Comments on: MSK transmitter and receiver

By: Krishna Sankar

Krishna Sankar — Tue, 31 May 2011 00:33:00 +0000

@Neumann: A moving average over a window of 2T samples

By: Neumann

Neumann — Mon, 30 May 2011 15:21:35 +0000

Hi Krishna,
thanks for your comments to msk modulation. I am busy to construct some software solutions for msk decoding.
Looking at your receiver block diagram, the integration over 2T, is this an moving average, sum over 2T samples after every sample, or only every n*2T samples?

Thank you and good luck.

Emil Neumann , Schwerte Germany

By: Krishna Sankar

Krishna Sankar — Mon, 21 Jun 2010 00:42:47 +0000

@Sarah: Well, if your sampling period is 1/Ts, you need to sum over 2T/Ts samples
I am unable to write the precise code as I do not follow your code well. But am sure that you will be able to figure this out. Good luck.

By: Sarah

Sarah — Tue, 15 Jun 2010 13:07:04 +0000

this was the time intv i was using. The solution you gave me previously is fine for the exp msk version but doesnt work for this one.

time_sampling = [0:1/fs:T];
time_sampling = time_sampling(1:end-1);% gets ride of the final value
%Produces the time intervals

By: Sarah

Sarah — Tue, 15 Jun 2010 13:01:35 +0000

Hi,

I used this equation to modulate MSK

xt = cos(2*pi*(fc+frequency_differance_repeat).*time_sampling_repeated + Phase_value_array_repeat );

This equation is from your website. It works fine.
The i multiplied the received (xt) signal by the following (first stage of the receiver as instructed on your website):
( i hvnt added noise yet)

%even arm
stage1_cos = xt.*(cos(2*fc*pi*time_sampling_repeated)).*(cos(pi.*time_sampling_repeated/2*T));

%odd arm
stage1_sin = xt.*(sin(2*fc*pi*time_sampling_repeated)).*(sin(pi.*time_sampling_repeated/2*T));

How do i recover the signal from here. I’m still having a problem with evaluating the intergtration from 0 to 2T for the above signal.

By: Krishna Sankar

Krishna Sankar — Sun, 18 Apr 2010 09:08:12 +0000

@Sarah: The convolution operation with taps as all ones for duration of 2T takes care of the integration of 2T
conv(real(yt).*ct,ones(1,2*T))

By: Krishna Sankar

Krishna Sankar — Sun, 18 Apr 2010 09:07:20 +0000

@Sarah: Well,
a) starting from 2T+1 ensures that the first 2T symbols are ignored (to account for the delay of the integration by 2T)
b) the step size of 2T ensures that we take one symbol from the even, then from odd, then even and so on…