Stability analysis of linear time-invariant dynamic systems using the matrix sign function and the Adomian decomposition method

Abstract

The assessment of the bounded-input bounded output (BIBO) stability of a dynamic system is of paramount importance in the process control theory and practice. In this paper, we have developed a BIBO stability analysis method for linear time-invariant systems on the basis of Howland’s eigenvalue separation theorem, which involves the matrix sign function, and the Adomian decomposition method. Our proposed method is conceptually convenient and merely requires matrix addition and multiplication. Furthermore, our method eliminates the need for the availability of the system’s characteristic equation, is devoid of any graphical representation, and does not involve the accustomed set of defined rules in the previous approaches. The method’s convergence analysis is presented, and its application is demonstrated through five real-world case studies. Based on a CPU-time analysis, it is demonstrated that the proposed method is computationally superior to the classical Routh–Hurwitz stability test and its efficiency is almost unaffected by the size of the system matrix.

Appendix

Theorem 1.

Let ${\mathbf{A}} \in M_{n} \left( {\mathbb{C}} \right)$, i.e., ${\mathbf{A}}$ is a square matrix of size $n$ whose entries can be complex, and assume that ${\mathbf{A}}$ is invertible. It holds that ${\mathbf{A}}^{2}$ and ${\mathbf{A}}^{ - 1}$ are invertible.

Proof

The proof is simple. Since ${\mathbf{A}}$ is invertible, it is nonsingular, and hence $\det \left( {\mathbf{A}} \right) \ne 0$. Elsewhere [44], we can find the proof of a theorem that states $\det \left( {{\mathbf{A}} {\mathbf{B}}} \right) = \det \left( {\mathbf{A}} \right)\det \left( {\mathbf{B}} \right)$ for ${\mathbf{A}},{\mathbf{B}} \in M_{n} \left( {\mathbb{C}} \right)$. Consequently, $\det \left( {{\mathbf{A}}^{2} } \right) = \left[ {\det \left( {\mathbf{A}} \right)} \right]^{{ 2}} \ne 0$. This concludes that ${\mathbf{A}}^{2}$ is invertible. Additionally, from the aforementioned theorem, it follows that $\det \left( {{\mathbf{A}}^{ - 1} } \right) = {1 \mathord{\left/ {\vphantom {1 {\det \left( {\mathbf{A}} \right)}}} \right. \kern-\nulldelimiterspace} {\det \left( {\mathbf{A}} \right)}}$ by considering ${\mathbf{A}} {\mathbf{A}}^{ - 1} = {\mathbf{I}}$ and taking ${\mathbf{B}} = {\mathbf{A}}^{ - 1}$, where ${\mathbf{I}}$ is the identity matrix. Now, because $\det \left( {\mathbf{A}} \right) \ne 0$, it is yielded that $\det \left( {{\mathbf{A}}^{ - 1} } \right) \ne 0$; Thus, ${\mathbf{A}}^{ - 1}$ is invertible.

Theorem 2.

Let ${\mathbf{A}} \in M_{n} \left( {\mathbb{C}} \right)$ be invertible. It holds that ${\mathbf{A}}^{\frac{1}{2}}$ is invertible.

Proof

If we take ${\mathbf{B}} = {\mathbf{A}}^{\frac{1}{2}}$, then we can write that

$$ \det \left( {{\mathbf{B}}^{2} } \right) = \det \left( {\mathbf{A}} \right). $$

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Then, it follows that

$$ \left[ {\det \left( {\mathbf{B}} \right)} \right]^{{ 2}} = \det \left( {\mathbf{A}} \right), $$

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or equrivalently,

$$ \det \left( {\mathbf{B}} \right) = \pm \sqrt {\det \left( {\mathbf{A}} \right)} . $$

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Now, since $\det \left( {\mathbf{A}} \right) \ne 0$, it is straightforward that $\det \left( {\mathbf{B}} \right) \ne 0$ or in other words, ${\mathbf{A}}^{\frac{1}{2}}$ is invertible.

Theorem 3.

Let ${\mathbf{A}} \in M_{n} \left( {\mathbb{C}} \right)$ be invertible. It holds that ${\mathbf{A}}$ has a square root, i.e., a matrix ${\mathbf{T}} \in M_{n} \left( {\mathbb{C}} \right)$ such that ${\mathbf{T}}^{{ 2}} = {\mathbf{A}}$.

Proof

Our proof is based on the fact that every invertible matrix can be expressed as the exponential of some other matrix, which is because the matrix exponential function is a surjective map.

$$ \exp :\,M_{n} \left( {\mathbb{C}} \right) \to {\text{GL}}\left( {n,{\mathbb{C}}} \right), $$

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from the space of all square matrices of size $n$ to the group of all $n \times n$ invertible matrices [45].

As a result, we write that

$$ {\mathbf{A}} = e^{{ {\mathbf{M}}}} = \left( {e^{{ \frac{{\mathbf{M}}}{2}}} } \right)\left( {e^{{ \frac{{\mathbf{M}}}{2}}} } \right). $$

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Clearly, by definition, ${\mathbf{T}} = e^{{ \frac{{\mathbf{M}}}{2}}}$ is one possible square root of ${\mathbf{A}}$.