Implementation of a linear approximated GMSK block scheme based on Laurent decomposition. My goal is the Implementation of the following block scheme:

enter image description here I think $v_k$ is an NRZ element ( 1 or -1).

ovs = 4;
L = 3 % laurent decomposition
Tr = 3;% truncation length
N = 20; 
a = randi([0,1],N,1);
vk = 2*a-1;       

vk is an input for the first branch of this block scheme and the first S/P converter and for the second. An input of the second S/P converter, vk_3 looks like:

vk_1 =  kron(ones(length(vk), 1), 0);  
vk_1(1) = vk(end);
for i = 1: length(vk)-1
    vk_1(i+1) = vk(i);
vk_2 =  kron(ones(length(vk),1), 0);  
vk_2(1) = vk(end-1);
vk_2(2) = vk(end);
for i = 1: length(vk)-2
    vk_2(i+2) = vk(i);
vk_3 = vk .* vk_1 .* vk_2;

S/P converter separate the input signal into 2: the first signal contains odd elements, the second contains the even elements. Before I apply C0 and C1 I take a product with (-1)^k and upsample the signals:

Tsym = 2* ovs; % ovs – oversampling ratio
ai_1 = vk(1:2:end);
aq_1 = vk(2:2:end);

for i = 1:length(ai_1)
    ai_1(i) = ai_1(i) .* (-1)^(i); 
    aq_1(i) = aq_1(i) .* (-1)^(i); 

ai_1 = [ai_1 zeros(length(ai_1),Tsym-1)]; 

aq_1 = [aq_1 zeros(length(aq_1),Tsym-1)];  

ai_1 = [ai_1(:) ; zeros(Tsym/2,1)].';       
aq_1 = [zeros(Tsym/2,1); aq_1(:)].'; 

The same implementation has the signal ``vk_3`.

I_1= filter(C0,1,ai_1);
Q_1 = filter(C0,1,aq_1);
IQ_1 = I_1 – 1i*Q_1

IF I take BT = 0.5, the second branch with C1 will not have a big influence of the result, so I skip its implementation.

I compute a phase of the IQ_1 : unwrap(atan2(Q_1,I_1)) and compare it with a phase of GMSK transmitter (I-Q representation) enter image description here

The result of the comparison:

enter image description here

Mr Dan Boschen explained that both produce identical results (not an approximation).

In my simulation you see I have a big difference more than 1 rad. I have mistaken in my simulation, but what the mistake is and how to find it I have no idea.

Did I implement GMSK Transmitter based on AMP wrong ( the first block)?

  • $\begingroup$ I'll have to repeat that Dan didn't say the Laurent approximation was identical to the nonlinear method for generating GMSK, he said that the appropriate linear method was identical to the nonlinear method. May I ask why you choose to approximate the Gaussian using the Laurent decomposition? This feels really strange, as your system has no advantages from it. What's the motivation for that? Why choose such a bad approximation? $\endgroup$ Commented Jun 21, 2022 at 13:29
  • $\begingroup$ Note that I'd argue that without a technical reason to use this very truncated and suboptimal approximation, yes, your choice of filter approximation was wrong. $\endgroup$ Commented Jun 21, 2022 at 22:26
  • $\begingroup$ @MarcusMüller My big limitation, a reason is a transceiver with linear filtering. If I want to implement nonlinear method as Dan suggested, I need to implement it in an addition board (ZYNQ/FPGA). Oversampling ratio will decrease my data rate of the transceiver and resultant data rate is too small for my research. I have found with Laurent I can linearized method and it will perfect match. So... transiever and desired data rate are my reasons for this choice. $\endgroup$
    – FrimHart64
    Commented Jun 23, 2022 at 9:36
  • $\begingroup$ @MarcusMüller "Note that I'd argue that without a technical reason to use this very truncated and suboptimal approximation, yes, your choice of filter approximation was wrong. ".... do you mean the matlab function ` filter` or something else? $\endgroup$
    – FrimHart64
    Commented Jun 23, 2022 at 9:37
  • $\begingroup$ I mean the fact that you're using a length 3 Laurent decomposition instead of actually implementing the Gaussian, which is not any more computationally hard. I'm really not sure you know what you're doing there. Why did you choose to do the Laurent decomposition? Why not simply use a reasonable-length Gaussian filter tap vector? We've established in a previous question of yours that 3 is too short. And you're not only too short, you're on top approximating the filter badly. I really don't understand why. $\endgroup$ Commented Jun 23, 2022 at 9:40


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