How Does the Positive Feedback Bring an Oscillator's Poles on the Imaginary Axis?
For oscillators you can either have linear or relaxation (nonlinear) ones.
For the nonlinear ones, usually you have some type of hysteresis behavior (although this is not a general definition!).
Two examples I can give are the Multivibrator and the 555 IC. The have two operational modes one &quot;charges&quot; and the other one &quot;discarges&quot;, and since the transition from charge-&gt;discharge and discharge-&gt;charge are not at the same point, say,
charge-&gt;discharge @ 5V and discharge-&gt;charge @ 2V. The system would oscillate between 5 and 2 V, with some waveform that probably is not a pure sinusoidal.
This type of oscillation is usually pretty &quot;self-sustainable&quot; because is is related to some property of the system (its dynamics) and even a lot of noise would have a hard time making it stop oscillating.
If you have the linear one you must have a system with conjugate poles with zero real part (along the imaginary axis), like,
\$ (s-j8)(s+j8),\$ or \$ (s-j8)(s+j8)(s-j20)(s+j20).\$
That means that any noise to the system, such as an impulse \$ \delta(t) \$, will lead to an oscillation that does not decay. For the system,
$$ H(s) = \frac{1}{(s-j8)(s+j8)(s-j20)(s+j20)}, $$
any impulse input will lead this system to have the following output,
$$ h(t) = a \cos(8t)+b \cos(20t).$$
But, regarding the Barkhausen criterion, it states that at the frequency of oscillation
|-------&gt;[ P ]--------&gt; Vout
|
|-------[ B ]&lt;-------- Vin

\$ |PB| = 1 \$

\$ \angle PB = 0 \$
So, for \$ P = \frac{s-1}{(s+1+j)(s+1-j)}\$ and \$ B = k\$ (the gain for a positive feedback), this would oscillate for,
$$  \text{for }\omega = 2, k \geq 0, ~~ \angle PB = 0, $$
$$  \text{we must have }\left| \frac{j2-1}{(j2+1+j)(j2+1-j)} k \right| = 1, $$
$$  \left| \frac{j2-1}{(j2+1+j)(j2+1-j)} k \right| = 1, $$
$$  \left| \frac{1}{2} k \right| = 1, $$
$$  k = 2.  $$


Now, if you analyze the closed-loop transfer function you will reach the same result but the algebra to find the necessary \$ k \$ might be a bit harder,
$$  H(s) = \frac{P}{ 1-kP} = \frac{ \frac{s-1}{(s+1+j)(s+1-j)}}{ 1-k \frac{s-1}{(s+1+j)(s+1-j)}} = \frac{s-1}{ (s+1+j)(s+1-j)-k(s-1)},  $$
$$  H(s) = \frac{s-1}{s^2+ (2-k) s +2+k},$$
With poles at,
$$  s_0 = \frac{k-2}{2}-\frac{\sqrt{k^2-8k-4}}{2},$$
$$  s_0 = \frac{k-2}{2}+\frac{\sqrt{k^2-8k-4}}{2},$$
And with \$ k = 2\$ you get a pole at the imaginary axis (zero real part).
$$  s_0 = 2j, s_0 = -2j.$$