Errata: sign errors in Example 5.18 (noise cancellation)

Chapter	Dynamic Behavior
Page	5-34
Line	8
Version	3.1.5
Date	26 Nov 2024

In Example 5.18 (noise cancellation), there are two sign errors in equation (5.26) that are propagated through the next several lines. The corrected text should read (with changes in red):

Assuming for simplicity that $S=0$, introduce $x_{1}=e=z-w$, $x_{2}=a-a_{0}$, and $x_{3}=b-b_{0}$. Then

${\frac {dx_{1}}{dt}}=a_{0}(z-w)+(a-a_{0})w+(b-b_{0})n=a_{0}x_{1}-x_{2}w-x_{3}n.$

(5.26)

We will achieve noise cancellation if we can find a feedback law for changing the parameters $a$ and $b$ so that the error $e$ goes to zero. To do this we choose

$V(x_{1},x_{2},x_{3})={\frac {1}{2}}{\bigl (}\alpha x_{1}^{2}+x_{2}^{2}+x_{3}^{2}{\bigr )}$

as a candidate Lyapunov function for equation (5.26). The derivative of $V$ is

${\dot {V}}=\alpha x_{1}{\dot {x}}_{1}+x_{2}{\dot {x}}_{2}+x_{3}{\dot {x}}_{3}=\alpha a_{0}x_{1}^{2}+x_{2}({\dot {x}}_{2}-\alpha wx_{1})+x_{3}({\dot {x}}_{3}-\alpha nx_{1}).$

Choosing

${\dot {a}}={\dot {x}}_{2}=\alpha wx_{1}=\alpha we,\qquad {\dot {b}}={\dot {x}}_{3}=\alpha nx_{1}=\alpha ne,$

we find that ${\dot {V}}=\alpha a_{0}x_{1}^{2}<0$, and it follows that the quadratic function will decrease as long as $e=x_{1}=w-z\neq 0$.