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
Only (a) is correct
Hartley's law ties information rate to two things and two things only: bandwidth and time. In its original form
\(H=k\,B\,t\)
where B is the channel bandwidth and t the duration of transmission — the total information that can be sent is proportional to their product. Statement (a) is therefore correct and statement (b) is not: the depth of modulation affects the amplitude of the transmitted signal and hence its signal-to-noise ratio, but it does not appear in Hartley's law at all. So the answer is option 1.
Why bandwidth is the limiting quantity. A channel of bandwidth B can carry at most \(2B\) independent symbol values per second — the Nyquist rate. Squeeze symbols closer together than that and they smear into one another as intersymbol interference, no matter how much power is used. Hence
\(C=2B\log_{2}M\ \ \text{bits/s}\)
for M-level signalling, which is Hartley's law in its modern form.
The obvious follow-up question — why not raise M without limit? — is what Shannon answered. Noise eventually makes the levels indistinguishable, and the true ceiling is the Shannon-Hartley theorem:
\(C=B\log_{2}\left(1+\dfrac{S}{N}\right)\ \ \text{bits/s}\)
| Law | Depends on | Says |
|---|---|---|
| Hartley | Bandwidth, time | Symbol rate is capped at 2B |
| Nyquist | Bandwidth | 2B independent samples per second |
| Shannon-Hartley | Bandwidth and S/N | Error-free capacity ceiling |
A worked example : a 3.1 kHz telephone channel with a 30 dB signal-to-noise ratio (S/N = 1000) gives
\(C=3100\log_{2}(1001)\approx30.9\ \text{kbps}\)
which is precisely why voice-band modems stalled around 33.6 kbps — they had reached a physical limit, not an engineering one.
The trade-off the two laws expose is that bandwidth and signal-to-noise ratio can be exchanged for one another: spread spectrum deliberately uses far more bandwidth than the data needs in order to work at a very low S/N, while a crowded band forces the opposite compromise.
Hence, only statement (a) is correct.
Similar Questions
The information capacity (bits/sec.) of a channel with bandwidth W and transmission time T is given by
According 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 information capacity (bits/sec.) of a channel with bandwidth W and transmission time T is given by
According 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:
