decibels english ver

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Decibels

The Decibel is the basis of the audio industry. All levels of power, voltageand sound pressure level (SPL) are expressed in dB form. 1 dB equals ontenth of a Bell. It is also a ratio, and to have a ratio you must have 2quantities. All this relates back to the ear, which can cope with values thatare vast and can operate over an energy range of 1 billion to 1. Therefore

we need something that is more manageable, yet realistic.

Logarithms

By expressing a number in power form we can make it smaller andtherefore more manageable, e.g:

100 = 10 x 10 = 102

1,000 = 10 x 10 x 10 = 103

1,000,000 = 10 x 10 x 10 x 10 x 10 x 10 = 106

1,000,000,000,000 = 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 1012

‘Log’ is the number by which you need to power the base to achieve another number, e.g:

log10 100 = 2

This 10 is known as the ‘base’. It is this number that is raised to the power that gives youanother number, in this case 100. i.e. 10 to the power of 2 gives 100.

With a base of 2 this is the number that gets raised to a power to produce another number.E.g.

Log2 8 = 3

i.e. 2 to the power of 3 gives 8 (or 2 x 2 x 2)

Note: All of our dB calculations use a base of 10.

Number Log

110100

1,00010,000

01234

When you want to multiply numbers which are powered, then you can simply add the powers.

102 ! 10

5 = 10

7

= 100 ! 100,000 = 10,000,000

Also, when you want to divide numbers that are powered, you subtract the powers.

105 ÷ 10

2 = 10

3

= 100,000 ÷ 100 = 1,000

Note: A number to the power of 0 is always equal to 1. If we had a –ve power, then thatwould represent a number that is less than 1.

This number is knownas the ‘power’



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Relating these values to Powers (i.e. Watts)

100 W = 102 W

10 W = 101 W

1 W = 100 W

0.1 W = 10-1

W0.01 W = 10

-2 W

0.001 W = 10-3

W

1 W = 1 W0.1 W = 100 mW0.01 W = 10 mW0.001 W = 1 mW

The dB

Power Ratios

The dB has no unit attached to it. Instead it expresses a ratio between 2 values. For power(Watts) use the following equation:

dB = 10 log Power1

Power2

The dB value is equal to ten times the log of power1 divided by power2. The result of thisformula is +ve if power1 is greater than power2, and –ve if power2 is greater than power1. E.g:

• We have 2 amplifiers, one is known as P1 with a power output of 60 Watts. The otheris P2 with a power output of 40 Watts. We can now express the power ratio betweenthem in dB form.

dB = 10 log P1

P2

= 10 log (60/40)

= 10 (log 1.5)

= 10 ! 0.176

= 1.76 dB (power ratio / logarithmic ratio)

Therefore a numerical ratio of 1.5 : 1 yields power or logarithmic ratio of 1.76 dB.

• If the values of the two amplifiers were changed to 90 W and 60 W:

P1 = 90 W, P2 = 60 W

dB = 10 log 90

60

= 10 log (1.5)

= 1.76 dB



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We can conclude that in both examples P1 is 1.76 dB more powerful thanP2.

Note: this dB value only expresses a ratio but does not give anabsolute value, since dB has no value on it own.

Find the ratios of these powers:

1. P1 = 2 W, P2 = 1 W2. P1 = 10 W, P2 = 5 W3. P1 = 200 W, P2 = 100 W4. P1 = 10 W, P2 = 1 W5. P1 = 1000 W, P2 = 100 W

1. dB = 10 log 2

1

= 10 log (2)

= 10 x 0.3

= 3 dB

2. dB = 10 log 10

5

= 10 log (2)

= 10 x 0.3

= 3 dB

3.

dB = 10 log 200

100

= 10 log (2)

= 10 x 0.3

= 3 dB

We can conclude that whenever one power is twice another there will always be a 3 dBdifference between them.

4. dB = 10 log 10

1

= 10 log (10)

= 10 x 1

= 10 dB



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5. dB = 10 log 1000

100

= 10 log (10)

= 10 x 1

= 10 dB

We can conclude here that whenever one power is ten times another it willalways be 10 dB greater.

Voltage, Current and SPL Ratios

However the formula:dB = 10 log Power1

Power2

Is only suitable for the ratio of one power to another.

When finding voltage, current or SPL (sound pressure level) ratios, a slightly different formulashould be used:

dB = 20 log Voltage1

Voltage2

Or

dB = 20 log SPL1

SPL2

Or

dB = 20 log Current1

Current2

Find the ratios of these voltages and SPL’s:

6. V1 = 2 V, V2 = 1 V7. SPL1 = 10 Pa, SPL2 = 5 Pa8. V1 = 200 V, V2 = 20 V

9. SPL1 = 1 mPa, SPL2 = 100 µPa

6. dB = 20 log 2 V

1 V

= 20 log (2)

= 20 x 0.3

= 6 dB



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7. dB = 20 log 10 Pa

5 Pa

= 20 log (2)

= 20 x 0.3

= 6 dB

We can conclude that whenever one voltage or SPL is twice another therewill always be a 6 dB difference between them.

8.

dB = 20 log 200 V

20 V

= 20 log (10)

= 20 x 1

= 20 dB

9. dB = 20 log 0.001 Pa

0.0001 Pa

= 20 log (10)

= 20 x 1

= 20 dB

We can conclude here that whenever one voltage or SPL is ten times another it will always be20 dB greater.

2 ! Power = 3 dB change

10 ! Power = 10 dB change

2 ! voltage, current, or SPL = 6 dB change

10 ! voltage, current, or SPL = 20 dB change

Remember to convert different

numbers i.e. mPa and µPa to the

same values. In this case theyhave both been converted to Pa.



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Useful Tables

The following tables can be derived from the above equations and havebeen included for easy reference.

Power Level of P1 Level Relationship to 1 W (i.e. P2 = 1W)

1 W 0 dB

10 W 10 dB

100 W 20 dB

200 W 23 dB

400 W 26 dB

800 W 29 dB

1000 W 30 dB

2000 W 33 dB

4000 W 36 dB

8000 W 39 dB

10 000 W 40 dB

20 000 W 43 dB

40 000 W 46 dB

80 000 W 49 dB

100 000 W 50 dB

The value of using dB to express relative levels should be apparent here, since a mere 50 dBrepresents a 100,000:1 ratio. For finding smaller dB values, the following may be helpful.

Power Level dB

1.00 0

1.25 1

1.60 2

2.00 3

2.50 4

3.15 5

4.00 6

5.00 7

6.30 8

8.00 9

10.0 10

The following table show dB levels for both power and voltage / SPL levels:

NumericalRatio

dBEquivalentfor Power

Ratio

dBEquivalent

for Voltage /SPL Ratio

NumericalRatio

dBEquivalentfor Power

Ratio

dBEquivalent

for Voltage /SPL Ratio

2:1 3 dB 6 dB 2.5:1 4 dB 8 dB

4:1 6 dB 12 dB 5:1 7 dB 14 dB

8:1 9 dB 18 dB 10:1 10 dB 20 dB

16:1 12 dB 24 dB 20:1 13 dB 26 dB

32:1 15 dB 30 dB 40:1 16 dB 32 dB

64:1 18 dB 36 dB 100:1 20 dB 40 dB

128:1 21 dB 42 dB 1000:1 30 dB 60 dB



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Expressing Gains and Losses in dB

Common Electronic Diagrams

1. INPUT OUTPUT

TRANSMISSION LINE

2.

INPUT OUTPUT

ATTENUATION PAD

3.INPUT OUTPUT

VARIABLE RESISTOR

4.

INPUT OUTPUT

OP AMP (Operational Amplifier)

-20 dBPAD

-20 dB

+20 dB



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The bell and the decibel are used as units of attenuation or gain, expressingthe decrease in the power caused by a transmission line or piece of audioequipment, or the increase in power caused by an amplifier.

Note – When using the calculator to determine dB gains and losses, dividethe output by the input.

dB = 10 log POUT

PIN

1.

When the input power to a transmission line is 100mW and the output power is 1mW.Calculate the loss this would represent.

dB = 10 log POUT

PIN

dB = 10 log 0.001

0.1= 10 log (0.01)

= 10 x - 2

= - 20 dB.If you calculate the output over the input, the calculator will automatically put a minus sign infront of the result, if it is a loss. In the example, the power ratio is 100:1, therefore theattenuation is -20 dB

2.

INPUT OUTPUT

OP AMP (Operational Amplifier)

Input: 0.1 W Output: 0.001 W

Attenuation?

+20 dB



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What is the gain of an amplifier in dB when the input is 2mW and the outputis 40mW.

dB = 10 log POUT

PIN

dB = 10 log 40

2= 10 log (20)

= 10 x 1.3

= 13 dB

3.

Calculate the attenuation in dB when an input is 10 mV and and output is 0.5 mV:

dB = 20 log VOUT

VIN

dB = 20 log 0.0005

0.01= 20 log (0.05)

= 20 x - 1.3

= - 26 dB

Relating Power and Voltage Levels

If we have to find the power in a circuit, the most common formula we use is:

P = V2

R

Attenuation?

Input: 10 mV Output: 0.5 mV



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1. If the Voltage in a circuit is 20 V and the resistance is 8 ", calculate

the power.

P = V2

R

P = 202

8

P = 400

8

P = 50 W

If the voltage was doubled what would the new output power be?

P = 402

8

P = 1600

8

P = 200 W

Since power is proportional to the voltage squared, if you double the voltage, you get fourtimes the power.

Therefore doubling the voltage results in a power dB increase of:

dB = 10 log 200

50

= 10 log (4)

= 10 x 0.6

= 6 dB

2. Calculate the actual power that is delivered into an 8 " load by 10 V and 100 V

outputs. Then find the power ratio of the two powers.

P = 102

8

P = 100

8

P = 12.5 W



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Using a tenfold increase in voltage:

P = 1002

8

P = 1000

8

P = 1250 W

Therefore a tenfold increase in voltage results in a power dB increase of:

dB = 10 log 1250

12.5

= 10 log (100)

= 10 x 2

= 20 dB

From the previous example we can see that 100 V is 10 ! 10 V and when we refer back to

power values from which the dB is delivered we find that this represents a 20 dB power ratio.This is why voltages use twice the multiplier before the log and the dB equation, and this alsoholds true for current relationships.

dB = 20 log I1

I2

Relative Vs Absolute Level

The key concept is that dB in itself has no absolute value, it can only describe the value ofone quantity relative to another. However, when one absolute value is known it is possible tofind another absolute value when given a suitable increase or decrease in level expressed asa dB value.

Expressing Gains and Losses

To find the power ratio of a circuit when the gain or loss is already known, a reversal of theprocedure is used.

Power Ratio = Antilog dB

10

For voltage and SPL ratios use:

Voltage / SPL Ratio = Antilog dB

20

Note the use of the Antilog (or Inverse log) function. This is used when converting dB valuesto numerical values.



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When the ratio is known the output power can be calculated for a giveninput or the input can be calculated for a known output by using thefollowing equations:

Input Value = Output Value

Ratio

Output Value = Input Value x Ratio

Example 1:

A signal provides 1 mW of power. If this is amplified by 23 dB then what isthe output power?

Power Ratio = Antilog dB

10

Power Ratio = Antilog 23

10

Power Ratio = Antilog (2.3)

Power Ratio = 200 : 1

Output Power = Input power x Power Ratio

Output Power = 1 mW x 200

Output Power = 200 mW

Example 2:

The output signal of an amp is measured at 0.4 V. If it was attenuated by 17 dB then what isthe input voltage?

Voltage Ratio = Antilog dB

20

+23 dB

Input: 1 mW Output:?

Input:? Output: 0.4 V

-17 dB



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Voltage Ratio = Antilog -17

10

Voltage Ratio = Antilog (-0.85)

Voltage Ratio = 0.141 : 1

Input Voltage = Output Voltage

Voltage Ratio

Input Voltage = 0.4

0.141

Input Voltage = 2.83 V

Calculating Overall Gain or Loss

When a circuit consists of several items of equipment the signal is subjected to bothpower/voltage losses and gains. The overall gain or loss of the circuit is expressed in decibelsand is the difference between the sum of the gains and the sum of the losses.

Overall power gain or loss = Total Gain – Total Loss

When the total gain exceeds the total loss, the circuit has an overall gain. When the total lossexceeds the total gain, the circuit has an overall loss. In many circuits the gain ofpower/voltage due to amplification is arranged to equal the losses due to the attenuation oflines, filters etc and the circuit is said to have 0 loss (unity gain). Under these conditions, theoutput power/voltage level is equal to the input power/voltage.

Example:

Total Gain = 26 + 13 + 16 + 15 = +70

Total Loss = -27 + -22 + -18 = -67

Overall Gain = 70 – 67 = 3 dB

Remember touse the minussign here as thesignal isattenuated.

-27

+16+13+26

-18-22

+15

Input Output



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0 dB Reference Levels

dB on its own without being qualified means nothing. In the previousexamples two power, voltage or SPL levels were compared and a ratiofound between them. This was achieved by using one of the amplifiers as areference and comparing the other to it.

If, however, a known reference level is used, dB values can then bechanged from a relative to an absolute level, i.e. a value expressed in watts,volts or SPL’s.

In order to achieve this a number of reference levels have beenimplemented by various organisations so that dB values can express exactquantities. Each of these reference levels are assigned a known quantity toexpress 0 dB and are required so that dB values can be compared withother readings and measurements.

The most common reference levels are as shown in the table below:

0 dB Reference Level Value

dBmdBWdBVdBvdBudBFS (Full Scale)

dBSPLdBA (A-weighted)

1 mW1 W1 V0.775 V0.775 V0 dB (max level) to –96 dB for 16bit digital audio

20 µPa or 2 ! 10-5

N/m2

dBSPL

SPL (Sound Pressure Level) is a measure of pressure so Pascals (Pa) or Newtons per meter2

(N/m2) are used.

1 Pa = 1 N/m2

SPL does relate to the perceived loudness of sound though SPL’s are an exact measurementof pressure whereas loudness is subjective and can vary between one person and another.

The ear is limited by the levels of SPL that it can cope with. These levels are known as:

• The Threshold of Hearing: This is the quietest sound we can perceive. It is theequivalent of sound intensity levels measuring 1 billionth of a Watt. The level wouldbe the equivalent of hearing the blood running through your veins after spending timein an anechoic chamber where your ears can adjust and make out low pressurelevels.

• The Threshold of Pain: This is the loudest sound we can stand. This may be referredto between 120 and 140 dBSPL. For the purposes of the course 140 dBSPL will beused though levels at 120 dBSPL and above will be unpleasant and can causeserious hearing damage. Levels of around 140 dBSPL may be caused by a gun shotat close range. Levels of 160 dB and above can easily burst an ear drum.

0 dBSPL Reference

0 dBSPL relates to the Threshold of Hearing and is equal to 20µPa or 0.00002 Pa, whereas

the threshold of pain is 140 dB SPL equivalent to 200 Pa.

0 dBSPL = 20µPa (or 0.00002 Pa)



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Example

For SPL equations we can use 20 dB increments as tenfold increases inSPL’s

0.00000002 Pa = -40 dBSPL0.000002 Pa = -20 dBSPL0.00002 Pa = 0 dBSPL0.0002 Pa = 20 dBSPL0.002 Pa = 40 dBSPL0.02 Pa = 60 dBSPL

Note: The –ve dB values represent SPL values lower than thereference level and not negative SPL’s

Sound Intensity

Sound intensity is the power of sound per unit area and is thereforemeasured in Watts per metre

2 (W/m

2).

Sound Intensity = Pressure2

Rho

Rho is the air resistance that remains constant at a value of 400.

