A antenna with 50dB gain transmits 0dBm power and the receiver antenna(gain 0db) at 10m distance receives 10dBm power. How does this happen? Path loss for 10m is taken as -40dBm. frequncy of operation 2400MHz. This violates the law of conservation of energy. anything to prove this?
I think that 0 dBm is the power measured at the input of the Tx antenna which must be highly directive and pointed toward Rx antenna. If this is the case, the equivalent isotropically radiated power (EIRP) is 50 dBm, therefore, the link budget is correct. The path loss is maybe higher than 40 dB, due to the cable losses and imperfect orientation of the Tx antenna toward Rx antenna. It seems that Tx antenna is very directive, with small half power beam width (HPBW).Therefore its effective gain can be smaller than 50 dB if the Tx antenna is not correctly pointed to Rx antenna.
Antenna is passive device and cannot generate power. If input power (Feed power) to the antenna is 'X', received power from the antenna is always less than 'X'
If you are getting received power more than expected, essentially, received power from the multiple sources/paths
The gain of antenna also does not increase received power more than transmitted power (it is not gain in absolute term, it is relative quantity and hence called directive gain)
Clearly your estimate of 10dBm for the received power is incorrect. I calculated the received power taking your data at face value and using the Friss equation and got -10dBm. However, your transmit antenna gain of 50dB implies an antenna receiving aperture of around 100sqm at 2.4GHz. Increasing this to get a physical aperture assuming it's an aperture antenna (the only practical way to get 50dB of gain at this frequency) gives a physical aperture of about 200sqm i.e. an antenna of about 14m diameter. At 2.4GHz the Rayliegh Range for such an antenna is 1.6km so using it a t a distance of 10m is not going to be valid anyway.
Yes, The 10m distance is in too near field region of the antenna. If we calculate the far field lower limit considering it as parabolic reflector antenna we get it as 5km. So the general friss equation is not valid. Is this the reason that the antenna is placed in the very near field and the friss equation for near field need to be applied?
Whatever be the scenario, Received power can not be more then transmitted power if no other source is contributing OR some active device is present.
Unless the transmit antenna is highly focussed on the receive antenna you can expect a loss of at least 50 dB in this arrangement. In any case you can never receive more power than you transmit (unless you are transmitting though a Lasing medium!). You must be making a mistake somewhere. Are you transmitting 0 dBW and receiving 10 dB microwatt?
The Friis equation does not hold for this close to such a big antenna. In round numbers an antenna with a gain of 50 dB has to have an aperture of 10000 square wavelengths, i.e. 100 wavelengths across. The beam from this is nearly parallel out to 10000 wavelengths, and doesn't reach its final shape until 20000 wavelengths, after which the Friis equation works (see Mr Marvin's comments). The receive antenna only captures power from an area of about 1/10 of a square wavelength, so it intercepts about 1/100000 of the radiated power, which at 10 m is going past the 1/2 wavelength long antenna in a beam 100 wavelengths wide .
If the antenna was focussed to a range of 10 m by increasing the curvature of the phase in the aperture, then you might get close to 0 dB link budget, i.e. no power loss between transmit and receive - Gaussian beam optics get clos to this.
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