I want to solve the following equations for $x_1$, $x_2$, and $x_3$

$$ K_0\, x_1(t) + K_1\, \frac{dx_1(t)}{dt} + M_0\, \frac{dx_2(t)}{dt} + x_3(t) =A_0\,\sin (\alpha_0 t) \tag{1} $$

$$ M_0\, \frac{dx_1(t)}{dt} + L_0\,x_2(t) + K_1\, \frac{dx_2(t)}{dt}+L_1\,\int x_2(t)\, dt =0 \tag{2} $$

$$ x_1(t) -\alpha_1\,B_1\,\cos(\alpha_1\,t + \phi) \, x_3(t)- \big[B_0+B_1\,\sin(\alpha_1\,t + \phi) \big] \frac{dx_3(t)}{dt} =0 \tag{3} $$

Here is my try using Laplace transform:

I can assume $x_1(0)=0$, and define $x_2(0)=x_{20}$, and $x_3(0)=x_{30}$

$$(1) \Rightarrow K_0\, X_1(s) + K_1\, sX_1(s)+ M_0\, sX_2(s)+ X_3(s) + M_0\, x_{20} = \frac{A_0\alpha_0}{s^2+\alpha_0^2} \tag{4} $$

$$(2) \Rightarrow M_0\, X_1(s) + L_0\, X_2(s) +K_1\, sX_2(s)+ L_1\, \frac{X_2(s)}{s}+ X_3(s)+ K_1\, x_{20} =0 \tag{5} $$ Next step brings me the trouble. I cannot simply take the Laplace transform of (3)

Update 1 (as suggested by @Cesareo)

Using the Euler's expansion of sine and cosine (3) can be transformed to

$$ X_1(s) - B_0 s X_3(s) -\\ \frac{B_1 (\alpha_1 +1) s}{2} \big(e^{i \phi} X_3(s-i \alpha_1) + e^{-i \phi} X_3(s+i \alpha_1)\big)+x_{30} (B_0 - B_1 \sin \phi)=0 \tag{6}$$

Now, my question is: How can I move forward from here? How can I handle $X_3(s \pm i \alpha_1)$ ifI want to solve for $X_3(s)$?


1 Answer 1



$$ \mathcal{L}(\cos(\alpha t)x(t)) = \mathcal{L}(\frac 12(e^{i\alpha t}+e^{-i\alpha t})x(t)) $$


$$ \int_0^{\infty}e^{i\alpha t}x(t)e^{-s}dt = \int_0^{\infty}e^{-(s-i\alpha)t}x(t) dt = X(s-i\alpha) $$

so after the Laplace transformation you should handle a system of difference equations ...

  • $\begingroup$ Thanks for the tip. I will try it and update here. isn't $\mathcal{L} \big(e^{i \alpha t} x(t) \big) = X(s-i \alpha) $? $\endgroup$
    – Pojj
    Commented Nov 20, 2018 at 19:10
  • $\begingroup$ @Pojj You are right. I had a little imagination left. $\endgroup$
    – Cesareo
    Commented Nov 21, 2018 at 1:40
  • $\begingroup$ any suggestion how I can move forward from here? (Edited the question) $\endgroup$
    – Pojj
    Commented Nov 21, 2018 at 18:43
  • $\begingroup$ @Pojj In the same line, eliminating $X_1(s), X_2(s)$ we arrive to a quasi polynomial equation for $X_3(s)$ with the structure $$ f_1(s)X_3(s)+f_2(s)\left(e^{i\phi}X_3(s-i\alpha_1)+e^{-i\phi}X_3(s+i\alpha_1)\right)= C_0 $$ with $f_1(s),f_2(s)$ polynomial fractions ... $\endgroup$
    – Cesareo
    Commented Nov 22, 2018 at 1:37

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