Getting the absolute max value using the calculator

Without having to write a custom function, you could do it this way:

value(VT("/out") xmax(abs(VT("/out"))))

xmax gives the x-value where the ymax occurs.

Andrew

Dear Kaspar,

KasperM said:

I am trying to find the global Y peak value of a graph. Basically the ymax but than also with the posibility of it being negative.
Using ymax(abs(expression)) will not cut it as I'd like to preserve the sign.

Is there a simple way of doing this?

I happened to think of this last night after I first read your post. The sign is preserved by including the second term. I think this might accomplish your objective as well as Andews's suggestion.

Shawn

Dear Kaspar,

As I looked at my expression, I realized I made an error! Sorry! I think the correct expression should read:

In this fashion if the maximum value is positive and the minimum value is negative and the positive value exceeds the negative value in absolute value, the term:

will be +1., If the minimum value is negative and exceeds the maximum value in absolute value, the value of ymax(expression) + ymin(expression) will be less than 0 and the added term will be -1.

Hence, I believe this should retain the sign of your maximum absolute value.

Sorry for my initial error Kaspar!

Shawn

Dear Kaspar,

As I looked at my expression, I realized I made an error! Sorry! I think the correct expression should read:

In this fashion if the maximum value is positive and the minimum value is negative and the positive value exceeds the negative value in absolute value, the term:

will be +1., If the minimum value is negative and exceeds the maximum value in absolute value, the value of ymax(expression) + ymin(expression) will be less than 0 and the added term will be -1.

Hence, I believe this should retain the sign of your maximum absolute value.

Sorry for my initial error Kaspar!

Shawn

Hi Shawn,

A small nit-pick on your proposed solution (which is a neat idea, although does involve more processing of waveforms than mine); it won't behave properly if the ymax and ymin are equal (but with different signs) - such as with an ideal sine wave centred around 0; then you'll have a divide by zero error.

Regards,

Andrew.

Dear Andrew,

Andrew Beckett said:

A small nit-pick on your proposed solution (which is a neat idea, although does involve more processing of waveforms than mine); it won't behave properly if the ymax and ymin are equal

You are absolutely correct!  I did realize the potential flaw when I thought of it.  I was going to include it in writing, but thought it might be obvious. And, yes, you are also correct it will require more processing. I thought I would just throw out the idea if it might spark some interest. Thank you, as always, for reading my post and including your comment and the caveat explicitly.

Shawn