Where is the mistake? A very simple feedback loop

A very simple feedback control loop:
input --> *(+-)* ---> 2/(s-1) ---*--> output
There is a feedback connection from -*
- to *(+-)*
Here's my question:
1. Is this system stable?
The output/input transfer function is 2/(s+1)
( H(s) = 2/(s-1), transfer function = H/(1+H) = 2/(s+1) )
The only pole of the transfer function is at -1, LHP,===> Stable
2. On the otherhand, if the input is a constant dc
then 2/(s-1) becomes just "-2" (s=jw, w=0, 2/(s-1) = -2)

Therefore, the total feedback is (-1)*(-2) = 2 (negative feedback)
The feedback is 2*
A positive feedback! ===> Not stable.
The bode plot shows the same thing: open loop transfer function
= 2/(s-1) ==> At w=0, phase = -180 deg and gain = positve =>
Not stable!
Where is the mistake?
Reply to
Harry Lin
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Hmm> The pole of the CLOSED-LOOP transfer function is at -1.
Closed-loop stability is typically examined by looking at the poles of the OPEN-LOOP system, in your case at +1 (RHP), therefore: the system is, itself, internally closed-loop unstable - regardless of signals you may or may not apply at the external input.
Kelvin B. Hales
Reply to
Kelvin Hales
snipped-for-privacy@aol.com (Harry Lin) wrote in news: snipped-for-privacy@posting.google.com:
You're assesing the steady state response of your feedforward filter funny. You can't assess the stability without accounting for the feedback. The input to that filter, while in the feedback loop, is not u, it is [u-2u/(s-1)], so your step 2 needs work.
The system is most definately negative feedback, and it is stable. Your feedforward filter has a RHP, and is unstable, but the overall system has none. Stick to your feedback equations to assess feedback stability.
Reply to
Scott Seidman
Hi Scott,
Then how do you explain the Bode Plot of the open loop function?
The Bode Plot of 2/(s-1) at w = 0 is: phase = -180 deg and gain > 0dB.
Is it true that in a BODE PLOT, a -180 deg phase and a positive gain => UNSTABLE?
Reply to
Harry Lin
Hi Harry,
One must always be careful about assessing stability of non-minimum phase (which includes unstable) systems from the open loop Bode plot. In such cases, the definitive answer is given by the Nyquist criterion which includes the encirclement component which a Bode plot omits as it is an evaluation on the imaginary axis only, with no "D contour" traversement.
Reply to
Fred Stevens
Yes, you've just proved it.
As Fred mentions you can use a Nyquist plot (and try to count unstable zeros), or you can just take it that the Bode plot shows potential stability _boundaries_ -- so a Bode plot of a system will show you gain and phase margins, but it won't show you whether the system is stable in the first place.
Try this -- consider a system that includes a double integrator, such as a speaker coil actuator with position feedback, H(s) = A/s^2. Now wrap it with PID control, and stabilize the loop with the PID (you can actually do this with pole placement -- your free parameters are k_i, k_p and k_d plus your derivative pole, and you have a four pole system).
Now examine your Bode plot -- with most decent tunings this system will have over 180 degrees of phase shift at low frequencies, and you have not only a high gain margin but a low one as well.
Reply to
Tim Wescott

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