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I have a doubt on the relationship between phase and frequency. We know that the definition of angular frequency is:

ω(t) = d(\$phi\$(t) / dt, where \$phi\$(t) is the istantaneous phase of our signal.

In this case ω(t) is a function of t. We can write:

enter image description here

If we pass to Laplace domain, we get:

enter image description here

Is it correct to say that, if we want to pass to Fourier Transform, s = jω, and so writing

\$phi\$(jω) = ω (jω) / jω

?

It seems quite strange to me (I think that s = jω represents the angular frequency of the signal "angular frequency" ω(t), and so not equal to jω), but somewhere I read this statement.

For instance, I saw this statement for proving that the Spectrum (power spectral density) of Phase noise is bounded to that of Angular frequency through the relationship:

enter image description here

Kinka-Byo
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  • Technically, \$s=\sigma + j\omega\$ ... If you assume there's no real component of \$s\$ then yes, it's just \$j\omega\$ . –  Jun 26 '19 at 15:03
  • But is that omega, i.e. the pulsation of w(t), exactly equal to w? – Kinka-Byo Jun 26 '19 at 15:08
  • (Redacted my last comment) Never mind, I found a similar article on Wikipedia. \$\omega(t)\$ is defined as the instantaneous angular frequency, not... whatever "pulsation" means. –  Jun 26 '19 at 15:16
  • Sorry, with "pulsation" I mean angular frequency (or radian frequency). I will correct the question. – Kinka-Byo Jun 26 '19 at 15:20
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    [This question](https://electronics.stackexchange.com/questions/316918/meaning-of-sigma-in-laplace-transform) may give you further insight of what \$s=\sigma +j\omega\$ means –  Jun 26 '19 at 15:27
  • There is no "w". You should be using lower-case omega (\$\omega\$) instead of w for angular frequency. – Elliot Alderson Jun 26 '19 at 15:49
  • What do you mean by the 'instantaneous phase' of a signal? Please give an expression for the signal. – Chu Jun 26 '19 at 16:05
  • For instance consider v(t) = sin (ωt). Its instantaneous phase is phi(t) = ωt, while its angular frequency is ω. Let's suppose that because of noise (for instance), the angular frequency is not stable in time, so ω = ω(t). This is the correct premise to my question – Kinka-Byo Jun 26 '19 at 16:08
  • Phase is the integral of f so the Spectrum is modified by 1/f and -90 phase shift – Tony Stewart EE75 Jun 26 '19 at 17:32
  • Can you explain me this passage more in detail? Phase is the integral of f in time, so the spectrum is modified by 1/frequency of the signal we called "phase". Are these two frequencies the same? – Kinka-Byo Jun 26 '19 at 17:46

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