Vco

[YAN-1]

Hello everyone. Can someone please tell me why the voltage-controlled oscillator is always modelled as an integrator when it is used as part of a closed-loop system (like PLL). Why is the relationship between its input (voltage) and its output (frequency) modelled as K/s, where K is its gain and s is the laplace operator?

Thanks a lot.

Nichola V. Abdo

 

[hjames]

PLL's use a phase comparator. A constant frequency offset will result in a VCO continuously "integrating" the difference in frequency, resulting in a linearly varying phase difference.

 

[Analog]

Hello everyone. Can someone please tell me why the voltage-controlled oscillator is always modelled as an integrator when it is used as part of a closed-loop system (like PLL). Why is the relationship between its input (voltage) and its output (frequency) modelled as K/s, where K is its gain and s is the laplace operator?

Thanks a lot.

Nichola V. Abdo

This is because the roots of K/s fall on the j axis. Recall your S plane from controls class. Any pole on the j axis is a pure oscillator. Poles to the right of the j axis are out of control and will saturate. Poles to the left of the j axis are damped sine waves that will decay exponentially. That is why.

 

[dknguyen]

This is because the roots of K/s fall on the j axis. Recall your S plane from controls class. Any pole on the j axis is a pure oscillator. Poles to the right of the j axis are out of control and will saturate. Poles to the left of the j axis are damped sine waves that will decay exponentially. That is why.

Coincidentally, irrefutable proof requires knowledge to understand that most people do not posess.

 

[ThermalRunaway]

Laplace transforms, ewwwww. That was my single most hated subject at college!

Brian