# db: Understanding Of ?

Status
Not open for further replies.

#### BOBRR

##### Member
Hello,

Have always had trouble in "understanding" db the right way.

e.g., when I look at attenuation charts for the various coax, the losses are always given in db.

Is this a db voltage, or db power loss ?
How would one know ?

e.g., if I see a loss at some freq. of 3 db, does this imply that I am "losing"
about 30 % of the signal, or,... ?

If someone might explain all of this, would be appreciative.

Thanks,
B.

#### dlwtrunked

##### Member
Hello,

Have always had trouble in "understanding" db the right way.
...e.g., if I see a loss at some freq. of 3 db, does this imply that I am "losing"
about 30 % of the signal, or,... ?

No. A loss of 3 dB (-3 dB, a loss is a negative) is a loss of 1/2 the signal so that 1/2 remains.
If you have a calculator, you calculate 10^(s/10) to see what the new signal is, where "s" is the loss or gain. A 3 dB loss is -3 dB so one calculates 10^(-3/10)=10^(-.3)=.50 (approx.) or 50% remains (a 50% loss) .
A 6 dB lost would be 10^(-6/10)=10^(-.6)=.25 or 25% of the signal remains (a 75% loss)
A 10 dB loss would be 10^(-10/10)=10^(-1)=.1 or 10% signal remains (a 90% loss)
A 3 dB gain would give 10^(3/10)=10^(.3)=2 (approx.) or twice the signal; while a 6 dB gain will give 4 times the signal and a 10 dB gain gives 10 times the signal (but 20 dB is not 20 times the signal, doing the calculation will see that it is 100 times the signal).

An advantage of using dB is one can add the dB of a series of gains and losses quickly to get a final loss or gain; for example a 10 dB gain followed by a 6 dB loss (-6 dB) will give 10+(-6)=4 dB so there will be 10^(4/10)=2.51or 2.51 times the original signal.

To change a loss or gain to dB one does 10*log10 (new/original), where log10 indicates the logarithm base 10, which is probably what your calculator has (do not use "ln" which is a logarithm in a different base). So for example, if a signal had a 1/3 loss (.33), then the new signal is .67 as much, so new/original=.67 and the calculation is 10*log10(.67)=-1.74 dB or 1.74 dB loss.

#### jwt873

##### Member
Is this a db voltage, or db power loss ?
How would one know ?

That can be confusing... Power dB is 10(log(P1/P2), but Voltage dB is calculated as 20(log(V1/V2). . So if you double the power you get a 3 dB increase and if you double the voltage you get a 6 dB increase.

But, the coax ratings are for power. (Transmitting). As dlwtrunked points out a -3dB loss per 100 feet of coax means that half the power will be lost in the cable.

#### dlwtrunked

##### Member
That can be confusing... Power dB is 10(log(P1/P2), but Voltage dB is calculated as 20(log(V1/V2). . So if you double the power you get a 3 dB increase and if you double the voltage you get a 6 dB increase.

It is complicated and the above is almost right...Actually, voltage dB is still 10(log(V1/V2) but if the impedance is held constant for this same example and you measure voltage, then the power (not voltage) dB is 20(log(V1/V2)). So where you said *Voltage dB* above, you should have said "Power dB".

https://en.wikipedia.org/wiki/Decibel
or
Decibel dB | Calculation Formula Definition | Radio-Electronics.com
and note he is talking bout power level throughout

This comes from the fact that P=EI=(E^2)/R (Where R is replaced by Z if non-resistive impedance). So
P1/P2=(E1^2)/(E2^2) (The R or Z's cancel). So
log(P1/P2)=log((E1^2)/(E2^2))=log(E1/E2)^2)=2log(E1/E2) (by the rules of logarithms) so
dB (power) =10*log(P1/P2)=20log(E1/E2)
And your 2nd sentence is "So if you double the power you get a 3 dB increase and if you double the voltage you get a 6 dB increase" should be "So if you double the power you get a 3 dB increase in power of if you double the voltage (and hold resistance or impedance constant you get a 6 dB increase in *power*".
Used to confuse me years ago and now I have a PhD in mathematics to fix that

Last edited:

#### lu81fitter

##### Member
GOOD GRIEF CHARLIE BROWN!!! (dlwtrunked) LOL! That just blew me away! It is interesting. I don't understand all of the mathematical jargon, but I understand coax loss better. Hope BOBRR got his question answered.

#### dlwtrunked

##### Member
GOOD GRIEF CHARLIE BROWN!!! (dlwtrunked) LOL! That just blew me away! It is interesting. I don't understand all of the mathematical jargon, but I understand coax loss better. Hope BOBRR got his question answered.

Nothing was above high school level algebra; but it may have been awhile. The main point in the second point is that the 20 appears when one calculates power dB from voltages but is a power dB and not a voltage dB

#### lu81fitter

##### Member
I don't remember any algebra like that in 1985, but you are correct.... It's been a while. Once again, hope BOBRR is better informed. Don't want to hijack his thread.

#### dlwtrunked

##### Member
I don't remember any algebra like that in 1985, but you are correct.... It's been a while. Once again, hope BOBRR is better informed. Don't want to hijack his thread.

The second post was really note for him but my first post was.

Status
Not open for further replies.

Replies
15
Views
1K
Replies
18
Views
829
Replies
9
Views
1K
Replies
24
Views
1K
Replies
1
Views
396
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}