By bandwidth, do you mean RF channel width? Because in WiFi, that's always 20 MHz, or multiples of 20 MHz if your WiFi supports carrier aggregation.

Since you mention specifically 54 Mb/s, and 2.4 GHz, it sounds like you're talking about 802.11g. That would not use carrier aggregation, so the answer is 20 MHz channel width.

To achieve 54 Mb/s, 802.11g would use the highest OFDM constellation, in this case 64-QAM. So what you might be able to calculate is the total symbol rate, given that each symbol carries 6 bits worth of data. But the channel width is predetermined.

In theory, bandwidth is related to data rate by:

1) Nyquist formula: data rate = 2 * bandwidth * log2 (M) ; where M is the modulation level (eg., M=4 for QPSK ).

2) Shannon formula: data rate = bandwidth * log2(1+SNR) ; where SNR is the signal to niose ratio.

Thank you all

Nikola. you are saying that the standard  OFDM is realized by 64 subcarriers . but what about pilot sub-carriers, is that taken into consideration???   And whether it will use the same modulation as that  total sub carrier???

Thank you sir

The required bandwidth is related to bit rate and the modulation order M.

It is so that the double sided bandwidth w = symbol rate= bit rate rb/ divided by the number of bit per symbol n. The number of bits per symbol is = log 2M with M is the M is the QAM modulation order.

So, the bandwidth w= rb/ log 2(M)

Best wishes

Please look at the book

Digital Communication

by

Proakis

at

Northeastern University

Boston, MA

USA

Hello,

This was interesting.

Could some one help me out with the following

Q) Given a bandwidth of 20kHz, what is the maximum symbol rate, according to the Nyquist theorem?

Given:

Passband: 𝐻(𝜔) = 1 for |𝜔| ≤ 0.3𝜋

Stopband: 𝐻(𝜔) ≤ 0.01 for 0.61𝜋 ≤ |𝜔| ≤ 𝜋.

Determine the range of the transition band and the width of the transition band Δω for the response above.

* what are difference of range of the transition band and the width of the transition band*

Dear Mekala,

The double sided bandwidth = Symbol rate

So the single sided bandwidth = symbol rate /2

Therefore 20 kHz as single sided bandwidth = Symbol rate /2

Therefore the symbol rate= 2x20= 40 symbol per second.

This is in agreement with the sampling theory of Nyquist such that the sample rate = 2 fmax with fmax = The bandwidth.

Best wishes

The bandwidth is proportional with the one over symbole duration T. Or BW=1/(2T)

interesting