Question about amplitude, frequency and energy in an audio signal

[33AndAThird]

Hi, I've been trying to wrap my head around why low frequencies require more power to generate and why in a complex audio signal the amplitude gets lower as the frequency gets higer. In performing some research I have confirmed the following 5 points, please correct them if they are wrong.

1. Given the same amplitude, generating a high frequency tone requires more energy over time than a low frequency tone because you are expaanding and contracting the air more often. eg. Generating a 2KHz tone requires 10 times the energy per second as a 200Hz tone using the same driver (ignoring parasitic losses).
2. Given the energy of a wave is proportional to the square of the amplitude, if we want to generate a 200Hz tone that uses the same energy over time as a 2KHz tone, we'd need to increase the amplitude by a factor of 3.16 (square root of 10).
3. Sound Intensity is proportional to both the square of amplitude and the square of frequency as demonstrated by this equation I=2^2^2^2.
4. The square of Sound Pressure Level is proportional to Sound Intensity.
5. Our perception of loudness proportional to SPL

Given the above points I have come to the following conclusion which doesn't seem to be the commonely held belief. So either I'm wrong, or a lot of other people are.

1. Amplitude does not determine SPL in isolation, frequency is also important. In order for a lower frequency to generate the same SPL as a higher frequency, it needs greater amplitude.

However I'm still left with the original question of why low frequencies require so much more power to generate. If we increase the amplitude of the 200Hz signal by a factor of 3.16 we'll impart the same energy as the 2KHz signal, and SPL would then be equal. I understand there are two other factors:

1. Humans are most sensitive to 1-5Khz and less sensitive to lower frequencies. I believe by about 15db.
2. The greater the excursion of a driver the less efficient it becomes

These two factors alone don't seem to make up for the fact we might need a 500W amp to power a subwoofer that matches to a tweeter that will consume <10W at the same SPL? What am I missing (I presume something logarithmic)?

Thanks for your help

 

[Keith_W]

I think you have not accounted for the fact that low frequencies are omnidirectional and high frequencies are directional. If your speakers are pointed at you, a high frequency transducer wastes less energy by directing sound into a beam. But a subwoofer wastes energy spraying sound all over the place and pressurizing the room.

 

[Hayabusa]

Hi, I've been trying to wrap my head around why low frequencies require more power to generate and why in a complex audio signal the amplitude gets lower as the frequency gets higer. In performing some research I have confirmed the following 5 points, please correct them if they are wrong.

1. Given the same amplitude, generating a high frequency tone requires more energy over time than a low frequency tone because you are expaanding and contracting the air more often. eg. Generating a 2KHz tone requires 10 times the energy per second as a 200Hz tone using the same driver (ignoring parasitic losses).
2. Given the energy of a wave is proportional to the square of the amplitude, if we want to generate a 200Hz tone that uses the same energy over time as a 2KHz tone, we'd need to increase the amplitude by a factor of 3.16 (square root of 10).
3. Sound Intensity is proportional to both the square of amplitude and the square of frequency as demonstrated by this equation I=2^2^2^2.
4. The square of Sound Pressure Level is proportional to Sound Intensity.
5. Our perception of loudness proportional to SPL

Given the above points I have come to the following conclusion which doesn't seem to be the commonely held belief. So either I'm wrong, or a lot of other people are.

1. Amplitude does not determine SPL in isolation, frequency is also important. In order for a lower frequency to generate the same SPL as a higher frequency, it needs greater amplitude.

However I'm still left with the original question of why low frequencies require so much more power to generate. If we increase the amplitude of the 200Hz signal by a factor of 3.16 we'll impart the same energy as the 2KHz signal, and SPL would then be equal. I understand there are two other factors:

1. Humans are most sensitive to 1-5Khz and less sensitive to lower frequencies. I believe by about 15db.
2. The greater the excursion of a driver the less efficient it becomes

These two factors alone don't seem to make up for the fact we might need a 500W amp to power a subwoofer that matches to a tweeter that will consume <10W at the same SPL? What am I missing (I presume something logarithmic)?

Thanks for your help

What is you definition of amplitude?

It looks like you take the speaker movement as amplitude in your reasoning?

If you would take the input voltage as definition of 'amplitude' the whole thing would change.
Then the same amplitude (of a 'perfect' speaker) at a different frequency would give the same SPL but the speaker movement would just be 1/f

 

[DVDdoug]

Given the same amplitude, generating a high frequency tone requires more energy over time than a low frequency tone because you are expaanding and contracting the air more often. eg. Generating a 2KHz tone requires 10 times the energy per second as a 200Hz tone using the same driver (ignoring parasitic losses).

I don't think that's true. 20V RMS into 8-Ohms is 50 Watts at any frequency. If the speaker's frequency response is flat and equally efficient throughout the frequency range, the acoustical energy would also be the same.

Humans are most sensitive to 1-5Khz and less sensitive to lower frequencies. I believe by about 15db.

