0
\$\begingroup\$

Consider the transfer function of a block being $$\frac{1}{s-0.5}$$ which is unstable in open loop. If we go for a feedback, magically it becomes stable even with the presence of the inherently unstable block. My thoughts on these are

  1. When an element is unstable , it doesn't mean it is unstable for all inputs. For certain inputs the block can still produce stable outputs.
  2. During feedback, the input given to the the block gets modified such that output doesn't blow up.

My questions are

  1. Are the above observations correct?
  2. The impulse response of the block is $$ e^{0.5t}u(t) $$ What the block does is, take the area under the signal at present instant and blow it up for subsequent instants. If so, a positive valued input should blowup the output exponentially. Then why is the output not blown up (exponentially) in following case, shown in figure?

The signal shown as input is the actual error signal when we use the block in a closed loop format with unit step input.enter image description here

\$\endgroup\$
  • \$\begingroup\$ What figure???? \$\endgroup\$ – Andy aka Dec 10 '16 at 10:16
  • \$\begingroup\$ @Andy aka: Figure added. \$\endgroup\$ – Divya K.S Dec 10 '16 at 10:39
  • \$\begingroup\$ "The impulse response of the block (open loop) is e^0.5t u(t)?": Yes. \$\endgroup\$ – Dirceu Rodrigues Jr Dec 10 '16 at 14:03
  • \$\begingroup\$ An integrator is unstable in open loop, but if the loop is closed around it and a step input is applied to the system, the integrator output stops increasing when it's input is zero, i.e. when the integrator output= system step input. Thus the closed loop system is a stable first order lag. \$\endgroup\$ – Chu Dec 10 '16 at 16:22
2
\$\begingroup\$

"When an element is unstable , it doesn't mean it is unstable for all inputs. For certain inputs the block can still produce stable outputs."

At first, an "element" (a part) cannot be unstable. It is better to use the term "active block" or "amplifier". Hence, an amplifier can be unstable if it has feedback, which produces a closed-loop pole (pair) with a positive real part. This is true for all inputs because it is the feedback loop which produces self-excitement of the system - not the input signal.

"During feedback, the input given to the the block gets modified such that output doesn't blow up."

An unstable system can be stabilized if the overall negative feedback overrides (dominates) the positive feedback which would cause instability (without negative feedback). In your case (example), the gain block has an inherent positive feedback (causing a pole at +0.5). This gain block ist stabilized using - as shown - 100% negative feedback causing a total closed-loop gain of "2 (and a pole at "-0.5")

\$\endgroup\$
  • \$\begingroup\$ "...active block" or "amplifier....". Or filter in frequency domain context. \$\endgroup\$ – Dirceu Rodrigues Jr Dec 10 '16 at 14:02
0
\$\begingroup\$

Not necessarily all feedbacks stabilize the system. In your pic, if it were positive feedback, the system would be unstable.

When an element is unstable , it doesn't mean it is unstable for all inputs. For certain inputs, the block can still produce stable outputs.

No. When an LTI system(like the one you have) is stable, it stable for all stable inputs. When an LTI system is unstable, the response is unstable for all stable and unstable inputs. Only for non-linear systems, a certain input range may stabilize and few will lead to an unstable response.

During feedback, the input given to the the block gets modified such that output doesn't blow up.

Yes. The reason why it is stable in this case is that for some input say x when given just to your unstable system, it generates a response(high value). This high value is given back to your unstable system but with reversed polarity/effect. So this feedback signal again when given back to the system, it generates high value(this time in opposite polarity) and the process continues and so on. So for any signal, your unstable system generates a response which is fed back negatively/counteractively to the system thereby stabilizing it.

Though the system is overall stable, internally it is unstable

\$\endgroup\$
0
\$\begingroup\$

enter image description here

\$V_{out} = \dfrac{1}{s-0.5}\times (V_{in} - V_{out}\$)

Rearranging: -

\$V_{out}(s-0.5+1) = V_{in}\$

and: -

\$\dfrac{V_{out}}{V_{in}} = \dfrac{1}{s+0.5}\$ i.e. a previously unstable system becomes stable with negative feedback.

Should the minus 0.5 term have been minus 1 (or more negative) then using the simple negative feedback in the question would not have produced stability.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.