Intensity (W/m2) Pressure (Pa) dB SPL

1 ! 10-12

0.00002pa 0dB SPL

1 ! 10-10

0.0002pa 20dB SPL

1 ! 10-8

0.002pa 40dB SPL

1 ! 10-6

0.02pa 60dB SPL

1 ! 10-4

0.2pa 80dB SPL

1 ! 10-2

2pa 100dB SPL

1 20pa 120dB SPL

100 200pa 140db SPL

As can be seen from the above table a 20 dBSPL increase results in a tenfold increase inpressure and a one-hundredfold increase in intensity.

dBA

dBA very similar to dBSPL. dBA is an A-weighted measurement. It still describes soundpressure levels but with reference to the perceived loudness of sounds to the human ear.

Example

A 50 Hz and 1 kHz sound wave are measured at 60 dB SPL. The two dBA measurementswould be different, for the 50 Hz wave it would be around 40 dBA whereas the 1 kHz wouldbe heard at 60 dBA. This means that we perceive the 50 Hz wave to be lower in loudnessthan the 1 kHz wave since the ear is less sensitive to lower frequencies (remember theFletcher-Munson Curves) even though they are at the same SPL.



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dBm

dBm was first set as a standard by the IRE (Institute of Radio Engineers).The term dBm expresses an electrical power level and has no directrelationship to voltage or impedance. The typical circuit in which dBm was

first measured was a 600 " telephone line.

P = V2

R

V = # (P x R)

V = # (0.001 x 600)

V = # (0.6)

V = 0.775 V

It so happens that 1 mW power is dissipated when a voltage of 0.775 V is applied to a 600 "

line. It is for this reason many people mistakenly believe that 0 dBm is equal to 0.775 Volts,

but this is only the case in a 600 " circuit.

0 dBm Reference

The 0 dBm is a power reference level and equates to 1 mW.

0 dBm = 0.001 Watt

Example

For power equations we can use 10 dB increments as tenfold increases in power levels

0.00001 W = -20 dBm0.0001 W = -10 dBm0.001 W = 0 dBm0.01 W = 10 dBm0.1 W = 20 dBm1 W = 30 dBm

Note: The –ve dB values represent powers lower than the reference level and notnegative power levels

dBu

Most modern audio equipment is sensitive to voltage levels. Power output is not a realconsideration except in the case of power amplifiers. As we know dBm expresses a powerratio. The voltage can be calculated if the impedance is known but this complicates things anddBu was created to simplify matters. dBu is a more appropriate term for expressing outputand input voltages (Note: the voltage represented by dBu is equivalent to that of dBm if the

dBm figure is derived with a 600 " load. However the dBu value is not dependant on the load

Telephone Line

0.001 W dissipated

600 "



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(hence the u for unloaded meaning it if free from the load impedance), therefore 0 dBu isalways referenced at 0.775V.

0 dBu Reference

The 0 dBu is a voltage reference and equates to 0.775 V.

0 dBu = 0.775 Volts

Example

For voltage equations we can use 20 dB increments as tenfold increases involtage levels:

0.00775 V = -40 dBu0.0775 V = -20 dBu0.775 V = 0 dBu7.75 V = 20 dBu77.5 V = 40 dBu775 V = 60 dBu

Note: The –ve dB values represent voltages lower than the reference level and notnegative voltage levels

dBu was specified as a standard in order to avoid confusion with another voltage related unit,dBv.

dBv

dBv (Note the lower case v) is used by NAB (National Association of Broadcasters) to denotethe voltage value corresponding to the power indicated in dBm – (0 dBv is referenced at0.775 V). This was convenient because dB values would tend to be the same as though dBm

was being used, provided that the output was specified to achieve 600 ". This made is easier

to compare dBv specifications with products that were specified in dBm.

0 dBv Reference

The 0 dBv is a voltage reference and equates to 0.775 V.

