the material below is copyright 1993 by M. Gore. used by permission. Use of this material forpersonal use only, is granted. Any other use is prohibited. And is also very BAD Karma. DECIBELS The decibel (which is one tenth of a BEL), is used to compare two levels of electric power in terms of the logarithm of their ratio. In somewhat simpler terms, a decibel is a RATIO between two powers, voltages or Sound Pressure Levels (SPLs). The advantage of using decibels (abbreviated as db) is that decibel units compress a huge range of RATIOS into numbers that are easy to understand. As an example, 10,000 watts vs. 1 watt is equal to a 40db difference between them, a bit easier to understand and talk about than to have said "a 10,000 to 1 ratio". Plus a couple of db is about the minimum change the normal person can detect in audio signals, and thus if we can discuss specifications and measurements in detectable units, these units then mean something to us. A 3 db change in noise is detectable. A difference between .001 volt and .00111 volt isn't. A good thing to remember is that a full BEL (ten deci-bels) is "TWICE AS LOUD" to most people. Since the central figure in dbs is POWER, the readings must be the same unit, you cannot compare watts and voltages unless you first convert the readings into the SAME UNIT. Also note that for correct comparisons, wattage readings (and voltage readings) must be made using the same load (resistance in most cases). It does little to compare an amplifier that can drive an 8 ohm speaker to one that can only drive a 400 ohm device. There are two formulas used to find a db ratio between units: The first is used ONLY FOR POWER (watts): P1 10x log ------ P2 In words this is "10 times the log of the product of Power #1 Divided by Power #2" First you find the RATIO by dividing the Power of one reading into the Power of the other reading, then find the Log of this result then multiply that figure by 10. You must use this formula when comparing wattages, or Sound Intensity which is also measured in watts. The second is used ONLY FOR VOLTAGES (and SPL levels): V1 20x log ------ V2 First find the ratio by dividing the two voltages, then find the Log of that figure, then multiply by 20. This formula is for voltage relationships and Sound Pressure Levels. If you know math very well, you will see that since voltage squared divided by the resistance equals watts, when you have two voltages that you are comparing across the SAME RESISTANCE, the resistances cancel each other out and 20 log V1 / V2 equals 10 log V1 squared divided by V2 squared. Thus the two formula really do give equal results when transformed to equal units!!! In other words, the two formulas are equal to one another, the voltage formula just cancels out the resistances and takes into account the squared voltages required in the power formula 2 Power = I R Often we use a already known standard and compare a power or voltage to this standard. Some of the standards are listed below: 0 dbm equals 1 milli Watt dissipated in a 600 ohm resistor. Any reading with dbm in it, MUST have a 600 ohm resistor in which the power is measured. NOTE: 1 mW is equal to .775 volts rms across 600 ohms. -128dbm equals the noise contained in a resistor without any external connections, at room temperature, due to the collisions of electrons. Sorry, you can't have less noise than this in the "real" world". +4dbm equals the standard for output at "0VU" from professional equipment. This is 1.228 Vrms across 600 ohms. This is "NORMAL Professional OUTPUT LEVEL" -10 db equals different readings when made across different resistance loads. Usually it is .3 Vrms across 10,000 ohms. This is the norm for SEMI-PRO gear. dbV is used for references of 1 Volt RMS across a "given" load, usually a high impedance more than 10,000 ohms NOTE WELL that a 3 db increase in POWER is twice the POWER, while a 6 db increase in VOLTAGE is twice the VOLTAGE, and a 10 db increase in POWER is about what the "normal" person considers "twice as loud". Note that for every 10 db increase in POWER you get a decade increase in Wattage: 0 db increase = same power 3 db increase = 2 x power 10 db increase = 10 x power 20 db increase = 100 x power 30 db increase = 1000 x power 40 db increase = 10,000 x power Note that for every 20 db increase in VOLTAGE you get a decade: 0 db increase = same voltage 6 db increase = 2 x voltage 10 db increase = 3.16 x voltage 20 db increase = 10 x voltage 40 db increase = 100 x voltage 60 db increase = 1000 x voltage 80 db increase = 10,000 x voltage Some "normal" Real-World db readings might be: noise of a tape recorder, in repro, but having no tape running: -58 dbm Overall noise level of a good Analog console: -85 dbm Max output level of a older Neve Console: +28 dbm Tape recorder S/N : 70 db Equivalent Mic Preamp noise: -122 dbm Analog Console crosstalk: -80 db Note some readings are in dbm (referenced with a 600 ohm resistor and related to .775 Volts rms) and others are just db, meaning the ratio is between two measurements made on the same piece of gear under different conditions (ie: no signal vs. signal). *******************************************************************

