Hi Don,
Assuming the Gaussian has the form: f(x) = a*exp{ -0.5*[(x-m)/s]^2}, then the total area under the curve (integral from -Inf to Inf) is exactly sqrt(2*pi)*a*s, and the fwhm is 2*sqrt(2*ln(2))*s. It becomes a little trickier to evaluate the integral for limits other than +/- Inf. The error function in LabVIEW is defined as:
erf(a) = [2/sqrt(pi)] * integral(from 0 to a) exp(-t^2) dt
let t = (x-m)/(sqrt(2)*s), and apply some algebra, and the integral(from 0 to a) f(x)dx = a*sqrt(2*pi)*s*erf[ (a-m)/(s*sqrt(2)]
Attached is a VI which compares results from Quadrature.vi (general purpose integrator) with the integral evaluated using erf() (error function.vi) and the closed-form expression sqrt(2*pi)*a*s. The integral from -Inf to Inf is approximated by doubling the integral from 0 to 100. As expected the erf-based integral is much faster than the adaptive quadrature algorithm.
If the Lorentzian has the form: f(x)=a/(1+((x-m)/c)^2), then total area (integral from -Inf to Inf) is pi*a*c, and the fwhm is 2*c
-Jim
