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The meter movement is linear but you are measuring a logarithmicIs the "needle device" (I think it is called a vu meter)
of a sound level meter is it linear, if so, how do we
get a non linear scale, namely -10 at extreme left,
zero in the middle, and +6 at extreme right?
It is the units of measurement, rather than the signal, that isfred wrote:
Is the "needle device" (I think it is called a vu meter)
of a sound level meter is it linear, if so, how do we
get a non linear scale, namely -10 at extreme left,
zero in the middle, and +6 at extreme right?
The meter movement is linear but you are measuring a logarithmic
signal, so the meter scale has to reflect this.
Can you explain this then:fred wrote:
Is the "needle device" (I think it is called a vu meter)
of a sound level meter is it linear, if so, how do we
get a non linear scale, namely -10 at extreme left,
zero in the middle, and +6 at extreme right?
The meter movement is linear but you are measuring a logarithmic
signal, so the meter scale has to reflect this.
"Michael A. Terrell" <mike.terrell@earthlink.net> wrote in message
news:<401AB69F.258AF24F@earthlink.net>...
fred wrote:
Is the "needle device" (I think it is called a vu meter)
of a sound level meter is it linear, if so, how do we
get a non linear scale, namely -10 at extreme left,
zero in the middle, and +6 at extreme right?
The meter movement is linear but you are measuring a logarithmic
signal, so the meter scale has to reflect this.
Can you explain this then:
why is -10db equal to +6db in terms of needle displacement ?
If 20 db is times 10 then 10 db is times root 10 i.e. 3.16
Hence, the voltage must increase times 3.16 to cause the needle
to move from -10 to 0 and from 0 to +6 the voltage doubles.
Since the needle is linear this does not make sense, what am
I doing wrong ?
In dB meters, because of the logarithmic scaling, the dB values near the
beginning of the scale become so dense that it isn't easy to read them. I
suspect the discrepancy in your calculations is because of this.
An increase of 6 dB is a doubling of voltage or pressure,I thought I understood this but now I am confused again:
When the needle moves from center to the +6 at the
extreme right, the displacement doubles, and so does
the voltage, so this makes perfect sense. However, when
the voltage doubles, sound pressure level quadruples,
so the reading of the sound level will be wrong, what
am I doing wrong here?
I thought I understood this but now I am confused again:
When the needle moves from center to the +6 at the
extreme right, the displacement doubles, and so does
the voltage, so this makes perfect sense. However, when
the voltage doubles, sound pressure level quadruples,
so the reading of the sound level will be wrong, what
am I doing wrong here?
"fred" <fred222@btinternet.com> wrote in message
news:2685a85f.0401310407.94db8a6@posting.google.com...
I thought I understood this but now I am confused again:
When the needle moves from center to the +6 at the
extreme right, the displacement doubles, and so does
the voltage, so this makes perfect sense. However, when
the voltage doubles, sound pressure level quadruples,
so the reading of the sound level will be wrong, what
am I doing wrong here?
OK, let's see what we have here. The meter scale is in dB, but the meter's
needle really measures voltage. We also have the following relation:
dB = 20 * log ( P / Po )
where P and Po are the two sound pressure levels we're comparing. So, if
you feed your meter with a voltage that is proportional to sound pressure
level (SPL), then the readings will be correct.
I think it simply works out that a 6db voltage increase and"Costas Vlachos" <c-X-vlachos@hot-X-mail.com> wrote in message
news:bvgkdo$hhb$1@sparta.btinternet.com...
"fred" <fred222@btinternet.com> wrote in message
news:2685a85f.0401310407.94db8a6@posting.google.com...
I thought I understood this but now I am confused again:
When the needle moves from center to the +6 at the
extreme right, the displacement doubles, and so does
the voltage, so this makes perfect sense. However, when
the voltage doubles, sound pressure level quadruples,
so the reading of the sound level will be wrong, what
am I doing wrong here?
OK, let's see what we have here. The meter scale is in dB, but the meter's
needle really measures voltage. We also have the following relation:
dB = 20 * log ( P / Po )
where P and Po are the two sound pressure levels we're comparing. So, if
you feed your meter with a voltage that is proportional to sound pressure
level (SPL), then the readings will be correct.
Hmmm, I shouldn't have used SPL in my reply above, as SPL has a fixed point
of reference. Please read the above as simply "sound pressure", *not* SPL.
So, if you feed your meter with a signal whose voltage is proportional to
sound pressure, then the readings will be correct. This is because of the
20*log relationship shown above. In addition to that, if your reference Po
is equal to a sound pressure of 20 microPascal (the threshold of hearing),
then the readings will be in dB [SPL].
Sorry for any confusion caused.
Costas
I think it simply works out that a 6db voltage increase and
a 6db sound level increase are the same think, which can be
concluded from the above mentioned formulas.