25th July 2014

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27th August 2014

#**33**

27th August 2014

#**34**

Run that signal directly into your VU meter. The meter should read then 0VU. Otherwise there should be a possibility to turn a knob or similar on the VU. Turn it until it reads 0 VU.

Maybe someone else can chime in and tell us how to do it the most accurate way.

30th October 2014

#**35**

Quote:

The dB is a logarithmic unit of ratios. What does this mean exactly? First, consider what a linear scale is: imagine you're charting your little cousin's height as he gets older. Every year you measure his height and write it down. Ten years go by and you decide to graph this as his birthday present. To create the graph, you put 10 equally-spaced tick marks on the horizontal axis and number them 0 through 9, each tick representing his age. On the vertical axis you put 10 equally-spaced ticks representing his height up to now. Finally, you add a dot for each height/age measurement you took and draw lines connecting them.

You've just created a linear graph: both the horizontal and vertical scales are linear. This means that between any two tick marks there is the same amount of "stuff" -- for example, between any two tick marks on the horizontal scale there is always one year's worth of time.

Now imagine that you're in audio engineering school and your professor has asked you to graph the relationship of loudness to certain events: a mosquito flapping its wing, a typical conversation, standing next to a jack hammer, and standing next to a nuclear explosion (if such a thing were possible). You do some research to find the pressure magnitude of these events and start building your graph, with the event on the horizontal axis and the pressure magnitude on the vertical axis. The horizontal axis is easy: four tick marks, each representing one of the events. But for the vertical axis, you realize you have a problem: the magnitude of a mosquito flapping its wing is on the order of 0.0001 pascals, the jack hammer around 2 pascals, and the nuclear explosion at about 100000 pascals. That means you need to show a range of 10^-5 to 10^5. If you make each tick equal to the smallest magnitude (10^-5), you'd need 10^10 ticks. That's 10*trillion* ticks! (If you could manage to average 10 ticks per second, it would take you a little under 32 years to finish drawing the vertical axis. That's a lot of paper.)

Obviously this won't do. You need something manageable, say 10 horizontal tick marks. But how do you break up such a huge range into 10 chunks? If you try to make it a linear scale -- where each chunk has the same amount of "stuff" in it -- you run into another kind of problem: If each tick is worth 100000/10 = 10000 pascals, then the very first tick contains all the events except the nuclear explosion. Worst still, the first tick has so much stuff in it that you won't have a visual indication of the difference in loudness between a mosquito flapping its wing and a jack hammer. Considering the very audible difference between the two, a linear scale is therefore a poor choice.

This is one situation where logarithmic scales shine. The range of stuff we want to graph is between 10^-5 and 10^5; as said before, this is 10^10 worth of stuff: you need 10^5 stuff to get from 10^-5 to 10^0, and then another 10^5 to get from 10^0 to 10^5. Iinstead of dividing this huge range into ten even chunks, why not use the ten exponents as the tick marks? This would give us, starting from the botton: -5, -4, -3, -2, -1, 0, 1, 2 ,3, 4, 5 (which is eleven ticks, but that's ok). The mosquito wing (0.00001 = 10^-5) would correspond to the -5 tick, the conversation (0.001 = 10^-3) with the -3 tick, the jack hammer (2 = a bit more than 10^0) with the 0 tick, and the nuclear explosion (100000 = 10^5) with the 5 tick. In this scale we can visually see that the jack hammer is halfway between the mosquito wing and the nuclear explosion, which makes pretty good sense.

So what exactly did we do there? If you notice, the stuff between ticks is not evenly distributed: the farther you go up (on the vertical scale; or to the right, on a horizontal scale) the more "stuff" there is between the ticks. In this case, the amount of stuff between the 0 tick and the 1 tick is ten units (10^0 = 1 and 10^1 = 10); the amount of stuff between 1 tick and 2 tick is almost a hundred units (10^1 = 10 and 10^2 = 100); the amount of stuff between the 2 tick and 3 tick is almost a thousand units (10^2 = 100 and 10^3 = 1000); and so on. The*magnitude* of each tick increases by a factor of 10; this nonlinearity allows us to represent huge numbers with small numbers. In this particular case we used factors of 10, which means our scale was *base-10 logarithmic*.

