Question about amplitude, frequency and energy in an audio signal

[33AndAThird]

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.

Thanks, that makes sense. So in order for a full range speaker to have a flat frequency response at a given input voltage, it needs to consume more power at lower frequencies to compensate for decreased efficiency. How does it do this? I recognise many speakers will have lower impedance at lower frequencies but usually not by more than half, and some speakers keep a fairly flat impedance. So if not via variable impedance, how does a speaker consume variable power for a given voltage to keep a flat SPL? The only other thing I can think of is purposefully wasting energy to heat in the tweeter or high pass crossover to artificially lower the efficiency of the higher range drivers.

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

This neatly brings us back to the second question in my original post! The answer being that musical signals have lower amplitude in the higher frequencies because that’s how they occur naturally, and that’s how our ears are tuned. This begs the question why nature has chosen this path?
Could it be for the same reason our low frequency drivers require more power to drive than our tweeters and if fed the same power our tweeters would fail? ie. To move something fast enough to produce a high frequency signal it naturally has to be light weight, which means it is likely fragile and would break up should high levels of energy be passed through it. Or put another way, only large heavy structures are strong enough to impart large amounts of energy through high amplitude waves, and by nature of their mass can only do it at low frequencies. Hence larger amplitude sound waves tend to occur at lower frequencies. Am I on the right track?

 

[kemmler3D]

So in order for a full range speaker to have a flat frequency response at a given input voltage, it needs to consume more power at lower frequencies to compensate for decreased efficiency. How does it do this?

The speaker doesn't do this per se, the amplifier outputs the voltage that corresponds to the intended audio output. It can do this as long as it can draw enough current without overloading to supply that voltage at the speaker's impedance at a given frequency.

It also bears mentioning that different speakers don't always respond the same to a given voltage - far from it. This is one reason why getting a speaker with a wide, flat frequency response is so hard.

The only other thing I can think of is purposefully wasting energy to heat in the tweeter or high pass crossover to artificially lower the efficiency of the higher range drivers.

Yes, this is done pretty commonly, but the drivers themselves are usually (hopefully) pretty close in sensitivity to begin with.

This begs the question why nature has chosen this path?

Way out of my area of expertise, but probably because very high frequencies are significantly attenuated in air, and very low frequencies require impractically large structures to produce those resonances. So the useful range of hearing (and vocalizing) for creatures of human size ends up being what we call "upper bass to midrange".

BTW, welcome to ASR, and I appreciate your effort to start from the very basics of how these things work. It was a long time into my audio journey before I even understood voltage ~ acceleration ~ SPL, I went around thinking voltage = cone position for a long time.

 

[NTK]

Thanks, that makes sense. So in order for a full range speaker to have a flat frequency response at a given input voltage, it needs to consume more power at lower frequencies to compensate for decreased efficiency. How does it do this? I recognise many speakers will have lower impedance at lower frequencies but usually not by more than half, and some speakers keep a fairly flat impedance.

For a single driver loudspeaker, the effective moving mass is constant, so the efficiency (if we kept the simplifying assumption and neglect the other effects) is also frequency independent.

People have a mistaken focus on "power". Amplifiers amplify voltages → output voltage = input voltage × gain, (ideally) regardless of load impedance and frequency. (The job of the amplifier is to provide the output voltage that equal to input voltage × gain, and to provide the current as determined by Ohm's law.) Loudspeakers are also designed to response to voltages, and ideally give the same SPL with the same input voltage, independent of frequency. Loudspeakers are thus not designed to respond to power, i.e. their efficiencies* vary with frequency as their impedance, and therefore the power demanded from the amplifier at a fixed input voltage, vary with frequency.

Note: * Efficiency is defined as ratio of (acoustic) power output to (electrical) power input. That's why the standard these days for loudspeakers is to rate "sensitivity" (= dB SPL @ 2.83 V), not "efficiency".

So if not via variable impedance, how does a speaker consume variable power for a given voltage to keep a flat SPL? The only other thing I can think of is purposefully wasting energy to heat in the tweeter or high pass crossover to artificially lower the efficiency of the higher range drivers.

Yes. That's why you often see "padding resistors" in the cross-over filter for tweeters. For example:

 

[Ken Tajalli]

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).

AFAJK Not true.

1. 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).

AFAIK not true.
Also he is very right. https://audiosciencereview.com/foru...-energy-in-an-audio-signal.53495/post-1935440
With music, even with Rock POP and Jazz, high frequencies are at 30/50dB from Bass..
Also bear in mind that applying the energy to a transducer at differing frequencies does not mean it can radiate the same amount of energy at every frequency, the viscosity of air will have something to do with it. In a thicker medium such as water, things change.
My 2 pence.

 

[33AndAThird]

The speaker doesn't do this per se, the amplifier outputs the voltage that corresponds to the intended audio output. It can do this as long as it can draw enough current without overloading to supply that voltage at the speaker's impedance at a given frequency.

It also bears mentioning that different speakers don't always respond the same to a given voltage - far from it. This is one reason why getting a speaker with a wide, flat frequency response is so hard.

Way out of my area of expertise, but probably because very high frequencies are significantly attenuated in air, and very low frequencies require impractically large structures to produce those resonances. So the useful range of hearing (and vocalizing) for creatures of human size ends up being what we call "upper bass to midrange".

For a single driver loudspeaker, the effective moving mass is constant, so the efficiency (if we kept the simplifying assumption and neglect the other effects) is also frequency independent.

People have a mistaken focus on "power". Amplifiers amplify voltages → output voltage = input voltage × gain, (ideally) regardless of load impedance and frequency. (The job of the amplifier is to provide the output voltage that equal to input voltage × gain, and to provide the current as determined by Ohm's law.) Loudspeakers are also designed to response to voltages, and ideally give the same SPL with the same input voltage, independent of frequency. Loudspeakers are thus not designed to respond to power, i.e. their efficiencies* vary with frequency as their impedance, and therefore the power demanded from the amplifier at a fixed input voltage, vary with frequency.

Note: * Efficiency is defined as ratio of (acoustic) power output to (electrical) power input. That's why the standard these days for loudspeakers is to rate "sensitivity" (= dB SPL @ 2.83 V), not "efficiency".

Yes. That's why you often see "padding resistors" in the cross-over filter for tweeters. For example:

Thanks for explaining, I think my original questions are now fully answered and my understanding is greatly improved!

BTW, welcome to ASR, and I appreciate your effort to start from the very basics of how these things work. It was a long time into my audio journey before I even understood voltage ~ acceleration ~ SPL, I went around thinking voltage = cone position for a long time.

Thank you! And thanks to you and everyone for taking the time to reply, having these pennies drop has been very satisfying!