matlab coding

I have a matrix A=[A0 A1]

A0 and A1 are [M x L] matrix and A is [M x LNt]

Another matrix B=[A0A0' A1A1']

B is M X M matrix and Ak' is [M x (p-L)] matrix

I have obtained A, but couldn't get B as dimensions vary

I have taken M=128,L=16,Nt=2,Nr=1,p=M/Nt=64

This is mimo ofdm project " Joint estimation of cfo,i/q imbalance and channel response of mimo ofdm sysytems"

[Jan]

If it is urgent, I suggest to improve the question. We have to guess to much details, e.g. what is an "M X L matrix"? I guess this means [M x L]. But I have no idea about "M X LNt" or "[A0A0' A1A1']" - does the quote mean a transponation here? And what is "Ak'"?

How did you obtain A and what is the problem at obtaining B?

Does the last sentence provide any necessary information?

Please use dots to separate the sentences. The more time you invest to make the qeustion as easy to understand as possible, the high is the motivation to create a meaningful answer.

[Janet]

Pls have a look at my base paper

http://www.scribd.com/doc/48495968/My-Base-Paper in 3rd page

I am beginner in matlab and doing a project in communication

Pls help with it

[Walter Roberson]

http://www.mathworks.com/matlabcentral/answers/29922-why-your-question-is-not-urgent-or-an-emergency

[Janet]

M xLNt means 128x32 matrix

A is a matrix having elements [ A0 A1.....ANt-1 ]

since my Nt is 2, A becomes A=[A0 A1]

B=[A0A0' A1A1']

Ak' means A0 to Ant-1

[Janet]

i am a newbie here Walter..sorry for using urgent word

[Honglei Chen]

You said you have A, so do you have A0 and A1? I don't think a lot of people will volunteer to read the paper for you so you may want to frame your question to be very specific and relevant to MATLAB.

[Janet]

A is a matrix with elements A0 and A1

A=[A0 A1]

[Janet]

I have formed matrix A:

A0=randn(M,L);

A1=randn(M,L);

A=horzcat(A0,A1);

Answer 1

[Honglei Chen]

OK, the question becomes more concrete now. Since you have A0 and A1, then you can do

A = [A0 A1];
B = [A0*A0' A1*A1'];

What is the dimension issue you mentioned in the question?

[Janet]

i have found out :)