This convention is especially appropriate for a sinusoidal function, since its value at any argument t {\displaystyle t} then can be expressed as ( t ) {\displaystyle \varphi (t)} , the sine of the phase, multiplied by some factor (the amplitude of the sinusoid). (The cosine may be used instead of sine, depending on where one considers each period to start.)
Usually, whole turns are ignored when expressing the phase; so that ( t ) {\displaystyle \varphi (t)} is also a periodic function, with the same period as F {\displaystyle F} , that repeatedly scans the same range of angles as t {\displaystyle t} goes through each period. Then, F {\displaystyle F} is said to be "at the same phase" at two argument values t 1 {\displaystyle t_{1}} and t 2 {\displaystyle t_{2}} (that is, ( t 1 ) = ( t 2 ) {\displaystyle \varphi (t_{1})=\varphi (t_{2})} ) if the difference between them is a whole number of periods.
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This concept can be visualized by imagining a clock with a hand that turns at constant speed, making a full turn every T {\displaystyle T} seconds, and is pointing straight up at time t 0 {\displaystyle t_{0}} . The phase ( t ) {\displaystyle \varphi (t)} is then the angle from the 12:00 position to the current position of the hand, at time t {\displaystyle t} , measured clockwise.
The phase concept is most useful when the origin t 0 {\displaystyle t_{0}} is chosen based on features of F {\displaystyle F} . For example, for a sinusoid, a convenient choice is any t {\displaystyle t} where the function's value changes from zero to positive.
Moreover, for any given choice of the origin t 0 {\displaystyle t_{0}} , the value of the signal F {\displaystyle F} for any argument t {\displaystyle t} depends only on its phase at t {\displaystyle t} . Namely, one can write F ( t ) = f ( ( t ) ) {\displaystyle F(t)=f(\varphi (t))} , where f {\displaystyle f} is a function of an angle, defined only for a single full turn, that describes the variation of F {\displaystyle F} as t {\displaystyle t} ranges over a single period.
In the clock analogy, each signal is represented by a hand (or pointer) of the same clock, both turning at constant but possibly different speeds. The phase difference is then the angle between the two hands, measured clockwise.
The phase difference is particularly important when two signals are added together by a physical process, such as two periodic sound waves emitted by two sources and recorded together by a microphone. This is usually the case in linear systems, when the superposition principle holds.
For arguments t {\displaystyle t} when the phase difference is zero, the two signals will have the same sign and will be reinforcing each other. One says that constructive interference is occurring. At arguments t {\displaystyle t} when the phases are different, the value of the sum depends on the waveform.
If the frequencies are different, the phase difference ( t ) {\displaystyle \varphi (t)} increases linearly with the argument t {\displaystyle t} . The periodic changes from reinforcement and opposition cause a phenomenon called beating.
The phase difference is especially important when comparing a periodic signal F {\displaystyle F} with a shifted and possibly scaled version G {\displaystyle G} of it. That is, suppose that G ( t ) = F ( t + ) {\displaystyle G(t)=\alpha \,F(t+\tau )} for some constants , {\displaystyle \alpha ,\tau } and all t {\displaystyle t} . Suppose also that the origin for computing the phase of G {\displaystyle G} has been shifted too. In that case, the phase difference {\displaystyle \varphi } is a constant (independent of t {\displaystyle t} ), called the 'phase shift' or 'phase offset' of G {\displaystyle G} relative to F {\displaystyle F} . In the clock analogy, this situation corresponds to the two hands turning at the same speed, so that the angle between them is constant.
Therefore, when two periodic signals have the same frequency, they are always in phase, or always out of phase. Physically, this situation commonly occurs, for many reasons. For example, the two signals may be a periodic soundwave recorded by two microphones at separate locations. Or, conversely, they may be periodic soundwaves created by two separate speakers from the same electrical signal, and recorded by a single microphone. They may be a radio signal that reaches the receiving antenna in a straight line, and a copy of it that was reflected off a large building nearby.
A well-known example of phase difference is the length of shadows seen at different points of Earth. To a first approximation, if F ( t ) {\displaystyle F(t)} is the length seen at time t {\displaystyle t} at one spot, and G {\displaystyle G} is the length seen at the same time at a longitude 30 west of that point, then the phase difference between the two signals will be 30 (assuming that, in each signal, each period starts when the shadow is shortest).
