Many papers involving ultrafast spectroscopy measure the instrument response function and model it as a Gaussian, but the sources on any of this are never cited and I have no intuition for what this could mean. What is the instrument response function intuitively and why is it represented as a convolution?

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    $\begingroup$ I Googled "What is the instrument response function" and got this hit. You might want to do some perfunctory research and amend your query to include what your efforts yielded. $\endgroup$ May 25, 2021 at 14:57
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    $\begingroup$ @ToddMinehardt Good hit! This query is really better suited to the signal processing stack exchange: they are all about linear, time invariant (LTI) systems, impulse responses, frequency domain transfer functions, and on and on. TLDR: in the time domain, the output of a LTI system is the convolution of the input with the impulse response function of the LTI system. In the frequency domain, the Fourier transform of the output is the product of the transfer function and the Fourier transform of the input. $\endgroup$
    – Ed V
    May 25, 2021 at 15:13
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    $\begingroup$ More generally, you want to look for an "impulse response" function, which dictates how the spectrometer (or any LTI system) responds to input. It's a bit long, but the Signals and Systems course on MIT OCW is a good place to learn these things. Of course, it raises the question of whether your spectrometer is truly LTI :-) $\endgroup$ May 25, 2021 at 15:15
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    $\begingroup$ Usually this a measured via autocorrelation of the laser pulse and there are various ways of doing this. Then you have to assume a pulse shape and fit to the autocorrelation. Actually what you measure is the convolution of the actual pump and probe pulses with the molecular response. If the response is very long, say nanoseconds then you can measure the rise time profile, perhaps as a few tens of femtoseconds depending on you laser system, and use this to obtain quite a good estimate of the pulse duration by fitting assumed pulse shape to the data. This ignores any chirp in the pulses. $\endgroup$
    – porphyrin
    May 25, 2021 at 16:16
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    $\begingroup$ There are great answers, I wish I could select both of them. Thank you! $\endgroup$ May 26, 2021 at 14:15

2 Answers 2


Let me provide the intuitive idea so that you take the lead from here and understand the notion behind "instrument response" function. Any analytical/physical chemist, who uses an intrument, should have some basic idea of the instrument response. After that you should look up formal concepts of convolution on Youtube.

Whenever you collect experimental data using an instrument, you would like to have no distortions induced by the instrument in your data. For example, if your eyesight is weak, and you are trying to read a road sign, but it appears blurred to you, your eyes and brain have "convoluted" the true image of the road sign. Therefore, your eye/brain response (=instrument response) is to blurr the image and this is what you do not want. At least, one should be aware of this fact that the true image is not a blurred picture.

Hence people who wish to study lineshapes, linewidths in a spectrum they would like to know how their spectrometer set-up is distorting the true experimental result. In an ideal case there should be no distortion but a real instrument can introduce its own effects which we call as convolution.

Let us see a example to illustrate what Gaussian broadening does to a "true" signal shape, using an light absorbance detector (of a chromatograph). I did this just to understand the response function of that instrument.

Imagine that a light source is turning on and off periodically. It should look like a square wave to a light detector (signal is up when the light is on and it is down when the light of off). However, if you have increasing Gaussian broadening done by the instrument, you begin to loose the true shape of the square wave. If you have too much of Gaussian broadening, the square wave, appears to be completely distorted. Look at the top and the bottom figure. The top shows the relatively true signal shape, and the bottom figure shows the effect of Gaussian broadening on the same signal. All true information is lost.

Hope that highlights the importance of instrument response function in analytical chemistry/instrumental analysis.

enter image description here


Suppose that the lifetime of electronically excited atoms or molecules is to be measured by exciting them with a pulse of light and their fluorescence measured as it decays with time. This fluorescence could be observed with a photodiode or photomultiplier, whose output voltage is measured with an oscilloscope. Before doing this experiment, two questions have to be answered;

Is the laser used to excite the molecules of short enough duration that the molecules or atoms can be excited quickly enough before any significant number can decay back to the ground state?

