Why would you say that VCO act as an integrator?

[A.Anand Srinivasan]

VCO

we were asked in our lab exam about what an VCO act as....
the correct answer that we were given is that it is an integrator...
i couldn't reason out why but we were informed that it is the answer from an gate paper...

 

[Euler's Identity]

Re: VCO

That one threw me for a loop when I first ran across it, but then it made sense.

A VCO receives an input voltage and then produces a frequency output proportional to Vin. My last experience with a VCO was by way of a Phase-Locked Loop, and there the Vin was a control voltage, Vc.

ω(t) = KVc(t) (Eq. 1)

where
ω(t): the output frequency in radians (rad) as a function of time
K: VCO constant (rad/V)
Vc: control voltage as a function of time

Now consider that ω is

ω = rad/s = dΦ/dt

and substitute in Eq. 1, which is

dΦ/dt = KVc(t) (Eq. 2)

Now here's where the integration happens. Assume that Φ(0) = 0 and take the Laplace Transform of Eq. 2.

L(dΦ/dt) = L(KVc(t))

Φ(s)s = KVc(s)

Now solve for Φ(s) -- whch is the variable of interest for a PLL.

Φ(s) = KVc(s)/s

Now, assuming you've followed this far, notice that Φ(s) is the constant K multiplied by the integral of Vc(s), as the Laplace Transform of integration is multiplication by 1/s.

It is likely that this is what was meant when it was said that the VCO is an integrator.

However, even if you didn't get the Laplace part (although it is still assumed you've had Calculus), notice that Eq. 2 is a derivative with respect to time equal to a function of time. Integrating both sides of Eq. 2 to solve for Φ(t) gives

Φ(t) = K∫Vc(t)dt

Once again, integration takes place. Specifically, the output phase is the product of the VCO's constant of proportionality and the integral of the control voltage with respect to time.

 

[electronics_kumar]

VCO

it is V to F converter...has integrator.. lowpass filter...