# How to model an antenna as a voltage source?

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\$$ ?

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.

• It follows from the last para that you can crudely model the antenna as a voltage source followed by a series RLC resonant circuit, with R around 75 or 300 ohms, and LC chosen to approximate the tuned frequency of the antenna (via their product and a well known equation for frequency) and its bandwidth (via their ratio, R, and a well known equation for Q). – user_1818839 Sep 28 '17 at 12:00
• I want to know how to model the voltage source as 100 uV. Actually, I am considering that received signal's power is transmit signal's powerKd^(-alpha), where K is a constant regarding to antenna and so on, d is distance between tx and rx, and alpha is attenuation factor, normally 2 to 8. In this case how to decide the voltage source is 100 uV! Can you explain this? :) – God Danny Sep 29 '17 at 2:59
• Actually, I am studying wireless communication, so I usually dealt with the received signal power which is calculated from channel and transmit power. However, recently, I am studying on simultaneous wireless information and power transmission (SWIPT) system, so I want to know how much voltage signal value the antenna makes. – God Danny Sep 29 '17 at 3:02

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.

I think that it's something like this (feel free to correct)

The load on the dipole is in blue. The first resistor is radiation resistance, second is loss resistance. The capacitor is the desired capacitive hats of the antenna, the smaller capacitor is the parasitic capacitance. The inductor is the parasitic inductance of the antenna. The generator models the variable received power and received frequency.

For a dipole/monopole I think you have to put the capacitor in series with the capacitor as above. In an actual dipole/monopole, the 'capacitor' is actually in parallel, but it cant be modelled like this with a real power source.