0 dBv = 0.775 Volts

Example

For voltage equations we can use 20 dB increments as tenfold increases in voltage levels:

0.00775 V = -40 dBv0.0775 V = -20 dBv0.775 V = 0 dBv7.75 V = 20 dBv77.5 V = 40 dBv775 V = 60 dBv

dBV

dBV (Note the capital V) was introduced to as a reference by the IEC (International Electro-Technical Commission). This is a term, which describes an output that uses no load, i.e. anopen circuit or an insignificant load, such as the typical high impedance outputs of modernconsumer audio equipment.



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0 dBV Reference

The 0 dBV is a voltage reference and equates to 1 V.

0 dBV = 1 Volt

Example

For voltage equations we can use 20 dB increments as tenfold increases involtage levels:

0.01V = -40 dBV0.1 V = -20 dBV1 V = 0 dBV10 V = 20 dBV100 V = 40 dBV1000 V = 60 dBV

dBW

dBm is useful when dealing with very small power levels, such as the output of microphones(that can be around one millionth of a watt) and the level of signal processors. But it is of nouse when dealing with the multi-hundred watt outputs of large power amplifiers typically foundin live sound environments. Hence dBW was introduced.

0 dBW Reference

The 0 dBW is a power reference and equates to 1 W.

0 dBW = 1 Watt

Example

For power equations we can use 10 dB increments as tenfold increases in power levels

0.01 W = -20 dBW0.1 W = -10 dBW1 W = 0 dBW10 W = 10 dBW100 W = 20 dBW1000 W = 30 dBW

Converting Numerical Values to dB

Since we now have the reference values for all the dB values we can now convert numericalvalues in to dB values that still maintain an absolute reference, since they are relative to aknown quantity.

The two equations required for this are:

dBm and dBW = 10 log P1

Ref

dBu, dBv, dBV and dBSPL = 20 log V or SPL

Ref



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Example 1

A console’s maximum operating power is 0.12 W. What is the maximumoperating level expressed in dBm?

dBm = 10 log P1

Ref

dBm = 10 log 0.12

0.001

= 10 log 120

= 10 ! 2.08

= 20.8 dBm

Example 2

A signal level is measured at 0.065 V. What is this expressed in dBu?

dBu = 20 log V1

Ref

dBu = 20 log 0.065

0.775

= 20 log 0.084

= 20 ! -1.08

= -21.5 dBu

Example 3

A SPL is measured at 0.37 Pa. What is this expressed in dBSPL?

dBSPL = 20 log SPL1

Ref

dBSPL = 20 log 0.37

0.00002

= 20 log 18500

= 20 ! 4.2

= 85.3 dBSPL

Remember to keepboth units the same,either in mW or inthis case W

Both valuesexpressed in V

1 mW

0.775 V

This is a negative value sincethe original voltage level waslower than the reference level

Both valuesexpressedin Pa

20 micro Pa



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Converting dB Values to Numerical Values

It is also possible to convert values from dB back to numerical values. Inorder to do this it is necessary to use the Antilog (or inverse log) function.The following formulas should be used:

Power = Antilog dBm / dBW x Ref

10

Voltage / SPL = Antilog dBu / dBv / dBV / dBSPL x Ref

20

Example 1

Express 13 dBm in mW:

Power = Antilog dBm x Ref

10

Power = Antilog 13 x 1 mW

10

Power = Antilog 1.3 x 1 mW

Power = 20 x 1 mW

Power = 20 mW

Example 2 Part A

A console’s maximum output level is +24 dB. This statement is meaningless since the zeroreference for dB is not specified.

Example 2 Part B

A console’s maximum output level is +24 dBm. This means that the output level is 20 dBhigher than the reference level (for dBm this is 1 mW).


10


10


Power = 251 x 1 mW

Power = 251 mW

This makes a specific claim. It tells us that the console is capable of delivering 251 mW intoan unknown resistance. Note that this is a typical peak level output of a desk.

