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I am wondering how to calculate transfer function $H(s)$ of system described by 3 coupled differential equations. The pourpose of work is to calculate "Bodedx" diagram ($|H(i\omega)|(\omega)$). My initial idea is to apply Laplace transform to the left and right side of the equation as it is done in the case of system described by only 1 differential equation. This includes expressing $H(s)=Y(s)/X(s)$, where $X$ and $Y$ are input and output signal. This approach works well for the equations of shape $$ay''+by'+cy=dx''+ex'+jx+k;$$

But I don't know how to adapt this tehnique to more complex system: $$ \mathbf{M}\ddot{x}+\mathbf{D}\dot{x}+\mathbf{K}x=\mathbf{L} $$

where $\mathbf{M, D,K, L}$ are 3 by 3 matrixes and x is the vector with 3 elements.

As a concrete example I have a following problem:

enter image description here

The picture shows force transducer with load mass, which is standing on the shaker table during the calibration. In reality, the force transducer can be modeled as 2 masses (upper and bottom part) connected with strings. During the experiment the shaker undergoes sinus induction of oscialtion and frequency response of the transducer is measured with comparason with reference sensor.

There are 4 different displacements of the masses $x_L, x_H, x_B, x_S$, with corresponding spring coefficients $k_L, k, k_B$. The output of the force transducer is proportional to deformation of spring with coefficient $k$, e.g. $output=x_H-x_B$. For every spring we also introduce damping coefficient $d$ with corresponding index.

Therefore I can rewrite matrices as $$ \mathbf{M}= \begin{pmatrix} m_L &0&0\\ 0&m_H&0\\ 0&0&m_B \end{pmatrix} \hspace{0.5in} \mathbf{D}=\begin{pmatrix} d_l&-d_l&0 \\ -d_L &d_L+d& -d\\0&-d&d+d_B \end{pmatrix}\hspace{0.5in} \mathbf{K}=\begin{pmatrix} k_l&-k_l&0 \\ -k_L &k_L+k& -k\\0&-k&k+k_B \end{pmatrix} $$ and $$ \mathbf{L}=\begin{pmatrix}0\\0\\d_B \dot{x_s}+ k_B x_s\end{pmatrix} \hspace{0.5in} x=\begin{pmatrix}x_L\\x_H\\x_B \end{pmatrix} $$

As I understand I should apply 3-Dimensional Laplace transform, but I do not know how exactly. The other option would be more likely to solve the system which I want to avoid, since I am interested only in frequency-dependancy of amplitude. Is there any better way to do so?

Figure and problem from: Michael Kobusch, Sascha Eichstädt, "A case study in model-based dynamic calibration of small strain gauge force transducers", Acta IMEKO, vol. 6, no. 1, article 2, April 2017, identifier: IMEKO-ACTA-06 (2017)-01-02

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  • $\begingroup$ Why would it be a 3-dimensional Laplace transform? You Laplace transform in time, not space. $\endgroup$
    – jacob1729
    May 30 at 10:52
  • $\begingroup$ @jacob1729 That makes sense... But how to apply it, since I have 3 equations? $\endgroup$
    – Vid
    May 30 at 10:53
  • $\begingroup$ Have you tried simply Laplace transforming (in time) each of the three equations and seeing what you end up with? $\endgroup$
    – jacob1729
    May 30 at 10:54

1 Answer 1

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form $$ \mathbf{M}\mathbf{\ddot{x}}+\mathbf{D}\mathbf{\dot{x}}+\mathbf{K}\mathbf{x}=\mathbf{L} \tag 1$$

with $$\mathbf x(t)\mapsto \mathbf x(s)~,\mathbf{\dot {x}}\mapsto s\,\mathbf x(s)~,\mathbf{\ddot{x}}\mapsto s^2\,\mathbf x(s)\\ \mathbf{L}(t)\mapsto \underbrace{\left[ \begin {array}{c} 0\\ 0\\ d_{{B}}\,s+k_{{B}}\end {array} \right]}_{\mathbf b(s)} \,x_s(s)$$

you transformed the ODE to Laplace space

thus

$$\mathbf H(s)=\left[s^2\mathbf M+s\,\mathbf D+K\right]^{-1}\,\mathbf b(s)$$

$$\mathbf x(s)=\mathbf H(s)\,x_s(s)$$

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