VU question pop quiz

Tim Ressel Tim_R1 at verifone.com
Fri Feb 25 18:00:03 CET 2000


Mike et al,

This is on-topic, unless you believe no one ever records their efforts.

As to log scales, there are two: voltave and power. A 1dB change in power is
ALWAYS a 1dB change in power, no matter what your reference is. There are
certain standard references, such as dBm (1mW into 600 ohms) or dBV (0dB=1Vrms).
When talking recorders, the 0dB mark is choosen to indicate maximum level, with
a little on top for headroom.

So you see, the 0dB point is totally arbitrary (but essential to know), but the
scales are indeed fixed.

Tim Ressel--Compliance Engineer
Hewlett-Packard
Verifone Division
916-630-2541  
tim_r1 at verifone.com                     



-----Original Message-----
From: Mike Ricos [mailto:mikeri at comm.mot.com]
Sent: Thursday, February 24, 2000 12:59 PM
To: synth-diy at mailhost.bpa.nl
Subject: RE: VU question pop quiz


Greetings,

A few comments on VU and logarithmic audio measurement:

0. While this thread is not strictly synth-diy, it seems enough on topic
to merit discussion. Please correct me if you disagree. I wonder if
others on the list agree with the following. 

1. Absolute VU readings: By all accounts I have read, a Volume Units
meter is only relative to a user defined point, there is no absolute
mapping between a VU scale and any logarithmic voltage scales (dBV, dBv,
dBu, dBm, etc.). Because an arbitrary voltage level can represent a
specific sound pressure level, it makes sense that VU is arbitrary and
independent of voltage levels and impedance. 

2. Are VU meters displaying voltage or power?: My answer is that they
display 20log[rms(Vin/Vref)]. In other words, they measure a voltage but
assume a logarithmic measurement commonly associated with voltage
signals representing power. I do not think it should be interpreted as a
power measurement with an assumed impedance. It is not a power
measurement. It is a voltage measurement of a signal that represents
pressure (force/area) impinging on ear drums. It is 20log and not 10log
so that dBspl = dBdecibel. (A doubling of voltage corresponds to a 6
dBspl increase.)

3. 10log() or 20log() for voltage measurements?: My answer is
20log(V1/Vref) because we typically measure a voltage and are ultimately
interested in sound pressure (force/area in N/m^2), not intensity
(power/area in W/m^2). In audio circles, voltage is typically used to
represent the instantaneous changes in air pressure, not air power
(force/time).  

4. dBm measurements: There are times when a measured voltage signal is
more appropriately interpreted as a power signal via the equation
Power=10log[(V^2/R)/ref_power]. This is true for any field of work where
impedance cannot be assumed constant or power is of primary interest.
Because 600 ohms is a common impedance for microphones and telephone
lines, it is often the assumed impedance for dBm calculations for audio
voltage measurements. A less ambiguous label for such a measurement is
dBv, but I see dBm (w/ 600 ohm assumed) used more frequently in operator
manuals and metering devices.
 
Regards,

Mike R.



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