This question is in reference to my earlier question:
Delay and Stablity in Negative Feedback Systems: Confusion
In that question I wanted to ask why the second order system is always stable. But I could not understand the issue well from the answers. So I am asking in a more definite way what doubt I have.
Delay block is ideal
When a step input is applied to the system with, say, ideal delay, the output would began to rise and after the specified delay Td (which is delay of the 'delay' block shown below in the figure) the sensed input would began to rise with the output. This is shown in the figure below, where the red curve shows the output, the blue curve is the sensed voltage which would be subtracted from the input (here the delay Td is shown to be 5s) and the black dotted line represents input step (Although, the figure is shown for speed but similar analogy can be drawn for voltages as well). Clearly, if this delay is too large then the error voltage (which is the difference between the input and the sensed voltage and is the input to the integrator) would remain high and the output of the integrator would continue to rise resulting in large overshoot. This should cause instability in the system. Is this correct? Delay Block is RC system
If this delay block is a first-order RC system (so the overall system is second order now) and if the product R*C is very large, again the blue curve would rise very slowly causing the error to remain large and the red curve, which is the output of the integrator, would again have a large overshoot above the input step. Shouldn't this again cause the system to become unstable?
In other words, for large product of R*C there should be large overshoot. But is the system still stable? If so, why?