please help me in my back projection algorithm.

Dear All, the attached file is for the Cross-correlation between the TX and RX. and my target is one rod. what i am trying to do in my code is backprojection in time domain not in frequency domain in order to get the target image. but it is not work. I do not know if it is from the data or the algorithm. if anyone has information please let me know. Thank you so much in advance.
load AAA
for Theta = 0:360
Current_Range = AAA(:,Theta+1);
Current_Image = zeros (41,41);
iii = 0; jjj = 0;
for xxx = -20:1:20
iii = iii +1;
jjj = 0;
for yyy = -20:1:20
jjj = jjj +1;
Tx_Pos = [sin(Theta.*pi./180) cos(Theta.*pi./180)].*3;
Distance = norm(Tx_Pos-[xxx yyy]);
Current_Image(iii,jjj) = Current_Range(round(Distance/(3e8/200e6/2))+1);
end
end
pcolor(abs(Current_Image));
pause

5 件のコメント

Walter Roberson
Walter Roberson 2016 年 9 月 18 日
What do you observe that leads you to suspect it is not working?
Mansour Aljohani
Mansour Aljohani 2016 年 9 月 23 日
I observe that the result is not at focused on the target, in other word, the result of the crosscorrelation gives you peak wich is the target and at each time of the correlation i want to find the pixel that contain the peak or the target.It might be from the Tx_pos. I need to look at each pixel for the target with respect to the TX_pos.
Thank You
Mansour Aljohani
Mansour Aljohani 2016 年 9 月 25 日
Dear Roberson,
I saw that you edited on some thing. I am not sure, what was this?
Thank you for trying helping me.
Walter Roberson
Walter Roberson 2016 年 9 月 25 日
I am not sure, but I think it was that I removed an "if true" around your code. The best way to enter code is to type (or paste it in) and then highlight it all and then click on the {} Code button.
Mansour Aljohani
Mansour Aljohani 2016 年 9 月 26 日
Oh, I see. i got confused, because my problem is about Image processing, or Signal processing. I don't know.

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