Quick and easy math lesson: a logarithm is an exponent. Next time you see the word logarithm or logarithmic, just think to yourself "exponent". For example, what is the base-10 logarithm of 100000? A logarithm is just an exponent, so what exponent of 10 gives you 100000? 5, because 10^5 = 100000. Therefore log(100000) = 5. Simple as that. We can change the*base* of the logarithm, but again, logarithms are all about the exponent. So what is log(8) in base-2? Well, what exponent of 2 gives us the number 8? The answer is 3, because 2*2*2 = 2^3 = 8. Now it should be a bit easier to understand what a logarithmic scale means: it's just a scale based on exponents. And because exponents are multiplicative, they allow us to map a huge ranges of numbers to a much smaller range. This is the essence of the dB.

The other important aspect of the dB is that it is a ratio; it tells us the magnitude of a measurement relative to*something*. So when you read that a preamp has 60 dB of gain, what they're telling you is that whatever level the input signal has, the output signal can have 60 dB more -- the 60 dB is relative to the input signal. We know that the dB is logarithmic (based on exponents), so what does 60 dB mean in linear terms?

By definition, a dB is a*decibel*, meaning one-tenth of a *bel*, which is an archaic unit that nobody uses. But the bel is the simplest representation of a logarithmic scale: it is the logarithm of the ratio of two quantities. For example, imagine I have a radio transmitter that has 1 watt of transmitting power and my friend has one with 100 watts of power. What is the difference in power between the two? The ratio of powers is 100:1, so his radio has 100 times the power (linear scale) of mine. In the logarithmic scale, we take the base-10 log of 100: what power of 10 gives us 100? The answer is 2: log(100) = 2. This tells us that his radio has 2 bels more power than mine. But the bel is a bit too big for comfortable practical use, so we defined the decibel. If one radio is 2 bels more powerful than the other, and there are 10 decibels per bel, then that radio is 20 dB stronger than the other.

What does this teach us? A couple of things: specifically, that 100 times the power equals 20 dB greater power; more generally, if P2 is the power produced by one device and P1 is the power produced by another device, then a dB is equal to 10 * log(P2/P1). But notice that I'm specifically using the word*power*, and when describing preamp gain, we're talking about magnitudes of voltage not power. So what's the distinction?

To understand this first we need a quick lesson in electronics. You probably have heard of Ohm's Law, which relates voltage, current, and resistance/impedance in electrical circuits. The fundamental expression for Ohm's Law is that the current through a resistor is equal to the quotient of the voltage across the resistor and its resistance value. Mathematically, I = V/R. Another electrical fact is that the power dissipated by a resistor is equal to the product of the voltage across the resistor and the current through the resistor, or, mathematically: P = V*I. Replacing the current term I with its Ohm's Law expression we get P = V*V/R, or equivalently, P = V^2/R. In English, this states that power is proportional to the square of the voltage.

Back to the linear understanding of 60 dB of preamp gain. Earlier I said that dB in power equals 10 * log(P2/P1). But now I just claimed that P is proportional to V^2. This means that dB in voltage must be equal to 10 * log(V2^2/V1^2) which we can rewrite as 10 * log[(V2/V1)^2]. A wonderful mathematical property of logarithms is that the exponentiation reduces to multiplication, e.g.: log(2^3) = 3 * log(2). This means that our definition of voltage dB can be changed from 10 * log[(V2/V1)^2] to 20 * log(V2/V1).

So if my preamp says it has 100 times (linear) gain, how many dB is that? 20 * log(100) = 40 dB. That means that whatever the input signal is, if I crank the volume on the preamp, it will be 40 dB = 100 times more voltage than the original signal. To go the other way, from dB to linear, we can use the antilog formula: V2 = V1 * 10^(dB/20). For example, if our preamp has 60 dB of gain and I input a 1 millivolt signal, the output will be 0.001 * 10^(60/20) = 1 volt. Normalizing V1 to 1 we see that 60 dB of gain is equal to 1000 times the amplitude.