For sinusoidal signals (and a few other waveforms, like square or symmetric triangular), a phase shift of 180 is equivalent to a phase shift of 0 with negation of the amplitude. When two signals with these waveforms, same period, and opposite phases are added together, the sum F + G {\displaystyle F+G} is either identically zero, or is a sinusoidal signal with the same period and phase, whose amplitude is the difference of the original amplitudes.
A real-world example of a sonic phase difference occurs in the warble of a Native American flute. The amplitude of different harmonic components of same long-held note on the flute come into dominance at different points in the phase cycle. The phase difference between the different harmonics can be observed on a spectrogram of the sound of a warbling flute.[4]
Phase comparison is a comparison of the phase of two waveforms, usually of the same nominal frequency. In time and frequency, the purpose of a phase comparison is generally to determine the frequency offset (difference between signal cycles) with respect to a reference.[3]
A phase comparison can be made by connecting two signals to a two-channel oscilloscope. The oscilloscope will display two sine signals, as shown in the graphic to the right. In the adjacent image, the top sine signal is the test frequency, and the bottom sine signal represents a signal from the reference.
If the two frequencies were exactly the same, their phase relationship would not change and both would appear to be stationary on the oscilloscope display. Since the two frequencies are not exactly the same, the reference appears to be stationary and the test signal moves. By measuring the rate of motion of the test signal the offset between frequencies can be determined.
Vertical lines have been drawn through the points where each sine signal passes through zero. The bottom of the figure shows bars whose width represents the phase difference between the signals. In this case the phase difference is increasing, indicating that the test signal is lower in frequency than the reference.[3]
Rajan, the issue you're facing is because you're using TBCTR for enabling the shift and the clocks for ePWMs are not in-sync. As I can see above TMCLKSYNC are individually configured. Try this change first:
Rajan Joshi said:
I was wondering if I could be specific about how much the offset would be ? Since i am using a driver to power a three phase motor, I need the waves to be 120 degrees out of phase, is it possible to do this?
You have several options--however the best is to devise a formula for the phase shift and then put this is the formula node in LabView to process the array data. Alternatively, you can use the Hilbert Transformation to apply a uniform phase shift to waveform data. Please refer to Echo Detector.vi or my simple example attached below.
What is the nature of the signal? For a sinusoid, a phase shift shifts the entire waveform in time by an amount -(phase shift in radians/(2 pi f), so a 1Hz sinusoid shifted 90 = pi/2 would be shifted in time -(pi/2)/(2 pi) = -0.25 seconds. The usual convention is a phase lead (positive numbers) means the waveform is advanced, so the "delay" is negative.
If your signal has a variety of frequencies in it, did you want to phase shift all of them by the same phase? Note that since the phase shift will shift-in-time depending on the frequency, the components of your composite signal will be shifted differing amounts, potentially changing the nature of the signal.
The problem I have now is making a third timer output that is synchronized to the first two but with a programmable phase shift. In reality I do have code that will make a synchronized, phase-shifted signal as shown in the image but there is an issue.
I don't know if this changes the situation, but when I set TIM2 CH1 (magenta line) CCR to 0 to have zero delay on the phase-shift signal (blue line), the signal stays high the whole 1st cycle so there is no rising edge on the 2nd cycle. The 3rd cycle on everything is good.
I made a little head way with this. I believe you are right that when I set CCR1=0 (0 phase shift) there's no trigger on the first cycle. If I set a delay of CCR1=1 then the phase-shift signal stops in the first cycle stops when it's supposed to instead of continuing into the second cycle.
I also figured out that for non-zero phase shifts the early high pulse (image in original post) is due to enabling the TIM3 capture compare channel. I tested this by setting the phase-shift trigger to have a delay of 500ms and started TIM1 and TIM2. I then did a HAL delay of 400ms and enabled TIM3 capture compare channel. At that point the signal went high, which is 100ms before the TIM3 trigger. This further suggests I need to look into your earlier suggestion of using CCxE preloading.
I have a working solution with which I will move forward. The key was to set the CR2.CCPC and CR2.CCUS bits to preload the capture compare enable. (I moved the phase-shift signal to an advanced timer that has this functionality.) This prevented the output from going high early by not enabling CCxE until the timer was triggered. Note in the code below I reset the bits and then set them again. This was required for when I stopped and restarted the timers. Without resetting the bits, the output would go high early again. 2351a5e196