Changing the amplitude (y-axis) of an FFT back to units of force.

Hello. I used Matlab to compute the FFT for my force vs time signals. The result was amplitude vs frequency, however, I would like to convert my amplitude signals back to units of force and plot force as a function of frequency. Does anyone have any idea on how I can do that with my current code? Thanks in advance.

Here is my code in a screen shot.

Answer 1

[Ameer Hamza]

Amplitude is just a general term used for signals. If your signal has a unit (in your case, units of force), then the output of fft() still has the same units. You can just replace the ylabel('P1(f) [Units of force]'), and it will be correct.

[Kev]

Intersting. I find that one hard to believe because my amplitude at 0 Hz should equate to the average force from my 100,000 force points from my entire signal , but it is not. For example, the average force from entire signal (100,000 points) is 16.73699 Newtons. But at 0 Hz the amplitude is reading 334.7372339. I'm not sure if amplitude is reading untis of force?

[Ameer Hamza]

Can you attach your data? Yes, you are correct, 0Hz value shoule be same as mean(F). I am not sure why you are getting something else.

[Kev]

Hi Hamza,

Here is an example, first column is time, second column is force.

Thanks,

Kev

[Kev]

Hi Hamza,

It seems that after changing my L to the correct length of 100,000 I was able to generate the correct FFT reflecting Force on the y-axis. I know it is correct because the average force from all my data points equates to the force at 0 Hz on my FFT. Thanks for your help with that!

Now I have just one question, Ive attached a resulting FFT for one of my data. If I choose for example a low-pass filter cut off of 5,000 Hz, does that mean I'm willing to accept a less than 1 Netwon difference between my original force signal and my filtered signal.

[Ameer Hamza]

Yes, L needs to be 100,000 for correct values.

I didn't understand your 2nd question. If you use a low-pass filter, then it will mean that there will be some differences between your original signal and the low-passed signal. As visible from your fft plot, the frequencies in your signal are concentrated near zero, and most frequencies are just contributing noise. Therefore, for your case, the low-passed signal will be smoother. Apart from that, you shouldn't see a significant change in amplitude.

[Kev]

I guess I'm confused with what the Force amplitude (y-axis) is telling me? If I look at a very low frequency, say 50 Hz and it corresponds to say 9 N on the y-axis based on my FFT, what does that 9 N mean?

[Ameer Hamza]

It just means that that the relative strength of 50 Hz frequency in your original signal. From the Fourier transform, we know that any arbitrary signal can be expressed as a sum of sinusoids. So the value of 9N at 50 Hz means that if you want to create the original signal, then you can need to add a 50Hz sinusoidal wave with an amplitude of 9. Similarly, you will multiply other sinusoids with their corresponding amplitudes and sum them all together to get the original signal back.

[Kev]

Thank you, Hamza. That was simply put, I understand now!

Answer 2

[David Goodmanson]

Hi Kev,

The first thing is, the time and frequency arrays have to be correct. You are not actually using the 'time' variable but presumably it is equally spaced and consistent wth your choice of Fs. The frequency grid calculation is correct except that L is supposed to be the number of points in the time array (and also the Fx1 array). L is set at 4999.94, close to 5000. But you say that there are 1e5 points. With 5000 the frequency array is going to be incorrect, as will the amplitude in the frequency domain.

For the amplitude, if the data reperesents a periodic function that keeps going on in time, then using Y/ L is correct. However, using 5000 for L instead of 1e5 gives an answer that is too large by a factor of 20, which is what you are off by.

In this case the units in the frequency domain are Nt.

If the data represents a nonrepeated pulse function of some kind. such as a decaying oscillation, then there are two possibilities. One is to decree that the decaying oscillation repeats as it would if you put many of your existing time windows side-by-side. Then (after fixing up L) you use Y/L as above. That keeps things simple, as long as you are aware of what is being represented.

The other possibility is to say it really is a one-time occurrence. In that case the natural assumption is that the fft represents an approximation to the fourier integral of a continuous function. Since you are approximating Integral [something] dt, dt is replaced by delta_t = 1/Fs, the spacing of the time array. The fft is multiplied by delta_t and the result is Y/Fs.

In this case the units in the frequency domain are Nt/Hz.

[Kev]

Hi David,

So for clarification, my L must equate to the the number of data points i collected in total. so 100,000? This correction should hopefully make the y-axis of my FFT truely in the units of Force?

Thanks for your thorough response.

Kev

[Kev]

Hi David,

It seems that after changing my L to the correct length of 100,000 I was able to generate the correct FFT reflecting Force on the y-axis. I know it is correct because the average force from all my data points equates to the force at 0 Hz on my FFT. Thanks for your help with that!

Now I have just one question, Ive attached a resulting FFT for one of my data. If I choose for example a low-pass filter cut off of 5,000 Hz, does that mean I'm willing to accept a less than 1 Netwon difference between my original force signal and my filtered signal.

[David Goodmanson]

HI Kev,

hard to say much without seeing the plot zoomed out near the origin and without knowing the nature of the low-pass filter. In particular, what is the value of the filter function down near zero frequency?

[Kev]

I guess I'm confused with what the Force amplitude (y-axis) is telling me? If I look at a very low frequency, say 50 Hz and it corresponds to say 9 N on the y-axis based on my FFT, what does that 9 N mean?

[David Goodmanson]

Hi Kev,

That means that there is a continuous wave fourier component with an amplitude of 9 N at that frequency, which contributes to the force in the time domain. If 50 Hz were an isolated frequency, you would see a wave of that size in the time domain. When there are a lot of nearby frequencies as well, things get mixed together and you don't see that.

When you have a lot of amplitudes that are dense in frequency, that's an argument for the second mehtod I mentioned in the answer. When the horizontal scale is V/Hz, suppose the fourier transform is fairly constant in a certain region, say around 50Hz and has a value 2 N/Hz. Roughly speaking, that tells you that frequency content between, say 40 and 60 Hz is worth 20 x 2 = 40N of response in the time domain.

[Kev]

Thank you, David. I understand now!