You are correct that the Fourier Transform will return the average peak amplitude of each of the frequency components. Remember that the Fourier Transform is speed over positive find negative frequencies. So if, as usual, we only look at the positive frequencies then the amplitude calculated by DATS will be half the amplitude. That is if the Fourier analysis shows say 0.4 then the amplitude in terms of an equivalent sinewave is 0.8.

In the examples I used the DATS Moph command. This computes values in terms of modulus and phase (this in the same as some people call amplitude and phase).

Generally within DATS if you use the default values for parameters you will get a sensible result. The Moph command asks three questions, all to do with how the phase is calculated. The first question asks about phase limits : zero to 2PI (0 to 360 degrees); +/-PI (-180 to +180 degrees); or phase unwrap. The phase unwrap mode, which I normally use, detects each time the phase crosses the 0 to 2PI (0 to 360 degree) boundary and either adds en subtracts 2PI (360 deg) appropriately. The second question is do you wish to see the phase in radians or degrees, with degrees as the default.

The third question is more technical and if you are unsure use the default of 80dB, as I do 99% of the time. What this question is about is handling the situation when the individual real and imaginary are small as the phase at each frequency comes from arctan(imaginary/real). Now with acquired data if there is no meaningful energy at a frequency then both the real and the imaginary parts will be small and basically act like random noise. This causes the phase to jump around like no tomorrow. Setting the value to zero will turn it off, but generally making it smaller than 80 will make it more severe (less phase noise).

Why restrict to first mode? Do you know its frequency?

Abiye Zerfu

The classical reason for a phase “jump” is that one is passing thpartrough a resonance in that frequency range. This should be accompanied by an increase in the amplitude. If you have truncated your transient then the sharp stop will cause some phase changes as the Fourier transform has to fit this really rapid amplitude change. Such a change way be small in amplitude but could be quite significant in phase change.

Now the phase at each frequency is determined by the inverse tangent of the imaginary part of the Fourier component at that frequency divided by the corresponding real part. For reference the real front is the cosine coefficient and the imaginary part is the sine coefficient. If over some frequency region the real and imaginary parts are small and are fluctuating around zero then because we use their ratio in finding the phase then clearly the phase will also fluctuate wildly. Basically then if the amplitude, which is the square root of the sum of the real part squared and the imaginary part squared, is small then the phase is not reliable in a real system where we have ‘noise’ present. In deed if it is only the real part that is small and is fluctuating do to noise, then one can have large phase changes.

It is away useful if in doubt to look at the real and imaginary parts. Also try looking at a vector plot of Real versus Imaginary as this can be most instructive.

Colin

I’m curious about the significance of the frequency (or range thereof) which the phase jump occurs. I have transients (not sinusoidal) showing a lot of phase jumps and I am trying to interpret what these frequencies mean.

Thanks,
James.

But beware with real life signals only use this as a simple model to aid understanding.

Colin

1. Frequency and time are inversely proportional. Thus if we have a signal of length T seconds then the raw frequency resolution is (1/T)Hz.

2. Fourier Transforms may be inverse transformed back from frequency to time. This means that we have no extra information, way just have a different representation which we may understand better. It also means that if we have a time signal that has some random data in it then the Fourier Transformed signal also has some random element in the Fourier coefficients. This usually means we have to average the Fourier transformed data, not directly but as a modulus squared. This leads us to spectral densities and spectrum levels.

3. The act of Fourier transforming a finite length of the time signal causes leakage of information into adjacent frequencies. This brings us into the domain of data windows that are designed to minimise these effects. Whilst minimising the leakage of information from one frequency to others, using a data window means that the frequency resolution is made larger. If in doubt use a Hanning window.

4. Frequency resolution is not the same as Frequency separations (or steps). That is we may form estimates of the Fourier coefficients at frequency steps of say df Hz but the frequency resolution may be dF Hz, where dF > df .

5. When we use a finite length of data and Fourier analyse it then each coefficient is an ‘average’ over the resolution dF. We may liken the Fourier transform to passing the time signal through a series of narrow band filters each of width dF Hz but spaced df Hz apart, that is the filters overlap each other.

Hope these help, there is no magic bullet – just a lot of work, time and effort trying to understand.

Colin

best regard