Hey, everyone,
Hope you are doing great!
I just started learning about digital phase locked loop. I have a lot of things that I cannot figure out. It will be great if I can get some answers from here.
In the textbook,
The gain of the voltage controlled oscillator is
Kvco = 2 x Pi x (fmax - fmin) / (Vmax - Vmin)
The VCO output frequency, f clock, is related to the VCO input voltage by
W clock = 2 x Pi x fclock = Kvco x Vinvco + W0
Where W0 is a constant. However, the variable we are feeding back is not frequency but phase ( hence the name of the circuit). The phase of the VCO clock output is related to fclock by
The integration of Wclock = Kvco x Vinvco / jw
Here, I do not understand that why the integration of the Wclock is equal to Kvco x Vinvco / jw.
If I do Laplace transform on the integration of the Wclock, then I will get 1/ jw x Laplace transform of Kvco x Vinvco. And since these two are constant, the laplace transform will be Kvco x Vinvco / s. Hence the Laplace transform of the integration of Wclock will be Kvco x Vinvco / square of s. Here in the book, I do not think they are talking about taking a Laplace transform. And even they are taking Laplace transform, my answer will be Kvco x Vinvco / square of s. Not Kvco x Vinvco /s.
Do you have any idea here, please?
- I can understand the equation about the natual frequency and damping factor. The book also just gives me the equations about pull in range, lock time, lock range. However, there is no deduction for these equations at all.
Specifically, the pull in range is
Pi/2 x square root of ( 2 x damping ratio x natual frequency x Kvco x Kpd - the square of natural frequency )
Here Kpd is the gain of phase detector.
The lock time is 2 x Pi / natual frequency.
The lock range is Pi / 2 x (1/ R x C ) , here R and C is the resistor and capacitor value of the low pass filter.
I have no idear how these equations are deducted.
Thanks for reading this post and any answers are greatly appreciated. Sarah