- #1

genxium

- 141

- 2

## 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 R

_{3}.

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. [itex]ClosedLoopTransferFunc = \frac{G(s)}{1 - G(s)H(s)}[/itex],

**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 [itex]\delta(t)[/itex] as input signal [itex]x(t)[/itex] when "switched on" in the block diagram. Then [itex]x(t)[/itex] adds with the feedback signal [itex]\mathcal{L}^{−1}\{H·Y\}(t)[/itex] both at [itex]t=0[/itex] and after [itex]t>0[/itex] in time-domain, where [itex]\mathcal{L}^{−1}[/itex] is the inverse Laplace transform, [itex]x(t \ne 0)=0[/itex] and [itex]\mathcal{L}^{−1}\{H·Y\}(t=0)=0[/itex]. Thus [itex]X(s)[/itex] and [itex]H(s)[/itex] 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 [itex]G(s)[/itex];
- the second figure shows how the "closed-loop", if exists, is opened, thus [itex]H(s) = \frac{R_3}{R_3 + R_4}[/itex].

All comments are appreciated.