Forum: GNU Radio Noisy source block

048ed9e907d8d46dbb96d0c73bba3261?d=identicon&s=25 猪猪猪猪猪猪头 (Guest)
on 2014-04-15 05:19
(Received via mailing list)
higuys
why the unit of amplitude in Noisy source block is volt?
how to explain this ?
thanks,
BZS
B4ffbc711addde4c649b1ed526df6157?d=identicon&s=25 Martin Braun (Guest)
on 2014-04-15 10:32
(Received via mailing list)
On 04/15/2014 04:46 AM, 猪猪猪猪猪猪头 wrote:
> hi,guys
> why the unit of amplitude in Noisy source block is volt?
> how to explain this ?
> thanks,
> BZS

Not sure what your question is. Signal amplitudes are usually in Volts,
that's just the right physical unit.

I assume you're asking because in the digital domain, there's really no
volts, and that's a good point. The unit could also just be 1. This is
something you sometimes see in (especially older) DSP textbooks.

There's a similar thing, where the mag-square of a signal is often
considered its instantaneous power (which, would assume a resistance of
1 \Ohm if you're still thinking in physical power).

So, just ignore this if it confuses you.

Martin
7d89a70df32c0ae27c1235016f9e5441?d=identicon&s=25 Marcus Müller (Guest)
on 2014-04-15 12:06
(Received via mailing list)
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Hi 猪猪猪猪猪猪头,

you've replied to the wrong thread; however:

dB is *always* a relative measure. You can't output dB; it always has
to be dB in relation to something.

The parameter of the noise source is amplitude.
Like Martin, if I *want* to stick a physical entity to it, it would be
voltage. Usually, I don't -- the noise source just outputs a
realization of a complex random process; it's just numbers without unit.

I'm under the impression you're mixing up absolute noise power, noise
variance, and SNR; for SNR, which is a ratio, dB does indeed make
sense. However, you don't get a noise source that generates a certain
SNR, because that depends on the signal power, and is something that
you/your application has to calculate.

Greetings,
Marcus

> hi ,Martin Braun well ,in my impression, the unit of is noisy is
> dB,so i feel fuzzy
about the volt,
> thanks for your reply!
>


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