# voltage controlled oscillator

#### suzuki

Joined Aug 10, 2011
119
Hi everyone,

looking for some help with finding the transfer function of a vco circuit. hopefully my explanations here are clear enough

on this (http://www.minicircuits.com/app/VCO15-10.pdf) and other websites, that the transfer function can be given by Ko/s, where Ko is the slope of the oscillator frequency over the voltage (with units rad/(s * V))

So what i have done so far is, by simulation of a VCO circuit, enter a control voltage and measured the corresponding oscillator frequency. I then multiplied the oscillator frequency by 2*pi to get the value of ω (rad/s).

I then plotted the control voltage vs ω to get the value of Ko. If it matters, the relationship between the frequency and the control voltage was found to be linear, which was something i expected. So with this plot, i found the slope (ω1-ω2)/(V1-V2) in order to find the value of Ko.

In order to check my work, and to verify that this transfer function is correct, i built a simulation in Simulink that accepts the control voltage as the input, multiplies it by the value of Ko that i found, integrates, and then spits out the output.

What I expected the output to be is the oscillator frequency for that particular control voltage. However, this did not seem to be giving the correct solution. Does anybody have some insight on where I might have gone wrong?

Thank you.

#### t_n_k

Joined Mar 6, 2009
5,455
The transfer function for the VCO of K0/s specifies the phase change rather than the frequency change - the PLL 's' domain is normally cast in terms of phase differences rather than frequency.

Plugging a non-zero value into an integrator would lead to an unbounded output increase - which is probably what you are observing in trying to simulate just the VCO [phase] transfer function.

#### suzuki

Joined Aug 10, 2011
119
Hi there,

Sorry, but can i have some more clarification here? i'm sort of confused by your first statement. If the K0 term specifies phase change wouldn't the units just be rad/V? I'm not sure about this part as the given units are in the form or rad/s/V. I guess this is also confusing since how would I be able to simulate/measure the phase without some reference signal?

and Yes, you are correct, i am seeing an unbounded output increase. So i guess my second question is, if this is indeed a phase transfer function, could i just take the derivative / slope of that line to get the frequency ω?

thank you

#### Hi-Z

Joined Jul 31, 2011
158
The output from your simulation is actually phase, not frequency. K0 is specified as a frequency - when you integrate it you get phase. So, as far as a phase-locked-loop is concerned, the vco is an integrator.

#### suzuki

Joined Aug 10, 2011
119
Okay, i think what I am taking from this is that, it is my Matlab simulation that is incorrect, which upon further review, you two are both correct, as the equation is given by theta = (K0*Vc)/s.

So, if i took the derivative of that equation, i would have ω = Ko*Vc. Therefore, can i say that the deviation in frequency is given by multiplying the Ko i found, with the control voltage?

I would then add this change in frequency to the centre frequency of my VCO (when the control voltage = 0) to get the actual output frequency i.e ω = ω0 + K0*Vc.

Does anybody see any flaws in this logic? I won't be able to re-simulate my system until next week, but i'd like to have some peace of mind over this problem. Can someone also confirm that my method of finding Ko is correct?

Thanks again.

#### Hi-Z

Joined Jul 31, 2011
158
Yes, the deviation in frequency is K0*Vc, so you're on the right track now.