Fourier Transform: Separate High & Low Frequency Signals

In summary, the conversation discusses the use of a Fourier transform to single out a low frequency signal from a continuous high frequency oscillating signal. One suggestion is to use a low pass filter in the frequency domain and transform back to the time domain. However, this may not work if the low frequency signal is "peaky" with sharp tips. Another suggestion is to use a band-pass filter with a narrow pass-band, but this may not be effective if the input and output signals are not simple superpositions of sine waves. The goal is to remove the effect of a small sine wave and extract the peak from the signal.
  • #1
ponjavic
225
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I have two signals one continuous oscillating at a high frequency and another one instantaneous at a lower frequency.

How can I use a Fourier transform to single out the low frequency one?

See at attached picture for what I am trying to do.

Edit:
Yeah by the way, data is collected in a two dimensional arrays (so discrete)
 

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  • #2
ponjavic said:
I have two signals one continuous oscillating at a high frequency and another one instantaneous at a lower frequency.

How can I use a Fourier transform to single out the low frequency one?

I am not quite sure what your graphs depict, but if your low-frequency signal is really low frequency and the other is high frequency, then obviously a low pass filter will do. A crude low pass is obtained by transforming to the frequency domain, set all frequencies that are too high to zero and transform back to the time domain.

If your 2nd graph is supposed to be the "low frequency" signal, you are in trouble, because it is very "peaky", i.e. the peak has a sharp tip. This indicates that you have rather high frequencies there. The low pass would round out the tip. If you cannot live with that, I would not know off hand how to proceed.

Harald.
 
  • #3
Why do you want to use a Fourier transform? Why not just run the data sets through a DSP lowpass filter of some order?
 
  • #4
The FFT technique given by birulami is actually a perfect digital filter, not a crude one. The problem is that you have to have the entire signal in memory before you can take the FFT. In some situations this is fine; in most, it's not.

- Warren
 
  • #5
birulami said:
I am not quite sure what your graphs depict, but if your low-frequency signal is really low frequency and the other is high frequency, then obviously a low pass filter will do. A crude low pass is obtained by transforming to the frequency domain, set all frequencies that are too high to zero and transform back to the time domain.

If your 2nd graph is supposed to be the "low frequency" signal, you are in trouble, because it is very "peaky", i.e. the peak has a sharp tip. This indicates that you have rather high frequencies there. The low pass would round out the tip. If you cannot live with that, I would not know off hand how to proceed.

Harald.

First graph is what I have, constant vibration due to hydraulics and a peak which is a measurement of force when breaking a specimen. 2nd graph shows what I need, removal of the vibration.

I thought it was possible to single out a frequency response using Fourier transform. I guess filters could be used but wouldn't they affect the peak? This is a very accurate experiment so if Fourier transform is better I'd like to do that. Using matlab.
 
  • #6
ponjavic,

What birulami was trying to say is that a low-pass filter cannot make a signal more peaky -- it can only smooth out peaks.

When you speak of isolating a specific frequency, you're talking about band-pass filter with a narrow pass-band. This is a well-understood sort of digital filter, but it cannot produce the output you claim to desire. The only output you can get from such a filter is a sine wave with arbitrary amplitude.

However, your second graph showed a "hump" in the input being transformed into a "peak" in the output. This is not going to happen with any normally-designed filter. The "peak" in your output is a cusp -- an instantaneously change in the derivatives of the signal, and actually contains frequency components all the way out to infinity.

The signals that you drew were not just a simple superposition of sine waves (the kind of well-behaved input people usually consider when thinking about filters). Perhaps you're not really looking for a linear system at all, and would be better off using something else.

Can you draw us a more realistic example of the input and output you're trying to achieve?

- Warren
 
  • #7
chroot said:
ponjavic,

What birulami was trying to say is that a low-pass filter cannot make a signal more peaky -- it can only smooth out peaks.

