Re: 3 dB bandwidth

Guy Macon wrote:

> Tony Williams wrote:
> >
> > keith <krw@xxxxxxxxxx> wrote:
> >
> >> You're as pig-ignorant as Guy! Decibels are a log ratio of
> >> *anything*. Crap, what are they teaching engineers these days?
> >
> > The Decibel is a Unit of Attenuation, defined in
> > terms of the ratio of power levels only.
> >
> > Attenuation = 10 * log10 (Pin/Pout) in Decibels.
> >
> > If (and only if) the input/output impedances are identical,
> > then the Attenuation (in Decibels) can be calculated as the
> > log10 ratio of the *square* of the currents or voltages.
> >
> > 2 2
> > Attenuation = 10 * log10 (Vin/Vout) in Decibels.
> >
> > Or, Attenuation = 20 * log10 (Vin/Vout) in Decibels.
> > /|\
> > |____ Notice where the squares went to?
> >
> > That calculation using (Vin/Vout) produces units of dB
> > only when the input/output impedances are identical.
> >
> > For example. An audio power amplifier requires an input
> > of 1mW into 600 ohms, for an output of 1W into 15 ohms.
> > The voltage/current ratios are 0.77/3.87 and 1.29/258.
> >
> > 10 * log10 (1mW/1W) is -30, Decibels of Attenuation.
> >
> > 20 * log10 (0.77/3.87) is -14, Units of Nothing.
> >
> > 20 * log10 (1.29/258) is -46, Units of Nothing.
> >
> > Only the power ratio produces legitimate units of dB.
>
> <applause>
>
> --------------------------------------------------------
>
> Adrian Tuddenham wrote:
> >
> >keith <krw@xxxxxxxxxx> wrote:
> >
> >> You're as pig-ignorant as Guy! Decibels are a log ratio of *anything*.
> >
> >Have a look at "Radio Designer's Handbook" by F. Langford-Smith (first
> >published 1934). In my 4th Edition copy (1953) the whole of Chapter 9
> >is devoted to the subject of decibels.
> >
> >It is quite clear that only the power ratio is directly described by
> >decibels, if you wish to express the ratio of other electrical
> >quantities in decibels you "must involve the resistance". (Page 807)
>
> <more applause>

So you managed to round up some clots to support your *cause* did you. You
can't *have* watts without it involving resistance you ignorant bugger ! No
resistance = no watts. Period. Then apply the VIR triangle to resolve the
equations in different units.

I'm sorry you're all *WRONG*. End of discussion.

Graham


.

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