According to Hartley's law
The maximum rate of information depends on the channel bandwidth
Hartley's law states that the amount of information that can be transmitted is proportional to the bandwidth and to the time available :
\(H=k\,B\,t\)
so the rate of information is set by the channel bandwidth — option 1.
Why bandwidth is the limiting quantity. A channel of bandwidth B can support at most \(2B\) independent symbol values per second — the Nyquist rate. Send symbols faster and adjacent pulses smear into one another as intersymbol interference, however much power is applied. In its modern form,
\(C=2B\log_{2}M\ \ \text{bits/s}\)
for M-level signalling.
Why depth of modulation (option 2) does not enter. Modulation index changes the amplitude of the transmitted signal and hence its signal-to-noise ratio, but it appears nowhere in Hartley's relation. Raising it improves noise performance, not the fundamental symbol rate the channel permits.
Options 3 and 4 belong to a different subject altogether. Redundancy is the province of coding theory — it is added deliberately for error detection and correction, and from an information standpoint it reduces the useful rate rather than being essential to it. And nothing in Hartley's law restricts the alphabet to two symbols; M-ary signalling is precisely how the rate is increased at fixed bandwidth.
| Law | Depends on | Statement |
|---|---|---|
| Hartley | Bandwidth, time | H = kBt |
| Nyquist | Bandwidth | 2B symbols per second |
| Shannon-Hartley | Bandwidth and S/N | \(C=B\log_{2}\left(1+\dfrac{S}{N}\right)\) |
The obvious follow-up — why not raise M indefinitely? — is what Shannon answered. Crowding more levels into the same amplitude range makes them harder to distinguish in noise, so the true ceiling involves the signal-to-noise ratio as well. For a 3.1 kHz telephone channel at 30 dB S/N,
\(C=3100\log_{2}(1001)\approx30.9\ \text{kbps}\)
which is why voice-band modems stalled near 33.6 kbps — a physical limit, not an engineering one.
Hence, according to Hartley's law the maximum rate of information depends on the channel bandwidth.
Similar Questions
The information capacity (bits/sec.) of a channel with bandwidth W and transmission time T is given by
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 modulation
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 information capacity (bits/sec.) of a channel with bandwidth W and transmission time T is given by
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 modulationMatch 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:
