>> So the reduction in SNR (assuming equal noise power) is
>> 3.01 dB, not 6 dB.
>That's the best case if the noise power is equal. If the
>"other" receiver has higher noise power (wider bandwidth,
>more interfering signals, etc.) the S/N reduction is greater.
>Even 3 dB reduction in S/N is a
Tayloe Dan-P26412 wrote:
However, if the situation is a weak signal situation where
the receiver noise floor is at least partially masking
the desired signal, we have a different situation. Band
noise and stations on the band will both be correlated coming
out of both receivers and thus get
On Nov 17, 2008, at 10:06 AM, Tayloe Dan-P26412 wrote:
Band noise from one receiver at
any instant in time will look exactly like band noise from
the second receiver.
That is true if the two receivers are tuned to the same passband and
you are using an identical antenna for the two receive
eases.
Receiver noise being a limiting factor happens either under
very quite band conditions, or when a very poor antenna is being
used.
- Dan, N7VE
-Original Message-
From: Kok Chen [mailto:[EMAIL PROTECTED]
Sent: Monday, November 17, 2008 12:55 PM
To: Elecraft Reflector
Cc: Tay
On Nov 17, 2008, at 1:01 PM, Tayloe Dan-P26412 wrote:
How can it hurt the SNR?
Because in split operation, you are adding the noise from two
different bandpasses, yet you are only hearing the signal from the
original single receiver.
Let receiver (1) hear s(t) + n1(t) and receiver (2) h
On Nov 17, 2008, at 3:01 PM, Kok Chen wrote:
Prob( (s+n1+n2)^2 ) = Prob(s^2 + n1^2 + n2^2 + n1.n2 + s.n1 +
s.n2). If s, n1 and n2 are uncorrelated, then Prob(n1.n2) =
Prob(s.n1) = Prob(s.n2) = 0.
Whoops, that should be
Prob( (s+n1+n2)^2 ) = Prob(s^2 + n1^2 + n2^2 + 2.n1.n2 + 2.s.n1 +
6 matches
Mail list logo