1
$\begingroup$

why if I lose 80% power in the output of a system, Does it mean that I had lost 10×log10(0.2) = -7dB ?

$\endgroup$
  • 3
    $\begingroup$ is it just me or does this question seem trivial? $\endgroup$ – CyberMen Apr 27 '12 at 18:44
  • 1
    $\begingroup$ @CyberMen Please see (dsp.stackexchange.com/faq) $\endgroup$ – Spacey Apr 28 '12 at 2:27
  • 1
    $\begingroup$ @0x90: I'm finding it hard to delete this question because the accepted answer has so many upvotes. That means lots of people thought the answer was interesting... so perhaps the question isn't as trivial as CyberMen suggests? $\endgroup$ – Peter K. Jul 30 '15 at 14:13
9
$\begingroup$

If you have a system that causes 80% of its input power to be lost, then 20% of the input power will remain at the output. The power gain of the network is therefore:

$$ G = \frac{P_{out}}{P_{in}} =0.2 $$ $$ G|_{dB} = 10 \log_{10}(G) = 10 \log_{10}(0.2) \approx -7 \text{ dB} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.