NaN issue and if statement

https://ww2.mathworks.cn/matlabcentral/answers/2143754-nan-issue-and-if-statement#comment_3232259

I'm having trouble with my code when I add an if condition to check if B_body(3) is less than 1e-5.

Without the if condition, everything works fine.

But when I add the if condition, I get a warning and B_body(3) values become NaN.

I've tried several solutions, but nothing works.

Any ideas on how to fix this?

epsilon=1e-5;
if abs(B_body(3))<epsilon
    T_rw=[0;0;0];
else
    T_rw = [0; 0; 1] * (B_body' * T_commanded) / (B_body(3));
end

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dpb 2024-8-7

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https://ww2.mathworks.cn/matlabcentral/answers/2143754-nan-issue-and-if-statement#comment_3232279

We don't have enough code to know what happens next, but almost certainly what then occurs after the T_rw vector is set to identically zero is that B_body(3) also becomes identically zero and then the first step after that you end up with a case of

0/0
ans = NaN

which is where the NaN comes from. Beyond that, all hope is lost.

You'll need some other special casing to deal with whatever is supposed to happend afterwards, but the model as written doesn't handle the case when whatever T_rw represents becomes identically zero.

Steven Lord 2024-8-7

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But when I add the if condition, I get a warning

What is the full and exact text of the warning message? Please show us all the text displayed in orange in the Command Window. The exact text may be useful and/or necessary to determine what's going on and how to avoid the warning.

Rik 2024-8-7

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If you put your code in your question or a comment, you can run it. That way you can confirm we will see the same error as you.

Sahas 2024-8-8

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https://ww2.mathworks.cn/matlabcentral/answers/2143754-nan-issue-and-if-statement#comment_3232854

Hi @Alexandra,

As others rightly mentioned, looks like the issue might be arising from the division by B_body(3) when it is very close to zero which leads to division by zero which in turn might be the reason for NaNs.

To avoid this, you can add a small value (like delta) to the denominator to prevent division by zero. You can try this out:

epsilon = 1e-5;
delta = 1e-6;
if abs(B_body(3)) < epsilon
    T_rw = [0; 0; 0];
else
    T_rw = [0; 0; 1] * (B_body' * T_commanded) / (B_body(3) + delta);
end

This approach slightly modifies the calculation, so make sure it still meets your requirements. If precision is crucial, you might need to handle the code more carefully.

I hope this is beneficial to you. If you need further assistance, please feel free to reach out.

Stephen23 2024-8-8

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编辑：Stephen23 2024-8-8

https://ww2.mathworks.cn/matlabcentral/answers/2143754-nan-issue-and-if-statement#comment_3232879

"when it is very close to zero which leads to division by zero which in turn might be the reason for NaN"

Dividing zero by zero leads to NaN. Even dividing zero by the nearest value closest to zero (i.e. the smallest subnormal) does not produce NaN:

0/eps(0)
ans = 0
0/0
ans = NaN

Being "very close to zero" is not sufficient.

Sahas 2024-8-8

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Thanks you @Stephen23 for clearing that out.

dpb 2024-8-8

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编辑：dpb 2024-8-8

https://ww2.mathworks.cn/matlabcentral/answers/2143754-nan-issue-and-if-statement#comment_3233334

And, just for good measure, anything nonzero divided by 0 produces a signed Inf, not NaN.

eps(0)/0
ans = Inf
realmax/0
ans = Inf

ADDENDUM:

(1)/(-0)
ans = -Inf
(-1/0)
ans = -Inf
(-1)/(-0)
ans = Inf

NOTA BENE: the sign is based on the sign of the quotient; the (-0) in the denominator is also significant. This can be helpful in some branch analyses...

Steven Lord 2024-8-8

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https://ww2.mathworks.cn/matlabcentral/answers/2143754-nan-issue-and-if-statement#comment_3233449

And, just for good measure, anything nonzero divided by 0 produces Inf, not NaN

To be technically correct it produces an infinite value, not necessarily positive Inf.

y = (-1)/0
y = -Inf

There are three non-finite values in MATLAB: Inf, -Inf, and NaN. Those are the values for which the built-in isfinite function will return false, but only the first two will have isinf return true.

x = [1; Inf; -Inf; NaN; 5];
fin = isfinite(x);
infinite = isinf(x);
results = table(x, fin, infinite, VariableNames = ["value", "is finite", "is infinite"])
results = 5x3 table
    value    is finite    is infinite
    _____    _________    ___________

       1       true          false   
     Inf       false         true    
    -Inf       false         true    
     NaN       false         false   
       5       true          false   

dpb 2024-8-8

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OK, Steven, you're correct...I didn't mean to imply all would return +Inf...I added the (I thought implied) "signed".... :)

Alexandra 2024-8-9

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Thank you all for your helpful suggestions and for taking the time to assist me with my question. I think I have finally found a solution that works for my needs.

dpb 2024-8-10

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