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: John (YA)

John has over 15 years of Research and Development experience in the field of Wireless Communications. He has worked for a number of companies around the world including Qualcomm Inc. USA. He has an MS in Electrical Engineering from Virginia Tech USA and has published his work in international journals and conferences.

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??