Unfortunately YOU CANNOT JUST ADD DECIBELS!!! Lets' say that we have a voltage of 7.75 volts rms (+20 dbm) across a 600 ohm resistor and add to this another voltage with a dbm rating of +20 dbm. Is the result +40 dbm? NO WAY !!!!! Since the two voltages are the same voltage (both are 7.75 volts) and we all know that doubling the voltage gains only a 6 db increase, the correct answer would be +26 dbm. What we have to do in order to add dbs together (and SPL readings in db's) is to reduce the db figures BACK to voltages (or SPL readings or watts), then add these figures together then apply the correct db formula..... so for 7.75 volts @ 600 ohms: 7.75 + 7.75 20x Log ------------- .775 (our reference voltage for 0 dbm) equals 15 20x Log ------ .775 equals 25.73 dbm !!! close enough to +26 dbm for me!!! another way to state this is that if we have 1 guitar amp putting out 70 db Sound Intensity (Sound INTENSITY is measured in WATTS, thus is power)... how many more amps to we need to sound twice as loud? (remember that we need a 10 db increase for the average listener to hear something "twice as loud") 1 amp = 70 db Sound intensity (in Watts per square meter) use the formula : (since we've got the sound in Watts per square meter) Intensity 1 10x LOG ------------ Intensity 2 (the threshold of hearing will be our reference here) for 1 guitar amp: -5 10 10x LOG ------------- -12 10 (threshold of hearing) which equals 70 db Acoustic Intensity note this is the level ABOVE the threshold of hearing which is what "Acoustic Intensity" is defined as... So for 2 guitar amps: -5 2 x 10 10x LOG ----------- -12 10 which equals: 73 db !!! ( Remember that if we add two equal wattages we get an increase of only 3 db!!!! ) now for 4 guitar amps: -5 4 x 10 10x LOG --------- -12 10 equals 76 db, not yet the required 10 db increase!!!! lets try 10 guitar amps: -5 10 x 10 10x LOG ------------ -12 10 which now equals 80 db !!! SO.... in order for the guitar player to sound TWOCE AS LOUD (a 10 db increase) she would need to have a total of 10 guitar amps (all the same) on stage! And quite a large stage if all the players needed 10 amplifiers each..... Then think if they wanted it "twice as loud" once again...... 100 amplifiers for each musician.... whew... So how do we get around this over crowed stage yet 'sound twice as loud" ??? We use more EFFICENT speakers... expensive ones. But we still have to have a lot of them... just look at any large performance space and you'll see lots of power amps and lots of speakers. Usually these are built into the venue's own PA system or rented as needed. ***************************************************************

The normal conversational speaker's voice is about 20 micro watts of SPL. ( talkers during the classes I used to teach always seemed to have a lot more power... even whispers sounded loud to me!) 20 million people all talking at once (and saying the same thing, in exact phase with one another) would generate just enough power to light up one single 400 watt light bulb!!!! *****************************************************************

There is confusion as to which of the two formulas for finding decibels one should use when dealing in SPL, and Acoustic Powers. as you remember: for Powers (wattages) P1 is Power measured for #1 unit P2 is Power measured for #2 unit... P1 10x LOG -------- P2 (our reference power) and for voltages: V1 20x LOG -------- V2 (our ref voltage) So when you measure ACOUSTIC POWER in WATTS PER SQUARE METER use the Power db formula But when you measure SPL in dbs: use the voltage db formula In truth one should square the SPL readings and use the Power db formula, but using the easier Voltage formula works since: SPL1 SPL1 squared 20x LOG ------- which equals 10x LOG -------------- SPL2 SPL2 squared ***************************************************************

We cannot simply add up db readings to get an overall noise level. To do correctly this we have to re- convert the db levels into watts, then add them, then find out the correct db ratios. (see note at the bottom of this page about noise phase !!) Since the first stage of a mic pre-amp amplifies the noise level (as well as the signal) from the mic by say about 65 db, it creates a noise level at the output of the pre-amp of let's say -55 db, plus the noise inherent in the mic-pre amp itself. We'll say the mic pre-amp has 2 db noise. Then the total would be P1 db = 10x log ------- P ref (our reference here will equal .001 watt) -55 db = .000 000 003 watts 2 db noise = P1 2 db = 10x log --------- P2 (P2 equals the absolute minimum noise level possible which is -128 dbm ) to find wattage of minimum noise P1 -128 dbm = 10x log ---------- .001 watt which equals .000 000 000 000 000 15 watts then P1 2 db noise = 10x log --------------------------- .000 000 000 000 000 15 2db noise = .000 000 000 000 000 24 watts .000 000 000 000 000 24 watts plus .000 000 003 watts equals .000 000 003 000 000 24 watts (whew!) THEN: noise in db is... .000 000 003 000 000 24 watts 10x log -------------------------- .001 watts which equals 10x log -.000 003 000 000 24 which then equals -55.0001 db next find out the power of the 5 db added in noise of the EQ amp: P eq 5 db = 10x log --------------------------------- .000 000 000 000 000 15 5 db noise = .000 000 000 000 000 475 watts add this to the total noise before the EQ amp: .000 000 003 000 000 000 24 + .000 000 000 000 000 475 = .000 000 003 000 000 000 715 watts then find the result: .000 000 003 000 000 000 715 noise = 10x log ---------------------------- .001 which equals 10x log -.000 003 000 000 000 715 which equals -55.0002 db and so on...... it isn't easy folks................

same as two in phase signals. While two in phase powers will add by

3 db, and two in phase voltages will add by 6 db, two noise voltages

will add by about 1.5 db to 2.5db.

This is due to the fact that each noise source has a different spectrum

of noise, and that noise spectrum is constantly changing, thus the phases of the

noise frequencies are always changing.

To find out the way two non-coherent signals add you can use this formula:

Total noise = square root of (signal 1 squared plus signal two squared)

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This only approximates the 'total noise' of 2 noise
signals combined together, but generally is close enough for use.
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