Notice that we've been using the "naked" dB; when the comparison is between two arbitrary things, such as the relative input and output levels of a preamp, we use dB without any suffix. However, certain standard measurements exist that we can compare things to. For example, you know that +4 dBu is the professional nominal operating level for line voltage. The little*u* is telling us that the measurement is relative to a standard measurement, namely 0 dBu = 0.775 volts rms. This needs to be perfectly clear: if you see a "naked" dB, it is the ratio between two arbitrary things; if you see a dB with a suffix, such as dBu (voltage) or dBm (power), it is the ratio of something to the standard defined by the suffix. So if I tell you that my preamp has 60 dB of gain and then ask you to tell me how much the voltage is at the output, you will correctly tell me "not enough information". All you can say is that whatever comes out will have 1000 times the amplitude of whatever goes in. But if I tell you that I have +4 dBu at the output of my preamp and I ask you what the voltage is, you *can* calculate a specific number: recalling the antilog formula, V2 = V1 * 10^(dB/20) we know that V2 is the output voltage, V1 is the reference voltage (0 dBu = 0.775 V rms), and we want +4 dB: V2 = 0.775 * 10^(4/20) = 1.23 V rms.

So in recap, a "naked" dB has no absolute meaning (though it can be worked out if enough information is known), while a "suffixed" dB always has an absolute meaning because it is in reference to some standard. The reason we have so many suffixes -- dBu, dBm, dBV, dBSPL, dBFS, etc. -- is because we have so many different kinds of things we want to measure. The only catch is to make sure you're referencing apples to apples. And for all that, it's hopefully clear that the simple reason we use logarithmic scales is because it makes the numbers easier to work with. With practice "thinking logarithmically" will become second-nature and you'll realize that it is indeed worth the effort.

Best of luck!

You've just created a linear graph: both the horizontal and vertical scales are linear. This means that between any two tick marks there is the same amount of "stuff" -- for example, between any two tick marks on the horizontal scale there is always one year's worth of time.

Now imagine that you're in audio engineering school and your professor has asked you to graph the relationship of loudness to certain events: a mosquito flapping its wing, a typical conversation, standing next to a jack hammer, and standing next to a nuclear explosion (if such a thing were possible). You do some research to find the pressure magnitude of these events and start building your graph, with the event on the horizontal axis and the pressure magnitude on the vertical axis. The horizontal axis is easy: four tick marks, each representing one of the events. But for the vertical axis, you realize you have a problem: the magnitude of a mosquito flapping its wing is on the order of 0.0001 pascals, the jack hammer around 2 pascals, and the nuclear explosion at about 100000 pascals. That means you need to show a range of 10^-5 to 10^5. If you make each tick equal to the smallest magnitude (10^-5), you'd need 10^10 ticks. That's 10

Obviously this won't do. You need something manageable, say 10 horizontal tick marks. But how do you break up such a huge range into 10 chunks? If you try to make it a linear scale -- where each chunk has the same amount of "stuff" in it -- you run into another kind of problem: If each tick is worth 100000/10 = 10000 pascals, then the very first tick contains all the events except the nuclear explosion. Worst still, the first tick has so much stuff in it that you won't have a visual indication of the difference in loudness between a mosquito flapping its wing and a jack hammer. Considering the very audible difference between the two, a linear scale is therefore a poor choice.

This is one situation where logarithmic scales shine. The range of stuff we want to graph is between 10^-5 and 10^5; as said before, this is 10^10 worth of stuff: you need 10^5 stuff to get from 10^-5 to 10^0, and then another 10^5 to get from 10^0 to 10^5. Iinstead of dividing this huge range into ten even chunks, why not use the ten exponents as the tick marks? This would give us, starting from the botton: -5, -4, -3, -2, -1, 0, 1, 2 ,3, 4, 5 (which is eleven ticks, but that's ok). The mosquito wing (0.00001 = 10^-5) would correspond to the -5 tick, the conversation (0.001 = 10^-3) with the -3 tick, the jack hammer (2 = a bit more than 10^0) with the 0 tick, and the nuclear explosion (100000 = 10^5) with the 5 tick. In this scale we can visually see that the jack hammer is halfway between the mosquito wing and the nuclear explosion, which makes pretty good sense.

