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How to calculate the noise figure of active antenna?

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fly_fish

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A dual-band antenna with one feed integrates with two LNA.

C/N C/N1
Antenna-------->Power Divider -----------> LNA1
-----------> LNA2
C/N2
Supposing :
Power Divider has no insert loss,
Antenna has equal gain for both band;
LNA1 and LNA2 have equal noise figure.
C/N and C/N1 is measured in f1 band ,C/N2 is measured in f2 band.
Question:
Does C/N equal to C/N1 and C/N1 equal to C/N2?
 

Please provvide more comprensive info. A picture will be appreciated.

Please explain if the Power splitter act as a diplexer!
 

Hi,
The noise figure may be calculted with respect two different frequencies. It shall be greater than the NF at the input of the antenna. An ideal splitter introduces 3dB of NF for each channel.
B R M
 

I'm sorry for not posting a picture to explain more comprensively. I just can't do it successfully. So I try to describe it in such:
The power divider doesn't act as a diplexer. It divides the signal from antenna into two ways,each fed into a LNA.
LNA1 and LNA2 work on two different frequencies.

An ideal splitter (insert loss is odB) will introduce 3dB power loss for noise as well as for signal. So an ideal splitter introduces 0dB of NF for each channel.
Is it right?
 

fly_fish said:
An ideal splitter (insert loss is odB) will introduce 3dB power loss for noise as well as for signal. So an ideal splitter introduces 0dB of NF for each channel.
Is it right?
Click to expand...

Right.

Take in consideration just a branch of the splitter, closing the third port on 50 Ohm.

If the loss of a branch (from port "in1" to port "out") of the splitter is 0 (ideally), the thermal noise added is 0, so the NF is 0.

Wrong ! :cry: Wrong ! :cry: Wrong ! :cry:
 

ok, now it's more comprensible, Are you speaking about an academic exercise or a real case?
In any case, the two way power splitter act sometime as a magic.

In general case, the path loss is almost 3 dB, so it act as an attenuator; if the splitter is placed at 290K of phisical temperature, it rise the LNA NF of 3 dB's.

The magic appear in the case you split signal, send to 2 identical LNA's and then sum the signals together; if the paths are identical in Gain, Delay and Phase and the splitter have no intrinsec loss (i.e. 3 dB only) the NF of the system is the same of one LNA.
 

hello,
marcomdd, I am afraid that taking only one branch of the splitter will increase the NF in 3dB (as sergio notices).

consider the following: put the splitter with the unused (match-loaded) branch in a black-box, so that you have only two accesible ports. then, the insertion loss of your setup is 3dB, and therefore the noise figure of the black-box is 3dB. when you connect this branch to the lna1, the NF of this path is increased in 3dB
 

dowjones said:
hello,

consider the following: put the splitter with the unused (match-loaded) branch in a black-box, so that you have only two accesible ports. then, the insertion loss of your setup is 3dB, and therefore the noise figure of the black-box is 3dB. when you connect this branch to the lna1, the NF of this path is increased in 3dB
Click to expand...

You are right !

I said a big bullshit ! :cry:
 

Thanks for all your explanation. The power spliiter will introduce at least 3dB NF for each channel.
To avoid this, the power splitter is deleted . LNA1 and LNA2 are united since f2 band and f1 band are very close. F1 band and f2 band will be seperated after LNA. So the new block is:
Antenna---->LNA---->filter1---->mixer1
----->filter2---->mixer2
 

The topic is less trivial than expected, read

Noise in lossless microwave multiports (p 99-110)
Q. García-García
Published Online: 16 Mar 2004
DOI: 10.1002/mmce.10122

at **broken link removed**.
 

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    fly_fish

    Points: 2
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hi,marcomdd
would you mind posting up the article you sited? I could not download this paper.
I would appreciate it very much!
 

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