Question 13.17: Obtain the response of the system of Example 13.16 by a freq...
Obtain the response of the system of Example 13.16 by a frequency-domain analysis but without using a normal coordinate transformation.
As in Example 13.16, we will determine the response for 3.0, so that the period of the forcing function is 3.0 \mathrm{~s}. As before, we select a sampling interval \Delta t=0.05 \mathrm{~s}. The complex frequency response function is obtained by solving the following equation
\left(-\Omega^{2} \mathbf{M}+\mathbf{K}\right) \mathbf{H}=\mathbf{f} \qquad (a)
Substitution for \mathbf{M}, \mathbf{K}, and \mathbf{f} gives
\left[\begin{array}{cc}192-2 \Omega^{2} & -64 \\-64 & 64-\Omega^{2}\end{array}\right] \mathrm{H}=\left[\begin{array}{l}1 \\\frac{1}{2}\end{array}\right] \qquad (b)
Table E13.17 Comparison of frequency domain displacement response obtained by the exponential window method and exact response in the physical coordinates.
On solving Equation b, we get
\mathbf{H}=\left[\begin{array}{l}H_{1} \\H_{2}\end{array}\right]=\frac{1}{D}\left[\begin{array}{c}96-\Omega^{2} \\160-\Omega^{2}\end{array}\right] \qquad (c)
where D=8192-320 \Omega^{2}+2 \Omega^{4}
The complex frequency functions are evaluated for \Omega=l \Delta \Omega, l=0,1,2, \ldots, 30, with \Delta \Omega=2 \pi / T_{0}=2.094. Because the system is undamped, \mathrm{H} is real. To generate the discrete version of frequency function, values of the function for l>30 are obtained by folding \mathrm{H} about a frequency of 30 \Delta \Omega. Thus
H_{j}(l \Delta \Omega)=H_{j}\{(L-l) \Delta \Omega\} \quad j=1,2 \quad l=31,32, \ldots, 59 \qquad (d)
where L=60.
The parameter a required in the exponential window method is selected as equal to 2 and the following steps are carried out to obtain the displacements in the two physical coordinates.
1 Obtain the discrete Fourier transform of the scaled function \hat{g}(k \Delta t)=g(k \Delta t) e^{-a k \Delta t}
\hat{G}(l \Delta \Omega)=\sum_{k=0}^{L-1} \hat{g}(k \Delta t) e^{-2 \pi i k l / L} \Delta t \qquad (e)
2 Calculate \hat{H}_{1} and \hat{H}_{2} from H_{1} and H_{2} using the relationships
\hat{H}_{j}(l \Delta \Omega)=H_{j}(l \Delta \Omega-i a) \quad j=1,2 \qquad (f)
3 Obtain the scaled displacements in the physical coordinates
\begin{aligned}&\hat{u}_{1}(k \Delta t)=\frac{1}{2 \pi} \sum_{l=0}^{L-1} \hat{U}_{1}(l \Delta \Omega) e^{2 \pi i k l / L} \Delta \Omega \qquad (g) \\&\hat{u}_{2}(k \Delta t)=\frac{1}{2 \pi} \sum_{l=0}^{L-1} \hat{U}_{2}(l \Delta \Omega) e^{2 \pi i k l / L} \Delta \Omega \qquad (h)\end{aligned}
where
\begin{aligned}&\hat{U}_{1}(l \Delta \Omega)=\hat{G}(l \Delta \Omega) \hat{H}_{1}(l \Delta \Omega) \\&\hat{U}_{2}(l \Delta \Omega)=\hat{G}(l \Delta \Omega) \hat{H}_{2}(l \Delta \Omega)\end{aligned}
4 Recover the true displacements in physical coordinates from the scaled displacements by using the following relationship.
u_{j}(k \Delta t)=\hat{u}_{j}(k \Delta t) e^{a k \Delta t} \qquad (i)
The final displacements calculated as above are shown in Table E13.17 at intervals of 0.1 \mathrm{~s}. For the purpose of comparison, exact values obtained by a mode superposition analysis are also shown. The same sets of data are presented in Figure E13.17. The two sets match very closely
