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It is known that the initial feedback signal is developed from the noise in the resitors and the transients of the power supply. The question here is : why the waveform at the output is sine wave particulary and not any other waveform?

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closed as unclear what you're asking by Dmitry Grigoryev, Bence Kaulics, PeterJ, Daniel Grillo, uint128_t Jun 14 '16 at 16:08

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    \$\begingroup\$ Because that's the solution of the equation describing the oscillator. Is there something specific you wanted to know about it? \$\endgroup\$ – Dmitry Grigoryev Jun 14 '16 at 7:50
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Barkhausens oscillation condition is applicable for LINEAR systems (with feedback) only. This condition requires a loop gain of unity (magnitude of "1" and zero phase). For most of the frequency-dependent networks this phase condition can be fulfilled for one single frequency only. Hence, there is only one single frequency, which can lead to self-sustained oscillations. In case of the WIEN oscillator a simple RC bandpass is used providing a magnitude maximum and a phase of zero deg at the same frequency.

As mentioned, this applies to linear systems only. On the other hand, a safe start of oscillations needs a loop gain slighter larger than unity - with continuously rising amplitudes. In order to make sure that the amplitudes are not hard-limited (clipped) at the supply voltage rail, we use an amplitude-control mechanism to enable soft-limiting which can maintain the sinusoidal form of the signal.

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