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How can we relate the measured data of speakers with its SIZE (speaker)? I have the data on white noise and pink noise (FFT and 1/3 octave), I'm a bit confused about how to show the relationship (connection) between them. Here I have 2 speakers, one is a car speaker and the other is a large one.

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    $\begingroup$ Size is only one of many factors affecting power output, not to mention spectral power curve $\endgroup$ – Carl Witthoft Nov 6 '20 at 12:58
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In general you cannot say anything because there is no direct relashionship between the size of the speaker and the output level. Think for example of a speaker box containing three speakers, the bass, the middle range and high frequency one. If you apply a high frequency signal (in the audio range) the big bass speaker will barely move whereas the most output comes from the thinny high frequency speaker. This is just one example. There are other things that makes the relationship complicated.

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  • $\begingroup$ thank you very much $\endgroup$ – bass_1u1 Nov 6 '20 at 13:59
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The answer can be found in any undergraduate acoustics text. I summarize it here:

The sound power radiated by any vibrating object (like a speaker cone) depends on how well matched the impedance of the object is to the characteristic impedance of the air into which it is radiating sound waves. A poor match means weak radiation. The radiated power "looks" like a resistive loss to the radiator and so its capability of radiating away sound energy is referred to as its radiation resistance.

The radiation resistance scales with the radiating area of the vibrating object: a large area means more radiation resistance. This means there is a size effect for sound radiation. Now I must leave for an MRI test which will determine whether or not my liver is cancerous. I'll edit this partial answer after I get back from the hospital.

EDIT: I'm back, and now living in a superposition of states (dead in 12 months and dead in 12 years) until my oncologist opens the box I am sitting in, makes a measurement, and collapses my wavefunction. Anyway:

So the larger the vibrating object is, the higher its radiation resistance. This means that we expect a 15" speaker to radiate more sound for a given power input than a 1" speaker, at the same driving frequency. BUT:

The radiator has its own impedance which depends on its mass and springiness. This means that the radiation resistance of a radiator will be a function of frequency, and when it is driven at its resonance frequency, the amplitude of its vibrations will go through a maximum and the amount of acoustic power it radiates will also go through a maximum, since the radiated power depends on the vibration amplitude.

Beyond that resonant frequency, the mass of the radiator will progressively clamp down the amplitude of its vibrations, so a radiator of significant size (and therefore mass) will become ineffective at high frequencies. In summary:

The heavier the speaker cone is, the less well it can radiate high frequencies. The larger the speaker cone is, the better it can radiate at all frequencies. But since cone mass scales with cone area, larger cones mean worse high frequency response.

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  • $\begingroup$ thank you for ur input. Amazing you came back again and answered it, really appreciate it. $\endgroup$ – bass_1u1 Nov 7 '20 at 6:24

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