Quick dB question

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.)

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

dB= 20*log E1/E2.

Don

Have a look at

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:

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

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.

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

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