Decibel Equation

[Dave H]

...and e is approximately equal to 2.7 something. This number has some elegant significance when using calculus.

That is probably because the exponential function is its own derivative. ie. The derivative of e^x = e^x. The rate of change (derivative) is proportional to the value of the function. eg. Bacteria grow at a rate proportional to the size of the population (exponential growth).

[Shadrock2]

Thank You R. G. for that explanation. I have looked through the Negative Feedback chapter in RDH4 and couldn't quite come up with that straight forward explanation. It makes more sense to me now

[trem]

OK...Gee Whiz...I am still having problems.
Another formula is:
Attenuation (dB) = 20 x log10 x R2/R1 + R2

The example given is confusing then (to me).
They say .....if R1 = R2 the signal is attenuated by 6dB
Well.....if R1 = 100 Ohms and R2 = 100 Ohms
100/100 + 100 = 100/200 = 0.50
Is that right so far.?
If the multiplier is 20db and the log10 (after doing the above math) = 0.50
Wouldn't the answer be:
20dB x 0.50 = 10dB

The book says:
R1 = R2
E(out) = 0.50 x E(in)
Then the signal is attenuated by 6dB.
What am I doing wrong.? Where do they get the 6dB.?
Thank You
(i am assuming i copied this correctly)

[eschertron]

OK...Gee Whiz...I am still having problems.
Another formula is:
Attenuation (dB) = 20 x log10 x R2/R1 + R2

The example given is confusing then (to me).
They say .....if R1 = R2 the signal is attenuated by 6dB
Well.....if R1 = 100 Ohms and R2 = 100 Ohms
100/100 + 100 = 100/200 = 0.50
Is that right so far.?

Yes.

If the multiplier is 20db and the log10 (after doing the above math) = 0.50

No.
we got to 100/(100+100) = 0.5
but then failed to take the Log(base10) of 0.5 which is equal to -0.301029995663981

and 20 * -0.301029995663981 will equal right around -6dB

Keep in mind the 6dB (for voltage) and the 3dB (for power) doubling rules are no more than handy engineer's thumbrules, and are are just approximations.

[loudthud]

I had to compute dB in a BASIC program for a computer that didn't do Log base 10. I used natural Log:

Log10 (X) = ln (X) / ln (10) ... I think that's right, correct me if I'm wrong.

Made the program run slow but it plotted the right data.

[trem]

10-4
Thank You
I will absorb these ratios.
Thanks So Much (again)

[eschertron]

I had to compute dB in a BASIC program for a computer that didn't do Log base 10. I used natural Log:

Log10 (X) = ln (X) / ln (10) ... I think that's right, correct me if I'm wrong.

Made the program run slow but it plotted the right data.

LT,
I hope you're taking a trip down memory lane with that recollection
I always enjoyed making old x86-based computers do computationally-intensive numerical methods in BASIC. My favorite was plotting different fractal patterns. Sometimes took a hefty chunk of an hour to run a sequence, then I'd tweak a constant and run the thing again...

[trem]

I had to compute dB in a BASIC program for a computer that didn't do Log base 10. I used natural Log:

Log10 (X) = ln (X) / ln (10) ... I think that's right, correct me if I'm wrong.

Made the program run slow but it plotted the right data.

Sorry.....can you guys Dick and Jane this for me.? I am not sure what all the nomenclature represents.
Log10(x) is that Log10(1/2) .?
I do not know what to do with the second half of the equation .....In(x) divided by In(10)
How do we come up with ..... Log10(0.50) = -0.301
Thanks

[loudthud]

Sorry.....can you guys Dick and Jane this for me.? I am not sure what all the nomenclature represents.
Log10(x) is that Log10(1/2) .?
I do not know what to do with the second half of the equation .....In(x) divided by In(10).
Thanks

X is just any number (except zero). Log10 is log base 10. When computers ran on what were essentially teletypes, there was no handy way to make subscripts and superscripts, so ways were devised to just use simple ASCII characters. This was used on many early computer languages. On my handy Windows calculator you just punch the key marked "log". The "Log base 10" is implied but sometimes when you want to be crystal clear you might say Log10. It's not In (eye en), its ln (EL en). The font makes them look almost the same. Ln is the "Natural Logarithm" or Log base e which is 2.7 something.

IF you want to know Log{base A} of X, it equals Log{base B} of X divided by Log{base B} of A. It's an easy way of computing Logs in any base.

Extra credit: What is the Log base 2 of the base 10 number 0.25 ???

[eschertron]

Sorry.....can you guys Dick and Jane this for me.?

In my high school algebra class, inside the back cover of our clay tablets , there were "log tables" inserted. We'd take one number (0.5 for example) and look up another number (-3.01 for example). The difference between a Log (or Log base 10) table and a Ln (or log base 'e') table is that the number thus looked up is a different number. Slide rules work the same way, allowing astronomically large (or microscopically small) numbers to be computed within the space of a few short "look ups". All of that is lost to me now, and I simply cling to the dB = 20*log(V2/V1) or dB = 10*log(P2/P1) which allow me to convert audio signals to dB relationships. Everything else is rocket science.

Just know that there is a difference between base 10 logarithms and base e (or any other base, although base 2 is becoming quite common ), so that you hit the correct key on the calculator.

[Shadrock2]

"Sorry.....can you guys Dick and Jane this for me.? I am not sure what all the nomenclature represents.
Log10(x) is that Log10(1/2) .?
I do not know what to do with the second half of the equation .....In(x) divided by In(10)
How do we come up with ..... Log10(0.50) = -0.301
Thanks"

Trem... You are have trouble understanding the properties of exponents or specifically negative exponents.

There is a web site "Kahn Academy" that can be extraordinarily helpful in understanding math concepts. His explanations are as straightforward as any I have seen. And ultimately thorough. I have used this site myself occasionally.
The site is https://www.khanacademy.org/math/ Or Google it.
I would recommend watching the videos on the properties of exponents in Algebra 1 and on logarithms in Algebra 2

[trem]

I appreciate the link, and will save it, thanks.
What I do not understand (at this point) is the math that equates Log10(1/2) with 0.301.
I also do not understand the equation:
IF you want to know Log{base A} of X, it equals Log{base B} of X divided by Log{base B} of A.
If Log {base A} = 10, and X = 0.50
what is, or how do I find ..... Log{base B} .?

[Dave H]

What I do not understand (at this point) is the math that equates Log10(1/2) with 0.301.

It's not 0.301 it's -0.301. Enter 0.5 (1/2) into the Windows calculator (say) and press the log key and it gives the answer -0.301. ie. log10(1/2) = -0.301.

It's that way because 0.5 = 10^(-0.301). If x is any non zero number then by definition x = 10^log10(x) for logs base 10. similarly x = 2^log2(x) for logs base 2 and x = e^ln(x) for logs to base e.

If you want to know the log to base A of a number (x say) but only have log tables to base B then you can use the formula loudthud posted.

LogA(x) = LogB(x) / LogB(A)

[trem]

Ah...OK...that is Where/Why that formula comes from.
Thanks

[Dave H]

Ah...OK...that is Where/Why that formula comes from.
Thanks

Then you should be able to get extra credit by using loudthud's formula to get log base 2 of 0.25. Look up log10 of 0.25 then divide that by log10 of 2. What do you get?