Tag Archives: LTE

Open Signal Coverage Maps for Pakistan

Open Signal is a mobile application that collects the data about your wireless network (2G/3G/4G) and generates coverage maps and host of other reports. The data is collected in the background while the user is busy in his daily routines. But data can also be collected on the request of the user. This is much better than drive testing since the data is collected in real life scenarios and on thousands of different devices that are in use.

The app works while the user is indoor or outdoor, at rest or in motion, on land or on water, at sea level or on a mountain, in dry weather or in rain. Basically anywhere and anytime there are wireless signals available. There are currently 20 million users of the app (both Android and iOS combined) and this number is increasing. In Pakistan all major networks are supported including Jazz, Telenor, Zong and Ufone (both 2G/3G and 4G networks are supported).

JAZZ Islamabad Coverage Map

Telenor Islamabad Coverage Map

Zong Islamabad Coverage Map

Ufone Islamabad Coverage Map

Shannon Capacity CDMA vs OFDMA

We have previously discussed Shannon Capacity of CDMA and OFMDA, here we will discuss it again in a bit more detail. Let us assume that we have 20 MHz bandwidth for both the systems which is divided amongst 20 users. For OFDMA we assume that each user gets 1 MHz bandwidth and there are no guard bands or pilot carriers. For CDMA we assume that each user utilizes full 20 MHz bandwidth. We can say that for OFDMA each user has a dedicated channel whereas for CDMA the channel is shared between 20 simultaneous users.

We know that Shannon Capacity is given as

C=B*log2(1+SNR)

or in the case of CDMA

C=B*log2(1+SINR)

where ‘B’ is the bandwidth and SINR is the signal to noise plus interference ratio. For OFDMA the SNR is given as

SNR=Pu/(B*No)

where ‘Pu’ is the signal power of a single user and ‘No’ is the Noise Power Spectral Density. For CDMA the calculation of SINR is a bit more complicated as we have to take into account the Multiple Access Interference. If the total number of users is ‘u’ the SINR is calculated as

SINR=Pu/(B*No+(u-1)*Pu)

The code given below plots the capacity of CDMA and OFDMA as a function of Noise Power Spectral Density ‘No’.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% CAPACITY OF CDMA and OFDMA
% u - Number of users
% Pu - Power of a single user
% No - Noise Power Spectral Density
%
% Copyright RAYmaps (www.raymaps.com)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

clear all
close all

u=20;
Pu=1;
No=1e-8:1e-8:1e-6;

B=20e6;
C_CDMA=u*B*log2(1+Pu./(B*No+(u-1)*Pu));

B=1e6;
C_OFDMA=u*B*log2(1+Pu./(B*No));

plot(No,C_CDMA/1e6);hold on
plot(No,C_OFDMA/1e6,'r');hold off
xlabel('Noise Power Spectral Density (No)')
ylabel('Capacity (Mbps)')
legend('CDMA','OFDMA')
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Shannon Capacity of CDMA and OFDMA
Shannon Capacity of CDMA and OFDMA

We see that the capacity of OFDMA is much more sensitive to noise than CDMA. Within the low noise region the capacity of OFDMA is much better than CDMA but as the noise increases the capacity of the two schemes converges. In fact it was seen that as the noise PSD is further increased the two curves completely overlap each other. Therefore it can be concluded that OFDMA is the preferred technique when we are operating in the high SNR regime.

Does Shannon Capacity Increase by Dividing a Frequency Band into Narrow Bins

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.