-4t) from 0 to infinity = 1/4, and P = lim (1 - e^(-2T))/(4T) = 0. (Key point: Energy of e^(-2t) u(t))

## Related in Communication Systems

- [Shannon-Hartley theorem; capacity limit 1.44S/N₀](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ashannon-hartley-1-44) — Communication Systems, chapter 6, asked 2×
- [Implications and limitations of Shannon-Hartley theorem](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ashannon-hartley-implications-limits) — Communication Systems, chapter 6
- [Shannon channel capacity theorem and its limitations](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ashannon-capacity-limitations) — Communication Systems, chapter 6, asked 2×
- [Shannon channel capacity theorem and its implication](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ashannon-capacity-implication) — Communication Systems, chapter 6
- [Nyquist channel capacity theorem](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Anyquist-channel-capacity) — Communication Systems, chapter 6
- [Rayleigh energy theorem; power from Fourier series](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Arayleigh-periodic-power-fourier) — Communication Systems, chapter 1

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