Somebody recently asked me this question “Does Shannon Capacity Increase by Dividing a Frequency Band into Narrow Bins”. To be honest I was momentarily confused and thought that this may be the case since many of the modern Digital Communication Systems do use narrow frequency bins e.g. LTE. But on closer inspection I found that the Shannon Capacity does not change, in fact it remains exactly the same. Following is the reasoning for that.

Shannon Capacity is calculated as:

C=B*log2(1+SNR)

or

C=B*log2(1+P/(B*No))

Now if the bandwidth ‘B’ is divided into 10 equal blocks then the transmit power ‘P’ for each block would also be divided by 10 to keep the total transmit power for the entire band to be constant. This means that the factor P/(B*No) remains constant. So the total capacity for the 10 blocks would be calculated as:

C=10*(B/10)*log2(1+P/(B*No))

So the Shannon Capacity for the entire band remains the same.

PS: The reason for the narrower channels is that for a narrow channel the channel appears relatively flat in the frequency domain and the process of equilization is thus simplified (a simple multiplication/division would do).

Note: ‘No’ is the Noise Power Spectral Density and ‘B*No’ is the Noise Power.

#### Author: Yasir Ahmed (aka John)

More than 20 years of experience in various organizations in Pakistan, the USA, and Europe. Worked as a Research Assistant within the Mobile and Portable Radio Group (MPRG) of Virginia Tech and was one of the first researchers to propose Space Time Block Codes for eight transmit antennas. The collaboration with MPRG continued even after graduating with an MSEE degree and has resulted in 12 research publications and a book on Wireless Communications. Worked for Qualcomm USA as an Engineer with the key role of performance and conformance testing of UMTS modems. Qualcomm is the inventor of CDMA technology and owns patents critical to the 4G and 5G standards.

But when we talk about individual channel then our capacity will increase……..as bandwidth decreases. Its live example is LTE-Advance.

Vikram,

Can you please elaborate.

John

(Please ignore the 1st comment)

Thanks for the Info,

So what if I have to calculate the capacity for 5 users of a femtocell. I am dividing the bandwidth B by 5 among the 5 user. Say the TX power of is 15dbm (30mW), do I have to divide it by 5, which is 7.78dbm (6mW) for each user? And I have to calculate the pathloss and SINR using 7.78 dbm??

Yes the power is divided by 5 (on linear scale). But the AWGN Power is also divided by 5, since Noise Power is Power Spectral Density into Bandwidth. So the SNR remains the same. You may want to take a closer look at what happens to the SINR…

Thanks for the Info,

So what if I am calculating a coverage of femtocell for 5 users. I am diving the bandwidth B by 5. Say the tx power of is 15dbm, do I have to divide it by 5, which is 3 dbm for each user? And I have to calculate the pathloss and SINR using 3dbm??