How to Model antenna as voltage source?

Question

I've already asked about this antenna system. However, I still have some questions and somethings 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 is $$s(t) = 10\cos(2\pi f_c t).$$ And we assume that the receiver receives the signal that is reduced a 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.