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That's a pretty basic one but for some reason I am stuck now:

Suppose I have a transmitter that transmits Ptx=4W_EIRP with an ideal, omni-directional antenna. The power applied to the antenna terminals is 4W.

Now Frii's equation states that the received power is Prx=Ptx * Gtx * Grx * L (Gtx, Grx antenna gains and L path loss).

Now assume a transmission 1m at 950MHz, the path loss is about 32dB. Assuming lossless components, I would receive 2.5mW ... enough to power a micro controller, sensors etc.

Now I assume I have perfect line-of-sight and create super high gain antennas like Gtx=20dBi, Grx=20dBi. Then the received power would be 4W * 10^((-32+20+20)/10) = 25W ! This is higher than the transmitted power and hence unphysical.

Where is the mistake?

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  • \$\begingroup\$ your (dish) antennas are huge compared to the 1meter spacing \$\endgroup\$ Oct 26, 2019 at 3:06

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You misspelt his name Friis, read his requirements and assumptions again.

The error is in geometry and the assumption this is far field not 1m near field.

Also with antenna gain, one measures the EIRP not the Pd of the Amp output signal.

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  • \$\begingroup\$ Could you elaborate on this a bit? 1.) Is it fair to say that for high gain we need a parabolic dish? I then have found the formula $$G\approx \frac{4\pi\eta A f^2}{c^2}$$ (A: physical size, eta: efficiency). The loss is given by $$L=16\pi^2d^2f^2/c^2$$. With A=r^2*pi and eta=1 G~2*pi^2*r^2*f^2/c^2. From this, G > L implies r > 2*d, so the radius would need to be larger than twice the distance RX<->TX. I do understand that Friis' equation is only valid in the far-field but since lambda drops out I still don't see the physical constraint. 1m would be far field but let's say 100m or 1km then \$\endgroup\$
    – divB
    Oct 28, 2019 at 16:21
  • \$\begingroup\$ (So why not build a hypothetical 10km dish for a 100m communication ... which is far-field for 1GHz) \$\endgroup\$
    – divB
    Oct 28, 2019 at 16:24
  • \$\begingroup\$ 2.) What do you mean with "Also with antenna gain, one measures the EIRP not the Pd of the Amp output signal."? When I was writing "The power applied to the antenna terminals is 4W" I implied an ideal isotropic antenna (no loss implying power at terminals 4W equals 4W EIRP). Did you refer do anything else? \$\endgroup\$
    – divB
    Oct 28, 2019 at 16:26
  • \$\begingroup\$ 1m is still near field but 10m would be far field. But antenna with gain is EIRP \$\endgroup\$ Oct 28, 2019 at 21:42
  • \$\begingroup\$ 1m is far field (far field: >2*lambda; lambda=31cm). But that's not my point. Let's do 100m for 1GHz. Why doesn't a hypothetic, ideal dish with 500m radius give me 6x more power than I put in? And "But antenna with gain is EIRP" I still do not understand. \$\endgroup\$
    – divB
    Oct 28, 2019 at 23:21

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