Attenuation measurement aginst frequency

Status
Not open for further replies.

yasine79

Member level 2
Joined
Nov 10, 2009
Messages
49
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,286
Location
germany
Visit site
Activity points
1,602
Hi,
I make a transmission (S21) measurement with two horn antennas. between the sender and VNA a make 30 db attenuation element. the result are in the attachement. the mesurement is taken in the air. the distance between the horn antennas is 50cm.

My Question:

The attenuation should increses with the Frquency, should not it?

thank you
 

Attachments

  • attenuation.jpg
    18.5 KB · Views: 80

In free air and far field conditions, the path-loss is given by the formula PL=-27+20*Log(d*f) with d distance in meter and f frequency in MHz. This is the attenuation due to the free space. It doesn't apply in near field. At 3Ghz the wavelength is about 10 cm. You are in far field if d > 2*D^2/wavelength where D is the dimension of the antenna i.e. if D < sqrt(0.5*0.1/2) = 16 cm. Also objects near the main lobe can distort the field and change the attenuation.
However did you take into account the gain of the antennas (I mean its variation with frequency) ? In fact the received signal will be given by:

RX = TX + AntennaGain(TX) - PL + AntennaGain(RX) - ExtraAttenuation (the 30 dB) - CableLosses

if the two antennas are the same type, then rougly:

RX = TX + 2*AntennaGain(TX) - PL - ExtraAttenuation - CableLosses

So to calculate the space attenuation, that is the path loss you have to apply:

PL =TX -RX + 2*AntennaGain(TX) - PL - ExtraAttenuation - CableLosses
 

-If the distance between high-gain antennas is so small, you cannot expect the free-space propagation to function. Your two horns form a resonant cavity where you can see funny things to happen. Try five meters of distance, then you can follow the theory.
 

Thank you all for your answer. the distance between bothantenns was not 50cmm, but 500 cm->for field. i have a question to you albbg: could u give me the name of the book or paper where you found the relation:
PL=-27+20*Log(d*f)

Thank you.
 

Hi again!

i think i have an understNDING PROBLEM.

assuming the S21(f1)=-15db, S21(f2)=-30db,f1<f2. We say the attenuation at f1 is greater or lower than at f2?

-15>-30->attenuation at f1 is greater than at f2.
BUT

-30db means that smaller amount of received power than in -15 db.

Could you help?

thank you
 

There are various formulations of free space path loss and Friis equation, depending on the involved units and assumed standarad antenna gain. The measurement shows that either the setup is involving multi-path transmission (e.g. ground reflections) or the antennas itself have non-monotonous gain versus frequency.

Confusingly, you told about 50 cm distance in post #1 and 500 cm in post #5.

https://en.wikipedia.org/wiki/Friis_transmission_equation
https://en.wikipedia.org/wiki/Free-space_path_loss

Accoridng to literature, horns have gain increasing with frequency, so the loss of the Friis equation would be at least compensated.
 

Status
Not open for further replies.
Cookies are required to use this site. You must accept them to continue using the site. Learn more…