Is the detection equipment (photodiode, oscilloscope) used able to respond quickly enough to measure the decaying fluorescence properly?


If either one or both of these conditions cannot be met, then the data will be distorted by the relatively slow response of the instrument. The convolution curve in the figure shows how this distortion affects data. In this figure, the top curve is the ideal decay of the excited state, but it could represent any ideal response. This behaviour would be observed if the molecules could be excited with an infinitesimally narrow laser pulse and measured with a photo-detector with an unlimited time response. The second curve is the actual shape of the laser pulse, and/or detector response, and is the 'instrument response' drawn on the same timescale. Clearly, this has a width and a rise and decay time that is not so very different to that of the ideal response. The lower curve is the convolution of the ideal response with the instrument response, and is what would be measured experimentally and clearly has characteristics of both curves. A log plot of the data would show that only at long times does the convoluted response have the same slope as the ideal one. It makes no difference if the instrument response consists of a slow 'driving force' for the experiment, in this case a long-lived light-pulse, or a slowly responding detector or both, because the effect producing the convolution is the same.

To understand how convolution works, suppose that the overall instrument response is made up of a series of $\delta$-function impulses. Suppose these impulses are made at ever shorter time intervals, then the effect is that of smoothly exciting the object. After a while these impulses stop. Each of the impulses elicits an ideal response but because there are many of them, their responses must be added together. The result is the convolution; the effect is shown below. It is always assumed in the convolution that the response is linear with the impulse, which simply means that doubling the impulse doubles the response and so forth.

Written as a summation, the convolution at point $k$ is

$$\displaystyle C(k) = \sum_{i=0}^k f(i)w(k - i ) \tag{32}$$

This sum evaluates just one point; to calculate the whole convolution, the index $k$ must now be varied from 1 to $n$, which is the number of data points, making a double summation. One reason Fourier transforms are used to calculate convolutions is that the fast Fourier transform algorithm, FFT, is far quicker on the computer than calculating the convolution as a double summation, particularly for a large number of data points. (However, FFT should not be used if the signal at beginning and end of data is not of the same magnitude. This caused severe distortion in the convoluted data and is a real drawback when doing experiments, i.e. it is safer to use summation, even though it is slower to calculate. )

convolution 2

Suppose that the pulse exciting the sample has a shape given by some function $f$, the ideal experimental response $w$, and the convolution $C$. The terms can be written down at each time if it is assumed, for the present, that the impulses are discrete and the data is represented as a series of points at times $1, 2, 3$, and so forth; $f$(6), for example, represents the value of $f$ at the sixth time position. The first point of the impulse is $f$(1) and this produces the response

$$\displaystyle f (1)[w(1) + w(2) + w(3) + \cdots]$$

The second and third impulses produce

$\displaystyle f(2)[w(1) + w(2) + w(3) + \cdots]$ and $f(3)[w(1) + w(2) + w(3) + \cdots]$.

The convolution is the sum of these terms at times 1, 2, 3, and so on therefore;

$$\displaystyle\begin{align} C(1)& = f (1)w(1)\\ C(2)& = f (1)w(2) + f (2)w(1)\\ C(3) &= f (1)w(3) + f (2)w(2) + f (3)w(1)\\ C(4)& = f (1)w(4) + f (2)w(3) + f (3)w(2) + f (4)w(1)\\ \end{align}$$

You can see how the series folds back on itself, hence the name convolution.

  • $\begingroup$ (+1) Excellent example and explanation! $\endgroup$
    – Ed V
    May 26, 2021 at 11:23
  • $\begingroup$ @Porphyrin, Nice illustration. You can add that DFT can produce the same linear linear convolution if we make sure the length of the DFT is N1+N2-1 (by zero padding). There are nice tricks to get rid of circular convolution effects. $\endgroup$
    – AChem
    May 26, 2021 at 15:50
  • $\begingroup$ @M. Farooq , thanks I had forgotten to mention that. $\endgroup$
    – porphyrin
    May 26, 2021 at 16:23

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