Delay and Stablity in Negative Feedback Systems: Confusion

Question

Below is a block diagram that shows an op-amp in negative feedback with feedback network made from resistive divider:

The encircled block is Op-Amp with 1 and 2 as its inverting and non-inverting pin respectively. I am assuming ideal conditions.

This would always be a stable system unless there is a delay in the feedback path. If we give a step input to the system and if the delay of the system is small then there would be no overshoot above the input step voltage but as we go on increasing the delay the system would start to overshoot and for higher delay it can become unstable. Below are the diagrams that show the system with delay and its response to the step input:

For this system with ideal delay, I read that if the delay \$T_d\$ is greater than \$\dfrac{\pi}{2} \cdot \dfrac{k}{\text{unity gain freq}}\$, the unity gain frequency is the \$\omega_u\$ in the diagram above, than the system would become unstable and its output would start diverging.

For the second order system the block diagram is shown below, here the delay element is first order, let's say it is some RC delay:

The delay element has a single pole at \$p_2\$. I read that this system is unconditionally stable with delay being \$1/p_2\$. From the above requirement of the delay, we can state that unconditional stability means the delay \$T_d\$ for this system always remains less than the above limit which is \$\dfrac{\pi}{2} \cdot \dfrac{k}{\text{unity gain freq}}\$.

Since the delay is of the order of \$R \cdot C\$ of the delay element and the \$R \cdot C\$ value can be made arbitrarily large, I think it is not always guaranteed that the delay will be restricted within this limit and hence system should go towards instability whenever the delay exceeds this limit.

Could someone please explain this apparent paradox about why the system is unconditionally stable even when the delay could be greater than \$\dfrac{\pi}{2} \cdot \dfrac{k}{\text{unity gain freq}}\$.

Answer 1

In your second example (2nd-order system with feedback) the phase of the loop gain function will approach the critical value of -180 deg for infinite frequencies only. Tha means: The phase shift will never reach -180deg at a fixed frequency - and the systenm will be stable.

In your first example (first order with fixed delay block) the loop gain phase will not be limited to any fixed value. Instead, the phase will rise with the frequency without any limitation. Hence, if the loop gain is larger (smaller) than 0 dB at a total phase of -180 deg, the closed loop system will be unstable (stable).

Answer 2

\Consider the open-loop with gain, \$\small K\$, integrator \$\frac{1}{s}\$, and delay \$\small e^{-sT}\$.

Phase angle (in radians) is: \$\small \phi=-\frac{\pi}{2}-\omega T\$, and gain is \$\frac{K}{\omega}\$.

For unity loop gain, \$\small K=\omega\$, which means, at this frequency \$\small \phi=-\frac{\pi}{2}-K T\$, and for a phase angle of \$-\pi\$, which would give a conditionally stable system (i.e. oscillatory):

\$\small -\pi=-\frac{\pi}{2}-K T\$ or \$\small T=\frac{\pi}{2K}\$, and including the feedback resistor network with gain, \$\frac{1}{k}\$, we have the condition: \$\small T=\frac{k\pi}{2K}\$