How do I pretend that my high resolution spectrometer only has low resolution? I have a spectrometer with resolution high enough to recognize the fine structure in alkali metal. Now if I add these two peaks together and ignore the non-resonance part between them, what do I get? Is it the right way to pretend that my high resolution spectrometer only has low resolution?
 A: The way to artificially decrease the resolution of your spectrum is to convolve it with a broadening function. Typically you'd use a gaussian.
Suppose your spectrum is the function $F(\lambda)$ i.e. the function $F(\lambda)$ gives the intensity measured at the wavelength $\lambda$. Take the function $g$ given by:
$$ g(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-x^2/(2\sigma^2)} $$
and calculate the new function:
$$ F'(\lambda) = \int_{-\infty}^\infty F(x)g(\lambda - x) dx $$
This is called the convolution of $F$ with $g$, and the effect is to smooth out the function $F$ and decrease the resolution. The parameter $\sigma$ determines the degree of smoothing. The larger the value of $\sigma$ the greater will be the smoothing. You will need to experiment to get the amount of smoothing you want.
This may seem a bit odd, but this is roughly what happens in real life. If $F$ is the perfect spectrum and $g$ describes the resolution of your spectrometer then the convolution of $F$ and $g$ is the measurement that you will get.
Finally, the odd prefactor of $1/\sigma\sqrt{2\pi}$ is just a normalising factor. It keeps the overall intensity of your smoothed spectrum constant.
