Does data rate increase with channel frequency?

Question

It is commonly said that a higher channel frequency implies a higher data rate. For example, in https://www.howtogeek.com/222249/whats-the-difference-between-2.4-ghz-and-5-ghz-wi-fi-and-which-should-you-use/ they say that 5 GHz wifi connection is simply faster than 2.4 GHz. However, Shannon–Hartley theorem states that the maximum channel capacity in bits/s depends only on the channel bandwidth and the SNR. Therefore, the data rate shouldn't really depend on the center frequency of the channel, only on the available bandwidth. Is it a misconception that higher frequency channels support a higher bit rate or is it true for some reason?

Answer 1

Is it a misconception that higher frequency channels support a higher bit rate or is it true for some reason?

It is not the (carrier) frequency itself that determines the supported bit rates but the available Bandwidth of the channel

Suppose I have 10 MHz of bandwith available at 100 MHz, for example 100 MHz to 110 MHz

or

I have 10 MHz of bandwith available at 1000 MHz, for example 1000 MHz to 1010 MHz

Then the highest bit rate I can achieve will be the same as there is 1 MHz available in both cases.

However note how at 100 MHz that channel bandwith is 10% of the carrier frequency but at 1000 MHz it is only 1 %.

At that 1000 MHz I could fit 10 of those 10 MHz channels to come to the same 10% (or use a wider channel of 100 MHz.)

If we want divide a certain frequency band between service providers, that's much easier to do at the higher frequencies. It can be done also at lower frequencies but that would result in narrow channels (small bandwidth) and therefore lower bit rates.

To put it more simply: there's more "space" at higher frequencies so it "costs less space" to implement higher bit rates.

Suppose I go to 10 GHz, that will give me another factor 10 more space.

The 2.5 GHz Wifi vs 5 GHz Wifi isn't completely fair because the 2.5 GHz band is about 100 MHz wide (that's all channels together) but the 5 GHz band has about 900 MHz available (depending on your country it might not be one continuous 900 MHz range though). See here. So there's simply lots more space (bandwidth) assigned to the 5 GHz standard.

Answer 2

I agree that sending the same bandwdith on different Carrier signal frequencies doesn't necessarily mean higher data throughput.

I guess what everyone means when they compare lower frequency applications like GSM900 with higher frequency applications like UMTS2100 or 2,4Ghz Wifi with 5Ghz Wifi is, that the available bandwidth in that particular band is just larger.

E.g. the bandwidth in GSM900 is very limited, it is a scarce resource which is usually regulated by a national authority. Same for any commercial mobile technology.

Wifi uses 2,4Ghz and 5Ghz non-regulated/licensed band. But the boundaries of the bands are specified, and the 5Ghz seems to be much "larger".

The 2,4Ghz WiFi band has 3 non overlapping channels. Whereas the 5Hgz band has 23!

(see https://www.electronics-notes.com/articles/connectivity/wifi-ieee-802-11/channels-frequencies-bands-bandwidth.php e.g.)

Answer 3

As mentioned in the other answer, the band around 5 GHz is less used in general which is advantageous.

In addition to this, it is often only practical to achieve very large bandwidths at higher frequencies. If you look at some basic texts in microwave engineering, such as "Microwave Engineering" by Pozar you find many of the building block circuits may have bandwidth listed as 30%-40% of the central frequency. At higher central frequencies this is obviously a larger bandwidth.

Intuitively, I think of this as being the case because the operation of say, a quarter-wave transformer, is dependent on the transmission line being exactly a quarter wavelength long at the central frequency. Frequencies that have wavelengths near this central frequency will also behave acceptably. At higher frequencies the wavelengths are spaced more closely together as:

\$ c=f\lambda \\ \lambda=c/f \\ \frac{\Delta\lambda}{\Delta f} = -c/f^2 \\ \Delta \lambda = \frac{-c}{f^2}\Delta f \$

Decreases with \$f^2\$

Therefore, practically it is often easier to achieve larger bandwidths at higher frequencies.

Ultimately, though, as you said the channel capacity is dependent only upon the bandwidth and SNR.