If there exists ω=φ such that |KG(jφ)| = 1 and ∠(KG(jφ)) = -180°, then we know that s=jφ must be a pole.

This is incorrect, and is leading to a misunderstanding.

The poles of the system are the roots of the characteristic equation of the **Open Loop** transfer function $$KG(s)=0$$

the roots are of the form $$s=\sigma+j\omega$$

These are the poles and zeros that are analysed in the Bode Plot.

Once the loop is closed the poles move location to be the roots of the characteristic equation $$1+KG(s)=0$$ You can view loop compensation as a method of moving the open loop poles into more suitable (stable) locations in the closed loop. This can even by performed directly using **pole placement**.

However, Bode stability analysis is based on the Nyquist Stability Criterion. So the condition for oscillation in a negative feedback system is unity gain and 180 degree phase shift :$$KG(s) = -1$$
and therefore the Bode plot illustrates the "stability condition" by rearranging $$KG(s)+1=0$$

This happens to be the same equation as the characteristic equation of the **Closed Loop**. But it is a misunderstanding to see this as relating to Bode stability plots, which are an Open Loop analysis.

The Nyquist Stability Criterion also tells us that in general (but there are exceptions), a closed loop system is stable if the unity gain crossing of the magnitude plot occurs at a lower frequency than the -180 degree crossing of the phase plot.

Given that your first equation has the correct signs, \$ a\$ must be negative otherwise it's an unstable pole and \$ s\rightarrow j\omega\$ is not a valid operation, i.e. there is no steady state, therefore a frequency response does not exist.

Hence it would be better (less confusing) to let the complex pole be: $$ H(s)=\dfrac{1}{s+(a-jb)}\:, \:a>0$$

Now, determining the amplitude Bode plot for this isolated complex pole is mathematically simple, but in practice it's meaningless. Rather, you need to consider the frequency response of the complex conjugate pair.

To illustrate the problem, the DC gain of the single pole (obtained by taking \$\small s=0)\$ would be \$\small \dfrac{1}{(a-jb)}\$. A complex gain is not something that is physically realisable.

However, including the complex conjugate pole in the formulation would give a real DC gain of \$\small \dfrac{1}{(a^2+b^2)}\$, which complies with the 2nd order TF: \$\small \dfrac{1}{s+(a-jb)}.\dfrac{1}{s+(a+jb)}=\dfrac{1}{s^2+2as+(a^2+b^2)}\$

## Best Answer

Stability when using negative feedback is an important thing in engineering and a simple example is an inverting op-amp; it naturally inverts (shifts by 180 degrees) and so any feedback acts to cancel the input signal to a lesser or greater extent i.e. that feedback is "negative" and stabilizing.

However, op-amps are not perfect and, at high frequencies, some op-amps shift there response so that negative feedback eventually becomes positive feedback i.e. another 180 degrees is added to the natural phase shift at lower frequencies.

This turns an amplifier into an oscillator and is one way of looking at "instability".

This is the open loop gain (before feedback is applied) for a typical op-amp (OP-77): -

If you study the graph and look at the frequency where the gain drops to unity (about 400 kHz) you can see that the op-amp (as an inverter) hasn't added enough phase shift to bring it to 180 degrees - this means that it will always be stable when simple resistor feedback is used. This isn't the case for some op-amps though.