Generally more information can be transmitted if the bandwidth of the signal is higher, but is there any advantage of keeping the bandwidth as low as possible?

  • 1
    \$\begingroup\$ Generates less noise. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 27 '17 at 22:33
  • 2
    \$\begingroup\$ @Creator: re-read your title. \$\endgroup\$ – Transistor Sep 27 '17 at 22:34
  • \$\begingroup\$ Y@IgnacioVazquez-Abrams less noise but does it have any advantage, how does less noise help us? \$\endgroup\$ – Creator Sep 27 '17 at 22:37
  • 1
    \$\begingroup\$ You scored 50% on your title fix. Completed. \$\endgroup\$ – Transistor Sep 27 '17 at 22:38
  • 2
    \$\begingroup\$ It keeps the local regulatory authorities from fining you for using bandwidth assigned to other users. \$\endgroup\$ – The Photon Sep 28 '17 at 0:38

The theoretical limit of the amount of data that a data channel can handle is given by the Shannon-Hartley theorem:`` $$ C = B \log_2 \left( 1+\frac{S}{N} \right) $$

So yes, Bandwidth (B) does have a proportional impact on data rate, but there is a catch. Approximatively, when you double the bandwidth, you also double the Noise (N) power by about 2. S will remain constant if you consider that you have a limited power budget (the power will be spread over the bandwidth, so S/B is lower). S and N are "averages" for signal and noise in a (symbol) time period.

Approximatively we have: $$C \approx 0.332 \ B \cdot \mathrm{SNR\ (in\ dB)},\ when\ SNR>>0dB $$ and $$C \approx 1.44 \ B \ {S \over N},\ when\ SNR << 0dB.$$

Let's look at this in two "practicle cases". In the first case we have a low SNR ratio (=0.1) and we will double the bandwitdh and see the maximum data rate. In the second case we have a higher SNR ratio (=1023) and we will do the same.

So in the first case, $$S_0=0.1 N_0$$ and therefore $$ C_0=0.138 B_0 $$. By doubling the bandwidth we get $$C_1=2B_0\log_2(1.05)=0.141B_0.$$ Doubling the bandwidth gives a very slight improvement. In fact if we suppose that noise density \$N_d\$ expressed in \$W/Hz\$ is constant, then the formula simplifies to \$C \approx 1.44 \ {S \over N_d} \$ for small SNR. You can see how the factor 1.44 is close to the factors in the example and that for small SNR the data rate essentially depends on signal power, not on bandwidth. So that is the "catch" mentionned earlier: while the theorem says that bandwidth has an impact, doubling the bandwidth also doubles the noise power, which mainly cancels the impact of the bandwidth for small SNR at constant signal power.

In the second case, $$S_2=1023N_2$$ then $$C_2=10B_2.$$ If we double the bandwidth with a limited power budget, we get $$S_3=1023N_2/2$$ and $$C_3=2B_2\log_2(1+1023/2)=18.002B_2.$$ Doubling the bandwidth gives an 80% improvement.

At an SNR of 0dB doubling the bandwidth gives an improvement of 17% with constant signal power.

So for a low SNR doubling the bandwidth is almost useless for constant signal power. While at high SNR it improves the data rate significantly. This can also be seen in the approximations for the Shannon formula given above.

So as soon as you reach about 0dB for your SNR at the receiver end, doubling the bandwidth is pretty much useless for the data rate.

Now all of this is the theoretical limit. The practical limit also depends on the modulation/demodulation that is used. If you use AM modulation, then you'll not be able to get a lot of data at a low SNR ratio, but with a spread spectrum technique you will. In other words, a higher SNR ratio can allow the use of simpler demodulation techniques than a low SNR ratio, but it does come at the expense of filtering (your must filter out unwanted signals).

There are other considerations such as jamming. High bandwidth and narrow bandwidth signals both have their techniques to counter jamming. It is difficult to say which is best. Jamming basically is "noise" in the channel.

By limiting the bandwidth for a channel, one can have more channels in the same total bandwidth and therefore increase the overall data rate if transmissions would interfere with each other in the same channel.

Some modulations require a "high" SNR - narrowing the bandwidth increases the SNR. For example, if your SNR for AM would be only 3dB you would not find your radio very pleasant [in that case, you as a human are part of the decoding process and you want to understand the words].

The power required for the decoder is also a limitation on the receiver end. Some modulations require computing power even if they allow a lower SNR. So if low power is the target, some modulations are not acceptable.

Considering all of this, if your acceptable SNR budget (which depends on modulation) at the receiver is more than 0dB, increasing the bandwidth allows you to significantly increase the theoretical data rate limit. The minimum SNR depends on your modulation, but 0dB is about the limit at which you would start to tradeoff with other design aspects such as jamming, filter quality factors, etc.

To reply to the question, the main advantages of narrow band are:

  • Can increase the SNR to the acceptable level;
  • Allows more channels in the same bandwidth.
| improve this answer | |

The primary advantage of using a narrow bandwidth is that it uses less of the available spectrum, thereby leaving more of it for other uses.

Since using wide bandwidth is advantageous to individual users, but detrimental everyone as a whole if used unnecessarily, there are regulations about what frequencies and bandwidths are allowed to be used for particular purposes. Some of these require explicit permission, in the form of a license.

Another advantage of using a narrow bandwidth is therefore that it keeps your transmitter within the law. Transmitting illegally risks confiscation of the equipment, fines, and in extreme cases, prison time.

| improve this answer | |

Narrow band will be more susceptible to phasenoise, unless you track the received phase and correct for that timing wander. If you do that, your distances become phenomenal with very low power.

Consider 1 milliWatt at 30MHz. With 10Hz bandwidth (you'll need crystal oscillators at TX and at RX; DDS may or may vary the closein phases too much to provide a reliable link), and 4dB budget for NoiseFigure and Antenna Matching, and 10dB for SNR, you'll need this much receiver input power:

-174 dBm/rtHz floor

+10dB 10Hz bandwidth [+20dB for 1ooHz, +30dB for 1khz, +40dB for 10KHz]

+4 dB NF/losses

+10dB SignalNoiseRatio

-174 + 24 = -150dBm signal noise needed.

0dBm/50 Ohm is 0.632vpp, -120dBm is 0.632 microvoltsPP, -160dBm is 0.00632uV or 6.32 nanoVolts. -150dBm/50 ohm is 6.32nv * sqrt(10) = 19 nanoVoltsPP.

We can have 0dBm - 150dBm = 150dB path loss.

Line-of-sight is +22dB + 10*log10[(distance/wavelength)*2]. We have 128dB as ratio of (distance/wavelength)^2 or 128/2 = 64dB as ratio of distance/wavelength or ratio of over a million. (2.5 million)

Thus at 30MHz, wavelength of 10 meters, distance can be 10meter * 2.5Million or

25 million meters, 25 thousand kilometers, 15 thousand miles.

Line of sight.

| improve this answer | |

Not the answer you're looking for? Browse other questions tagged or ask your own question.