So what exactly did we do there? If you notice, the stuff between ticks is not evenly distributed: the farther you go up (on the vertical scale; or to the right, on a horizontal scale) the more "stuff" there is between the ticks. In this case, the amount of stuff between the 0 tick and the 1 tick is ten units (10^0 = 1 and 10^1 = 10); the amount of stuff between 1 tick and 2 tick is almost a hundred units (10^1 = 10 and 10^2 = 100); the amount of stuff between the 2 tick and 3 tick is almost a thousand units (10^2 = 100 and 10^3 = 1000); and so on. The

Quick and easy math lesson: a logarithm is an exponent. Next time you see the word logarithm or logarithmic, just think to yourself "exponent". For example, what is the base-10 logarithm of 100000? A logarithm is just an exponent, so what exponent of 10 gives you 100000? 5, because 10^5 = 100000. Therefore log(100000) = 5. Simple as that. We can change the

The other important aspect of the dB is that it is a ratio; it tells us the magnitude of a measurement relative to

By definition, a dB is a

What does this teach us? A couple of things: specifically, that 100 times the power equals 20 dB greater power; more generally, if P2 is the power produced by one device and P1 is the power produced by another device, then a dB is equal to 10 * log(P2/P1). But notice that I'm specifically using the word

To understand this first we need a quick lesson in electronics. You probably have heard of Ohm's Law, which relates voltage, current, and resistance/impedance in electrical circuits. The fundamental expression for Ohm's Law is that the current through a resistor is equal to the quotient of the voltage across the resistor and its resistance value. Mathematically, I = V/R. Another electrical fact is that the power dissipated by a resistor is equal to the product of the voltage across the resistor and the current through the resistor, or, mathematically: P = V*I. Replacing the current term I with its Ohm's Law expression we get P = V*V/R, or equivalently, P = V^2/R. In English, this states that power is proportional to the square of the voltage.

Back to the linear understanding of 60 dB of preamp gain. Earlier I said that dB in power equals 10 * log(P2/P1). But now I just claimed that P is proportional to V^2. This means that dB in voltage must be equal to 10 * log(V2^2/V1^2) which we can rewrite as 10 * log[(V2/V1)^2]. A wonderful mathematical property of logarithms is that the exponentiation reduces to multiplication, e.g.: log(2^3) = 3 * log(2). This means that our definition of voltage dB can be changed from 10 * log[(V2/V1)^2] to 20 * log(V2/V1).

So if my preamp says it has 100 times (linear) gain, how many dB is that? 20 * log(100) = 40 dB. That means that whatever the input signal is, if I crank the volume on the preamp, it will be 40 dB = 100 times more voltage than the original signal. To go the other way, from dB to linear, we can use the antilog formula: V2 = V1 * 10^(dB/20). For example, if our preamp has 60 dB of gain and I input a 1 millivolt signal, the output will be 0.001 * 10^(60/20) = 1 volt. Normalizing V1 to 1 we see that 60 dB of gain is equal to 1000 times the amplitude.

Notice that we've been using the "naked" dB; when the comparison is between two arbitrary things, such as the relative input and output levels of a preamp, we use dB without any suffix. However, certain standard measurements exist that we can compare things to. For example, you know that +4 dBu is the professional nominal operating level for line voltage. The little

So in recap, a "naked" dB has no absolute meaning (though it can be worked out if enough information is known), while a "suffixed" dB always has an absolute meaning because it is in reference to some standard. The reason we have so many suffixes -- dBu, dBm, dBV, dBSPL, dBFS, etc. -- is because we have so many different kinds of things we want to measure. The only catch is to make sure you're referencing apples to apples. And for all that, it's hopefully clear that the simple reason we use logarithmic scales is because it makes the numbers easier to work with. With practice "thinking logarithmically" will become second-nature and you'll realize that it is indeed worth the effort.

Best of luck!

