We can represent oscillator as

y(t)= A cos (wt + Φ(t)), where Φ(t) is phase noise.

If Φ(t)=cos(w_nt)  then, we can show using Narrowband approximation that

y(t)≈ A cos (wt) + A sin (wt) sin ( w_nt)

Which one of the figures attached (a or b) is the actual spectrum of the oscillator?

I used to think it's (b) and the entire carrier power is spread throughout, but I can't explain it mathematically. Only (a) makes sense mathematically since we have a carrier term (highlighted above) and then a phase noise term.

Question 2:

Also when we measure SSB phase noise it's overlaying both left and right of the carrier, resulting in 3 dB more phase noise compared to DSB phase noise, right? In the plot (a), the dBc that I am plotting is DSB phase noise, which is 3 dB lower compared to SSB phase noise, right?

Hi Ken,

Thanks for the reply, I got it now. I was plotting Sv, I understand S_phi goes to infinity due to flicker but not Sv.