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?
9
- 2$\begingroup$ By "pretend" you mean reproduce exactly the measurement result from a lower resolution spectrometer? $\endgroup$stafusa– stafusa2018-02-08 14:34:02 +00:00Commented Feb 8, 2018 at 14:34
- 2$\begingroup$ One could always convolute the data with a lower resolution function. $\endgroup$Jon Custer– Jon Custer2018-02-08 15:30:05 +00:00Commented Feb 8, 2018 at 15:30
- 8$\begingroup$ @JonCuster Convolve $\endgroup$BlackThorn– BlackThorn2018-02-08 15:54:54 +00:00Commented Feb 8, 2018 at 15:54
- 4$\begingroup$ @DoktorJ modelling whether a low-resolution measurement can give sufficient information for an experiment based on very good data is useful with FTIR, where there's a trade-off between resolution and time (and signal-to-noise). You may not simply be able to wait longer if you can't treat you sample as constant. Or as dmckee says, getting simulation data into an experimental scale, as I did in my thesis: thermal simulations have higher resolution than Raman thermography, so convolution is how you compare simulation to experiment (in 3D) $\endgroup$Chris H– Chris H2018-02-08 16:53:20 +00:00Commented Feb 8, 2018 at 16:53
- 3$\begingroup$ @BlackThorn - Mea culpa, probably because the result is a convolution, it was not uncommon for 'convolute' to be used colloquially several decades ago. $\endgroup$Jon Custer– Jon Custer2018-02-08 17:05:55 +00:00Commented Feb 8, 2018 at 17:05
| Show 4 more comments
1 Answer
$\begingroup$ $\endgroup$
9 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.
- $\begingroup$ Thank you, sir. I basically understand your answer but still don't understand the effect of non-resonance part if the resolution is really high, could you please explain it? $\endgroup$Bettertomo– Bettertomo2018-02-08 09:51:22 +00:00Commented Feb 8, 2018 at 9:51
- 2$\begingroup$ Uh, I suddenly don't understand my own question, forgive me! $\endgroup$Bettertomo– Bettertomo2018-02-08 10:30:08 +00:00Commented Feb 8, 2018 at 10:30
- 1$\begingroup$ Convolving with a Gaussian will be essentially a form of low-pass filtering. From there, you could (if you wanted) down-sample to actually get lower resolution data. This combination of low-pass filtering followed by down-sampling is called decimation. The (basics of the) theory of low-pass filtering and decimation may be worth looking at to, e.g., avoid aliasing by not low-pass filtering aggressively enough (not an issue if you don't down-sample). $\endgroup$Derek Elkins left SE– Derek Elkins left SE2018-02-08 15:43:17 +00:00Commented Feb 8, 2018 at 15:43
- 2$\begingroup$ A cruder but simpler approach (often used for noddling around analyses) is to simple re-bin the histogram into fewer bins. $\endgroup$dmckee --- ex-moderator kitten– dmckee --- ex-moderator kitten2018-02-08 15:58:29 +00:00Commented Feb 8, 2018 at 15:58
- 1$\begingroup$ Computationally, you'd take your data and make it discrete (in magnitude and frequency) at sufficient resolution. Then you'd apply a FFT (fast fourier transform), bandlimit, and inverse-FFT. Or, if you are married to a guassian, you apply a FFT, pointwise multiply with guassian kernel, then inverse-FFT. Actually convolving is computationally expensive! Or you'd throw the signal and the kernel at an engine that (under the hood) does the FFT, pointwise multiplication, inverse-FFT without telling you. $\endgroup$Yakk– Yakk2018-02-08 16:03:49 +00:00Commented Feb 8, 2018 at 16:03