3A FFT Analysis

8/2/2019 3A FFT Analysis

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FFT Analysis

Understanding the limitations of

synchronisation and frequency

resolution


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Definitions

N = number of samples taken T = total sampling time = N/Fs

T= time increment between samples =T/N Fs = sample frequency = 1/ T

F= frequency resolution = 1/T = Fs/N

Fmax = Folding frequency =Fs/2

F = frequency


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What is an FFT

The FFT, or Fast Fourier Transform is a method of

calculating harmonics not one at a time, but as a group,

using a special algorithm.

Advantage

The FFT requires much less processing power than a DFT

for the same number of harmonic results.

Disadvantage

Requires that the number of samples being analysed are N

to a power of 2, such as 64, 128, 256 etc


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FFT Disadvatages

Frequency Resolution Harmonic Leakage

Requires synchronisation


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10Hz Waveform sampled at 200Hz

Consider a 10Hz sine wave with an amplitude of 1

As a starting point the sampling frequency (Fs)is 200Hz

The number of samples taken (N) was 256

Sample time = N/Fs =256/200 equals 1.28 seconds


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Frequency SpectrumFrequency Resolution = Fs/N = 200/256 equals 0.781Hz

Fmax( folding frequency) =Fs/2 equals 100Hz

Solution of (f) 10Hz = f/ F equals 12.8 samples


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Frequency Spectrum

Because the FFT resolution does not allow the

measurement of 10 Hz

Sample 12 = 9.372 Hz

Sample 13 = 10.153 Hz

Sample 14 = 10.934 Hz


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Comments

Although the frequency spectrum correctly shows a spike at 10Hz, the spike is notinfinitesimally narrow. In fact, it appears from the frequency spectrum that there is asignificant component of the signal at frequencies near 10Hz (specifically within about5Hz to either side of 10Hz). This unphysical error in the FFT is called leakage, whichappears when the discrete data acquisition does not stop exactly in the same phase of thesine wave as it started, it is not synchronised.

In principle if an infinite number of discrete samples are taken, leakage would not be aproblem. However, any real data acquisition system performing FFTs uses a finite(rather than infinite) number of discrete samples, and there will always be some leakage.

The maximum amplitude of the FFT is not exactly 1 (in fact it is less than 1), eventhough the amplitude of the original signal was exactly 1. This is another consequenceof leakage namely some of the energy at 10Hz is distributed among frequencies near10Hz, thereby reducing the calculated amplitude at 10 Hz

This can be reduced using a correction window function such as a Hanning windowwhich modifies the result of the FFT to mathematically

The frequency range is from 0 to 100Hz, half the sampling frequency (Nyquist criterion)


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How can leakage be reduced?

Will a higher sample rate help ?

Lets use an increased sample rate of 1000Hz

Consider a 10Hz sine wave with an amplitude of 1As a starting point the sampling frequency (Fs)is 1000Hz




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Increase sample rate to 1000Hz

Higher sampling frequency has vastly improved the resolution

of the sine wave, there are now 100 data points per wavelength.


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Fmax( folding frequency) =Fs/2 equals 500Hz



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Frequency Spectrum



Sample 2= 7.82 Hz

Sample 3 = 11.73 Hz

Sample 4 = 15.64 Hz


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Comments

Increasing the sampling frequency did not improve the frequency spectrum, in fact the

output frequency resolution is worse approximately 3.91Hz whereas at a sampling

frequency of 200Hz the resolution was approximately 0.781Hz

Furthermore, the peak amplitude of the frequency spectrum is now 0.66 significantly

smaller than the known signal amplitude of1. This is due to leakage, which is even

worse in this case because of poor frequency resolution.

Even worse, the peak amplitude of the frequency spectrum occurs at a frequency of11.72Hz instead of the known 10Hz. This is the result of poor frequency resolution.

The only thing gained from increasing the sampling frequency of an FFT is the

corresponding increase in the maximum frequency ( folding frequency ) of the

frequency spectrum but at the loss of frequency resolution.


