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Scenario (plant model subject to process variations)

Consider a plant \$P(s)\$ in series with a controller \$D(s)\$ with a negative feedback path around it, where the closed loop is given by the complementary sensitivity function \$T(s)\$. $$T(s)=\frac{P(s)D(s)}{1+P(s)D(s)}$$

Now consider that the plant model is subject to process variations \$\Delta P\$.

Question (\$ \Delta P \$ relation to complementary sensitivity function for non-minimum phase plant)

I read in a book\$^*\$ that to avoid system instability, the maximum allowable process model variation \$\Delta P(s)\$ must satisfy the following condition\$^*\$ derived from the Nyquist plot of the loop transfer function: $$\frac{|\Delta P(j\omega)|}{|P(j\omega)|}<\frac{1}{|T(j\omega)|}$$ This is derived for a Nyquist contour with no loops around \$-1+0j\$, i.e. a minimum phase system. I was wondering how this condition would apply to a non-minimum phase system, i.e. one with RHP poles or zeros, and hence with loops around \$-1+0j\$ on the Nyquist plot.

Footnotes for clarification

\$^*\$ This is derived from the Nyquist plot of the loop transfer function \$L(s)=P(s)D(s)\$, and comes from the condition that the magnitude of process variation \$|C\Delta P|\$ is not larger than the magnitude of the vector from \$L(s)\$ to the critical point \$-1 + 0j \$ on the Nyquist plot, or: $$|C\Delta P|<|1+L|$$

\$^*\$ page 192 of https://www.cds.caltech.edu/~murray/courses/cds101/fa02/caltech/astrom-ch5.pdf

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    \$\begingroup\$ The first time you mention \$P(s)\$ you appear to be saying it's the negative feedback; later, you say it's the plant transfer function. I'm pretty sure that's not what you meant; could you please edit your question for clarity there? \$\endgroup\$
    – TimWescott
    Commented Dec 28, 2023 at 23:02
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    \$\begingroup\$ Hello Tim, I've just edited my question for clarity, hope it is easier to follow now \$\endgroup\$ Commented Dec 29, 2023 at 10:59

1 Answer 1

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Yes, this is correct for non-minimum phase plants or controllers. As far as I know (and I'm quite certain this statement is true) it is correct for any linear time-invariant system, including ones that cannot be modeled by rational polynomials in \$s\$.

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  • \$\begingroup\$ Could you give me an argument for this? \$\endgroup\$ Commented Dec 30, 2023 at 14:38

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