How Does Antenna Gain Influence Signal Power Calculations?

[H_man]

Antenna Gain??

Hi all,

I am really confused by the subject of antenna gain.

I have no problem with the idea of the idea that an antennas performance should be related to how welll it focusses radiation relative to an isotropic emitter. Bus as this gives a figure greater than one I really don't understand how it can be used in power calculations.

Signal Power = Transmitted Power * Transmitting Antenna Gain * Recieving Antenna Gain / Losses in cables and through the medium

This equation to me suggests that if you placed two antennas opposite each other they will generate energy. Obviously this is wrong, but what am I not seeing?

 

[Born2bwire]

The antenna link equation is only valid in the far-field, this place restrictions on the range of values of R that you can have in relation to both the wavelength and physical size of the antenna. As such, this provides that any high gain is countered by the free space loss.

 

[H_man]

Thanks... It makes more sense if this equation can only be used under certain circumstances (far field).

Although it still seems/feels like there is some constant missing from the equation.

 

[Wannabeagenius]

Hi all,


Signal Power = Transmitted Power * Transmitting Antenna Gain * Recieving Antenna Gain / Losses in cables and through the medium

This equation to me suggests that if you placed two antennas opposite each other they will generate energy. Obviously this is wrong, but what am I not seeing?

I am currently studying antenna theory and I'm very confused also.

Why does this equation suggest that?

Thank you,

Bob

 

[Born2bwire]

I am currently studying antenna theory and I'm very confused also.

Why does this equation suggest that?

Thank you,

Bob

It makes more sense when you consider the equation as Friis originally stated it. He originally wrote it in terms of the effective aperture of the antennas. That is,

P_r = \frac{A_rA_tP_t}{d^2\lambda^2}

The effective aperture is the effective area of the antenna over which it captures the incident electromagnetic energy. It is related to the physical aperture by an efficiency coefficient, so the largest effective aperture that is possible is the physical aperture of the antenna. And the physical aperture is limited by the physical size of the antenna.

The easiest way to visualize this is with a horn antenna. The horn antenna is a rectangular waveguide that has an attached horn that acts as a transformer, going from the impedance of the waveguide to that of free space. We can think of the opening of the horn as our physical aperture, if the horn is a perfect transformer than all radiation incident on the mouth of the horn will be captured and converted into a guided wave inside the horn's waveguide. Thus, the amount of power that can be received is limited by the physical size of our horn antenna. Since the far-field condition is dependent upon the electrical size of the antenna, then the larger the antenna is then the farther away we need to setup our link antenna for the Friis antenna equations to be valid. By increasing the distance though, we offset the large effective aperture by the space loss factor.

The far-field limit is such that

d > \frac{2D^2}{\lambda}

where $D$ is the largest dimension of the radiator. So we can think of $D^2$ as a physical aperture. Thus, we can see that at the edge of the far-field, we already have a maximum relationship for the power received to be

P_r \approx \frac{P_t}{2}

So we cannot have a received power greater than the transmitted power as long as we pay proper attention to the requirements of the Friis equation.