Frequency to temporal period conversion

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SUMMARY

The discussion focuses on the conversion between frequency and time domains in signal processing, specifically analyzing the relationship between Free Spectral Range (FSR) and Full Width at Half Maximum (FWHM). The calculated values are FSR = 2.05 × 106 Hz and FWHM = 6.6 × 104 Hz, leading to a ratio of 30.6 in the frequency domain. However, when converted to the time domain, the FWHM appears larger than the FSR, which contradicts the expected relationship. The key takeaway is that FSR in one domain corresponds to FWHM in the other, necessitating careful inversion of calculations during conversions.

PREREQUISITES
  • Understanding of signal processing concepts
  • Familiarity with Free Spectral Range (FSR) and Full Width at Half Maximum (FWHM)
  • Knowledge of frequency and time domain analysis
  • Proficiency in mathematical conversions between domains
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  • Study the mathematical principles of signal transformation
  • Learn about the implications of FSR and FWHM in optical systems
  • Research time-frequency analysis techniques
  • Explore simulation tools for signal processing, such as MATLAB or Python libraries
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This discussion is beneficial for signal processing engineers, optical system designers, and researchers analyzing frequency-time domain relationships in signals.

roam
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I am working with signals that look like the following. The curve has the same form in both frequency and time domain.

I am trying to calculate the ratio of two distances:
The distance between two adjacent red "x"s (known as the free spectral range (FSR)).
The distance between two adjacent green "x"s (FWHM of the dips).

243563


For the simulation above I get ##\text{FSR} = 2.05 \times 10^6 Hz##, and ##\text{FWHM} = 6.6 \times 10^4 Hz##. So a calculation in frequency domain gives the correct answer:

$$2.05 \times 10^6 / 6.6 \times 10^4 = 30.6,$$

whereas with conversion to time domain I get:

$$(1/2.05 \times 10^6) / (1/ 6.6 \times 10^4) = 0.0326.$$

Note that FWHM of a curve must be smaller than FSR in both time and frequency domain. But when I take the inverse of these values (i.e., shift to time domain), FWHM appears to be larger than FSR!

Here is an actual time domain measurement. As one can see, it has the same general shape as it would in the frequency domain:

243564

Again, if I calculate the ratio in time domain I get the right answer, but conversion to frequency domain does not give the right answer. Why is that?

Any explanation would be appreciated.
 

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1/30.6=0.0326

roam said:
Note that FWHM of a curve must be smaller than FSR in both time and frequency domain.
That is sort of true. If you were considering a single domain then yes. But the FWHM in the frequency domain corresponds to something with a long time in the time domain, and he FSR in the frequency domain corresponds to something with a very small time in the time domain.

So the thing that corresponds to the FSR in one domain corresponds to the FWHM in the other domain. That is why you need to invert your calculation in the time domain
 
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