I would like to find the frequency response of a spring mass system of multiple degrees of freedom by using the Laplace transform. I think I know how to do this with one mass oscillating, however I find difficulty in extending the approach to multiple masses.

My approach for one mass

The equation of motion for one damped spring mass is as follows: $$m\ddot{x}(t) + c\dot{x}(t) + kx(t) = f(t) \tag{1}$$ To solve this in the frequency domain, I employ the Laplace transform on Equation (1) above: $$ms^2X(s) + csX(s) + kX(s) = F(s) \\ \Rightarrow H(s) = \frac{X(s)}{F(s)}=\frac{1}{s^2m + sc + k} \tag{2}$$ Substituting $j\omega$ in the place of $s$, and taking the modulus of the transfer function $H(s)$ above, we obtain the frequency response for the single mass for all $\omega$: $$|H(j\omega)|=\left| \frac{1}{-\omega^2m + j\omega c+ k} \right| \tag{3}$$

My approach for multiple masses

I pick up where I left of from Equation (2), however this time I replace the masss, damping and stiffness coefficients with their respective matrices, and make my substitution for $s$ as in Equation (3) above: $$[-\omega^2\mathbf{M} + j\omega\mathbf{C} + \mathbf{K}]\mathbf{X}=\mathbf{F}$$

However this is where I get stuck. Intuitively I can tell that my transfer function will be something like $\mathbf{H} =[-\omega^2\mathbf{M} + j\omega\mathbf{C} + \mathbf{K}]^{-1}$, however firstly I don't know how to prove it, and secondly I don't know how from this matrix, I will end up getting a vector for every value of $\omega$ that I insert in this equation, such that I get the frequency response for each degree of freedom


2 Answers 2


In frequency space a one dimensional linear system is characterized by the dynamical equation

$$ A(s)X(s) = F(s) $$

Which can typically be found from the equations of motion and a Laplace transform. Inverting this expression we find

$$ X(s) = A(s)^{-1}F(s) = H(s)F(s) $$

Where I've defined $H(s) = A(s)^{-1}$. $H(s)$ is the system transfer function or susceptibility. It tells us the response of the linear system (characterized by variable $X(s)$) due to excitation $F(s)$ at a particular frequency.

The system becomes more physically complicated when multiple degrees of freedom are included but the math is not much more complicated. The dynamical equation (from equations of motion and Laplace transform) is now

$$ \boldsymbol{A}(s) \boldsymbol{X}(s) = \boldsymbol{F}(s) $$

Here $\boldsymbol{A}$ is an $n\times n$ matrix To find the susceptibility matrix we invert

$$ \boldsymbol{X}(s) = \boldsymbol{A}(s)^{-1}\boldsymbol{F}(s) = \boldsymbol{H}(s)\boldsymbol{F}(s) $$

The susceptibility matrix $\boldsymbol{H}$ is an $n\times n$ matrix. The $(i,j)$ component of this matrix, $H_{ij}$ tells us the response of variable $X_i$ to "force" $F_j$.

If we want to "visualize" the susceptibility matrix we, in general, must plot all $n\times n$ components $H_{ij}(s)$ versus frequency $s$. This is the analog of the plot of $H(s)$ in the 1d case.


I don't think that you have to transfer your differential equation to first order to get the frequency response (solution of @Jagerber)

Time domain

$$\mathbf M\,\mathbf{\ddot{x}}+\mathbf{K}\mathbf x+ \mathbf C\,\mathbf{\dot{x}}=\mathbf f(t)$$

transformed to Laplace domain and solve for $\mathbf{X}(s)$

$$\mathbf X(s)=\underbrace{\left[s^2\,\mathbf M+\mathbf K+s\,\mathbf C\right]^{-1}}_{\mathbf H(s)}\,\mathbf F(s)$$

$~\mathbf H(s)~$ is a matrix $n\times n$ where n is the number of the ODE's


$$m\,\ddot x_1+k\,(x_1-x_2)+c\,\dot x_1=f_1(t)\\ m\,\ddot x_2-k\,(x_1-x_2)+c\,\dot x_2=f_2(t)$$

$$\mathbf M=\left[ \begin {array}{cc} m&0\\ 0&m\end {array} \right]\\ \mathbf K= \left[ \begin {array}{cc} k&-k\\ -k&k\end {array} \right] \\ \mathbf C=\left[ \begin {array}{cc} c&0\\ 0&c\end {array} \right] $$


$$\mathbf H=\left[ \begin {array}{cc} {\frac {m{s}^{2}+cs+k}{s \left( {m}^{2}{s}^ {3}+2\,m{s}^{2}c+2\,msk+{c}^{2}s+2\,ck \right) }}&{\frac {k}{s \left( {m}^{2}{s}^{3}+2\,m{s}^{2}c+2\,msk+{c}^{2}s+2\,ck \right) }} \\ {\frac {k}{s \left( {m}^{2}{s}^{3}+2\,m{s}^{2}c+2 \,msk+{c}^{2}s+2\,ck \right) }}&{\frac {m{s}^{2}+cs+k}{s \left( {m}^{2 }{s}^{3}+2\,m{s}^{2}c+2\,msk+{c}^{2}s+2\,ck \right) }}\end {array} \right] $$

