On Dec 8, 2:01 pm, Jim Black <fmla...@organization.edu> wrote:
> On Fri, 7 Dec 2007 16:50:31 +0000 (UTC), Efthimios wrote:
> > On Dec 7, 4:01 am, "Rich L." <ralivings...@sbcglobal.net> wrote:
> >> On Dec 6, 12:37 pm, Efthimios <eangelopou...@gmail.com> wrote:[/color][/color]
>
> >>> Hallo to all,[/color][/color]
>
> >>> I have been reading a paper regarding electric pulse attenuation and
> >>> the values are quoted in dB. The conversion to dB was done by
> >>> multiplying the value with (20log10e) where "e" is the base of the
> >>> natural logarithm.[/color][/color]
>
> >>> I see this conversion to dB fist time. I can not find any information
> >>> in my math books or the web.[/color][/color]
>
> >>> Does any of you have any information regarding this that I can read so
> >>> I can understand it better?[/color][/color]
>
> >> "dB" is short for "decibel". It is a logarithmic power scale relative
> >> to some reference power. The most general formula is:[/color][/color]
>
> >> dB = 10 Log10( P1/P0)[/color][/color]
>
> >> Where P1 is the power being described and P0 is the reference power
> >> level. On this scale zero dB means the power is equal to the
> >> reference power level. 10 dB means the power is 10 times the
> >> reference level. 20dB means it is 100x the reference level. etc.[/color][/color]
>
> >> When used often the reference level is not stated explicitely. Really
> >> it ought to be. In low power RF (radio frequency) electronics they
> >> often use dBm meaning decibels wrt 1 milliwatt. You also sometimes
> >> see dBuW (wrt to 1 micro Watt) and dBW (wrt 1 Watt). As electrical
> >> signals are often measured in volts, you often see decibels defined
> >> as:[/color][/color]
>
> >> dB = 20 Log10 (V1/V0)[/color][/color]
>
> >> This is correct assuming that the impedance of the circuit with V1 and
> >> V0 are the same, and that the power is related to V^2/R. If the
> >> impedances are not the same (e.g. between the input and output of an
> >> amplifier) then this formula is not correct. Decibels are always a
> >> power ratio expressed as a log.[/color][/color]
>
> >> There is more on this in the Wikipedia article on
> >> "decibel" (
www.wikipedia.org)[/color][/color]
>
> >> Rich L.[/color][/color]
>
> > Thanks Reich, but have you seen the expression (20log10e) where "e" is
> > the base of the natural logarithm.?
> > This Log is multiplied with the value of the attenuation to convert it
> > to the logarithmic dB.[/color]
>
> > I am trying to see where this particular strange (20log10e) expression
> > of comes from. This does not look like the normal dB expression that
> > you mentioned above.[/color]
>
> > Brgds[/color]
>
> > Efthimios[/color]
>
> Most likely it comes from the fact that
>
> 20 log10(V1/V2) = 20 log10(e) * ln(V1/V2).
>
> Note that if V1 is very close to V2, ln(V1/V2) is approximately (V1-V2)/V1.
>
> --
> Jim E. Black- Hide quoted text -
>
> - Show quoted text -[/color]
Efthimios,
I got the paper you sent, and Jim Black has the correct answer. The
parameter they are working with is the propagation constant, gamma =
alpha +i*beta. The voltage attenuation they are calculating is:
V = V0*e^(alpha*x)
So the voltage ratio is:
V/V0 = e^alpha
To calculate the dB attenuation per unit length:
dB = 20*log10(V/V0) = 20*Log10(e^alpha) = 20*Log10(e)*alpha
Which is what is shown in the paper. The factor Log10(e) comes about
because dB is defined with Log10, but alpha in the paper was defined
for an exponential of the natural base (e).
Rich L.