Question 9.4: Find the response of a damped single-degree-of-freedom syste...

The unit impulse response function h(t) is given by

h(t)=\frac{1}{m \omega_{d}} e^{-\xi \omega t} \sin \omega_{d} t \qquad (a)

where \omega_{d}=\omega \sqrt{1-\xi^{2}}.

Function h(t) is shown in Figure E9.4b. It is noted that although g(t), the forcing function, is of finite duration, h(t) is not. However, the magnitude of h(t) decreases fairly rapidly with time, so that we can truncate the function at a sufficiently large value of t and assume that it is zero for all greater t.

Let the sampling interval be chosen as 0.1 \mathrm{~s} and h(t) be truncated at 19 \times 0.1=1.9 \mathrm{~s}. We have p=4 and q=19, so that the period should be selected as N=p+q+1=24. The sampled functions with a period of 24 \times 0.1=2.4 \mathrm{~s} are shown in Figure E9.4c and d, respectively.

The discrete convolution of the two sampled functions is obtained through the frequency domain by using the convolution theorem of Equation 9.58.

g(k \Delta t) * h(k \Delta t) \Longleftrightarrow G(n \Delta \Omega) H(n \Delta \Omega)          (9.58)

An appropriate computer program is used to obtain the direct and inverse Fourier transforms involved. The resulting convolution values are shown in Figure E9.4e and Table E9.4.

The exact response of the system to the rectangular pulse function of amplitude p_{0} is given by the following expressions

The exact response, which is the same as the response obtained from a continuous convolution, is also shown in Figure E9.4e and Table E9.4. The discrete convolution results closely match the exact values. The difference is because of the rectangular summation used in discrete convolution. The results can therefore be improved by using a smaller value for \Delta t, say 0.05 \mathrm{~s}.

It will be noted that the match between the exact and discrete convolution values is not as good on samples 20 through 23 . This is because of the truncation applied to the unit

Table E9.4 Response of a single-degree-of-freedom system to a rectangular pulse function.
\begin{matrix}\hline & & u(k \Delta t) & u(k \Delta t) \\k & t(s) & \text{Exact }& \text{Frequency domain} \\\hline 0 & 0.0 & 0.0000 & 0.0000 \\1 & 0.1 & 0.0190 & 0.0183 \\2 & 0.2 & 0.0673 & 0.0647 \\3 & 0.3 & 0.1252 & 0.1208 \\4 & 0.4 & 0.1708 & 0.1650 \\5 & 0.5 & 0.1689 & 0.1635 \\6 & 0.6 & 0.1047 & 0.1017 \\7 & 0.7 & 0.0055 & 0.0059 \\8 & 0.8 & -0.0896 & -0.0862 \\9 & 0.9 & -0.1458 & -0.1408 \\10 & 1.0 & -0.1446 & -0.1399 \\11 & 1.1 & -0.0899 & -0.0873 \\12 & 1.2 & -0.0053 & -0.0056 \\13 & 1.3 & 0.0761 & 0.0733 \\14 & 1.4 & 0.1244 & 0.1202 \\15 & 1.5 & 0.1237 & 0.1197 \\16 & 1.6 & 0.0772 & 0.0750 \\17 & 1.7 & 0.0050 & 0.0052 \\18 & 1.8 & -0.0647 & -0.0622 \\19 & 1.9 & -0.1062 & -0.1025 \\20 & 2.0 & -0.1058 & -0.1021 \\21 & 2.1 & -0.0663 & -0.0733 \\22 & 2.2 & -0.0047 & -0.0383 \\23 & 2.3 & 0.0550 & -0.0105 \\ \hline \end{matrix}

impulse function. In a later section we discuss a method of applying suitable correction to improve these values.