# vu meter, is it linear ?

F

#### fred

Jan 1, 1970
0
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?

M

#### Michael A. Terrell

Jan 1, 1970
0
fred said:
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.

P

#### Peter Bennett

Jan 1, 1970
0
The meter movement is linear but you are measuring a logarithmic
signal, so the meter scale has to reflect this.

It is the units of measurement, rather than the signal, that is
logarithmic.

The units indicated on a VU meter are decibels, which are
20 log(Vin/Vref).

If you were to add a voltage scale to the VU meter, that scale would
be linear.

F

#### fred

Jan 1, 1970
0
Michael A. Terrell said:
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 ?

C

#### Costas Vlachos

Jan 1, 1970
0
fred said:
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 ?

Are you sure the -10dB point is *exactly* at the beginning of the scale?
Look carefully. If your meter has the +6dB at extreme right and the 0dB
exactly at the centre, then the extreme left *has* to be -oo (-infinity). If
there was an offset in the scale to bring the -10dB at extreme left, then
either the 0dB wouldn't be at the centre, or the +6dB wouldn't be at extreme
right... Plus you'd need to subtract a voltage from the measured signal to
have a correct measurement in this case.

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.

Costas
_________________________________________________
Costas Vlachos Email: [email protected]

F

#### fred

Jan 1, 1970
0
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.

yes, you are right.
thanks

F

#### fred

Jan 1, 1970
0
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?

B

#### Bob Masta

Jan 1, 1970
0
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?

An increase of 6 dB is a doubling of voltage or pressure,
but a quadrupling of *power*. The formula for dB is
20 * log (V / Vref)
or
20 * log (p / pref)
or
10 * log (P / Pref)
where p = pressure and P = power.
SPL = 20 * log (p / 20^10-6)
where p is pressure in Pascals.
(A Pascal = 1 Newton / meter^2)

Hope this helps!

Bob Masta

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com

C

#### Costas Vlachos

Jan 1, 1970
0
fred said:
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.

cheers,
Costas
_________________________________________________
Costas Vlachos Email: [email protected]

C

#### Costas Vlachos

Jan 1, 1970
0
Costas Vlachos said:
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
_________________________________________________
Costas Vlachos Email: [email protected]

F

#### fred

Jan 1, 1970
0
Costas Vlachos said:
Costas Vlachos said:
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.

B

#### BobGardner

Jan 1, 1970
0
The meter reads avg voltage, the scale is marked in dB

R

#### [email protected]

Jan 1, 1970
0
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.

It's easier than you think, if you double the voltage then twice the
current will be forced to flow (through the same load) quadrupling the
power:-

V
|
| power
|
|________ I

2V
|
|
|
| 4 * power
|
|
|
|________________ 2I

Robin

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