BAD REFERENCE SIGNALS ERROR IN TDOA

WARNING: BAD REFERENCE SIGNALS: ref delays differ by more than 2 samples!

taking ref with higher reliability, i.e. ref check (reliability: 0.00017386(ref) < 0.0054402(ref check)

[Walter Roberson]

What code are you using? https://www.mathworks.com/help/lte/ug/time-difference-of-arrival-positioning-using-prs.html ?

[john corner]

  %% Calculate Correlation Results
    
    if (interpol <= 1)
        delay1 = delay1_native;
        delay2 = delay2_native;
        delay3 = delay3_native;
    else
        delay1 = delay1_interp;
        delay2 = delay2_interp;
        delay3 = delay3_interp;
    end 
    
    
    if abs(delay1 - delay3) <= 2
        avg_delay13 = (delay1 + delay3) / 2; % this delay includes: 1) different reception start time 2) ref signal delay due to different distances to ref transmitter
    else
        disp('<strong>WARNING: BAD REFERENCE SIGNALS: ref delays differ by more than 2 samples! </strong>');
        if corr1_reliability > corr3_reliability
            avg_delay13 = delay1;
            disp(['<strong>taking ref with higher reliability, i.e. ref (reliability: ' num2str(corr1_reliability) '(ref) > ' num2str(corr3_reliability) '(ref check) </strong>']);
        else
            avg_delay13 = delay3;
            disp(['<strong>taking ref with higher reliability, i.e. ref check (reliability: ' num2str(corr1_reliability) '(ref) < ' num2str(corr3_reliability) '(ref check) </strong>']);
        end
        
    end

Answer 1

[Walter Roberson]

It looks like you are using a system that has three signals total, two of which are intended to be the same reference signal, and the third is probably the signal whose distance is to be judged. But the correlation between the two reference signals is more than 2 samples. If your sampling frequency is 22100 Hz, that would correspond to 1/11050 seconds; at 343 meters per second in air, that would be

format long g
Fs = 22100;  %samples/second
SoS = 343;  %m/s
Nsamp = 2;  %samples
distance_m = Nsamp / Fs * SoS

distance_m = 
        0.0310407239819005

distance_cm = distance_m * 100

distance_cm = 
          3.10407239819005

But if your sampling frequency is 8000:

Fs = 8000;  %samples/second
SoS = 343;  %m/s
Nsamp = 2;  %samples
distance_m = Nsamp / Fs * SoS

distance_m = 
                   0.08575

distance_cm = distance_m * 100

distance_cm = 
                     8.575

I would guess that you are intended to place your microphone at the same distance relative to two speakers that are emitting the same sound, in order to calibrate, but that you have not done that placement accurately enough.