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I have a simple cruise control closed-loop system with negative feedback and have no problem using the Nyquist stability criterion to find its stability. However, I have another function that describes the fuel consumption of the throttle in the closed loop. There is no feedback from the fuel consumption to the throttle or the speed.

In other words, I have an additional transfer function outside the loop from the plant to the fuel consumption and want to assess the stability of the fuel consumption.

  • Since the fuel consumption is just a function of the throttle which is inside the feedback loop, am I guaranteed stability if the closed loop system is stable?
  • Is it possible for the fuel consumption function to essentially add a gain that makes the consumption unstable even if the speed control is stable?
  • Should I treat the fuel consumption as a purely open loop problem from the throttle to the fuel?

Thank you

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Since you're worried about an open-loop system, in theory all you need to do is perform standard stability analysis on the open-loop system alone. The source of the input signal to the open loop system is irrelevant if you can prove that it is stable for a known/expected set of input signals. For example, for a linear system you would need to demonstrate its stability for an expected range of input magnitudes and frequencies. A nonlinear system would be more complicated, but methods such as Lyapunov stability analysis and passivity can be applied.

However, I have a counter-question for you: what does it mean to have "stable" fuel consumption? Stability usually implies that given a certain initial state, the state does not move beyond a certain distance from the equilibrium point for all time (Lyapunov stability). Since the state in a fuel system would be the volume remaining in the tank, and we know that the tank has a maximum volume ($V_{max}$) and a minimum volume (0, no fuel) then for any initial state,

$$0 \leq V(0) \leq V_{max} \implies 0 \leq V(t) \leq V_{max} \text{.}$$

Note this conclusion is independent of the input. If we further consider that, in the absence of refueling, $\frac{dV}{dt} \leq 0$, we can conclude that $0 \leq V(t) \leq V(0)$. Either result leads to the conclusion that the fuel consumption is stable (in the Lyapunov sense) regardless of input.

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  • $\begingroup$ open loop stability is not equivalent to "prove that it is stable for any input", with a bad controller a closed loop system can be instable even if the open loop is stable $\endgroup$ – OpticalResonator May 23 '18 at 19:59
  • $\begingroup$ @OpticalResonator as the OP states, there is no controller $\endgroup$ – BarbalatsDilemma May 23 '18 at 19:59
  • $\begingroup$ I worded my comment badly, just because the open loop is stable doesn't mean that it will stay stable for any input $\endgroup$ – OpticalResonator May 23 '18 at 20:00
  • $\begingroup$ I'd say it's technically right how you wrote it, but right now it can be misunderstood as "standard stability analysis on the open-loop" automatically implying "stable for any input", which is not the case $\endgroup$ – OpticalResonator May 23 '18 at 20:03
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    $\begingroup$ I'll update the answer to be more clear $\endgroup$ – BarbalatsDilemma May 23 '18 at 20:04

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