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Signals, physics, and instrumental response
In many cases the signal you obtain contains the physics of the phenomena as well as the
physical characteristics of the instruments used for measurement. Part of your job is to separate
the physics part of the signal from the instrumentation part.
The problem: Let’s look at this in the context of an optical measurement. Suppose you
send light from a source, through a monochromator (only certain wavelengths pass), through a
sample, then you detect the intensity as a function of the wavelength.
You would hope that the signal characteristics (amplitude, phase, polarization, etc.) are
telling you something about the physics of the sample. However, it could also be telling you
about the physics of the instrumentation. For example, the light source might not have the same
intensity at all wavelengths, the mirrors or gratings in the monochromator might have different
efficiencies at different wavelengths or change the polarization, or the detector sensitivity might
depend on wavelength. Any of these could cause your signal to vary as a function of wavelength
even though the physics of the sample was constant over that range. Let’s restrict our discussion
to the effect of wavelength.
The signal you measure is actually the product of the physics of the sample and the
instrumental response.
Signal = (Physics of sample)(Instrumental response)
S( " ) = P( " )I( " )
In order to get just the physics we need to divide out the instrumental response at each
wavelength, how do we figure out the instrumental response?
!
The solution:
The way to do this is to run your experiment without the sample. This will
allow you characterize just the instrumental response as a function of wavelength. To see how
this works let’s look at a simple example. On the left is the signal obtained with the sample in
place. The right shows the instrumental response with no sample in the system.
S
I
2
6
1
!
1
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2
3
4
2
!
1
2
3
4
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Looking at the signal you might think there is a minimum between 2 and 3 in wavelength that is
about 1 unit wide…perhaps this is interesting physics. However, the instrumental response is not
uniform over the range of wavelengths measured. Also notice that the signal without the sample
is overall larger than with the sample. (Apparently the sample reduces the overall signal size
along with any other physics.)
So we can find the physics of the sample by dividing the measured signal by the
S (! )
instrumental response at each wavelength, P(! ) =
. This produces the following:
I (! )
From this we see that the minimum is actually between 1 and 2 and that it is wider than we
would have thought from just looking at the signal, hence there is some different physics.
Mjm SignalsInstrumentsW10.doc
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