Simple singularly perturbed systems

102 Views Asked by At

I'm an engineer looking for math insights into an engineering problem. I have a simple first order ODE: $$\dot x(t)=k\Big(-x(t)+f(t)\Big).$$ As you can see, this system behaves like a low-pass filter with $k$ acting as the filter bandwidth. The higher the filter bandwidth, the higher frequency can pass through the filter. So by hand waving, I thought when the bandwidth is very very high, almost everything passes through the filter and $x(t)$ will behave like $f(t)$. I did many Simulink runs as well as experiments with an RC circuit and a function generator that produces different signals $f(t)$. My hand-waving statement seems to be true. My questions are:

1) How do I rigorously show that as $k\to \infty,$ $x(t)$ behaves like $f(t)$?

2) I did some literature review and figured that the statement is true if $f(t)$ is constant in light of Tikhonov's result [1]. I wanted to dig deeper into the literature, but my background in maths is too limited to understand deeper results. If the statement is not true in general, what is the condition on $f(t)$ for the statement to be true?

Any ideas or discussions are greatly appreciated.

Thank you!

[1] A.N. Tikhonov, Systems of differential equations containing a small parameter multiplying the derivative, Mat. Sb. 31 (1952) pp. 575-586

1

There are 1 best solutions below

8
On BEST ANSWER

Answer to 1): A singular perturbation is a degeneration of the systems order, i.e. the derivative is perturbed.

Write the ODE in standard form for singularly perturbed systems:

$\frac{1}{k} \dot x = -x + f(t)$

or with $\varepsilon=1/k$

$\varepsilon \dot x = -x + f(t)$ .

Now, let $k \to \infty$:

$0=-x+f(t)$

$x=f(t)$ is called a quasi steady state.

It follows: $x=f(t) + O(\varepsilon)$. The last expression is a result of Tikhonov. It means that it is an approximation with error of order $ O(\varepsilon)$.

2) $f(t)$ must be a unique root of the degenerate system. This is the case in this example. Furthermore Tikhonov’s Thm requires some stability conditions of the so called boundary layer model. To consider this, set $y=x-f$ and insert it in the ODE:

$\varepsilon \dot y = -y - \varepsilon \dot f(t)$ and set $\tau= t/\varepsilon$:

$\frac{d}{d\tau}y=-y -\varepsilon \dot f(t)$ and let $\varepsilon=0$:

$\frac{d}{d\tau}y=-y $.

This is the boundary layer system which describes the fast transient before reaching the quasi steady state.

It is exponentially stable. Hence, due to Tikhonov‘s Thm the approximation $x=f(t) + O(\varepsilon)$ is valid.