The Barkhausen criterion for system with positive feedback (being \$B(s)\$ the feedback network transfer function and \$A(s)\$ the gain network transfer function) afirms that for a system to keep oscillating without input signal and without loss, we need the poles of:

\$H(s) = \dfrac{A(s)}{1 - B(s)A(s)}\$

in the imaginary axes at the complex plane.That means there should be a solution to:

\$B(j \omega_o) \cdot A(j \omega_o) = 1\$

and ωo would be the oscillating frequency.

On the other hand, if i have a system with negative feedback (being B(s) the feedback network transfer function and A(s) the gain network transfer function), the overall transfer function will be:

\$H(s) = \dfrac{A(s)}{1+B(s)A(s)}\$

So, in order to have poles in the imaginary axis (no loss), I need a frequency ωd that solves:

\$B(\omega_d).A(\omega_d) = -1\$

That loop transfer function being -1, should mean that the sinusoidal input with frequency ωd wil be shifted by 180º and will keep the same amplitude after passing through the loop \$A(s).B(s)\$, then will be shifted by 180° again because of the negative feedback and will return to its original form to keep oscillating indefinitely.

It will also mean that the transfer function \$H(s)\$ will have infinity gain at d which is a requirement to keep a system living without any input signal. Isn't it enough to implement a oscillating system that will keep oscillating any sinusoidal with frequency ωd that enters the loop?

Why do we need positive feedback to implement an oscillating system ?

  • \$\begingroup\$ An unusual oscillator like the "Bridged T", filter out the fundamental frequency and feed any distortion back to the Negative input (reducing output distortion). Any filter should work in this manner. Positive feedback can be had from a simple voltage devider used as an AGC control. \$\endgroup\$ Commented Feb 28, 2013 at 2:06
  • \$\begingroup\$ There is no such thing as infinite gain. Infinity is not a number. The value of \$f(x)/(x - a)\$ is not infinite at \$x = a\$, but rather undefined. \$\endgroup\$
    – Kaz
    Commented Feb 28, 2013 at 2:39
  • \$\begingroup\$ So what that means is that the transfer function has an undefined point. It is undefinedness that corresponds to oscillation, not "infinite gain". Oscillation will not occur with a transfer function which is defined everywhere. \$\endgroup\$
    – Kaz
    Commented Feb 28, 2013 at 2:44
  • \$\begingroup\$ The criterion applies specifically to linear feedback systems, meaning you don't necessarily need linear feedback for oscillation, just to be able to apply the criterion (which is by the way a necessary yet not sufficient condition for oscillation). \$\endgroup\$ Commented Feb 28, 2013 at 9:48

2 Answers 2


The answer is relatively simple: Each linear oscillator needs a loop gain of (at least) unity (unity magnitude and phase shift of zero deg) at one frequency only.

That means: We need a frequency-selective circuitry which can meet this condition at the desired frequency.

If we want to use a fixed gain stage (which is not always the case, we also can use integrators) we have two choices: inverting or non-inverting.

1.) Non-inverting (phase shift zero): The passive network must produce a phase shift of zero at the desired frequency (example: bandpass).

2.) Inverting (180 deg phase shift): The passive network must produce -180 deg at the desired frequency (example: three RC lowpass stages).

Answer to your final question: Why do we need positive feedback to implement an oscillating system ?

In order to avoid misinterpretations, I think that we should say: We need always a positive loop gain (of unity or - in practice - slightly larger) at the desired frequency. More than that - at the same time we need negative loop gain (negative feedback) for DC (stable bias point).


What you are saying is exactly correct.

Oscillation can happen even when the output of an amplifier is fed back negatively, provided either the amplifier or the feedback network, not both, should invert the signal (ie, \$A\beta = -1\$).

And negatively feedbacking an inverted signal is equivalent to positive feedbacking.

According to Barkhausen, oscillation needs the input and the fedback signal be in phase. You can use negative or positive feedback to attain this condition. But don't forget that feedback topology that results in augmenting the input is positive feedback.

This question should have been discussed under terminology. :-)


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