Put theappropriatereference valuein here. FordBm it is 1 mW



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Example 3

Calculate the actual voltage represented by –26 dBu:

This can be worked out using tenfold and twofold reductions in voltage

0 dBu = 0.775 V

-20 dBu = 0.0775 V

-26 dBu = 0.03875 V

or using the Antilog formula:

Voltage = Antilog dBu x Ref

20

Voltage = Antilog -26 x 0.775 V

20

Voltage = Antilog -1.3 x 0.775 V

Voltage = 0.05 x 0.775 V

Voltage = 0.03875 V

Example 4

Calculate the sound pressure level represented by 94 dBSPL:

SPL = Antilog dBSPL x Ref

20

SPL = Antilog 94 x 0.00002 Pa

20

SPL = Antilog 4.7 x 0.00002 Pa

SPL = 50119 x 0.00002 Pa

SPL = 1 Pa

Note that 94 dBSPL represents an actual SPL of 1 Pa. This SPL level is used to measuremicrophone and loudspeaker sensitivity.



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Standard Operating Levels

In the world of audio there are different operating levels for differentapplications. The three standard operating levels are:

Operating Level ReferenceLevel

Calibration Voltage CalibrationReference Level

Semi-Pro (Consumer audio i.e. TV’s,

HiFi’s etc)

- 10 dBm

- 10 dBV

0.245 V

0.316 V

0 VU

0 VU

Professional(Live PA and studio)

+ 4 dBm

+ 4 dBu

1.23 V

1.23 V

0 VU

0 VU

Broadcast(TV and Radio)

+ 8 dBm

+ 8 dBu

1.94 V

1.94 V

0 VU

0 VU

These standard operating levels are used for compatibility of equipment with other like-ratedequipment and also to establish a set of calibration levels so that equipment can be calibratedto ensure the correct gain structure throughout a studio for example. Therefore a meterreading of 0 VU should equate to an actual signal level of 1.23 V.

Note: The Pro Audio and Broadcast reference levels equate to the same calibrationvoltage even though they are rated in Watts (for dBm) and Volts (for dBu) as it is

assumed that an impedance of 600 " is used.

Pro Audio Operating Levels

0Vu

+ 4 dBu = 1.23 V = 0 VU Recording

0Vu

+8 dBu = 1.94 V = 0 VU Broadcast

0Vu

-10 dBV = 0.316 V = 0 VU Semi-pro



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Example

Calculate the actual voltage represented by + 4 dBu:


20

Voltage = Antilog 4 x 0.775 V

20

Voltage = Antilog 0.2 x 0.775 V

Voltage = 1.58 x 0.775 V

Voltage = 1.23 V

Calculate the actual voltage represented by + 4 dBm:


10


10


Power = 2.51 x 1 mW

Power = 2.51 mW

Assuming a Load resistance of 600 ":

V = # (P x R)

V = # (0.00251 x 600)

V = # (1.506)

V = 1.23 V

Calculate the actual voltage represented by +8 dBu:


20

Voltage = Antilog 8 x 0.775 V

20

Voltage = Antilog 0.4 x 0.775 V

Voltage = 2.51 x 0.775 V

Voltage = 1.94 V



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Calculate the actual voltage represented by -10 dBV:


20

Voltage = Antilog -10 x 1 V

20

Voltage = Antilog –0.5 x 1 V

Voltage = 0.316 x 1 V

Voltage = 0.316 V

The higher the operating level, the more amplification you need for less gainsince higher voltage and (hence signal levels) are present. You will find onthe rear panel of many of today’s audio equipment, a selector switch, which will normallyallow a choice of between –10 dBV and +4 dBu, such as consoles, FX units, amplifiers etc.Most semi-pro equipment runs at -10 dBV, such as semi-pro consoles, FX units, hi-fiamplifiers etc. Some semi-pro equipment may also use unbalanced circuitry throughout.

Therefore the hotter the level you run, i.e. +4dBu, then the less gain you need to achieve thedesired level (more gain = more noise).

Relating dBV, dBu and dBm to Specifications

In most consumer audio (semi-pro) products you will see phono (RCA) inputs and outputsrated in dBV. Typical line level phono jack inputs and outputs are for use with high impedanceequipment which is more sensitive to voltage than to power, and so these outputs arespecified as -10dBV. This is a standard used in the consumer audio market.

