How to model an antenna as a voltage source?

Question

I've already asked about this antenna system, but I still have some questions and some things I do not understand.

I heard that the antenna only receives POWER, not voltage. From this fact, I thought that the current will be 1 / (antenna resistance + load resistance.) That is, the input power is constant.

Let us define the transmit signal with power 50 watt as \$s(t) = 10\cos(2\pi f_c t)\$.

We also assume that the receiver receives the signal that is reduced to half of the transmit signal, i.e., the power of the received signal is 25 watt.

Then, is the voltage source defined as \$v(t)=A \cos(2\pi f_c t)\$ such that \$\frac{\frac1T\int_T |v(t)|^2 dt}{R_{antenna}+R_{load}}=25\$ ?

Answer 1

Antennas have something known as their characteristic impedance. You can think of a antenna as a Thevenin or Norton source at this impedance.

A common dipole has around 75 Ω impedance at its intended operating frequency. A folded dipole has about 300 Ω impedance.

For example, if a 75 Ω antenna picks up a 100 µV signal, then you can think of it in two equivalent ways from the point of view of the circuit. It could be a 100 µV source with 75 Ω in series, or a 1.33 µA source with 75 Ω across it.

Note that the above only applies at the antennas intended operating frequency. The impedance can be very different at other frequencies. For example, a dipole as infinite impedance at DC, and a folded dipole 0. The impedance also become reactive (no longer purely resistive) at off frequencies.

Answer 2

An antenna impedance is key to being able to extract the best signal given the incident power. For instance a monopole looks like an impedance in series with a resistance of about 37 ohms when it is a quarter wave long. At other frequencies it changes its impedance dramatically: -

The "height" mentioned in the diagram is in fact monopole length and not the distance a particular antenna is placed above ground.

So, at about 0.05 wavelengths, the antenna resistance is virtually zero but the impedance is -j1000 ohms (capacitive). At one quarter \$\lambda\$ the impedance looks resistive 37 ohms (although in this picture it is hard to see).

The voltage received at the terminals is proportional to antenna length and the E-field present. This is one of the reasons why an antenna at higher frequencies receives less power - the length of the antenna (assumed to be at a constant ratio to \$\lambda\$) defines how much power it can capture and this reduces with increasing frequency.