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Figure 1 Sine wave signal waveform
It can be expressed as ; (Figure 1).
Here, A is defined as amplitude, ω as angular frequency, f as frequency, T as period (= 1/f), ωt+φ as phase, and φ as initial phase.
Thus, the phase corresponds to the angle of the sine wave function, but since the above value (ωt+φ) only increases with time, the initial phase is often referred to as the phase. As shown in Figure 1, when the horizontal axis is time, the time difference from the coordinate origin corresponds to the phase.
In general, the relationship between phase θ and time difference τ is:
Therefore, if you know the time difference, you can calculate the phase. The reverse is also true.
So, how is the phase of channel 1 represented in an FFT analyzer?
The phase of channel 1 corresponds to the phase of the complex Fourier spectrum in an FFT analyzer. The complex Fourier spectrum C(k) is as follows:
Then the phase spectrum θ(k) is:
It is required as follows:
As can be seen from this equation, FFT analyzers use a cosine waveform as their reference. That is, in a cosine waveform like the one in Figure 2 below, the phase is defined as 0 degrees, and in a sine waveform like the one in Figure 3 below, the peak appears 90 degrees later than the cosine, so the phase is -90 degrees.
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Figure 2 Cosine wave <Phase φ = 0°>
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Figure 3 Sine wave (phase φ = -90°)
In this way, the complex Fourier spectrum of one channel is obtained, and the phase for each frequency k is calculated to obtain the phase spectrum θ(k). Generally, the phase spectrum of one channel is obtained by applying a trigger (internal or external trigger signal) and determining the phase from the timing of its acquisition. A major application of this is "balancing measurement of rotating bodies".
The phase of channel 1 is the phase relative to the data acquisition point, as explained above, but the phase of channel 2 is clearly the phase difference between channels. For example, consider the sine waves of two channels for input and output;
In this case, the gain of the transfer function is A2/A1 and the phase is φ2-φ1, which, like the 1-channel, is a function of frequency f. It is important to note here that the phase difference is the difference between identical frequency components contained in the 2-channel signal. If the frequencies are different, the phase cannot usually be determined. Also, the phase difference between the 2 channels is determined as the phase of the transfer function (= phase of the cross spectrum), but in principle, it is equivalent to the phase difference of the complex Fourier spectra of the 2 channels sampled simultaneously. The phase difference between channels represents the phase characteristics between the input and output of the system and is widely used in measuring transfer functions in electrical measurement, vibration analysis, servo analysis, etc.
Finally, let's consider the phase of a square wave. As you know, a square wave consists of a sine wave with its period as the fundamental frequency and sine waves of its harmonic frequencies, so the phase of each harmonic component can be determined. It is important to note here that even when the rising edge of the square wave pulse is taken as the origin, the phase of the fundamental wave differs depending on the duty cycle of the square wave. As shown in the example in Figure 4, when the duty cycle is 50%, only odd-order harmonics are present, and the phase is exactly -90 degrees. However, when the duty cycle is other than 50% (15.6% in this example), even-order harmonics also appear. In general, if the duty cycle of a square wave is d (100 d%), the phase of the fundamental wave of the square wave when the rising edge is taken as the origin can be expressed as -180 d (degrees). Figure 5 shows the phase of a square wave and a square wave that has been further delayed by a certain amount of time (78.125 μs) with the rising edge as the origin. You can see that the fundamental wave of the square wave on the right is delayed by an additional 22.5 degrees.
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Figure 4: Phase difference of a square wave due to duty cycle.
Left side: When the duty cycle is 50%
Right side: When the duty cycle is 15.6%
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Figure 5: Phase comparison of waveforms with one side delayed from the same square wave.
Left side: Starting point
Right side: Waveform with the left waveform delayed by approximately 78 μs.
(Note: At θ=ωτ, τ=78.125e-6, ω=360/1.25e-3)
Thus, even when comparing the rising phases of square waves with the same frequency, if the duty cycles are different, it will not represent the phase difference of the fundamental wave. Generally, if we take the transfer functions of square waves with duty cycles d1 and d2, and determine that the phase difference of the fundamental wave is θ1, then the rising phase difference θ of the square wave is:
It will be.
(Excerpt from the email newsletter issued on April 24, 2008)
