Question 10.12: It is desired to monitor the vibration level of a test struc......
It is desired to monitor the vibration level of a test structure by measuring auto power spectral density functions, at frequencies up to 1000 Hz, using accelerometers.
The analog outputs from the accelerometers are sampled digitally at f_{s} = 2500 samples/s, after passing through analog anti-aliasing filters, designed so that the response is nearly flat up to 1000 Hz, but then falls sharply, so that it is negligible at the Nyquist frequency, f_{N}, which is f_{s}/2 = 1250 Hz. The samples are stored in a computer containing a DFT program, which evaluates Eq. (9.45):
X_{k} = \frac{1}{N} \sum\limits_{j=0}^{N-1}{x_{j} e^{-\mathrm{i}\left(\frac{2\pi j k}{N} \right) }} (A)The value of N is fixed at 2048. The digital samples from the accelerometers are represented by x_{j} ( j = 0, 1, 2, 3, . . . , 2047). The program outputs 2048 complex values of X_{k} (k = 0, 1, 2, 3, . . . , 2047).
Derive expressions enabling acceleration PSD plots, without refinements such as data windows, overlap or sequential averaging, to be produced by the computer.
The output of a single accelerometer is considered. Since the sampling rate is f_{s} = 2500 samples/s, and the computer takes the data in a batch of N = 2048 values, the time duration of the sample is T where
T = N / f_{s} = 2048/2500 = 0.8192 sThe frequency spacing of the 2048 values of X_{k} \mathrm{is} 1/T = f_{s}/N = 2500/2048 = 1.2207 Hz. Values of X_{k} above the Nyquist frequency, 1250 Hz, will be spurious, and the useful frequency range will be further limited by the anti-aliasing filter cut-off at 1000 Hz. So, of the 2048 values of X_{k} output by the DFT program, only those below 1000 Hz will be useful, that is (1000/1.2207) or about 819 discrete frequencies, spaced at intervals of 1.2207 Hz.
From Eq. (10.82), X^{*}_{k}X_{k} = \frac{1}{4}(a^{2}_{n} + b^{2}_{n}), the power at each frequency, \frac{1}{2}(a^{2}_{n} + b^{2}_{n}), is given by:
The PSD at each frequency, f_{k}, will be the power at that frequency, \frac{1}{2}(a^{2}_{n} + b^{2}_{n}),divided by the frequency interval, \delta f = 1 / T, i.e.,
\frac{\frac{1}{2}(a^{2}_{n} + b^{2}_{n})}{\delta f} = \frac{\frac{1}{2}(a^{2}_{n} + b^{2}_{n})}{1 / T} = \frac{2X_{k}X^{*}_{k}}{1 / T} = \frac{2X_{k}X^{*}_{k}}{1.2207} (C)If the original samples, x_{j}, are scaled in g units, the PSD values will be in g²/Hz. The values of frequency will be
f_{k} = k / T (k = 0, 1, 2, 3, \cdot \cdot \cdot , 819) (D)that is from 0 to 1000 Hz (approximately) in steps of 1.2207 Hz.