# LMIs in Control/pages/Small Gain Theorem

LMIs in Control/Matrix and LMI Properties and Tools/Small Gain Theorem

The Small Gain Theorem provides a sufficient condition for the stability of a feedback connection.

## Theorem

Suppose ${\displaystyle B}$  is a Banach Algebra and ${\displaystyle Q\in B}$ . If ${\displaystyle \|Q\|<1}$ , then ${\displaystyle (I-Q)^{-1}}$  exists, and furthermore,

                   ${\displaystyle (I-Q)^{-1}=\sum _{k=0}^{\infty }Q^{k}}$


## Proof

Assuming we have an interconnected system ${\displaystyle (G,K)}$ :

${\displaystyle y_{1}=G(u_{1}-y_{2})}$  and, ${\displaystyle y_{2}=K(u_{2}-y_{1})}$

The above equations can be represented in matrix form as

{\displaystyle {\begin{aligned}{\begin{bmatrix}I&0\\0&I\end{bmatrix}}{\begin{bmatrix}y_{1}\\y_{2}\end{bmatrix}}&={\begin{bmatrix}\ 0&-G\\\ -K&0\\\end{bmatrix}}{\begin{bmatrix}y_{1}\\y_{2}\end{bmatrix}}+{\begin{bmatrix}G&0\\0&K\end{bmatrix}}{\begin{bmatrix}u_{1}\\u_{2}\end{bmatrix}}\end{aligned}}}

Making ${\displaystyle {\begin{bmatrix}y_{1}&y_{2}\end{bmatrix}}^{T}}$  the subject, we then have:

{\displaystyle {\begin{aligned}{\begin{bmatrix}y_{1}\\y_{2}\end{bmatrix}}&={\begin{bmatrix}\ I&G\\\ K&I\\\end{bmatrix}}^{-1}{\begin{bmatrix}G&0\\0&K\end{bmatrix}}{\begin{bmatrix}u_{1}\\u_{2}\end{bmatrix}}&={\begin{bmatrix}\ (I-GK)^{-1}G&-G(I-KG)^{-1}K\\\ -K(I-GK)^{-1}G&(I-KG)^{-1}K\\\end{bmatrix}}{\begin{bmatrix}u_{1}\\u_{2}\end{bmatrix}}\end{aligned}}}

If ${\displaystyle (I-GK)^{-1}}$  is well-behaved, then the interconnection is stable. For ${\displaystyle (I-GK)^{-1}}$  to be well-behaved, ${\displaystyle \|(I-GK)^{-1}\|}$  must be finite.

Hence, we have ${\displaystyle \|(I-GK)^{-1}\|<\infty }$

${\displaystyle \|G\|\|K\|=\|Q\|}$  and ${\displaystyle \|Q\|  for the higher exponents of ${\displaystyle \|Q\|}$  to converge to ${\displaystyle 0.}$

## Conclusion

If ${\displaystyle \|Q\|<1}$ , then this implies stability, since the higher exponents of ${\displaystyle Q}$  in the summation of ${\displaystyle \sum _{k=0}^{\infty }Q^{k}}$  will converge to ${\displaystyle 0}$ , instead of blowing up to infinity.