genxium
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Homework Statement
By Wien bridge oscillator (abbreviated as WBO below) I refer to this circuit where the op-amp is assumed to be ideal and the output is the voltage drained from the upper end of R3.
What confuses me is that I hardly find a reference which explicitly states that a WBO has a closed-loop transfer function. By closed-loop transfer function I refer to that of a negative feedback block diagram as follows in s-domain, e.g. ClosedLoopTransferFunc = \frac{G(s)}{1 - G(s)H(s)}, if the WBO could be modeled like so.
Homework Equations
Will be shown in the following section.
The Attempt at a Solution
The difficult part for me is to define/find the input to the WBO. As told to me by an esteemed person in a personal email, the WBO has no input, and thus its closed-loop transfer function shouldn't be defined. However this is kind of weird to me. I understand that in terms of energy conservation the energy is drained from a DC pull and converted to sinusoidal waves and possibly heat, that's reasonable, but the magical word "convert" is so outstanding here.
What I think is that there should be an abrupt impulse, or sharp edge of voltage/current amplitude change when the oscillator is "switched on", and that impulse contains all kinds of frequency components to be amplified or attenuated w.r.t. individual closed-loop transfer function of each.
My attempt to quantify this intuition, especially how signals are "addible in s-domain", is to assume that in time-domain there's an ideal impulse which is proportional to a Dirac delta \delta(t) as input signal x(t) when "switched on" in the block diagram. Then x(t) adds with the feedback signal \mathcal{L}^{−1}\{H·Y\}(t) both at t=0 and after t>0 in time-domain, where \mathcal{L}^{−1} is the inverse Laplace transform, x(t \ne 0)=0 and \mathcal{L}^{−1}\{H·Y\}(t=0)=0. Thus X(s) and H(s) adds in the s-domain as well due to the linearity of Laplace transform.
Moreover, my attempt to relate the 2 figures above is like the following.
where
- the first figure shows the ideal op-amp together with its negative feedback path being a finite-gain amplifier contributing to G(s);
- the second figure shows how the "closed-loop", if exists, is opened, thus H(s) = \frac{R_3}{R_3 + R_4}.
All comments are appreciated.