FWHM of Lorentzian

Do the algebra to find where the level is at one half of the peak amplitude (neglect the baseline y0):

(x-x0)^2 + B = 2*B
|x-x0| = sqrt(B) (at HWHM)
FWHM = 2*sqrt(B)

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Thanks S.r.chinn,

I'm glad i'm not going too mad. The reason why i ask is that I did a quick lorentzian fit on my data and got this as an output:

Coefficient values ± one standard deviation
y0 =1.0451 ± 0.544
A =94831 ± 1.19e+004
x0 =654.5 ± 1.97
B =1893.6 ± 278

So if B=(1/2 * FWHM)^2 then A=1/2 * FWHM
When i look at my peak have a FWHM at ~87 and an amplitude/height A~43 makes perfect sense. Where then does the A=94 831 come from?

Apologies if i am being completely obtuse. I can't see how these numbers fit...but they seem to be doing the correct things...

Thanks,
N

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"Coefficient values ± one standard deviation
y0 =1.0451 ± 0.544
A =94831 ± 1.19e+004
x0 =654.5 ± 1.97
B =1893.6 ± 278

So if B=(1/2 * FWHM)^2 then A=1/2 * FWHM
When i look at my peak have a FWHM at ~87 and an amplitude/height A~43 makes perfect sense. Where then does the A=94 831 come from?"

Your FWHM calculated from 2*sqrt(B) seems to be correct. It should occur at a level of 47416, which is 1/2 of your measured (or fit) peak amplitude. The ratio of 43 you cite (whatever it means) is irrelevant and confusing. There is no general intrinsic relation between peak amplitude and FWHM (they are independent parameters from a curve-fitting point of view). Your question was how to convert from B to FWHM. Note that the algebra I provided did not invoke 'A'.

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Hi s.r.chinn,

Thanks for the feedback. I was confused because the equation for a lorentzian distribution can be expressed as follows:
f(x)= [(1/2)*FWHM]/[(x-x0)^2+((1/2)FWHM)^2]
which is the closest to the form IGOR adopts. (There are other ways to express it that have pi's and other factors involved)

In this case B=((1/2)*FWHM)^2 which leaves is with 2squrt(B) which makes sense.

But A=1/2 *FWHM should equal the amplitude.

Now when I look at my data the B is correct and my calculation of A~43 using the equation above is the height of the curve. This is not the same 'A' IGOR outputs. This is where I'm confused.

But as you say the FWHM=2squrt(B) makes perfect sense and I got rid of half my doubts!

Thanks =)

N

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