atomic spectroscopy - having trouble understanding why source modulation eliminates interference by emitted radiation (basically, why does fluctuating the source correct for interference?)

Question

Vocaloid · Answer

nuts · Answer

The key to it is probably in "an alternating one in the source and a continuous one in the flame"

We don't know the effects of the flame on the source. Let's say the flame results in a constant 1 mV electrical signal while the source is 9mV. Your electrical circuit would receive 9+1=10mV, but wouldn't have any reference for this (no way to know what 10mV means).

However, if you modulated the source to say, 9sin(240*pi*t) (frequency chosen to be 120Hz), your signal would now be 1+9sin(240*pi*t) mV. Now, with your RC low pass filter with a cutoff frequency of say 60Hz, (blocking common powerline frequency), you will actually remove the 1mV DC (continuous) component leaving just the 9sin(240*pi*t), allowing you to determine emission from the source.

Summary: modulating the source is a method that relies on a key feature of electrical systems being able to deal with different FREQUENCIES separately.

An analogy might be, if you want to measure the mass of a reactant in an aqueous solution without knowing the concentration, you might be able to just precipitate it and filter (see what i did just there) and weigh that instead.

nuts · Answer

@Vocaloid

Vocaloid · Answer

ah. thank you.