[Nezer]

Now, Tangent you have my ass all confused on the dB to G relationship. I thought it was :

dB=20 log(G)

Where G=5 is about 14dB and where G=10 is 20? This gives a differenc of 6dB.

The formula that matchs your values (in the other thread I decided to keep this discussion out of) is

dB=10 log(G)

which is only useful when G is a watts value.

Uggh! Now I'm so confused again!!

So, in our case, calculating *voltage* gain, which of these formulas is accurate:

dB=10 log(Vout/Vin)

or

dB=20 log(Vout/Vin)

*when dealing with amps the Vin is assumed as 1 becasue we are more intrested in the gain ratio. So a gain of 5 would mean 5 volts out to every 1 volt in.

Would one of you creatures that is infinity more intelligent than I please clear this mess up for me!?

And while your at it, could you also invert the equation so that I can calculate voltage gain based on a dB?

 

[Nezer]

From another thread:

Quote:

Originally posted by tangent The inverse of a logarithm is exponentiation. If: X = log_subY_(Z) then: Z = Y^X The subY part is the "base" of the logarithm. Since decibels are figured using base-10 logs, the inverse function is: dB = 10 ^ (G / 20) That is, divide the gain value you calculated by 20, and then raise 10 to that power. Any calculator with a base-10 log function will have a 10^x function as well .... and now you know why!

Don't you mean:

G = 10^(dB/20)

As I've already got the formula with dB on the LHS.

So, log (x) is just shorthand for (assuming base-10 log): 10^(x)?

Sorry for having to have the math lesson on this board.

 

[slindeman]

Here you go: http://et.nmsu.edu/~etti/fall96/comm...ons/db/db.html

So the correct equations are:

dB = 20log(G) [20 multiplied by the base-10 log of G]
G = 10^(dB/20) [10 raised to the power of dB/20]

where G = gain or Vout/Vin

 

[tangent]

Sorry for the confusion, Nezer: I wrote that out on the fly without checking it. Slindeman's got the right equations, and I fixed that other thread.

 

[Nezer]

Quote:

Originally posted by tangent Sorry for the confusion, Nezer: I wrote that out on the fly without checking it. Slindeman's got the right equations, and I fixed that other thread.

I'm just glad to get it cleared-up and have an understanding of what a log is (mathmatically speaking). Like I've said 899+101 times before, math is NOT my strongest subject.