=?UTF-8?B?Um95IFcuIFJpc2luZw==?= <
rwrising@dslextreme.com> wrote:
On Mon Feb 26 13:43:28 2024 liz@poppyrecords.invalid.invalid (Liz
Tuddenham) wrote: > geoff <geoff@nospamgeoffwood.org> wrote: >
... - supposed to be great on
electric guitar speakers.
The bass tip-up effect should not be apparent close to a loudspeaker because the wavefront at that distance is plane, not spherical. It is rather counter-intuitive and difficult to explain without going too
deeply into theory.
--
~ Liz Tuddenham ~
(Remove the ".invalid"s and add ".co.uk" to reply)
www.poppyrecords.co.uk
Liz - Please go "too deeply into theory".
OK, here we go.....
Assume a 'conventional' ribbon mic with a strong magnet and some narrow
pole pieces with a thin aluminium ribbon tensioned loosely in the gap
between them. The voltage [technically the E.M.F.] developed across the
ends of the ribbon is directly proportional to the rate at which it cuts through the magnetic field in the gap, the 'velocity' of the ribbon.
This movement is caused by a difference in air pressure between the two
sides of the ribbon as the result of air vibration caused by sound.
The tension in the ribbon is very low, so its resonance is below the
audio spectrum and its mass is more important than its compliance at the frequencies we are interested in. The density of aluminium is a lot
greater than air, so the tiny volume of the ribbon has more mass than
the tiny volume of air it displaces. Therefore it does not move as
freely as the air around it, but its mass is accelerated by the
difference in air pressure between its two sides. At high frequencies
the ribbon has little time to be accelerated to a given velocity and the
force needed to do this is high -- but, as the frequency falls, it has
time to be accelerated to the same velocity by smaller and smaller
forces. The relationship is 6dB per octave between force and velocity,
so, for a constant force caused by a constant pressure difference, the
output voltage increases at 6dB per-octave-of-frequency-fall.
Theoretical Plane Wave Response
We can think of the pressure difference across the ribbon as being
caused by plane sound pressure waves passing it. The distance the wave
has to travel between the front of the ribbon and the back, means that
each side is exposed to a different point on the (assumed sinusiodal)
pressure waveform. If we take an example where the half-wavelength of
the sound is exactly equal to the path length between the two side of
the ribbon measured around the pole pieces, one side will be at peak
positive pressure while the other side is at peak negative pressure.
As the frequency falls and the wavelength of the sound becomes greater,
the fixed path length will be a smaller and smaller proportion of the increasing wavelength and the difference between the two pressures will
reduce. The reducing pressure with falling frequency very neatly
compensates for the increasing sensitivity of the ribbon with falling
frequency and the overall result is a flat frequency response. [At
least up to the frequency where the half-wavelength equals the path
difference and down to the point where resonance takes over.]
Theoretical Spherical Wave Response
If sound waves are issuing from a point source, they are spherical, not
plane; this introduces another source of pressure difference across the
ribbon. The spherical waves expand into a greater and greater volume of
the surrounding air, so the energy density in each wave drops as the
square of the distance from the source (the 'inverse square law'). This
effect is independent of the wavelength. When an expanding sperical
wave passes a ribbon mic, there will be an additional pressure
difference across the ribbon caused by the drop in pressure of the wave
itself, as it expands. This drop depends on the ratio of the path
difference of the mic to the radius of curvature of the spherical wave;
so it is greater nearer the source.
However... Unlike the plane wave effect, the spherical wave effect does
not change with frequency or wavelength and the pressure difference on
the ribbon does not reduce with falling frequency. This means that a
lower frequency gives the ribbon a greater velocity, so the voltage
across the ribbon caused by spherical waves increases at 6dB per octave
as the frequency falls.
Practical Response
In reality, most sources of sound give wavefronts which are spherical to
some degree near the source and become more and more plane at a
distance. Thus, as a ribbon mic approaches the source, its distant
plane wave response (flat) will gradually be overpowered by its
spherical wave response (bass tip-up). If the wavefront near the
source is mostly plane because the source is large and flattish, for
instance at the back of a double-bass or close to the cone of a large loudspeaker, the bass tip up effect will be weak or absent.
Other Types of Microphone
Cardioid and hypercardioid microphones achieve their directivity by
combining a bidirectional [ribbon] response with an omnidirectional
[pressure] response. They exhibit bass tip-up from the bidirectional component, but this is only part of the whole response and the effect is
less than it would have been in a ribbon mic.
Response Correction
The transition from a flat response to a 6dB per octave one is exactly
matched by a single-section RC high-pass filter and can be corrected by
nothing more complex than that ...if the correct time constant is
chosen. A variable time constant can be set by ear to correct for a
range of mic distances. Multi-section sharp-cut filters for removing
wind noise are unsuitable, so are shelving 'tone controls' of the
Baxandall type.
Hope I have gone deep enough , but if you want to go even more deeply
into this, at a mathematical level, the best book I have found is a BBC Engineering Training Manual called "Microphones" by A.E. Robertson
(London - Iliffe Books, New York - Hayden Book Co. Inc. ,1951)
--
~ Liz Tuddenham ~
(Remove the ".invalid"s and add ".co.uk" to reply)
www.poppyrecords.co.uk
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