More, depending on how deep you go, and it's non-linear in a way that it depends on how loud you're listening, Equal Loudness Curves

 

[33AndAThird]

I think you have not accounted for the fact that low frequencies are omnidirectional and high frequencies are directional. If your speakers are pointed at you, a high frequency transducer wastes less energy by directing sound into a beam. But a subwoofer wastes energy spraying sound all over the place and pressurizing the room.

That is at most 6db increasing from the baffle step right? So that is one doubling, but there needs to be many more to account for the change in energy used to drive a tweeter vs a woofer.

What is you definition of amplitude?

It looks like you take the speaker movement as amplitude in your reasoning?

If you would take the input voltage as definition of 'amplitude' the whole thing would change.
Then the same amplitude (of a 'perfect' speaker) at a different frequency would give the same SPL but the speaker movement would just be 1/f

Is the amplitude in the voltage not directly / linearly proportional to the driver excursion and hence interchangeable?
This statement also conflicts with the equation showing sound intensity being proprotional to f^2 as well as a^2 - how do we explain that?

I don't think that's true. 20V RMS into 8-Ohms is 50 Watts at any frequency. If the speaker's frequency response is flat and equally efficient throughout the frequency range, the acoustical energy would also be the same.

So I understand the frequency response argument, but how does that align with the reality that energy can neither be created or destroyed, and the same driver at the same amplitude at 2Khz is doing 10 times as much work as at 200Hz?

More, depending on how deep you go, and it's non-linear in a way that it depends on how loud you're listening, Equal Loudness Curves

This is a greater differential than I thought particularly at low volumes, but if this were the only effect we'd also need to be using higher amplitute signals above 5KHz as well as below 1KHz, and that is not the case to my understanding.

 

[Hayabusa]

Is the amplitude in the voltage not directly / linearly proportional to the driver excursion and hence interchangeable?

No, that's the fundamental difference.
A speaker at 2000Hz at 100dB SPL has halve the excursion of the same speaker at 1000Hz/100dB

look at: http://www.baudline.com/erik/bass/xmaxer.html

 

[janbth]

More, depending on how deep you go, and it's non-linear in a way that it depends on how loud you're listening, Equal Loudness Curves

Isn't it less, like the thread starter says? The way I read the equal-loudness contours is that a signal at 100 Hz with 70 dB SPL will sound equally loud as a 1000 Hz signal at 50 dB SPL (roughly). Is this interpretation wrong?

 

[terryforsythe]

This may be of some help: https://www.baudline.com/erik/bass/xmaxer.html

In the equation given, "Ro" is air density and "c" is the speed of sound.

EDIT: I just noticed that Hyabusa posted the same link above. I should have read that post more carefully.

 

[33AndAThird]

No, that's the fundamental difference.
A speaker at 2000Hz at 100dB SPL has halve the excursion of the same speaker at 1000Hz/100dB

look at: http://www.baudline.com/erik/bass/xmaxer.html

Wow! Mind blown. This is new information for me thank you! (note it actually has a quarter of the excursion as frequency appears to be inversely proportional to the square of exursion)

How does this all come together though? I now know that:

• To maintain a flat SPL from a speaker across frequencies the driver excursion needs to get larger as the frequency gets lower
• We ideally expect to get the same SPL out from from a loudpseaker across all frequencies given a fixed amplitude voltage signal - otherwise known as a flat frequency response

Therefore can we conclude that speaker driver excursion is not directly / linearly proportional to voltage amplitude of the input signal?
(This would be in condradiction to my understanding to date and a lot of what I can see online).
If so how does this work? What is voltage proportional to? Velocity or acceleration rather than excursion (distance)?

 

[NTK]

Therefore can we conclude that speaker driver excursion is not directly / linearly proportional to voltage amplitude of the input signal?
(This would be in condradiction to my understanding to date and a lot of what I can see online).

The force produced by the speaker driver motor which moves the driver diaphragm is the voice coil current multiplied by the Bl factor (force factor, one of the Thiele/Small parameters).

If we take the highly simplified (zeroth order approximation) that the driver impedance is constant with frequency, which means current is directly proportional to voltage, then the force acting on the diaphragm is also directly proportional to voltage. Since F = ma, the diaphragm acceleration is directly proportional to voltage.

The sound pressure produced by the motion of the diaphragm is proportional to the acceleration of the diaphragm, not displacement nor velocity (see this blog post by Purifi). This is also why for the same SPL, the displacement amplitude of the diaphragm goes down by the square of the frequency, since acceleration goes up by the square of frequency:

For x(t) = sin(ω t); v(t) = ω cos(ω t); and a(t) = -ω² sin(ω t)​

 

[33AndAThird]

The force produced by the speaker driver motor which moves the driver diaphragm is the voice coil current multiplied by the Bl factor (force factor, one of the Thiele/Small parameters).

If we take the highly simplified (zeroth order approximation) that the driver impedance is constant with frequency, which means current is directly proportional to voltage, then the force acting on the diaphragm is also directly proportional to voltage. Since F = ma, the diaphragm acceleration is directly proportional to voltage.