Most pro-audio gear with XLR connectors and balanced 1/4-inch jack inputs and outputs arerated in dBm or dBu. Typical line level XLR output and input connectors are intended for usewith both low and high impedance equipment. Therefore their levels are specified at +4 dBu

or dBm over 600 " that is the standard reference for studio recording and live sound. Note

however that broadcast studios operate at +8 dBu)

Relating dB to Acoustic Power Levels

The term sound level generally refers to SPL although it may refer to sound power (dBPWL).However dBPWL is a very rarely used term. Sound power is the total sound energy radiatedby a loudspeaker in all directions while sound pressure is a level measured per unit area at aparticular distance relative to the sound source.

The measurement for dBPWL is normally made using 1 Watt at a distance of 1 metre (1 W @1 m).

Question

Two transistor radios combined produce an SPL of 96 dBSPL. If one radio is turned off, whatis the new SPL?

Since one radio is turned off, the power producing the SPL will drop by half. Therefore a 3 dBdrop in level will be experienced so the answer is 93 dBSPL.



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Decibels And The Human Ear

The intensity of a sound and its loudness are two distinct aspects of sound.The intensity of a sound is related to the amount of energy present per unitarea, whilst loudness is a physical sensation that varies betweenindividuals. Because of this subjective nature of loudness it can neither beaccurately measured nor precisely defined.

A well-known law in psychology (Webber-Fechner Law) states that themagnitude of any sensation is proportional to the logarithm of the stimulus.What this means is that if a certain power is expended to produce a givenloudness, then doubling the power will not double the loudness.

It has been found that the human ear responds to logs of power ratios andso an increase in loudness can be found using a logarithmic function aswell.

Increase in Loudness = log Power Ratio

Example

Doubling an amp power results in a Power Ratio of 2:1

Therefore the expected loudness increase will be:

Increase in Loudness = log 2

= 0.3

A 0.3 increase in loudness is the smallest change that the human ear can perceive in acomplex waveform.

Therefore if the power is doubled again, the loudness will be further increased by 0.3.

Example

A tenfold increase in amp power results in a Power Ratio of 10:1

Therefore the expected loudness increase will be:

Increase in Loudness = log 10

= 1

Therefore to double the apparent loudness of a sound (i.e. going from a loudness of 1 to 2) a10 dB power increase is required.

Example

An amplifier delivers 5 W of power into a room. It is found that the sound level produced ismuch too low. It is suggested that the amp is changed for a 10 W model. Is this changejustified?

The increase in sound is proportional to the power ratios involved. In this case the power

output has been doubled (i.e. 5 W $ 10 W), which represents a gain of 3 dB or an increase in

loudness of 0.3. Although the power output has been increased by 5 W, the increase inloudness would only be very slight. This modification would not be worth the cost and effort.For the 5 W amp to sound twice as loud, the increase should be tenfold:

5 W ! 10 = 50 W



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Volume, Level and Gain

Volume, level and gain are the three most commonly used terms in audio.Therefore it is important to have a good understanding of them and to knowwhen they should be used.

Volume

This is defined as power level, so in terms of audio equipment if you turn upthe volume you are increasing the power. However, it can also describesound intensity or the magnitude of an electrical signal, or even the spaceinside an enclosure. It is better to avoid using the term ‘volume’ in the studiosince it does not have a specific meaning.

Level

Level is defined as the magnitude of a quantity in relation to an arbitrary reference value. For

example, SPL is expressed as dB SPL relative to 20 µPa. The audio level in a signal

processor would be expressed in dBm over 600 " or dBu, both of which are referenced to an

exact figure.

Gain

This has several definitions. If it is not specified it normally implies transmission gain from anamplifier (i.e. a microphone or line pre-amp found on every console), which is the powerincrease of a signal expressed in dB. Sometimes an increase in voltage is expressed as avoltage gain, but be careful because a voltage gain may represent a power loss depending onthe impedance of the circuit.

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Documents

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