In this measurement column, we address frequently asked questions received by our customer support center and provide answers. This time, continuing from the previous article, we will introduce a method for measuring the frequency response function using a sine sweep signal from external equipment such as a vibration exciter controller.
In methods using sine sweep signals from external devices, the frequency of the sine wave changes independently of the analysis device's FFT calculation. Therefore, if the sine wave sweep speed is too fast, the correct results will not be obtained. This time, we will show how the power spectrum values for each channel become smaller than their actual amplitude when the sweep speed is too fast.
Measurement of frequency response function using signals from external devices
Figure 1 shows an example of a system configuration for measuring the frequency response function using a sine sweep signal from an external device, as introduced in the previous article. In this example configuration, a controller with an oscillator outputs a sine sweep signal, causing an exciter to vibrate. Accelerometer are attached to both the exciter and the object under test, and the frequency response function (natural frequency) of the object under test is measured from these signals.
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Figure 1: Example of a system configuration for vibration measurement using signals from a vibration exciter controller.
Guideline for sweep speed
As introduced last time, frequency range, number of samples, FFT time length, and ideal sweep speed
The relationship is given by the following equation. Examples are shown in Table 1.
Number of lines [points] = Number of sample points [points] ÷ 2.56
Frequency resolution [Hz] = Frequency range [Hz] ÷ Number of lines [points]
FFT time length [seconds] =1÷frequency resolution [Hz]
Sweep speed [Hz/sec] = Frequency resolution [Hz] ÷ Time length [sec]
=(Frequency resolution [Hz]) 2
Table 1: An example of frequency range, number of samples, FFT time length, and ideal sweep speed.
|
Frequency range (Hz) |
Sample score (points) |
Frequency resolution (Hz) |
FFT time length (sec) |
Sweep speed (Hz/second) |
|
40000 |
4096 |
25.0 |
0.04 |
625.0 |
|
20000 |
4096 |
12.5 |
0.08 |
156.3 |
|
10000 |
4096 |
6.25 |
0.16 |
39.06 |
|
5000 |
4096 |
3.13 |
0.32 |
9.77 |
|
2000 |
4096 |
1.25 |
0.80 |
1.56 |
|
1000 |
4096 |
0.63 |
1.60 |
0.39 |
|
20000 |
1024 |
50.0 |
0.02 |
2500.0 |
|
10000 |
1024 |
25.0 |
0.04 |
625.0 |
|
5000 |
1024 |
12.5 |
0.08 |
156.3 |
|
2000 |
1024 |
5.00 |
0.20 |
25.00 |
|
1000 |
1024 |
2.50 |
0.40 |
6.25 |
Example of measurement using a sine sweep signal from an external device
Figures 2-1, 2-2, and 2-3 show the power spectra when a sine sweep signal input from an external device was measured. The frequency range was 400 Hz, the number of sample points was 1024, and the frequency resolution was 1 Hz, with settings that resulted in an ideal sweep speed of 1 Hz/second. A sine sweep signal with an amplitude of 0.1 Vrms and a frequency range of 20 Hz to 200 Hz was output from a function generator. The sweep time and sweep speed were 180 seconds (1 Hz/second), 45 seconds (4 Hz/second), and 11.25 seconds (16 Hz/second), which are 1, 4, and 16 times the ideal sweep speed determined from the measurement conditions, respectively.
The top row of each measurement result shows the power spectrum when a sine sweep signal is measured.
The lower section shows the instantaneous spectrum at a frequency of 100 Hz, with an enlarged view of the range from 50 Hz to 150 Hz.
The amplitude of the input signal is 0.1 Vrms (-20 dBVrms), but the amplitude of the power spectrum is approximately -21.6 dBVrms at a sweep speed of 1 Hz/sec, approximately -22.1 dBVrms at 4 Hz/sec, and approximately -26.3 dBVrms at 16 Hz/sec, decreasing as the sweep speed increases.
Furthermore, when examining the instantaneous spectrum, we can see that as the sweep speed increases, the width of the 100 Hz component peak widens, and consequently, the height of the peak decreases.
Under these measurement conditions, the FFT time length is 1 second. To obtain the instantaneous spectrum at 100 Hz, a time waveform of 1 second is required. If the sweep speed is 16 Hz/second, the time waveform includes time waveforms from 92 Hz to 108 Hz. As a result, the instantaneous spectrum obtained from this time waveform will be spread horizontally, and the peak height (amplitude) will be lower.
To avoid this phenomenon, it is necessary to either slow down the sine wave sweep speed or change the measurement conditions so that the ideal sweep speed is faster.
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Figure 2-1 Example of measurement results for a sine sweep signal (Sweep time: 180 seconds, Sweep speed: 1 Hz/second)
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Figure 2-2 Example of measurement results for a sine sweep signal (Sweep time: 45 seconds, Sweep speed: 4 Hz/second)
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Figure 2-3 Example of measurement results for a sine sweep signal (Sweep time: 11.25 seconds, Sweep speed: 16 Hz/second)
Example of measurement using a sine sweep signal from an external device (log sine sweep)
The above describes measurement results using a "linear sine sweep signal" where the frequency change rate per second is constant. In measurements using a sine sweep signal, a "log sweep signal" is sometimes used, where the sweep speed increases as the frequency increases.
The sweep speed of a "log sweep signal" is indicated by how many seconds it takes for the frequency to change by one octave, such as 1 octave/10 seconds. Here, one octave is the frequency range over which the frequency doubles.
Figure 3 shows the power spectrum when a log sine sweep signal input from an external device was measured. The frequency range is 400 Hz, the number of sample points is 1024, and the frequency resolution is 1 Hz. A log sine sweep signal with an amplitude of 0.1 Vrms, a frequency range of 20 Hz to 320 Hz, and a sweep time of 40 seconds was output from a function generator. The sweep speed was 1 octave/10 seconds. It took 10 seconds for the frequency to double from 20 Hz to 40 Hz, 10 seconds from 40 Hz to 80 Hz, 10 seconds from 80 Hz to 160 Hz, and 10 seconds from 160 Hz to 320 Hz, for a total of 40 seconds to sweep from 20 Hz to 320 Hz.
The sweep rate per unit time increases with increasing frequency. The sweep rate is 1.4 Hz/second at 20 Hz, 2.8 Hz/second at 40 Hz, 5.5 Hz/second at 80 Hz, 11.1 Hz/second at 160 Hz, and 22.2 Hz/second at 320 Hz.
Figure 3 shows that the amplitude of the power spectrum decreases as the frequency increases. This is because the sweep rate per unit time increases as the frequency increases. Also, there is a gap in the spectrum around 313 Hz. This is because the sweep rate was too fast for the FFT calculation to keep up.
Because of this phenomenon, log-sine sweep signals are not suitable for measuring frequency response functions using FFT analysis. If only log-sine sweep signals can be used, it is necessary to narrow the frequency range being measured or change the measurement conditions to increase the ideal sweep speed.
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Figure 3. Example of measurement results for a sine sweep signal (1 octave/10 seconds).
summary
In this article, we introduced measurement examples using sine sweep signals from external devices, specifically focusing on linear sine sweep signals and log sine sweep signals.
In either case, if the sweep speed of the sine sweep signal is too fast, accurate measurements cannot be obtained.
In that case, you will need to slow down the sweep speed or change the measurement conditions.
Furthermore, if you have the option to choose between a linear sine sweep signal and a log sine sweep signal, please use the linear sine sweep for measurement.
(Excerpt from the email newsletter issued on April 21, 2016)
