14
\$\begingroup\$

My question is very primitive :) I've read multiple articles today and answers here, on Stack Exchange, but still don't understand one thing. Why doesn't frequency itself affect data rate in mobile networks? 3G/4G networks use QAM modulation, that includes changes to amplitude and phase of signal. Let's say, we have 900 MHz signal with 10 MHz bandwidth and 2600 MHz signal with 10 MHz bandwidth. At the same "snippet" of incoming signal we'll have much more "cycles" to modulate on higher frequency (per same time), don't we? So why doesn't it matter?

\$\endgroup\$
11
  • 4
    \$\begingroup\$ Perhaps some reading on the basics of modulation will be useful... try reading up on AM modulation and analyze it in the time and frequency domains. That should paint a picture on why the answer to the question "At the same 'snippet' of incoming signal we'll have much more 'cycles' to modulate on higher frequency (per same time), don't we?" is "No, we don't." The key factor is bandwidth, not carrier frequency. \$\endgroup\$
    – Shamtam
    Jun 8 '20 at 15:41
  • 2
    \$\begingroup\$ If both signals have a 10 MHz bandwidth, than both have 10 million cycles per second. \$\endgroup\$ Jun 8 '20 at 15:49
  • 10
    \$\begingroup\$ @user1850479, When OP says, 900 MHz or 2600 MHz "signal" with 10MHz "bandwidth," they're almost certainly talking about a radio transmission with a carrier frequency somewhere within a nominal 900 MHz band or a 2600 MHz band, and occupying a "channel" that is 10MHz wide. \$\endgroup\$ Jun 8 '20 at 16:15
  • 1
    \$\begingroup\$ @user1850479, OK, I guess I didn't understand what you meant by a channel "having 10 million cycles per second." I would expect to talk about the number of digital bits per second one could send through a 10MHz-wide channel, or maybe talk about the maximum analog frequency which could be imposed on it. That max modulation frequency might be significantly less than 10MHz depending on the modulation method. \$\endgroup\$ Jun 8 '20 at 16:21
  • 2
    \$\begingroup\$ @Rob Don't they? ;) \$\endgroup\$
    – DKNguyen
    Jun 9 '20 at 5:19
36
\$\begingroup\$

It's not a bad question and shows quite a common misunderstanding of how Radio Frequency (RF) Systems work. The 900MHz and 2600MHz signals are called Carrier Frequencies. The actual information is contained in the 10MHz bandwidth. The original signal is a baseband signal that extends up to 10MHz. This is used to modulate the carrier signal. The reason we do this is so that we can have several channels sharing the same medium.

When received by a radio, the signal is downconverted back to the baseband (up to 10MHz in this case). The reason we do this, instead of sampling the signal directly, is the RF electronics are very complicated and relatively expensive, while baseband electronics is not.

So to answer your question, when both signals are downconverted, they are both 10MHz wide signals, so will transfer data at that rate (probably greater for QAM due the symbol rate but that's another story).

\$\endgroup\$
1
  • 12
    \$\begingroup\$ There is a slight relation though: it is easier to get a wider block of bandwidth with consistent propagation properties and fewer interferers at a higher center frequency, so you can be reasonably sure that the 2 GHz bandwidth from 58 to 60 GHz are usable for wireless networking, while 0 to 2 GHz are not. \$\endgroup\$ Jun 9 '20 at 12:09
18
\$\begingroup\$

Because the bandwidth of the channel determines how rapidly the symbols on that channel can change, not the carrier frequency. It is the bandwidth of the channel which determines how fast the channel "rings down" from a sudden change in phase or amplitude.

So your two 10MHz bandwidth channels can each sustain a symbol rate of no more than \$10 \cdot10^6\$ symbols per second, regardless of whether the carrier frequency is 0Hz or 1THz or anywhere in between.

\$\endgroup\$
5
  • \$\begingroup\$ But the higher the the carrier frequency, the higher the channel frequency can be. Because you can sample the same data in less time. What would be the point of higher carrier frequency then? \$\endgroup\$
    – Fredled
    Jun 8 '20 at 23:21
  • 4
    \$\begingroup\$ @Fredled Nope. Bandwidth and carrier frequency are two different things. A 10MHz bandwidth signal on a 100MHz carrier can range from 95MHz to 105MHz, no more, no less. A 10MHz bandwidth signal on a 1THz carrier can range from 999995MHz to 1000005Mhz -- no more, no less. \$\endgroup\$
    – TimWescott
    Jun 9 '20 at 0:14
  • \$\begingroup\$ @Fredled: If two channels each had a bandwidth that was a fixed percentage of the carrier frequency, which may be how you're viewing things intuitively, then the channel with the higher carrier frequency could handle more data. If the bandwidth is fixed, however, then the ratio of bandwidth to carrier frequency will drop as the carrier frequency, offsetting the aforementioned tendency. \$\endgroup\$
    – supercat
    Jun 9 '20 at 2:18
  • 1
    \$\begingroup\$ @Fredled: Note that if one were using old pre-heterodyne tuners, it would be hard to avoid having the bandwidth increase with the carrier frequency, but modern radios work by taking the incoming signal, rough filtering it, and subtracting a tuning oscillator's output frequency from all of the signals present in the rough-tuned source to yield an "intermediate" frequency which can be tuned with a fixed-frequency narrow bandwidth filter. The bandwidth of the IF filter is unaffected by the carrier frequency. \$\endgroup\$
    – supercat
    Jun 9 '20 at 2:22
  • \$\begingroup\$ @Fredled If you try to use the entire frequency as-is you will cause interference in other frequencies. This is OK if you are the only people on the planet using radio transmission (early Marconi days) but someone will complain to the regulators if you do this now. Bandwidth is indirectly a measure of how much of the frequency you can use before causing interference or how much you can extract real signal from your channel before it becomes too noisy (interference from other frequency channels) \$\endgroup\$
    – slebetman
    Jun 9 '20 at 3:28
11
\$\begingroup\$

