# SCIENCE CHINA Information Sciences, Volume 61, Issue 9: 092203(2018) https://doi.org/10.1007/s11432-017-9185-6

## Achievable delay margin using LTI control for plants with unstable complex poles Peijun JU1,2,
• AcceptedJun 20, 2017
• PublishedJan 4, 2018
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### Abstract

We consider the achievable delay margin of a real rational and strictly proper plant, with unstable complex poles, by a linear time-invariant (LTI) controller.The delay margin is defined as the largest time delay such that, for any delay less than this value, the closed-loop stability is maintained. Drawing upon a frequency domain method, particularly a bilinear transform technique, we provide an upper bound of the delay margin, which requires computing the maximum of a one-variable function. Finally, the effectiveness of the theoretical results is demonstrated through a numerical example.

### Acknowledgment

This work was partially supported by Taishan Scholar Construction Engineering by Shandong Government and National Natural Science Foundation of China (Grant Nos. 61573220, 61573221, 61633014).

### Supplement

Appendix

Proof of Lemma 1

We first write $P_{\alpha}(s)=P_0(s)B_{\alpha}(s)$ as the ratio of two polynomials, which have no common zeros in $\bar{\mathcal{C}}^+$ for small $\alpha\geq~0~$.

If i) holds, write \begin{eqnarray}P_0(s)=g\cdot \frac{\bar{n}_0(s)}{[(s-a)^2 +b^2]\cdot\bar{d}_0(s)} \end{eqnarray} with $\bar{n}_0(s)$ and $[(s-a)^2+b^2]\cdot\bar{d}_0(s)$ monic and coprime. Let \begin{eqnarray}\left\{ \begin{aligned}n_{\alpha}(s):=&[(s+a)^2+b^2-\alpha]\cdot(1-c_{\alpha} s)\cdot\bar{n}_0(s), \\ d_{\alpha}(s):=&[(s+a)^2+b^2]\cdot[(s-a)^2+b^2-\alpha]\cdot(1+\bar{c}_{\alpha}s)\cdot\bar{d}_0(s). \end{aligned} \right. \end{eqnarray}

If ii) holds, write \begin{eqnarray}P_0(s)=g\cdot \frac{\bar{n}_0(s)(z-s)}{[(s-a)^2 +b^2]\cdot\bar{d}_0(s)} \end{eqnarray} with $\bar{n}_0(s)(z-s)$ and $[(s-a)^2+b^2]\cdot\bar{d}_0(s)$ monic and coprime. Let \begin{eqnarray}\left\{ \begin{aligned}n_{\alpha}(s):=&[(s+a)^2+b^2-\alpha]\cdot(1-c_{\alpha} s)\cdot[z+s]\cdot\bar{n}_0(s), \\ d_{\alpha}(s):=&[(s+a)^2+b^2]\cdot[(s-a)^2+b^2-\alpha]\cdot(1+\bar{c}_{\alpha}s)\cdot\bar{d}_0(s). \end{aligned} \right. \end{eqnarray}

If iii) holds, write \begin{eqnarray}P_0(s)=g\cdot \frac{\bar{n}_0(s)(z-s)}{\bar{d}_0(s)} \end{eqnarray} with $\bar{n}_0(s)(z-s)$ and $\bar{d}_0(s)$ monic and coprime. Let \begin{eqnarray}\left\{ \begin{aligned}n_{\alpha}(s):=&[a+b\text{j} -\alpha s]\cdot[a-b\text{j} -\alpha s]\cdot[z+s]\cdot\bar{n}_0(s), \\ d_{\alpha}(s):=&[a-b\text{j} +\alpha s]\cdot[a+b\text{j} +\alpha s]\cdot\bar{d}_0(s). \end{aligned} \right. \end{eqnarray}

Then, combing the above three cases, we can write \begin{eqnarray}P_{\alpha}(s)=P_0(s)B_{\alpha}(s)=g\cdot \frac{n_{\alpha}(s)}{d_{\alpha}(s)}. \end{eqnarray}

It is straightforward to verify that $n_{\alpha}(s)$ and $d_{\alpha}(s)$ have no common zeros in $\bar{\mathcal{C}}^+$ for small $\alpha\geq0$. But this property would be lost at $\hat{\alpha}>0$ for there is an unstable pole-zero cancellation in $P_{\alpha}(s)$. Based on the continuity of the zeros of the characteristic polynomial as a function of the parameter $\alpha$, there exist an $\alpha^*\in~[0,~\hat{\alpha})$ so that the polynomial has a zero on the imaginary axis, i.e., $s=\text{j}\omega^*$, which means that $B_{\alpha^*}(\text{j}\omega^*)P_0(\omega^*)C(\omega^*)=-1$. Thus, the proof is completed.

Proof of Claim 1

Following the notation of $A=a^2+b^2$, and \begin{eqnarray}Q_\alpha(s)=[a+b\text{j} -\alpha s]\cdot[a-b\text{j} -\alpha s]\cdot[z+s], \end{eqnarray} at $s=\text{j}\omega$, we have \begin{eqnarray}\begin{aligned}Q_\alpha(\text{j}\omega) =&[(\text{j}\omega-a)^2+b^2][(\text{j}\omega+a)^2+b^2-\alpha][1-\text{j}\omega c_\alpha] \tag{11} \\ =&[A-\omega^2-2a\omega \text{j} ][A-\omega^2+2a\omega \text{j} -\alpha][1-\text{j}\omega c_\alpha] \tag{12} \\ =&[(A-\omega^2)^2-\alpha(A-\omega^2)+4a^2\omega^2+2a\alpha\omega \text{j} ](1-\text{j} c_\alpha \omega) \tag{13} \\ =&[\omega^4+\omega^2(4a^2+\alpha-2A+2a\alpha)+A^2-\alpha A] \tag{14} \\ &-\text{j}[c_\alpha\omega^5+c_\alpha\omega^3(4a^2+\alpha-2A)+c_\alpha (A^2-\alpha A)-2a\alpha]. \end{aligned} \end{eqnarray} Since \begin{eqnarray}c_\alpha=\frac{2a\alpha}{A^2-\alpha A}, \end{eqnarray} so $c_\alpha~(A^2-\alpha~A)-2a\alpha=0$, and note that $\Lambda(\alpha,\omega)={\omega}^2+4a^2+\alpha-2A$, then from the above equation, we have \begin{eqnarray}\begin{aligned}Q_\alpha(\text{j}\omega)=&[\omega^4+\omega^2(4a^2+\alpha-2A+2a\alpha)+A^2-\alpha A] -\text{j}[c_\alpha\omega^5+c_\alpha\omega^3(4a^2+\alpha-2A)] \tag{15} \\ =& [ \Lambda(\alpha,\omega){\omega}^2 +2a\alpha c_{\alpha}{\omega}^2+A^2-\alpha A]-\text{j} c_{\alpha}\Lambda(\alpha,\omega){\omega}^3 \tag{16} \\ =&{\rm Re}Q_{\alpha}(\text{j}\omega)-\text{j}{\rm Im}Q_{\alpha}(\text{j}\omega), \end{aligned} \end{eqnarray} as claimed.

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