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Found in: Page 439

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# feedback Loops:Suppose some quantities ${{\mathbit{k}}}_{{\mathbf{i}}}$ ${{\mathbit{x}}}_{{\mathbf{1}}}\left(t\right){\mathbf{,}}{{\mathbit{x}}}_{{\mathbf{2}}}\left(t\right){\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{,}}{{\mathbit{x}}}_{{\mathbf{n}}}\left(t\right)$ can be modelled by differential equations of the form localid="1662090443855" style="max-width: none; vertical-align: -109px;" $|\begingathered\frac{d{x}_{1}}{dt}=-{k}_{1}x -b{x}_{n}\hfill\phantom{\rule{0ex}{0ex}}\frac{d{x}_{2}}{dt}={x}_{1}-{k}_{2 } \hfill\phantom{\rule{0ex}{0ex}}.\hfill\phantom{\rule{0ex}{0ex}}.\hfill\phantom{\rule{0ex}{0ex}}\frac{d{x}_{n}}{dt} = {x}_{n-1}-{k}_{n}{x}_{n}\hfill\phantom{\rule{0ex}{0ex}}\endgathered|$Where b is positive and the localid="1662090454144" style="max-width: none; vertical-align: -9px;" ${{\mathbit{k}}}_{{\mathbf{i}}}$ are positive.(The matrix of this system has negative numbers on the diagonal, localid="1662090460105" ${\mathbf{1}}{\mathbf{\text{'}}}{\mathbit{s}}$ directly below the diagonal and a negative number in the top right corner)We say that the quantities localid="1662090470062" style="max-width: none; vertical-align: -9px;" ${{\mathbit{x}}}_{{\mathbf{1}}}{\mathbf{,}}{{\mathbit{x}}}_{{\mathbf{2}}}{\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{,}}{{\mathbit{x}}}_{{\mathbf{n}}}$ describe a (linear) negative feedback loop Describe the significance of the entries in this system inpractical terms. Is a negative feedback loop with two components localid="1662090478095" style="max-width: none; vertical-align: -4px;" $\left(n=2\right)$ necessarily stable? Is a negative feedback loop with three components necessarily stable?

a. The significance of the entries in the system are asymptotically stable

b. The negative feedback loop with two components are necessarily stable

c. The negative feedback loop with three components are necessarily stable

See the step by step solution

## Step 1: Explanation of the stability of a continuous dynamical system

For a system $\frac{\stackrel{⇀}{dx}}{dt}=A\stackrel{\to }{x}$, here A is the matrix form.

The zero state is an asymptotically stable equilibrium solution if and only if the real parts of all eigen values of A are negative

## Step 2: Description for the significance of the entries in the above system in practical terms

Suppose some quantities ${x}_{1}\left(t\right),{x}_{2}\left(t\right),...,{x}_{n}\left(t\right)$ can be modelled by differential equations of the form:

$\left|\\mathrm{begingathered}\frac{d{x}_{1}}{dt}=-{k}_{1}x -b{x}_{n}\\mathrm{hfill}\phantom{\rule{0ex}{0ex}}\frac{d{x}_{2}}{dt}={x}_{1}-{k}_{2 } \\mathrm{hfill}\phantom{\rule{0ex}{0ex}}.\\mathrm{hfill}\phantom{\rule{0ex}{0ex}}.\\mathrm{hfill}\phantom{\rule{0ex}{0ex}}\frac{d{x}_{n}}{dt} = {x}_{n-1}-{k}_{n}{x}_{n}\\mathrm{hfill}\phantom{\rule{0ex}{0ex}}\\mathrm{endgathered}\right|$

Here b is positive and ${k}_{i}$ are positive.

The matrix of the above system has negative numbers on the diagonal and first directly below the diagonal and a negative number in the top right corner.

The entries of this system are negative in the top right corner.

So the system is asymptotically stable.

Thus the solution

## Step 3: Negative feedback loop with two components

Consider the negative feedback loop of the system $\left|\\mathrm{begingathered}\frac{d{x}_{1}}{dt}=-{k}_{1}x -b{x}_{n}\\mathrm{hfill}\phantom{\rule{0ex}{0ex}}\frac{d{x}_{2}}{dt}={x}_{1}-{k}_{2 } \\mathrm{hfill}\phantom{\rule{0ex}{0ex}}.\\mathrm{hfill}\phantom{\rule{0ex}{0ex}}.\\mathrm{hfill}\phantom{\rule{0ex}{0ex}}\frac{d{x}_{n}}{dt} = {x}_{n-1}-{k}_{n}{x}_{n}\\mathrm{hfill}\phantom{\rule{0ex}{0ex}}\\mathrm{endgathered}\right|$

Here feedback loops are negative.

So according to the stability condition, the real part of the eigen values of $n=2$ are necessarily stable.

Thus the solution.

## Step 4: Negative feedback loop with three components

Consider the negative feedback loop of the system $\left|\\mathrm{begingathered}\frac{d{x}_{1}}{dt}=-{k}_{1}x -b{x}_{n}\\mathrm{hfill}\phantom{\rule{0ex}{0ex}}\frac{d{x}_{2}}{dt}={x}_{1}-{k}_{2 } \\mathrm{hfill}\phantom{\rule{0ex}{0ex}}.\\mathrm{hfill}\phantom{\rule{0ex}{0ex}}.\\mathrm{hfill}\phantom{\rule{0ex}{0ex}}\frac{d{x}_{n}}{dt} = {x}_{n-1}-{k}_{n}{x}_{n}\\mathrm{hfill}\phantom{\rule{0ex}{0ex}}\\mathrm{endgathered}\right|$

Here feedback loops are negative.

So according to the stability condition, the real part of the eigen values of $n=3$ are necessarily stable.

Thus the solution.