EP says 2*H/h, I could not interpret correctly.
In order to get S/n ratio, do I have to multiply Height by two times and then divide by noise?

Hi Anila,

I don't know the formula stands for but to me the S/N ratio could be calculated very simple. You divide the signal (=height in cm) of a peak by the noise (=height in cm). The noise height is the height from the highest to the lowest absorbance over a range of some minutes.

Durk

what software are you using.
most of them has an inbuilt calculator.

In European Pharmacopoeia under heading of "Chromatographic separation techniques" 2.2.46, given a formula for signal to noise ratio under "Precision of quantification" as 2*H/h.

I'm using Waters millinieum and Agilent Chemstation.

Hope understood my problem.

Anila, I think you understood correctly the formula.
I perform the same calculation (that is multiply peak height by 2 times and then divide by noise). Using Chemstation, you have to choose the "Point-To-Point" method to get the noise value.

With this calculation, for obtaining S/N ratio of 2 or 3, the height of the peak is almost equal to noise. How can we integrate then.

Integrate is always possible, more difficult is to be reproducible... and it is not compulsory to reach SNR=2/3!
But normally I only have to give results until the LOQ, so I consider it more important than LOD. For LOQ a SNR=10 gives (with DAD-UV) reasonable CV values. Moreover, it is also possible to use other criteria to set LOD.

Is the formula for signal to noise ratio means

Signal/noise or 2*signal/noise.?

If criteria = 2,
Then 2*Signal/noise =2
==> Signal/noise = 1
==> Signal = Noise
So, the response due to our peak = noise.
So, Even if we add some sample , there will not be any reponse.

Is my iterpretation is correct?

can anybody clarify me!

anila: Just forget about this factor of 2. Report your S/N by dividing your peak hight by the baseline width. If this S/N is below a certain number consider your peak to be uncertain. (Usually a figure of 2 to 4 is used for the limiting criteria, that means that some people are confident that they see an analyte peak when they get something 2x higher than their baseline width, other people are not sure until the have something 3x or 4x higher)

Thanks Mueller.

But, why the Ph.Eur saying to multiply by 2, under the section for Precision of quantification.

Any how my doubt is cleared now regarding the formula for S/N.

Thanks to all.

The EP factor of 2 was introduced because noise is deviates from the baseline up- and downwards and peaks are almost always positive.
If your noise is 4 (µV or whatever) 2 will bee above mean and 2 below. A peak height of 4 will be above (mean baseline). In that case you have a S/N = 2H/h = 2*4 /4 = 2. The peak is certainly higher than the noise.
LOD is required be ICH validation guideline. I personally do not understand the reason, as the important criterion is the LOD (often set to S/N = 10).

"If criteria = 2,
Then 2*Signal/noise =2
==> Signal/noise = 1
==> Signal = Noise "
Wrong! Signal is above baseline, noise is 0.5 above baseline, 0.5 below baseline, so ppeak height from baselin is 2*noise from baseline.

Alex

Thanks Alex,

Then what does the term " Precision of quantificaiton" meant for..., under which the said formula given.

Without looking into the EP: The "Precision of quantification" depends on S/N.
Prepare solutions at five levels from - lets say - dection limit to 50*detection limit. Inject each solution six times. Plot the RSD of peak areas against the concentration and you will see what I think its meant. And then tell us the results.

Alex

Thank you very much. I also looked at some text books and found the similar explanation as what you are telling.

If I don't see peak eitheir to left or right of my main peak in that case how to calculate limit of detection or (S/N ratio) . In other words my sample is 99.9 % pure.