3
$\begingroup$

Let $y(t)$ be a function that consists of a sum of terms, each of which is in general a product of an exponential function $e^{\sigma t}$, and a sine function $\sin{\omega t}$ or cosine function $\cos{\omega t}$ with some amplitude (it could also have a potential function $t^n$, but for the sake of simplicity let's assume $y$ has no terms with such factor). So $y$ has the general form

$y(t) = 2 e^{\sigma_1 t} (a_1 \cos{\omega_1 t} + b_1 \sin{\omega_1 t}) u(t) + 2 e^{\sigma_2 t} (a_2 \cos{\omega_2 t} + b_2 \sin{\omega_2 t}) u(t) + \cdots$

In other words, $y$ is the output of a continuous LTI system with no repeated poles.

My question is, how do we get the greatest time constant of the output, assuming the system is stable with a bounded input?

My proposed solution: Since we know the expression of $y(t)$, one way I think of tackling this problem is by following these steps:

  1. Get each term of $y$. This results in a list $\{ 2 e^{\sigma_1 t} (a_1 \cos{\omega_1 t} + b_1 \sin{\omega_1 t}) u(t), 2 e^{\sigma_2 t} (a_2 \cos{\omega_2 t} + b_2 \sin{\omega_2 t}) u(t), \dots \}$
  2. For each term, identify/get the exponential factor. This results in a list $\{ e^{\sigma_1 t}, e^{\sigma_2 t}, \dots \}$.
  3. For each exponential factor, get the coefficient of the exponent (i.e. the Neper frequency). This results in a list $\{ \sigma_1, \sigma_2, \dots \}$.
  4. For each Neper frequency, get the negative of the inverse of it. This results in a list of $\{ \tau_1, \tau_2, \dots \}$, where $\tau_i = -1/ \sigma_i$.
  5. Get the greatest time constant. This results in $\tau_{\text{max}}$.

While this algorithm is pretty descent for the kind of systems I'm studying, my problem is I don't know how to implement/code this. Do you know how? Or do you know another way? For the last step though, we could use the Max[] function of Mathematica.

EDIT: As an example, consider the output

$y(t) = [1 - 0.14 e^{-2.59 t} + 0.16 e^{-0.31 t} - 2 e^{-0.05 t} (0.51 \cos{0.95 t} + 0.20 \sin{0.95 t})] u(t)$

All of the coefficients of the exponents of exponential factors are non-positive, so the output is bounded and we can use the algorithm. After applying steps 1 to 3, Mathematica should get the list {-2.59, -0.31, -0.05}. For step 4, it gets the inverse of each element of the list and multiplies it by -1, getting the list {0.39, 3.23, 20}. For step 5, the Max[] function applied to the previous list should return 20, which is the greatest time constant of the output.

By the way, I just realized we can make the algorithm a bit more efficient. Change step 4 as: "Get the Min[] of the list obtained in step 3." And change step 5: "Get the inverse of the minimum element obtained and multiply it by -1."

$\endgroup$
  • $\begingroup$ it will be easier to answe if you show a MWE of an actual expression to use to test with with any concrete values they might have. $\endgroup$ – Nasser May 2 at 5:52
  • $\begingroup$ @Nasser I've added an example. $\endgroup$ – Alejandro Nava May 2 at 6:30
3
$\begingroup$
ClearAll[y, t, x, u];
y[t_] := (1 - 0.14*Exp[-2.59*t] + 0.16*Exp[-0.31*t] - 
        2*Exp[-0.05*t]*(0.51*Cos[0.95*t] + 0.2*Sin[0.95*t]))*u[t]; 

getPatterns[expr_, pat_] := Last[Reap[expr /. a:pat :> Sow[a], _, Sequence @@ #2 & ]]; 

(*r = getPatterns[y[t], Exp[x__]];*)
r = getPatterns[y[t], Exp[_. t + _.]]; (*may be better*)

Mathematica graphics

If[Length[r] > 0, Max[Cases[r, Exp[(x_.)*t] :> -x^(-1)]]]
(*20*)

The function getPatterns above is thanks to Carl Woll

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.