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Decrease sample rate to 25HzFor a 10Hz sine wave , a 1000Hz sample rate is overkill.

By the Nyquist criterion, since the maximum frequency is

10Hz, the minimum required sample rate to avoid aliasing is

20Hz, lets consider a sampling frequency of 25Hz

10Hz sine wave with an amplitude of 1

Sampling frequency (Fs)is 25Hz




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Fmax( folding frequency) =Fs/2 equals 12.5Hz



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Frequency Spectrum



Sample 102= 9.996 Hz

Sample 103 = 10.06 Hz

Sample 104 = 10.16 Hz


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Comments

The frequency resolution is better at 0.0977Hz While there is still some leakage, it is only significant

within about 1Hz to the right or left of the peak at 10Hz

The maximum amplitude is about 0.75, and occurs at

9.96Hz- much closer to the true frequency of 10Hz, due to

greater resolution

This illustrates a very significant, yet unappreciated

fact: Higher sampling rate of an FFT is not alwaysbetter.


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Increase the number of Samples

Can the FFT results be improved ?The answer is yes but at the cost of more computing time.

The previous case was repeated except the number of samples

are doubled from 256 to 512.


Sampling frequency (Fs)is 25Hz



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Frequency Spectrum



Sample 204= 9.96 Hz

Sample 205 = 10.01 Hz

Sample 206 = 10.06 Hz


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Comments

The frequency resolution has improved by a factor of two(for the same sampling frequency)

The peak in the spectrum at 10Hz is much narrower, with

reduced leakage.

The peak amplitude is now at 0.934 at a frequency of

10.01Hz

The improvement is due to greater frequency resolution

and reduced leakage.


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SynchronisationIt is theoretically possible to obtain a perfect FFT from

sampled data.

However this is only possible if the sampled data begins

and ends in the same phase of the signal


Number of samples = 512

Sampling frequency (Fs)is 25.6HzSample time = N/Fs =512/25.6 equals 20 seconds

This T corresponds to exactly 200 complete cycles of the

10Hz sine wave.


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Frequency SpectrumFrequency Resolution = Fs/N = 25.6/512 equals 0.05Hz


Solution of (f) 10Hz = f/ F equals sample 200


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Comments

The peak amplitude is exactly 1V, and occurs at

10HZ

The FFT is perfectly correct because there is no

leakage because of the correct frequency resolution


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DFT Analysis

The principle of the Fourier Analysis is to test for the

presence of each frequency component by multiplying the

wave form by sine and cosine waveform of the same test

frequency and average the results over one or more cycles.

In the case of DFT analysis, the first step is to determine

the fundamental frequency of the wave form to be

analysed.

The disadvantage of the DFT technique is that it requires

each harmonic to be calculated separately which requires

more computing power


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Advantages of the DFT

The DFT does not require that the number ofsamples N to be to a power of 2, such as 64, 128,

256 etc

The DFT harmonic frequency resolution isindependent of the sample rate.

The DFT does not suffer from harmonic leakage

because the DFT uses sine and cosine tablesfrequency referenced to the fundamental

frequency measurement.


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Conclusions

An FFT only returns the correct result thesampling frequency is matched to give theresulting number of samples (2 to the power N)over a complete number of waveform cycles. The

measurements are said to be synchronised to thewaveform .

In real life the sample time is an integer of the

instrument clock frequency and this means that allFFT measurements will be in error due toharmonic leakage


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Conclusions

Increased sample rate does not increase frequencyresolution.

Window functions such as the Hanning window

reduce the harmonic leakage error but cannot fullycompensate for it over a wide frequency range.


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Conclusions

An FFT has the advantage of allowing aoverview of a wide range of harmonic

frequencies as used on a Spectrum analyser

or Scope but is limited in accuracy. For an accurate Power Analyser the correct

harmonic measurement solution is to use a

DFT as specified by IEC61000-4-7 , Testing

and measurement techniques

3A FFT Analysis

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