Quick dB question

[Grey]

How do I calculatre if I have a 30v signal and I reduce it by -6dB?

Graham

 

[John Miller]

Grey wrote:

How do I calculatre if I have a 30v signal and I reduce it by -6dB?

To reduce it by -6 dB, multiply times four. (Although you probably intended
to reduce it by 6 dB, which would mean dividing by four.)

--
John Miller
Email address: domain, n4vu.com; username, jsm

Punishment becomes ineffective after a certain point. Men become
insensitive.
-Eneg, "Patterns of Force", stardate 2534.7

 

[Dbowey]

Grey posted:

<< How do I calculatre if I have a 30v signal and I reduce it by -6dB? >>

You will have a 15V signal after you do that.

dB= 20*log E1/E2.

Don

 

[Graham Knott]

Grey wrote:

How do I calculatre if I have a 30v signal and I reduce it by -6dB?

Graham

Have a look at

http://homepage.ntlworld.com/g.knott/elect498.htm

 

[Gareth]

John Miller wrote:

Grey wrote:


How do I calculatre if I have a 30v signal and I reduce it by -6dB?


To reduce it by -6 dB, multiply times four. (Although you probably intended
to reduce it by 6 dB, which would mean dividing by four.)

That is, of course, you divide the POWER by four. The voltage will be
reduced by a factor of two since power is proportional to voltage squared.

Gareth.

--
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To reply to me directly:

Replace privacy.net with: totalise DOT co DOT uk and replace me with
gareth.harris

 

[John Miller]

Gareth wrote:

That is, of course, you divide the POWER by four. The voltage will be
reduced by a factor of two since power is proportional to voltage squared.

Yes, exactly right. I shouldn't post before waking up.

--
John Miller
Email address: domain, n4vu.com; username, jsm

I was the best I ever had.
-Woody Allen

 

[Jamie]

lets see if my math will pop to my head.
(20 * log(30))-6 = 15 DB
of course you can put that back into the V signal after.
V = InvertLog(15/20) = 5.6; etc///////////////

which should give you something to work with..
to the best of my knowledge that should be close
enough.


Grey wrote:

How do I calculatre if I have a 30v signal and I reduce it by -6dB?

Graham

 

[john jardine]

"Grey" <grahame9@btinternet.com> wrote in message
news:cheldu$3uj$1@sparta.btinternet.com...

How do I calculatre if I have a 30v signal and I reduce it by -6dB?

Graham


You need to find what "-6dB" means in a real world so ...

Divide "-6" by the magic number of "20"
Get "-0.3"

Find the antilog of "-0.3" (ie press the 10^x button)
Get "0.501187"

This is how much the 30V needs reducing.
So ... 30V X .501187 = 15.035V

This works for volts and amps. People normally just regard the -6dB as "a
half" or "+6dB" as "2 times".

regards
john

 

[Tim Auton]

jeffm_@email.com (JeffM) wrote:

To reduce it by -6 dB, multiply times four.
(Although you probably intended to reduce it by 6 dB,
which would mean dividing by four.)
John Miller

Professor: Two negatives give a positive result,
but two positives don't result in a negative.

Voice from back of classroom: Yeah, right.

That's a (mis) quote of the late Sidney Morgenbesser. OK, perhaps he
wasn't the only person to say it, but he's arguably the most famous.

"The most celebrated Morgenbesser anecdote involved visiting Oxford
philosopher J. L. Austin, who noted that it was peculiar that although
there are many languages in which a double negative makes a positive,
no example existed where two positives expressed a negative. In a
dismissive voice, Morgenbesser replied from the audience, "Yeah,
yeah..."

(from the New York Sun website, but it's elsewhere too)


Tim
--
Guns Don’t Kill People, Rappers Do.