In the course of modal analysis, the terms in the equation of motion are transformed using the (Modal Analysis:modal matrix). As the eigenvectors are orthogonal with respect to the mass and stiffness matrix, this transformation leads to a diagonal mass \mathbf{M^*} and stiffness matrix \mathbf{K^*}. Hence, the equations of motion for all n degrees of freedom are decoupled.

For an arbitrary damping matrix \mathbf{C}\neq 0, the transformation

\mathbf{C^*}=\pmb{\Phi}^T\mathbf{C}\pmb{\Phi}

using the undamped eigenmodes does not yield a diagonal generalized damping matrix \mathbf{C^*} and the n equations of motion do not decouple.

Solving for an arbitrary damping matrix

Recall the damped equation of motion

(1) \mathbf{M} \ddot{\mathbf{u}} (t) + \mathbf{C} \dot{\mathbf{u}} (t) + \mathbf{K} \mathbf{u} (t) = \mathbf{f} (t)

In the general case, the following ansatz holds (compare (Modal Analysis:2) with \lambda = i \omega)

\mathbf{u}(t) = \pmb{\phi} \mathrm{e}^{\lambda t} , \label{eq:mdof_free_damped_ansatz_w}

and thus the first two derivatives with respect to time are

\begin{gather} \dot{\mathbf{u}}(t) = \lambda \pmb{\phi} \mathrm{e}^{\lambda t} \label{eq:mdof_free_damped_ansatz_wdot} \\ \ddot{\mathbf{u}}(t) = \lambda^2 \pmb{\phi} \mathrm{e}^{\lambda t} . \label{eq:mdof_free_damped_ansatz_wddot} \end{gather}

We know insert the ansatz into the equation of motion (1) and obtain

\left( \mathbf{M} \lambda^2 + \mathbf{C} \lambda + \mathbf{K} \right) \pmb{\phi} = \mathbf{0} ,

which is commonly referred to as a quadratic eigenvalue problem. A nontrivial solution for \pmb{\phi} is found for the case that the matrix \left( \mathbf{M} \lambda^2 + \mathbf{C} \lambda + \mathbf{K} \right) becomes singular and the determinant is zero

\det \left( \mathbf{M} \lambda^2 + \mathbf{C} \lambda + \mathbf{K} \right) = 0 .

This is considered as the characteristic equation of the problem. It is a polynomial with order 2 n and gives 2 n solutions for \lambda. Thus, we obtain 2 n eigenvalues \lambda and corresponding eigenvectors \pmb{\phi}. If all component matrices are symmetric, the solutions are either real or occur as complex conjugate pairs. As for the undamped case, the eigenvectors are scaled using an arbitrary factor.

The quadratic eigenvalue problem has to be linarized in order to be solved. The linearization results in a generalized eigenvalue problem, which can be solved using e.g. a computer algebra system. The linearization is not unique.

It can, for example, be constructed using a substitution \mathbf{x}=\lambda\pmb{\phi} in the quadratic eigenvalue problem

\left( \mathbf{M} \lambda^2 + \mathbf{C} \lambda + \mathbf{K} \right) \pmb{\phi} = \mathbf{0} ,

which results in

\lambda\mathbf{M}\mathbf{x} + \lambda\mathbf{C}\pmb{\phi} + \mathbf{K}\pmb{\phi} = 0.

Rewriting this in matrix notation yields the generalized eigenvalue problem

\begin{alignat}{2} \underbrace{ \begin{bmatrix} \mathbf{-K} & \mathbf{0} \\ \mathbf{0} & \mathbf{I} \end{bmatrix} }_{:= \mathbf{A}} % \begin{bmatrix} {\pmb{\phi}} \\ {\mathbf{x}} \end{bmatrix} - \lambda \underbrace{ \begin{bmatrix} \mathbf{C} & \mathbf{M} \\ \mathbf{I} & \mathbf{0} \end{bmatrix} }_{:= \mathbf{B}} % \begin{bmatrix} {\pmb{\phi}} \\ {\mathbf{x}} \end{bmatrix} = \begin{bmatrix} \mathbf{0} \\ \mathbf{0} \end{bmatrix} . \label{eq:DGL_mdof_damped_ext} \end{alignat}

The 2n eigenvalues can now be found solving

\det \left(\mathbf{A}-\lambda \mathbf{B} \right) = \mathbf{0} .

Response in time domain

Using the eigenvalues and eigenvectors, the response for given initial conditions $\mathbf{u}_0$ at any time $t$ can be computed. First, the initial conditions have to be transformed to modal space by

\mathbf{q}_0=\pmb{\Phi}^{-1}\mathbf{u}_0.

Now, using the exponential ansatz, the system response is given by

\mathbf{u} (t) = \sum_{i=1}^{2n} \, \pmb{\phi}_i q_{0,i} \;\, \mathrm{e}^{\lambda_i t} = \pmb{\Phi} \pmb{q}_0 \mathrm{e}^{\pmb{\lambda}t}