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Home/ Questions/Q 6857467
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T02:01:23+00:00 2026-05-27T02:01:23+00:00

I am doing a mini project and i am stuck with a code. My

  • 0

I am doing a mini project and i am stuck with a code. My project is based Positioning.

I have used a GUI to get the original and estimated position of a target.

I have placed 8 nodes to test.

In this the first four are reference nodes and the next four are random position of the unknown nodes.

cors = [6.62650602409639 194.444444444445;
        6.62650602409639 10.6837606837607;
        192.168674698795 7.83475783475797;
        192.168674698795 191.595441595442;
        70.4819277108434 171.652421652422;
        129.518072289157 168.803418803419;
        24.6987951807229 144.586894586895;
        42.7710843373494 79.0598290598291];

I want to plot the graph between x and y co-ordinates.

I have used TOA to estimate the position of target and I got the following output.

cors_est = [6.62650602409639 194.444444444445;
            6.62650602409639 10.6837606837607;
            192.168674698795 7.83475783475797;
            192.168674698795 191.595441595442;
            70.7600705547484 171.847603055024;
            129.443055817301 168.734648868329;
            25.01956026761   144.890243978875;
            42.6058125534278 79.1446327727804];

From this, I have calculated the difference between cors and cors_est, let it be le(x) and le(y) where le is the localization error.

And with the data I have calculated mean(x)= summation of le(x)/8;

similarly,

mean(y)=le(y)/8;
ΔX = LE(X) − Mean(X)
ΔY = LE(Y) − Mean(Y)

Now i need some guidance in doing the following tasks :

  1. To calculate the stdx=square of (ΔX) and stdy=square of (ΔY) and then,

  2. stdxy=summation of ((ΔX)+(ΔY))/N

Actually if I take the square root of the above data, I will get the standard deviation.

Please guide me with this.

Waiting for some good guidance 🙂

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  1. Editorial Team
    Editorial Team
    2026-05-27T02:01:24+00:00Added an answer on May 27, 2026 at 2:01 am

    I think you’re making this too hard. How about:

    meanXandY = mean(cors - cors_est, 1);
    

    And

    stdXandY = std(cors - cors_est, 0, 1);
    
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