thus you obtain 4 transfer functions and



solution of @Jagerber


$$x_1=y_1~,x_2=y_2~,\dot x_1=y_3~,\dot x_2=y_4$$ \begin{align*} &\mathbf{\dot{y}}=\mathbf{A}\,\mathbf{y}+\mathbf{b}\\ &\left[ \begin {array}{c} {\dot y}_{{1}}\\ {\dot y}_ {{2}}\\ {\dot y}_{{3}}\\ {\dot y}_ {{4}}\end {array} \right] =\left[ \begin {array}{cccc} 0&0&1&0\\ 0&0&0&1 \\ -{\frac {k}{m}}&{\frac {k}{m}}&-{\frac {c}{m}}&0 \\ {\frac {k}{m}}&-{\frac {k}{m}}&0&-{\frac {c}{m}} \end {array} \right] \,\left[ \begin {array}{c} y_{{1}}\\ y_{{2}} \\ y_{{3}}\\ y_{{4}}\end {array} \right] + \left[ \begin {array}{c} 0\\ 0\\ {\frac {f_{{1}}}{m}}\\ {\frac {f_{{2}}}{m}} \end {array} \right]\\\\ &\text{transfer to Laplace and solve for $~Y(s)$}\\\\ &\mathbf{Y}(s)=\underbrace{\left[s\,I_4-A\right]^{-1}}_{\mathbf{H}(s)}\,\mathbf{b}(s) \end{align*} with

the differential equations transfer to first order differential equations

\begin{align*} &\mathbf{H}(s)=\left[ \begin {array}{cccc} {\frac {m{s}^{2}+cs+k}{ \left( cs+m{s}^{2 }+2\,k \right) s}}&{\frac {k}{ \left( cs+m{s}^{2}+2\,k \right) s}}&{ \frac {m \left( m{s}^{2}+cs+k \right) }{s \left( {m}^{2}{s}^{3}+2\,m{s }^{2}c+2\,msk+{c}^{2}s+2\,ck \right) }}&{\frac {km}{s \left( {m}^{2}{s }^{3}+2\,m{s}^{2}c+2\,msk+{c}^{2}s+2\,ck \right) }} \\ {\frac {k}{ \left( cs+m{s}^{2}+2\,k \right) s}}&{ \frac {m{s}^{2}+cs+k}{ \left( cs+m{s}^{2}+2\,k \right) s}}&{\frac {km} {s \left( {m}^{2}{s}^{3}+2\,m{s}^{2}c+2\,msk+{c}^{2}s+2\,ck \right) }} &{\frac {m \left( m{s}^{2}+cs+k \right) }{s \left( {m}^{2}{s}^{3}+2\,m {s}^{2}c+2\,msk+{c}^{2}s+2\,ck \right) }}\\ -{\frac {k}{cs+m{s}^{2}+2\,k}}&{\frac {k}{cs+m{s}^{2}+2\,k}}&{\frac {m \left( m{s}^{2}+cs+k \right) }{{m}^{2}{s}^{3}+2\,m{s}^{2}c+2\,msk+{c}^{2}s+2 \,ck}}&{\frac {km}{{m}^{2}{s}^{3}+2\,m{s}^{2}c+2\,msk+{c}^{2}s+2\,ck}} \\ {\frac {k}{cs+m{s}^{2}+2\,k}}&-{\frac {k}{cs+m{s} ^{2}+2\,k}}&{\frac {km}{{m}^{2}{s}^{3}+2\,m{s}^{2}c+2\,msk+{c}^{2}s+2 \,ck}}&{\frac {m \left( m{s}^{2}+cs+k \right) }{{m}^{2}{s}^{3}+2\,m{s} ^{2}c+2\,msk+{c}^{2}s+2\,ck}}\end {array} \right] \end{align*}

thus the transfer function for example $~Y_3~$

\begin{align*} Y_3(s)=\sum_{i=1}^{4} H_{3,i}\,b_i= {\frac { \left( m{s}^{2}+cs+k \right) F_{{1}}}{{m}^{2}{s}^{3}+2\,m{s}^ {2}c+2\,msk+{c}^{2}s+2\,ck}}+{\frac {kF_{{2}}}{{m}^{2}{s}^{3}+2\,m{s}^ {2}c+2\,msk+{c}^{2}s+2\,ck}} \end{align*}

must be equal to the transfer function $~X_1(s)$ (my solution) this is the case so my approach is correct!!

who ever vote it down please let me know why ??

  • $\begingroup$ Please don't delete and repost your answers. If you want to make changes, use the edit button instead. $\endgroup$
    – Chris
    Aug 20, 2021 at 16:14
  • $\begingroup$ You can also undelete your own answers, if you delete an answer but later decide you want it to come back. $\endgroup$
    – Chris
    Aug 20, 2021 at 16:15
  • $\begingroup$ @Chris this is a new answer . if you vote it down please let me know why ? I think that my answer is still correct !!!!!! $\endgroup$
    – Eli
    Aug 20, 2021 at 16:25
  • 1
    $\begingroup$ I didn't downvote your answer and I don't know who did. Asking downvoters to explain themselves is rarely helpful- whoever downvoted your answer is unlikely to even see the edit. $\endgroup$
    – Chris
    Aug 20, 2021 at 19:05

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