# Why do Harmonics Decay Faster than the Fundamental?

• I
When looking at the FFT spectrum of a sonometer, I noticed that the harmonics decayed faster than the fundamental. Why is this?

anorlunda
Staff Emeritus

When looking at the FFT spectrum of a sonometer, I noticed that the harmonics decayed faster than the fundamental. Why is this?

Do you mean they decay faster in time? Looking at a FFT will not show you that, so your question confuses me.

Do you mean they decay faster in time? Looking at a FFT will not show you that, so your question confuses me.

Yes, I mean decaying faster in time. We kept the FFT spectrum running while we plucked the string and watched peaks form then decay, and the harmonics decayed faster. If this is still confusing (probably due to my inability to explain it well), then don't worry about the FFT part. The point of my question is that I want to know why the harmonics decay faster than the fundamental mode in vibrating string.

Gold Member
Do you mean they decay faster in time? Looking at a FFT will not show you that, so your question confuses me.

Sure it would, if you had the FFT at various points in time. This is sometimes called a short-time Fourier transform (STFT).

Yes, I mean decaying faster in time. We kept the FFT spectrum running while we plucked the string and watched peaks form then decay, and the harmonics decayed faster. If this is still confusing (probably due to my inability to explain it well), then don't worry about the FFT part. The point of my question is that I want to know why the harmonics decay faster than the fundamental mode in vibrating string.

Out of curiosity here, did you actually measure the decay rates of the fundamental and harmonic and the decay rates were actually different, or did you just notice that the harmonic disappears first?

Sure it would, if you had the FFT at various points in time. This is sometimes called a short-time Fourier transform (STFT).

Out of curiosity here, did you actually measure the decay rates of the fundamental and harmonic and the decay rates were actually different, or did you just notice that the harmonic disappears first?

I just eyeballed it, and noticed that the harmonics seem to decay faster. I then googled it, and read that this is usually the case, but I'm not sure why.

berkeman
Mentor
I just eyeballed it, and noticed that the harmonics seem to decay faster. I then googled it, and read that this is usually the case, but I'm not sure why.
Do you understand where the harmonics come from on a guitar string?

And could you post a link to where you read that it us usually the case? There are plenty of systems where this is not the case...

anorlunda
Staff Emeritus
I'm old fashioned, when I think of an FFT, I think of it printed on paper.

A general answer is that most physical systems are low pass filters. They attenuate high frequencies more than low frequencies. There are exceptions, but most things are like that.

DanielMB and Joella Kait
Dr. Courtney
Gold Member
2020 Award
I've looked at Fourier transforms (spectra) of lots of plucked strings, even published papers on it. I never noticed a trend one way or another. Also, thinking of the actual sound, usually the overtones (harmonics) persist in the sound and seem to decay at about the same rate most of the time.

I think I'd need to see the data to be convinced otherwise.

Dale and Joella Kait
sophiecentaur
Gold Member
2020 Award
and read that this is usually the case, but I'm not sure why.
The energy in a simple free oscillator will decay exponentially and the Q factor is the number of cycles for the energy to drop to 1/e of an initial value.
If the various modes on the string all have the same Q factor (a bit of an assumption perhaps but go with it) then the energy in the modes will all take the same number of cycles of oscillationto fall to 1/e of the original value. That would take half as long for a second harmonic and one third as long for a third harmonic.

DanielMB, Asymptotic, Tom.G and 2 others
Dr. Courtney
Gold Member
2020 Award
The energy in a simple free oscillator will decay exponentially and the Q factor is the number of cycles for the energy to drop to 1/e of an initial value.
If the various modes on the string all have the same Q factor (a bit of an assumption perhaps but go with it) then the energy in the modes will all take the same number of cycles of oscillationto fall to 1/e of the original value. That would take half as long for a second harmonic and one third as long for a third harmonic.

It is an errant assumption that the different modes all start with a peak energy and decay independently. In reality, energy is traded back and forth between modes and amplitudes of a given spectral peak will both rise and fall over time. Sure, in the long term, they are all decreasing, but the decrease of some modes is not even monotonic, much less exponential.

marcusl
Gold Member
Can you explain how energy is exchanged between modes in a string? Modes are, by definition, orthogonal.

davenn
Gold Member
I'm old fashioned, when I think of an FFT, I think of it printed on paper.

