Dear Suhas,

welcome,

The bit rate fb is related to the bandwidth by the bandwidth efficiency eta. Eta itself depends on the modulation technique. For QAM modulation the bandwidth efficeincy = log2M, where M is the modulation order. The modulation order rages from M=2 at the edge of the cell to M= 64 at the center of the cell.

So the minimum bandwidth required= fb/ log2 M= 100Mb/s divided by 6= 18.33 MHz. In case of one 1Gb/s, one needs BW= IG/6= 183.3 MHz.

Best wishes

Dear Suhas,

welcome,

The bit rate fb is related to the bandwidth by the bandwidth efficiency eta. Eta itself depends on the modulation technique. For QAM modulation the bandwidth efficeincy = log2M, where M is the modulation order. The modulation order rages from M=2 at the edge of the cell to M= 64 at the center of the cell.

So the minimum bandwidth required= fb/ log2 M= 100Mb/s divided by 6= 18.33 MHz. In case of one 1Gb/s, one needs BW= IG/6= 183.3 MHz.

Best wishes

There's also another way to approach this problem, in addition to Dr Zekry's approach, and that is to find the maximum bit/sec capacity theoretically possible. Without assuming any particular modulation scheme, but instead by using the desired signal to noise ratio as a constraint. The general idea is this. For a given RF channel bandwidth, and a specification of the maximum credible signal to noise ratio I can expect, what is the maximum theoretical bit rate?

Shannon's equation is used. But note that Shannon's equation applies to using a single antenna at the transmitter and receiver. If you allow for multiple antennas in the transmitter and multiple antenna in the receiver, in that one RF channel, and you have multiple de-correlated propagation paths, you can do better. For instance, in a 2X2 antenna example, with ideal decorrelation of the signal paths, you can double the channel capacity without requiring a higher SNR.

Anyway, back to basics. Here is Shannon's equation:

Capacity (b/s) = Channel Bandwidth (Hz) * logbase2(1 + S/N)

Note that S/N is the power ratio of signal and noise, expressed as a ratio and not in dB!

So for example, with a bandwidth of 20 MHz (single LTE channel) and a marginal SNR let's say 8 dB (small antenna at the mobile device), the maximum possible capacity comes out 57.4 Mb/s.

You will get better if you assume that higher SNR is okay. Same 20 MHz channel, allow 15 dB of marginal SNR, and now you get 100.6 Mb/s capacity.

You can see in the equation that doubling the RF channel bandwidth, keeping the rest the same, and you double the capacity.

So this approach allows you to get an answer that is the theoretical best, without taking modulation or FEC schemes into account. To be realistic, though, you should probably degrade the SNR by at least 3 dB, to get results that can be achieved in the real word. So, instead of 8 dB SNR, you might actually assume you have 5 dB of SNR, which reduces the answer to 41.1 Mb/s. This sort of thing.

You can do the algebra to answer your exact question directly.

Channel BW = Capacity (b/s) / logbase2(1 + S/N)

Then, if you can assume that you have multiple de-correlated paths available, you can try 2X2, 4X4, or similar MIMO, to get twice or four times the capacity, ideally.

Than you sir Albert Manfredi and sir Abdelhalim Zekry !

I mathematically calculated the parameters for an antenna using Shannon Hartley theorem for a data rate of 1 Gbps (for 4G LTE application) with an SNR of 5 dB. I obtained a bandwidth of 240 MHz.

1. Why is it necessary to use 2X2 or 4X4 MIMO antennas or can I use a single antenna for this purpose?

Dear Suhas,

What do you calculated is the band width of the maximum channel capacity as Dr,. Manfredi hinted. This is the absolutely minimum required bandwidth which can be realized under very large constellations according to Sharon theorem. This point was previously discussed in a a forum on the research gate. You can follow it at the link:https://www.researchgate.net/post/The_shanon_capacity_limit_of_a_communication_channel_how_to_approach_it_practically

- According to the 4G LTE standards you can aggregate the 5 x20 MHz bands to transmit data. So, the total available bandwidth is 100 MHz.

- In order to decrease the required bandwidth you increase the S/N ratio of the link to maximum possible value which is limited by allowed intercell interference. It is greater than the assumed 5 dB say 15 dB as suggested by Dr. Manfredi. If you need band width larger than 100 Mhz then you have to switch to MIMO systems. The MIMO cam increase the channel capacity by a multiple of the minimum of (Nt, Nr). Which will be 2 incase of 2x2 MIMO and 4 in case of 4x4MIMO. This is valid under the constraint of equal signal to noise ratio for all signal paths of the signals.

In case of constraining the total radiated power from all antennas, then

C= w N log2 (1+ s/ n N), with Nt=Nr=N,

Using this expression you can get the highest possible rate of your link

Best wishes