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Full range speaker not able to deliver all frequencies with perfection.

  1. Aug 29, 2009 #1
    Why is it that a full range speaker is not able to deliver all frequencies at once with perfection as compared to the standard tweeder, midrange, woofer and optionally subwoofer arrangement?
  2. jcsd
  3. Aug 29, 2009 #2
    In order to accurately reproduce high frequencies, the tweeter has a very light diaphram. There too much mass in an all-in-one speaker diaphram to reproduce high frequencies.
  4. Aug 29, 2009 #3


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    Also the exact opposite. For a woofer to produce low wavelengths especially at high volumes, the diaphragm needs to move very large distances. That would prove a very difficult thing for a very thin diaphragm and to not tear or fatigue.
  5. Aug 29, 2009 #4
    So it's more of a material issue.

    But why is it that a heavy cone will have difficulty to produce high frequencies?...it can't move that fast?
  6. Aug 29, 2009 #5
    Ok...I get it......
  7. Aug 31, 2009 #6


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    I'll go into a little overkill. Let's consider a circular piston of radius z, mounted in an infinte rigid baffle. The noise radiated by this oscillating piston can be modeled in terms of numerous monopoles radiating together. Each monopole is radiating from a rigid, fully-reflecting plane, not from free space. Therefore, the sound pressure due to any one of the baffled monopole is twice that of an equivalent monopole in free space[30].
    p'(r,t) = \frac{ik\rho_0c}{2\pi r}Q_p e^{i(\omega t - kr)}[/tex]
    In this equation, Q represents the source strength of the monopole on the surface and is equal to [tex]U_p \delta S[/tex] where U is the peak surface velocity of the monopole and [tex]\delta S[/tex] is an elemental surface area. We can then integrate over the whole surface to get the resultant pressure fluctuation due to all the monopoles vibrating in phase.
    p'(r,\theta,t) = \frac{ik\rho_0 c \pi z^2 U_p e^{i(\omega t - kr)}}{2\pi r}\left[\frac{2J_1(kz\sin\theta)}{kz\sin\theta}\right][/tex]
    So, basically I'm just putting equations to words. The Bessel term on the right is a directivity term and puts in those nasty "lobes" that make certain seats at the opera house the "ideal" place to sit.

    So, lower frequencies give us lower reduced frequencies, k. This gives us two variables to play with, U (the maximum velocity) and z (the piston, {e.g. speaker} diamter). In order to make the low frequency noise at a level that is equal to high frequency, one of these needs to go up. Increasing the velocity of the speaker can potentially cause damage, while increasing the piston diameter seems simple. In addition to this, because its exponential, we can get a greater effect simply by increasing the size.
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