How can I find a turning point or inflection point on the curve?

[J.Huang]

I have encountered a problem, is seeking a turning point or inflection point on the curve.

The curve is monotonically increasing (slope is essentially the same); the slope tends to 0 after the turning point. Because of the curve is not actually a straight line, so the derivative of point by point seems comparison difficult.

First, I intercepted two new curves from the curve.  One with monotonically increasing slope and another with remained constant slope. And I transform the two new curves to two absolutely straight with “regression” filter. Then find the intersection point of two absolutely straight. Through the X coordinate of the intersection point to find the corresponding Y value on the curve.Turning point is (x, y).

However, this method can only be used for a fixed curve. Because the range of the intercept values ​​need to give the range, if the new changes in the overall length of the curve, the value of this range may no longer apply.

My question is:  Is there a better way, or Diadem functions associated itself can solve this problem?

Any of your ideas and suggestions is important to me, thanks very much.

Best Regards

Jie

[Brad_Turpin · Accepted solution]

Hi J. Huang,

I would recommend the following generic approach, based on the pictures you submitted:

1)  Create a smoothed channel from the raw data curve to remove random variations

2)  Calculate a derivative/slope channel from the smoothed channel

3)  Use the ChnFind() function to find the points where...

        the slope is close to zero

        the slope is positive

        the raw data is positive

        the raw data is close to the maximum channel value

Brad Turpin

DIAdem Product Support Engineer

National Instruments

[J.Huang]

Hi Brad ,

thank you for your answer.

But can you tell me how to creat a new curve with script.  The new curve is synthesized by two another curves.  You can find the two curves in the picture ofin the attachment "inflection point.JPG 131 KB ".

thanks

Jie

[Brad_Turpin]

Hi Jie,

The approach I outlined above would avoid the need to calculate the asymptote channels by finding the inflection point directly based on the conditions I listed, which would all be included with " AND " conjunctions in a long string parameter and passed to the ChnFind() function.

If you'll send me the data set, I can show you what I mean,

Brad Turpin

DIAdem Product Support Engineer

National Instruments

brad.turpin@ni.com