23rd November 2014

#**37**

Quote:

You've just created a linear graph: both the horizontal and vertical scales are linear. This means that between any two tick marks there is the same amount of "stuff" -- for example, between any two tick marks on the horizontal scale there is always one year's worth of time.

Now imagine that you're in audio engineering school and your professor has asked you to graph the relationship of loudness to certain events: a mosquito flapping its wing, a typical conversation, standing next to a jack hammer, and standing next to a nuclear explosion (if such a thing were possible). You do some research to find the pressure magnitude of these events and start building your graph, with the event on the horizontal axis and the pressure magnitude on the vertical axis. The horizontal axis is easy: four tick marks, each representing one of the events. But for the vertical axis, you realize you have a problem: the magnitude of a mosquito flapping its wing is on the order of 0.0001 pascals, the jack hammer around 2 pascals, and the nuclear explosion at about 100000 pascals. That means you need to show a range of 10^-5 to 10^5. If you make each tick equal to the smallest magnitude (10^-5), you'd need 10^10 ticks. That's 10

Obviously this won't do. You need something manageable, say 10 horizontal tick marks. But how do you break up such a huge range into 10 chunks? If you try to make it a linear scale -- where each chunk has the same amount of "stuff" in it -- you run into another kind of problem: If each tick is worth 100000/10 = 10000 pascals, then the very first tick contains all the events except the nuclear explosion. Worst still, the first tick has so much stuff in it that you won't have a visual indication of the difference in loudness between a mosquito flapping its wing and a jack hammer. Considering the very audible difference between the two, a linear scale is therefore a poor choice.

This is one situation where logarithmic scales shine. The range of stuff we want to graph is between 10^-5 and 10^5; as said before, this is 10^10 worth of stuff: you need 10^5 stuff to get from 10^-5 to 10^0, and then another 10^5 to get from 10^0 to 10^5. Iinstead of dividing this huge range into ten even chunks, why not use the ten exponents as the tick marks? This would give us, starting from the botton: -5, -4, -3, -2, -1, 0, 1, 2 ,3, 4, 5 (which is eleven ticks, but that's ok). The mosquito wing (0.00001 = 10^-5) would correspond to the -5 tick, the conversation (0.001 = 10^-3) with the -3 tick, the jack hammer (2 = a bit more than 10^0) with the 0 tick, and the nuclear explosion (100000 = 10^5) with the 5 tick. In this scale we can visually see that the jack hammer is halfway between the mosquito wing and the nuclear explosion, which makes pretty good sense.

So what exactly did we do there? If you notice, the stuff between ticks is not evenly distributed: the farther you go up (on the vertical scale; or to the right, on a horizontal scale) the more "stuff" there is between the ticks. In this case, the amount of stuff between the 0 tick and the 1 tick is ten units (10^0 = 1 and 10^1 = 10); the amount of stuff between 1 tick and 2 tick is almost a hundred units (10^1 = 10 and 10^2 = 100); the amount of stuff between the 2 tick and 3 tick is almost a thousand units (10^2 = 100 and 10^3 = 1000); and so on. The

Quick and easy math lesson: a logarithm is an exponent. Next time you see the word logarithm or logarithmic, just think to yourself "exponent". For example, what is the base-10 logarithm of 100000? A logarithm is just an exponent, so what exponent of 10 gives you 100000? 5, because 10^5 = 100000. Therefore log(100000) = 5. Simple as that. We can change the

The other important aspect of the dB is that it is a ratio; it tells us the magnitude of a measurement relative to

By definition, a dB is a

What does this teach us? A couple of things: specifically, that 100 times the power equals 20 dB greater power; more generally, if P2 is the power produced by one device and P1 is the power produced by another device, then a dB is equal to 10 * log(P2/P1). But notice that I'm specifically using the word

To understand this first we need a quick lesson in electronics. You probably have heard of Ohm's Law, which relates voltage, current, and resistance/impedance in electrical circuits. The fundamental expression for Ohm's Law is that the current through a resistor is equal to the quotient of the voltage across the resistor and its resistance value. Mathematically, I = V/R. Another electrical fact is that the power dissipated by a resistor is equal to the product of the voltage across the resistor and the current through the resistor, or, mathematically: P = V*I. Replacing the current term I with its Ohm's Law expression we get P = V*V/R, or equivalently, P = V^2/R. In English, this states that power is proportional to the square of the voltage.

