The information capacity (bits/sec.) of a channel with bandwidth W and transmission time T is given by
WT
This is Hartley's law, which makes the information carried proportional to bandwidth and to time — the product WT, option 1.
\(H=k\,W\,T\)
Why the two multiply rather than divide. Both factors increase what can be sent, so neither can appear in a denominator. Widen the channel and more symbols per second are possible; keep the channel open longer and more seconds are available. Doubling either doubles the total; doubling both quadruples it. That alone eliminates options 2, 3 and 4, each of which would make the capacity fall as the transmission time or the bandwidth grows.
Where the proportionality to bandwidth comes from. By the Nyquist criterion a channel of bandwidth W supports at most \(2W\) independent symbol values per second. Push symbols faster and adjacent pulses smear into one another as intersymbol interference, however much power is applied. So the symbol rate is capped by the bandwidth, and over a time T the number of symbols sent is proportional to \(WT\).
The quantity WT is called the time-bandwidth product and appears throughout communication theory. It is dimensionless — hertz multiplied by seconds — and counts the number of independent degrees of freedom a signal possesses. A signal occupying bandwidth W for time T can be described completely by about \(2WT\) samples, and no more; everything else about it is determined.
| Law | Depends on | Statement |
|---|---|---|
| Hartley | Bandwidth, time | H = kWT |
| Nyquist | Bandwidth | 2W symbols per second |
| Shannon-Hartley | Bandwidth and S/N | \(C=W\log_{2}\left(1+\dfrac{S}{N}\right)\) |
What Hartley's law leaves out is the constant k, which hides the number of distinguishable amplitude levels. Shannon supplied it: crowding more levels into the same amplitude range makes them harder to separate in noise, so the true ceiling involves the signal-to-noise ratio. For a 3.1 kHz telephone channel at 30 dB S/N the limit is about 30.9 kbps — which is why voice-band modems stalled near 33.6 kbps.
Hence, the information capacity is WT.
Similar Questions
The Hartley law states that :
(a) the maximum rate of information depends on the channel bandwidth
(b) the maximum rate of information depends on the depth of modulationAccording to Hartley's law
Important Questions from Channel Capacity
The capacity of band-limited additive white Gaussian Noise (AWGN) channel is given by \(C = W{\log _2}\left[ {1 + \frac{P}{{{\sigma ^2}w}}} \right]\) bits per second (bps), where W is the channel Bandwidth, P is the average power received and σ2 is the one-sided power spectral density of the AWGN.
For a fixed \(\frac{P}{{{\sigma ^2}}} = 1000\), the channel capacity (in kbps) with infinite Bandwidth (W → ∞) is approximately
- Let the relevant bandwidth ($B$) of a digital communication system be 1 MHz and $kT = -174\text{ dBm/Hz}$, where $k$ is Boltzmann's constant and '$T$' is equivalent noise temperature of the receiver. The power ($S$) of signal received through an additive Gaussian channel is $-80\text{ dBm}$.
Which of the following options is/are TRUE about Shannon capacity ($C$) of the channel? The Hartley law states that :
(a) the maximum rate of information depends on the channel bandwidth
(b) the maximum rate of information depends on the depth of modulationAccording to Hartley's law
Match List I with List II:
List I List II (A) Shannon's theorem (I) Capacity of Gaussian Noise channel (B) Shannon-Hartley theorem (II) Rate of Information (C) Bayes theorem (III) Energy of a signal (D) Parseval's theorem (IV) Conditional probabilities Choose the correct answer from the options given below:
