Re: Re: something like dB

• To: mathgroup at smc.vnet.net
• Subject: [mg56631] Re: [mg56629] Re: something like dB
• From: Chris Chiasson <chris.chiasson at gmail.com>
• Date: Sun, 1 May 2005 00:46:27 -0400 (EDT)
• References: <200504300528.BAA23976@smc.vnet.net>
• Reply-to: Chris Chiasson <chris.chiasson at gmail.com>
• Sender: owner-wri-mathgroup at wolfram.com

```The dB measurement, as you are most likely aware, comes from the
logarithm of a ratio. One of the terms in the ratio might be a
standard (20 micro Pascal, or whatever). You could try keeping that

Try using the Notation package.

Maybe something like (pseudo code .. don't forget to use the palette):

Notation[x_ dB <==> dB[x_]]

then, assuming you're not working with equations... just expressions
(again, pseudo code):

FromdBRule=dB[x_]->reference*10^(x/20)

FromdBRule=num_->20 Log[10,num/reference] (*be careful with this rule...*)

Regards,

> On 4/29/05 at 3:20 AM, djmp at earthlink.net (David Park) wrote:
>
> >It is only sloppyness to say that something is x dB. (Maybe I'll
> >hear differently from other responders.) In any case, if you use
> >0.0 dB then the 0.0 will be retained, but if you write exact 0
> >anything Mathematica always returns 0.
>
> No, it isn't just sloppyness to use dB. For example, it is perfectly logical to talk of an attenuation or gain of say 3 dB which would mean for attenuation half of the input power is lost. Linear amplifiers increase power by a fixed ratio for a given setting and attenuators decrease power by a fixed ratio.
>
> This is made even more useful by measuring power levels in units like dBm. Here the m tells me the power level is referenced to 1 mW. So, 0 dBm would be 1 mW of power. And with an input of 0 dBm and an amplifer with a gain of 30 dB, I can easily determine the output power is 30 dBm or equivalently 1 Watt.
> --
> To reply via email subtract one hundred and four
>
>

--
Chris Chiasson
http://chrischiasson.com
Kettering University
Mechanical Engineering
1 810 265 3161

```

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