The sound pressure produced by the motion of the diaphragm is proportional to the acceleration of the diaphragm, not displacement nor velocity (see this blog post by Purifi). This is also why for the same SPL, the displacement amplitude of the diaphragm goes down by the square of the frequency, since acceleration goes up by the square of frequency:

For x(t) = sin(ω t); v(t) = ω cos(ω t); and a(t) = -ω² sin(ω t)​

Thank you @NTK - this has explained it perfectly, my understanding of the relationship of the driver to the voltage signal has been corrected!

This leads me back to the original question - if we go back to that simplified driver with constant impedance, shouldn't a fixed amplitude signal sweep of 20Hz to 20Khz require the same power regardless of frequency? Why do subs require so much more power than tweeters? Is it all down to the size of driver required to increase the displacement per cycle and the additional weight that brings - ie. its down to efficiency?

 

[KSTR]

No, that's the fundamental difference.
A speaker at 2000Hz at 100dB SPL has halve the excursion of the same speaker at 1000Hz/100dB

look at: http://www.baudline.com/erik/bass/xmaxer.html

Correction: 1/4th. Excursion goes with 1/f².

 

[KSTR]

Why do subs require so much more power than tweeters? Is it all down to the size of driver required to increase the displacement per cycle and the additional weight that brings - ie. its down to efficiency?

Yes.

 

[NTK]

Thank you @NTK - this has explained it perfectly, my understanding of the relationship of the driver to the voltage signal has been corrected!

This leads me back to the original question - if we go back to that simplified driver with constant impedance, shouldn't a fixed amplitude signal sweep of 20Hz to 20Khz require the same power regardless of frequency? Why do subs require so much more power than tweeters? Is it all down to the size of driver required to increase the displacement per cycle and the additional weight that brings - ie. its down to efficiency?

It is because woofer drivers are usually a lot less efficient than tweeters. It has a lot to do with F = m a. The higher the m, for equal F, the lower the a, and therefore lower SPL.

Just for example, the SB Acoustics SB34SWPL-76-4 12" subwoofer has a moving mass (Mms) of 266 g; while the SB Acoustics SB12MNRX2-25-04, 4" midrange has a moving mass (Mms) of 4.2 g, a 63x difference (OK, the magnets are different too, so the actual difference may be smaller).

 

[MaxwellsEq]

Although not a function of mass acceleration, you also need to remember that higher frequencies are considerably quieter in practice than lower frequency. Pretty much all the musical energy is in the bottom few octaves meaning more energy is demanded from the amplifiers. This is from one of @Jean.Francois images:

 

[kemmler3D]

Although not a function of mass acceleration, you also need to remember that higher frequencies are considerably quieter in practice than lower frequency. Pretty much all the musical energy is in the bottom few octaves meaning more energy is demanded from the amplifiers. This is from one of @Jean.Francois images:

View attachment 361136

Right, this is a pretty typical music spectrum, and remember that this is plotted in decibels, (logarithmic) so the linear values for actual wattage going into the high frequencies are really low.

This lack of energy in high frequencies is not an inherent property of audio, but because of how music and human hearing work, we typically don't need much power at all for treble.

 

[Sonic icons]

I found a good reference on basic physical properties of sound waves, including examples from our shared obsession (audio systems):

"Sound", chapter 17 of "University Physics Volume 1", published online by openstax.org, a division of Rice University, under a Creative Commons license.
https://openstax.org/books/university-physics-volume-1/pages/17-introduction
sections: 17.1 Sound Waves, 17.2 Speed of Sound, 17.3 Sound Intensity, 17.4 Normal Modes of a Standing Sound Wave [room modes!], 17.5 Sources of Musical Sound, 17.6 Beats [not the brand lol], 17.7 Doppler Effect, 17.8 Shock Waves, Review with Homework Problems

Depending on your physical science background, reading the preceding chapter "16. Waves" first may help a lot with understanding.

If you finish a chapter AND do the homework, you'll get credit on your Permanent Record (pinky promise )

 

[Sonic icons]

because of how music and human hearing work, we typically don't need much power at all for treble.

I wonder what the power spectrum for a string quartet recording looks like, probably not quite as weighted toward the bottom octaves.

 

[kemmler3D]

I wonder what the power spectrum for a string quartet recording looks like, probably not quite as weighted toward the bottom octaves.

You might be surprised. Here's the spectrum for a random string quartet I downloaded by Boccherini. If you listen to it, the mids are actually quite hot to the point of distorting, so this is actually showing more energy at/above 1khz than you'd normally see. It still drops off very quickly after 3khz.

 

[AdamG]

Does in not boil down to “E=MC²”. The lower the frequency the more mass is moved, and that requires more energy?

Sound Waves Carry Mass

Even if you ignore general relativity, sound waves transport a small amount of mass, according to theory.

physics.aps.org

I always marvel at the energy of a close and loud Thunder clap. It can rattle the door, the windows even the glasses in the cupboard. That’s very low frequency sound energy.