Back to the linear understanding of 60 dB of preamp gain. Earlier I said that dB in power equals 10 * log(P2/P1). But now I just claimed that P is proportional to V^2. This means that dB in voltage must be equal to 10 * log(V2^2/V1^2) which we can rewrite as 10 * log[(V2/V1)^2]. A wonderful mathematical property of logarithms is that the exponentiation reduces to multiplication, e.g.: log(2^3) = 3 * log(2). This means that our definition of voltage dB can be changed from 10 * log[(V2/V1)^2] to 20 * log(V2/V1).

So if my preamp says it has 100 times (linear) gain, how many dB is that? 20 * log(100) = 40 dB. That means that whatever the input signal is, if I crank the volume on the preamp, it will be 40 dB = 100 times more voltage than the original signal. To go the other way, from dB to linear, we can use the antilog formula: V2 = V1 * 10^(dB/20). For example, if our preamp has 60 dB of gain and I input a 1 millivolt signal, the output will be 0.001 * 10^(60/20) = 1 volt. Normalizing V1 to 1 we see that 60 dB of gain is equal to 1000 times the amplitude.

Notice that we've been using the "naked" dB; when the comparison is between two arbitrary things, such as the relative input and output levels of a preamp, we use dB without any suffix. However, certain standard measurements exist that we can compare things to. For example, you know that +4 dBu is the professional nominal operating level for line voltage. The little

So in recap, a "naked" dB has no absolute meaning (though it can be worked out if enough information is known), while a "suffixed" dB always has an absolute meaning because it is in reference to some standard. The reason we have so many suffixes -- dBu, dBm, dBV, dBSPL, dBFS, etc. -- is because we have so many different kinds of things we want to measure. The only catch is to make sure you're referencing apples to apples. And for all that, it's hopefully clear that the simple reason we use logarithmic scales is because it makes the numbers easier to work with. With practice "thinking logarithmically" will become second-nature and you'll realize that it is indeed worth the effort.

Best of luck!

24th November 2014

#**38**

22nd April 2015

#**39**

23rd April 2015

#**40**

Quote:

-1 on the fader means it lowers the signal by 1dB, compared to the signal before the fader. So your +4 signal (measured before the fader, or the fader at zero) will become +3 afterwards (let's say measured at the master buss).

23rd April 2015

#**41**

Quote:

Zero on the fader just means, that you don't increase or decrease the signal's volume. It has no meaning to the actual volume in matter of measuring.

-1 on the fader means it lowers the signal by 1dB, compared to the signal before the fader. So your +4 signal (measured before the fader, or the fader at zero) will become +3 afterwards (let's say measured at the master buss).

-1 on the fader means it lowers the signal by 1dB, compared to the signal before the fader. So your +4 signal (measured before the fader, or the fader at zero) will become +3 afterwards (let's say measured at the master buss).

23rd April 2015

#**42**

Quote:

i think i get it enough now.....0 dB just means that it's like neutral....0 dB is like the zero point or equilibrium point..but when you lower the dB it means that your attenuating the power....so you reducing the amps/volts whatever, this actual number of volts/dB could be anything depending on the power source, as soon as you push it past 0 dB you are now raisng the power level by x amount of real dB....the dB on the fader is just a marker...like a fishing floater that goes up and down....so when you raise the fader by 1 dB its actually raising the real dB by 1.3333 or something dB in reality....and sucking more energy

You change the value logarithmically, that's dB.

Very simplified said: The amount dB alone doesn't mean much, it must always correlate to something else. In the case of the fader: Pre-fader volume.

That's why all those dB-values out there, like dBu, dBVU, dBFS, etc. have a real measurable value like e.g. Volt. They correlate to this fixed point. In case of the fader, the fixed point is the pre-fader volume. So the new signal behind is pre-fade signal + fader (very simplified)

PS: At least that's how my dB-rookie-mind looks at it heh

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