Loop stabilty and bode plots. - Page 1
Forum section: electronics / beginners

author: Kevin.D
author_rank: Frequent Contributor
timestamp: January 13, 2013, 01:54:24 pm
content: Am studying loop stability at the moment ,doesn't seem theres many good tutorials on this subject for the enthusiast on the web (found lots of awful stuff written which just goes off in a tirade of mathematical formula's and yet explains nothing in practical terms.)I think I made some progress in understanding though after a hard time.But can someone explain something though please . I understand Two conditions necessary for oscillations are Phase shift > 180 deg and loop gain = 1 , which is why we have to look at the 0 db gain and phase shift intersection point on a bode plot to see if loop will be stable . Am I correct here in my understanding ?So my question is why a loop can only oscillate at a frequency where it's loop gain =1 .Why cant it oscillate ,say at some lower frequency if it's phase shift >180, but it's loop gain is still greater than 1. ?I CAN understand why it wont oscillate if it's phase shift is 180 deg and it'sloop gain is LESS than one ,but it wont if it's >1 either ?.

author: Kremmen
author_rank: Super Contributor
timestamp: January 13, 2013, 03:05:05 pm
content: Quote from: kevotronic on January 13, 2013, 01:54:24 pmYes, that is correct, although to be painfully precise, this is the Bode stability criterion which is not the last word on the subject. A more comprehensive one is the Nyquist criterion and then finally, for non-LTI systems you need to engage the Lyapunov which is not for the faint hearted. But there we are deep in the math that you don't want. So, at this level you are corrcect; if the loop gain at 180 degree phase shift is < 0db, the system is stable.QuoteThis oscillation is a resonance phenomenon. Resonances are self-sustaining or slowly decaying oscillations that occur at the natural frequency of the system. The natural frequency of a feedback system is the one where the feedback signal has shifted 180 degrees, because then the difference of the driving signal and the feedback signal turns into summation instead of subtraction. (Or more accurately, the summation effect reaches a maximun at that shift). When the feedback signal acts the same way as the driver signal, then the feedback has turned from negative to positive and the system is guaranteed to slew to one limit. At the natural frequency the feedback path delays the signal just enough to cause the system drive itself to the maximun opposite state next. And so on & so on. In this state the system does not need any input to start the oscillations, it will start anyway from any little glitch, thermal noise or whatever. Actually the math does not assume any initiating signal at all; instability is a fundamental property of a system like that.

author: Mechatrommer
author_rank: Super Contributor
timestamp: January 13, 2013, 06:40:34 pm
content: Quote from: kevotronic on January 13, 2013, 01:54:24 pmyup correct! and thats all you want to know about stability analysis, while doing open loop gain plot. the idea is that you dont want the phase shift to go beyond 180 deg at any signal gain (out >= in)Quote from: kevotronic on January 13, 2013, 01:54:24 pmi guess you are in confused condition, so... dont think too much, back to my above post... from the plot if you see the phase is out of 180 deg boundary while the gain is greater or equal 1, then you risk (certainly will and certainly murphy) oscillation/resonans of the system while in operation. why? i dont know thats what engineers and scientists found out long time ago probably from experience, exhaustive testings and experimentation (maybe this is also the reason why i was clueless while studying this subject long time ago).Yup! Kremmen explained that better!