When you speak of isolating a specific frequency, you're talking about band-pass filter with a narrow pass-band. This is a well-understood sort of digital filter, but it cannot produce the output you claim to desire. The only output you can get from such a filter is a sine wave with arbitrary amplitude.

However, your second graph showed a "hump" in the input being transformed into a "peak" in the output. This is not going to happen with any normally-designed filter. The "peak" in your output is a cusp -- an instantaneously change in the derivatives of the signal, and actually contains frequency components all the way out to infinity.

The signals that you drew were not just a simple superposition of sine waves (the kind of well-behaved input people usually consider when thinking about filters). Perhaps you're not really looking for a linear system at all, and would be better off using something else.

Can you draw us a more realistic example of the input and output you're trying to achieve?

- Warren
Yeah sorry I see what the problem is now. The peaks in the first and second pictures are supposed to be the same (except for the fact the the first peak is influenced by the small sine wave). So all I want to do is to remove the effect of the sine wave, extruding the peak from the signal.
 
  • #8
Then you're looking for a basic low-pass digital filter.

- Warren
 
  • #9
ponjavic said:
Yeah sorry I see what the problem is now. The peaks in the first and second pictures are supposed to be the same (except for the fact the the first peak is influenced by the small sine wave). So all I want to do is to remove the effect of the sine wave, extruding the peak from the signal.

If the sine wave you want to remove is itself really narrow-band, i.e. in the Fourier transform only very few frequencies are non-zero, you may want to use a carefully crafted band-stop filter. Just setting these frequencies to zero and transforming back may help. Someone said that zeroing frequencies is a perfect filter, while I called it crude. I say 'crude', because this will usually generate nasty sidelines when transformed back. I am not an expert in digital filter design, but that much I do remember:-)

Harald.
 
  • #10
birulami said:
If the sine wave you want to remove is itself really narrow-band, i.e. in the Fourier transform only very few frequencies are non-zero, you may want to use a carefully crafted band-stop filter. Just setting these frequencies to zero and transforming back may help. Someone said that zeroing frequencies is a perfect filter, while I called it crude. I say 'crude', because this will usually generate nasty sidelines when transformed back. I am not an expert in digital filter design, but that much I do remember:-)

Harald.

Well, you mean a perfect brick-wall filter in the frequency domain looks like a sinc function in the time-domain. It has a pretty nasty impulse response.

- Warren
 
  • #11
Thank you guys that should do it =)

I thought of a filter as well but didn't realize that you could do it digitally (stupid) thought of making an actual filter for the signal :P

Anyways my coordinator definitely confused me with the Fourier transform but I'm on track now
 

FAQ: Fourier Transform: Separate High & Low Frequency Signals

What is a Fourier Transform?

A Fourier Transform is a mathematical tool used to analyze and decompose a complex signal into its individual frequency components. It converts a signal from its original domain (such as time or space) to a representation in the frequency domain.

How does a Fourier Transform separate high and low frequency signals?

A Fourier Transform separates high and low frequency signals by breaking down the original signal into its individual frequency components. High frequency components will have a larger amplitude in the frequency domain, while low frequency components will have a smaller amplitude. This allows us to isolate and analyze specific frequency ranges within a signal.

What is the difference between high and low frequency signals?

High frequency signals have a higher number of oscillations per unit of time, while low frequency signals have a lower number of oscillations per unit of time. In other words, high frequency signals have a shorter period and low frequency signals have a longer period. High frequency signals are associated with fast-changing phenomena, while low frequency signals are associated with slower-changing phenomena.

How is a Fourier Transform useful in scientific research?

A Fourier Transform is useful in scientific research because it allows us to analyze complex signals and identify specific frequency components. This can help us better understand the underlying patterns and relationships within a signal, and can also be used for noise filtering, signal enhancement, and feature extraction.

Can a Fourier Transform be used on any type of signal?

Yes, a Fourier Transform can be used on any type of signal, as long as the signal is finite and has a well-defined frequency spectrum. This includes signals from various fields such as physics, engineering, economics, and biology.

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