A general answer is that most physical systems are low pass filters. They attenuate high frequencies more than low frequencies. There are exceptions, but most things are like that.

agreed. and "most of the time " the harmonics are also lower amplitude than the fundamental and as their frequency increases, their amplitude also decreases

this, along with the low pass filter function, means the harmonics "usually" die out sooner

Dave

rcgldr
Homework Helper
Link to an article with graphs. Figure 3 shows an unusual case where the fundamental harmonic decayed first, but the rest of the harmonics decayed quicker for higher harmonics.

https://courses.physics.illinois.ed.../Fall02/RLee/Ryan_Lee_P398EMI_Main_Report.pdf

Note it's also possible to place a finger on the node of a harmonic to prevent the fundamental frequency from being generated (or at least greatly reduce it).

davenn
Dr. Courtney
Gold Member
2020 Award
Can you explain how energy is exchanged between modes in a string? Modes are, by definition, orthogonal.

How is a theoretical question. One need not answer the how to look at the data and understand that it definitely does happen.

https://courses.physics.illinois.ed.../Fall02/RLee/Ryan_Lee_P398EMI_Main_Report.pdf

For example, look at the amplitude of the harmonics in yellow and green in Figure 1 of the above report. The amplitude is definitely rising and falling rather than decreasing monotonically in time. Likewise, look at the 3rd, 5th, and 6th harmonics in Figure 2 of the above report (also yellow and green): once again amplitudes rise and fall.

Modes are orthogonal in the theory. But real strings do not completely obey the theory.

Dr. Courtney
Gold Member
2020 Award
Link to an article with graphs. Figure 3 shows an unusual case where the fundamental harmonic decayed first, but the rest of the harmonics decayed quicker for higher harmonics.

https://courses.physics.illinois.ed.../Fall02/RLee/Ryan_Lee_P398EMI_Main_Report.pdf

Note it's also possible to place a finger on the node of a harmonic to prevent the fundamental frequency from being generated (or at least greatly reduce it).

Actually, Figure 4 also shows a case where the 2nd harmonic decays more slowly than the fundamental. Given that two of four cases shown in that paper have the fundamental decaying more slowly than the 2nd harmonic, I'm not sure you have a strong case for a trend.

sophiecentaur
Gold Member
2020 Award
It is an errant assumption that the different modes all start with a peak energy and decay independently. In reality, energy is traded back and forth between modes and amplitudes of a given spectral peak will both rise and fall over time. Sure, in the long term, they are all decreasing, but the decrease of some modes is not even monotonic, much less exponential.
I made a broad assumption - true. But the modes would be orthogonal on an ideal string. A guitar has a complex set of resonances and the ends of the strings are not well defined or constant throughout the ranger of angles made by the string. This can result in frequency / mode length modulation of one mode by another mode. In fact, the fact that your effect is observed actually implies that there must be some non linearity. Also, of course, the overtones are not perfect harmonics of the fundamental and you could perhaps expect some beating between modal frequencies and harmonics of the fundamental due to non linearity. Quite a can of worms, in fact.

I looked at the graphs in your l ink and I notice that the 'trading' of energy between modes tends to die out as the amplitudes reduce after the initial attack and I would suggest that the system behaves more linearly at low amplitudes. The near 2:1 ratio of the decay times of 2nd and 4th order modes sort of justify my comment about decay being about numbers of cycles when the Q factors are similar.
PS That's an interesting paper. An idea topic for a computer [Edit - I meant Guitar enthusiast] enthusiast / nerd. A busman's holiday as the English say.

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Gold Member
Whether the modes are orthogonal or not really only covers the question of whether they interact via linear processes, e.g. superposition. Orthogonal modes can still interact with each other nonlinearly, which is typically the case when one observes harmonics (assuming they are real harmonics and not spontaneously-excited modes that happen to be at twice the frequency of another mode).

This most commonly takes the form of quadratic phase coupling, i.e. there are quadratic terms in the dependent variable in the governing equations. That allows energy exchange between otherwise normal modes.

While a guitar strings is likely pretty close to ideal for small amplitudes, it would be fairly unusual for a real-world problem with all the aforementioned complications to not include some degree of nonlinearity at higher amplitudes.

sophiecentaur
Mister T
Gold Member
I just eyeballed it, and noticed that the harmonics seem to decay faster.

What did you see that led you to this conclusion? When I pluck a string and look at the FFT what I usually notice is that the higher harmonics have a smaller amplitude. This is not always the case, as it depends to some extent on how the string is plucked. In any case, the information I get about the relative amplitude of the harmonics doesn't necessarily tell me which of them are decaying faster.

Gold Member
What did you see that led you to this conclusion? When I pluck a string and look at the FFT what I usually notice is that the higher harmonics have a smaller amplitude. This is not always the case, as it depends to some extent on how the string is plucked. In any case, the information I get about the relative amplitude of the harmonics doesn't necessarily tell me which of them are decaying faster.

This is why I asked my original question about how OP was reaching this conclusion, and whether he or she actually measured the decay rate or simply based it on which peak disappeared first.

sophiecentaur