Noise figure of a 10 dB spliter (or x-dB splitter)

nf = 10 log (si/ni/so/no)

I always hear that noise figure (at the coupled -10 dB port) of a splitter is NOT 10 dB!!!!
I understand if the splitter is not a resistive splitter (no sources of noise, like branch line splitter as an example), then additive noise should be lower. Which lead to LOWER noise figure.
BUT, this contadicts the basic fact that a lossy element has a noise figure equal to its loss (at room temp).

Is not this contradicting?
Could it be that the termination resisotr adds noise to the couple port that makes the port have a noise figure of 10 dB?

This was a long time debate with a friend...

Any comments are welcome...

splitter noise figure

In fact,the transport and coupled port has the same noise floor, the different is the signal amplifude decrease 10dB in the coupled port, and when using noise figue instrument measure the coupled port noise, it shows the noise figue 10dB.

splitter noise

Ok. But if it is not resistive, how come the noise floor be the same on both ports?
Is not the noise coming IN at the input will get attenuated by the 10dB of the coupled port? same as the signal? How come the signal gets attenuated, but not the noise?

The basic noise figure equations show that the noise figure represents the ADDITIVE noise due to the device itself, and it adds to the input noise, thus the output noise is always GREATER than the input noise*Gain.
NF=(Si/Ni)/(So/No)=(Si/Ni)/(G*Si/(G*Ni+Na)).

In short, if the device itself does not add noise due to internal noise sources, then the noise figure SHOULD be 1, unless we treated the coupler as a resistive device that has the nominal thermal noise power of a resistor.

Sorry for the long comment, but any explanations will be helpful..

Thanks in advance..

3 db coupler noise figure

Dear,

if the noise sources in the splitter are thermal noise sources then spliiter NF is equal to splitter loss, 1\L (L<1), (at T=T0).
The noise power at the splitter input port is k*T0*B (thermal noise): this is attenuated (by the losses) to K*T0*B*L at the output port, but the splitter adds itself k*T0*B*(1-L) at the output port. The total output noise power is K*T0*B. Calculating the NF, as (Si/Ni)/(So/No), you obtain NF=1/L (>1) !

This is true only for thermal noise sources; if the noise sources are not of thermal type the output noise power added by the network is k*T*B*(1-L), where T may be greater or less than that of the network itself.


Bye

coupleur noise figure

when the input signal noise power is very small, the noise make up of half AM noise and half FM noise,the AM part can be attenuatted,FM can not be. thus the attenuation of signal and noise is diffrent.
in addition, when the input signal noise power is the minimal -174dBm/Hz(room temperture),it can not be attenuate.

noise figure of coupled port on coupler

I propose another view.
1.The 10 dB coupler is matched at all ports (Sii=0). If it is a one arm coupler then it is also lossy ( there is at least one resistive element in it ).
2. If the direct output port and the coupled line's not coupled port are terminated, then (seeing it from outside) there cannot be seen any difference between a 10 dB attenuator and a 10 dB coupler (loss 10 dB, rellection factor zero), so the noise properties / noise figures should be the same.

noise figure couplers

marcomdd said:

Dear,

if the noise sources in the splitter are thermal noise sources then spliiter NF is equal to splitter loss, 1\L (L<1), (at T=T0).
The noise power at the splitter input port is k*T0*B (thermal noise): this is attenuated (by the losses) to K*T0*B*L at the output port, but the splitter adds itself k*T0*B*(1-L) at the output port. The total output noise power is K*T0*B. Calculating the NF, as (Si/Ni)/(So/No), you obtain NF=1/L (>1) !

This is true only for thermal noise sources; if the noise sources are not of thermal type the output noise power added by the network is k*T*B*(1-L), where T may be greater or less than that of the network itself.


Bye

Click to expand...

I want to know how to understand "but the splitter adds itself k*T0*B*(1-L) at the output port"

thanks

how noise is 10db

sjyatou said:

marcomdd said:

Dear,

if the noise sources in the splitter are thermal noise sources then spliiter NF is equal to splitter loss, 1\L (L<1), (at T=T0).
The noise power at the splitter input port is k*T0*B (thermal noise): this is attenuated (by the losses) to K*T0*B*L at the output port, but the splitter adds itself k*T0*B*(1-L) at the output port. The total output noise power is K*T0*B. Calculating the NF, as (Si/Ni)/(So/No), you obtain NF=1/L (>1) !

This is true only for thermal noise sources; if the noise sources are not of thermal type the output noise power added by the network is k*T*B*(1-L), where T may be greater or less than that of the network itself.


Bye

Click to expand...

I want to know how to understand "but the splitter adds itself k*T0*B*(1-L) at the output port"

thanks

Click to expand...

This could be demonstrated, but not possible here in the forum.

Now I don't remember the title of the book I read some years ago.