How do I apply a uniform phase shift to a waveform?

In LabVIEW 6.0, I am attempting to apply a uniform phase shift to a non-periodic waveform but have run into a problem. Theoretically, if I take the Fourier transform of the signal, apply an offset to the phase and then take the inverse transform of that, the output of this process should be my original waveform with all of its frequency components delayed by the same number of radians (different amounts of time). However, I am not getting this result from LabVIEW. In the attached example code I generate a sine wave burst, take the FFT, convert from cartesian to polar, apply an offset to the phase, convert back to cartesian and take the inverse FFT. However, instead of getting a phase-shifted, same-amplitude versio
n of the input at the output, I get an amplitude shifted, same-phase signal. What am I doing wrong?
Attachments:
phase-shift.vi ‏51 KB

This only works (partially) if your array size is an integral power of 2 (e.g. 512, 1024, etc.), i.e. in cases where the fast fourier transform can be used.
You need to use the complex FFT (and the hilbert transform)
Please see my example posted HERE
LabVIEW Champion . Do more with less code and in less time .

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Phase shift a signal data

hi all
i want to phase shift a signal of 90 degree. i am having the signal data as 1d array,using the data i want to shift the phase of the the signal .....any one....
Indrajit
| [email protected] | [email protected] .

Hi Indar,
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.
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Attachments:
Hilbertwitharray.vi ‏90 KB

Phase shift

Dear all,
i have create a simple model to find the phase shift of 2 waveform using the dual channel spectral measurment model. My waveforms for both signal r of transient type, ie their phase shift is different from each cycle and as such i not sure abt the phase shift result.
refer to attachment:my first graph is the signals of both waveform and the first cycle phase different is 11.28 degree and the second cycle onwards is 28.2 degree. the 2nd graphs shows the phase response of the both signal and the result is 24.2 degree.can some 1 pls advise me how those the dual channel spectral measruement model give the result of both waveform phase response as 24.2 degree.I am bluffed.
urgent request.
Cheers
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Attachments:
PHASESHIFT.JPG ‏539 KB

I am assuming that negating the result is not going to give you the
right answer. Then, you can try the inverse Hilbert Transform.
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Hi,
I have a simple RLC circuit consisting of no more than 4 components. If i hook up the network analyser function to the circuit and read the S11 values, i can see a change in the input impedance and a phase shift when i run it across a range of frequencies.
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Attachments:
M1.png ‏3155 KB
m2.png ‏618 KB

Hi 08Ultrasound,
You need to measure the voltage phase difference over a load as the voltage source is ideal so will always be in phase.
Regards,
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National Instruments UK

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IF you were to take two +identical signals+, sum them in equal amounts and flip the phase on one of them 180º to the other you will get +complete cancellation+.
According to the manual on page 221 the Process>Invert will do this.
Invert
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+display, the wave’s crests become troughs and vice versa.+
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Hi Jinio,
Please take a look at the following KnowledgeBase article:
How Can I Determine Phase Mismatch Between Two Channels?

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Hi AI S
I have gone through the example of phase diff e.g and i have written the following code but when i run it , it gives the max value of 1st sine wave (outarray1) nearly equal (i-e 7.7 volt ) but it should be 8 V pk.
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    int count;
static double wave1maxval = 0.0;
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Thanks very much , that's really helpful,Cathy
Date: Thu, 23 Dec 2010 11:18:21 -0700
From: [email protected]
To: [email protected]
Subject: How do I apply the blur filter to all my layers at the same time?
Hello!
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Thanks a lot for your help. Maybe I explain more about my current structure
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Edited by: Daniel_problem on Aug 15, 2008 10:34 AM
Edited by: Daniel_problem on Aug 15, 2008 10:36 AM

Phase shift of a sine wave and a square wave

Hi,
I would like to simply compute the phase shift between a sinusoidal waveform and a square one. Between two sin-s I can compute, but I have no idea how to do that with a square waveform.
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Ussr123

Hi there,
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Phaseshift_Sqr_Sin.vi ‏89 KB

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I admit this is not 100% on mark but may give you some ideas. In this thread I suggested using a zero-phase shift filter. Ignore this post if it does not help.
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Hello,
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http://forums.ni.com/ni/board/message?board.id=170&message.id=6956&query.id=101197
These examples work great with theoritcal signals, but there is a problem (see figure below) I am having in practice.. Since the phase shift can occur in the range [-360,+360], i believe the index of the peaks are getting lost, causing it to pick up the next peak in the sequence, thus, causing a phase shift (see figure below, plot of phase shift) to be measured.  Another way of saying this is that after a small amount of phase shifting occurs, the lead peak begins to references the next peak in the comparison signal.
Im at loss on how to identify the peaks and track them so i can accurately measure phase shift. Any help is appreciated. Im sure this is soemthing that has been done before, esp since there is a vi just for measuring phase shift. I guess i just dont know the key words to find it...
Below is a screen shot showing the two signals. The bottom plot is showing there is a relative small shift in phase which periodically jumps.
many thanks
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Message Edited by JimmyinCT2 on 01-23-2008 07:17 PM

dahhh, okay, i think i finally figured out what is going on here, and i think im just being somewhat dumb or maybe just assuming to much.
Looking at matts example helped a lot as i see why.
with one of the old VIs for measuring phase shift, we were getting a phase shift range of [-360, 360] or more.    We were just assuming that was the range here with this new VI.
With the new dual channel measurement (or the example you posted), when phase is measured, there is an inflection point at some angle of phase difference.. The way your example works (and the express vi) is with an inflection at 180 degrees. So, lets imagine we have a measurement of phase shift that is 178 degrees, as this difference increases, the next is 179, and then 180.. Now as it continues to move, the next phase measurement will not bet 181 but instead -179.   As the signal continues to move in the same direction, its then 178, 177, 176.. So there is an inflection point at 180 degrees with two sinusoids, in simple terms, its because the phase difference of one sinusoid to the other has shifted and is now being measured in reference to the next peak in the series.
Its troubling for measurements are if you turn on a machine that is measuring phase, and it happens to be right near -/+180, it can cross that boundary often, creating substantial noise.   Also, that for certain types of peak detection, such as for motion, its desirable to have a larger range such as [-360,360] which should be theoretically possible (like the first vi we used) but i guess [-720, 720] could be done too assuming there is enough difference between the peaks. It would have been awesome to have a VI that measures peak movement outward to infinity.   We were originally using a VI that calculated out these larger differences, think it was doing the calcs in frequency domain (imaginary nums) rather then cross correlation like the example.
I think for now, we are going to have to work inside the range [-180,180] and initialize our hardware to 0 phase so we dont start near the inflection point and see if that works.
Message Edited by jimmyinCT on 01-26-2008 07:54 PM

How do I apply a uniform phase shift to a waveform?

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