baseline resolution

[push]

Hi,
When I try to figure a so-called baseline resolution I get Rs =1. How do all courses report Rs=1.5 for baseline resolution. Where is an error?

[tom jupille]

Resolution is the ratio of center-to-center separation (the difference in retention times) to the average baseline width.

Start by assuming two perfectly Gaussian peaks of the same width and same height.

Baseline width is equal to 4 sigma (sigma is the square root of the variance; think of it as the standard deviation in statistical terms). To a good approximation, that accounts for approximately 95% of the peak area (+/- 2 sigma on either side of the mean).

So, at Rs = 1.0, what you have is a "4-sigma" separation, which means you have a 5% mutual overlap.

Rs = 1.5 is a "6-sigma" separation. Because +/- 3 sigma accounts for 99% of the area, that means that at Rs = 1.5, you have just less than 1% mutual overlap. In fact, Rs = 1.5 was originally called "99% baseline resolution". Over the years, the "99%" bit has been forgotten.

[FranziskaW]

Hello Tom Jupille,

I have a question regarding the calculation of the mutual overlap. Maybe this is totally basic, but I could not find an appropriate explanation anywhere.

How do you calculate (starting from the fact, that at Rs=1 w=4-sigma, equalling < 5 % mutual overlap)
1) that Rs=1.5 equals w=6-sigma. Obviously this is just 1.5 x 4-sigma, but why can you just multiply this?
2) that 6-sigma equals < 1% overlap?

The formula to calculate the resolution (Rs = (tR2-tR1)/(0.5 x (w1+w2), does not seem to be of use, because I cannot insert numbers for the retention times, because I want to have a general calculation?

I would like to be able to follow the calculation.

Thank you!

[vmu]

Rs = (tR2 - tR1)/(0.5*(w1 + w2)).
For Gaussian peaks, w1 = 4*sigma1, w2 = 4*sigma2.
Assuming w1 = w2 = 4*sigma (i.e. sigma1 = sigma2 = sigma) for close peaks, one gets Rs = (tR2 - tR1)/(4*sigma).

Rs = 1.0 when tR2 - tR1 = 4*sigma; hence, one speaks about "4-sigma resolution".
Rs = 1.5 when tR2 - tR1 = 6*sigma; hence, one speaks about "6-sigma resolution".

A Gaussian peak is described by
I(t) = [A/(sigma*(2*Pi)^0.5)]*exp(-((t - tR)^2)/(2*sigma^2)),
where A - is the peak area.
The minimum between two partially separated peaks with identical parameters A and sigma is found at t = (tR1 + tR2)/2 = T.

The area of the part of the first peak that is on the right from T is A minus integral of I(t) from 0 to T. For the 4-sigma separation, this area is 2.275 % of A. The area of the part of the second peak that is on the left from T is the same, i.e. 2.275 % of A. One can say that the overlap is 2*2.275 % = 4.55 % of A (less than 5 %).

Similar calculations for the 6-sigma separation result in the overlap of 2*0.135 % = 0.27 % of A (less than 1 %).

[tom jupille]

. . . and because the overlap at 6 sigma is just under 1%, people started calling 6-sigma separation as "99% baseline resolution". Over time, the "99%" bit got dropped off and it was just called "baseline resolution".