By quick analogy (the other answers are sufficient in their detail):

If you own a hotel where all the rooms are the same size (bandwidth), it doesn't matter how high a floor (frequency) they're on--they all hold the same amount of stuff. If you have some larger rooms (more bandwidth), they'll hold more stuff, whether they're on the 2nd floor or the 26th.

\$\endgroup\$
6
\$\begingroup\$

Consider Morse Code modulation applied to human audio as one of many possible examples.

With input

- . . .   - . . .   - . - .

First, 880 Hz carrier frequency:

biyip bi bi bi  biyip bi bi bi  biyip bi biyip bi

Next, 55 Hz carrier frequency:

gruur gr gr gr  gruur gr gr gr  gruur gr gruur gr
\$\endgroup\$
1
  • \$\begingroup\$ This is actually a cool and easy to grasp analogy, just needs further explanation. Finger speed is limited and no matter if they are beeps or gru's, finger speed is the same. \$\endgroup\$
    – Ariel M.
    Jun 11 '20 at 23:31
4
\$\begingroup\$

10 MHz of bandwidth contains the same amount of information, regardless of whether it's centered at 900MHz or 2600MHz, or any other center frequency.

This is easy to show -- the center frequency of a signal can be shifted digitally or electronically without destroying it. You can move the 2600MHz signal down to 900MHz, and it will have the same bandwidth. Then you can move it back to 2600MHz and get exactly the original signal back, so obviously whatever information you can carry at one frequency, you can carry at the other.

By the the Nyquist-Shannon sampling theorem, we know that any < 10MHz signal at any center frequency can be reconstructed from samples taken at a 20MHz rate.

https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem

\$\endgroup\$
2
\$\begingroup\$

If one is using a single carrier to send information through an unshared, interference-free, communications medium, bandwidth will generally scale with carrier frequency. If, however, one modulates a signal with a 901,500,000hz center frequency and sends it through a communications medium which, while free of unwanted frequency content in the range 901,490,000Hz to 901,510,000Hz, may have unpredictable amounts of frequency content below 901,490,000Hz and above 901,510,000Hz, then even if one demodulates the signal perfectly, the demodulated signal may have unpredictable amounts of noise at any or all frequencies above 10,000Hz(*).

If one used a perfect filter to remove from the demodulated signal all content above 9,999Hz, one could faithfully recover all frequency content below that frequency, but would of course lose any content one had transmitted at frequencies higher than that. If one doesn't filter out everything above 10,000Hz, it might totally drown out everything else in one's signal. The usable frequency range will depend upon the distance between the carrier frequency and the nearest unwanted frequency content.

(*) Techniques called single sideband modulation or quadrature amplitude modulation may be used, respectively, to shield a channel from interference on one side (allowing one to use a carrier frequency which is off center toward that side, and thus further from the other side), or use a channel that is clear on both sides of the carrier frequency to send two signals, each of whose bandwidth would match the difference between the carrier and the nearest interference. The general problem of interference, however, remains.

\$\endgroup\$
1
\$\begingroup\$

https://en.wikipedia.org/wiki/Noisy-channel_coding_theorem

First, all communication channels are noisy.

The maximum data rate depends on the signal/noise ratio and the bandwith. The particular frequencies of the band are of no importance.

Signal level depends on the transmitter power and the propagation conditions. Noise depends on the temperature and the presence of other transmitters.

Propagation conditions and thermal noise both DO depend on the frequency in more or less predictable manner, as well as the presence of other transmitters, but all of these factors have only a second-order relation to your question. But yes, they are related.

The bandwidth is allocated by the regulator authority (when using a common media), some standard, media properties or other considerations when you control the media.

\$\endgroup\$
0
\$\begingroup\$

A higher bandwidth dictates higher data rate which we all know. Bandwidth refers to a frequency range and not just a single frequency. If your wireless system is operating say at 5.85 GHz to 5.95 GHz, a bandwidth of 100 MHz is available. Also, when your wireless operating system is operating at say 2.4 GHz to 2.45 GHz, the bandwidth available is 50 MHz. Thus, data rate is higher in the former case.

I mentioned the former band as ranging from 5.85 GHz to 5.95 GHz. Bandwidth is the difference between the highest frequency of the band and the lowest frequency of the band. If regulatory authorities allow us to expand the existing bandwidth to say 150 MHz, then our highest band frequency is now 6 GHz. Thus, bandwidth has risen and hence the data rate has also risen.

Whenever the highest frequency of the band is increased, the bandwidth increases.

\$\endgroup\$
0
\$\begingroup\$

In order to speed up the data rate 1)increase the bandwidth 2)increase the RF power 3)use data compression 4)use more effective modulation ...

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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