# How Antenna as RLC circuit works

Antenna is an important transducer and important component in RF communication circuits such as AM transmitter, AM receiver, FM transmitter and cell phones. Antenna are used to transmit signal and receive signal whose size is selected according to the frequency of the radio wave being transmitter. A real physical wire can be modeled as RLC circuit. RLC circuit is a circuit that is made up of resistor, inductor and capacitor. Here it is explained how antenna works when it is modeled as an RLC circuit.

Consider RF signal $$s(t)$$ which strikes an antenna that is modeled as RLC circuit as shown below.

Let $$s(t) = V_0 cos(wt)$$ and using KVL in the circuit we have,

$$\frac{d^2 q}{dt^2} + \frac{R}{L}\frac{dq}{dt} +\frac{1}{LC}q = V_0 cos(wt)$$   ---->(1)

The voltage across the capacitor is,

$$v_c(t) = V_c(w) cos(wt - \delta)$$        ---->(2)

or, $$V_c(w) = \frac{\frac{V_0}{LC}}{\sqrt{(w_0^2 - w^2)^2+(\frac{wR}{L})^2}}$$  ---->(3)

where $$\delta$$ is the phase difference between voltage source and the oscillator frequency.

At resonance $$w_0 = w$$ equation (3) becomes,

$$V_c(w_0) = \frac{V_0}{w_0 RC} = Q V_0$$  ---->(4)

If the RLC antenna circuit designed to resonant at resonance frequency $$w_0$$, the Q is high which means that there is high amplification of the incoming RF signal.

Consider an RF signal s(t) is applied to an antenna with ac frequency of 10MHz and ac voltage of 10mV. Let the value of the capacitor C be 1nF and the value of inductor H be 1uH. We can find out the capacitive reactance and inductive reactance and the voltage across the capacitor and inductor.

capacitive reactance, $$X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \pi \times 10MHz \times 1nF}=-15.915\Omega$$

inductive reactance, $$X_L =2 \pi f L = 2 \pi \times 10MHz \times 1uH=62.832\Omega$$

The total impedance is,

$$X_t = X_C+X_L=46.916\Omega$$

The inductive reactance at this total reactance is,

$$L=\frac{X_t}{2\pi f }=746.697nH$$

If the LC is replaced by L of inductance of 746.697nH then we will get the same result without disturbing rest of the circuit.

The total current in the circuit is,

$$I=\frac{V}{X_t}=\frac{10mV}{46.916\Omega}=0.213mA$$

Thus the voltages across the capacitor and inductors are:

voltage across capacitor, $$V_C=I X_C = 0.213mA\times-15.915\Omega=3.39mV$$

and voltage across inductor, $$V_L=I X_L = 0.213mA\times62.832\Omega=13.39mV$$

And since $$V = V_L-V_C = 13.39mV-3.39mV=10mV$$ as expected.

The oscillation in the RLC circuit antenna losses energy every cycle and the oscillation voltage decays over as explained in the previous article How to find Q-factor of Guitar

In the RLC circuit, the total impedance is,

$$Z = R+ jwL + \frac{1}{jwC} = R + j(wL - \frac{1}{wC})$$

The magnitude of Z is,

$$|Z| = \sqrt{R^2 + (X_L-X_C)^2}=\sqrt{R^2 + X_t^2}=\sqrt{R^2 + (wL-\frac{1}{wC})^2}$$

At resonance the magnitude of total impedance Z becomes purely resistive when the inductance reactance and capacitive reactance are equal.

$$X_C=X_L$$

which means,

$$\frac{1}{wC}=w L$$

which gives,

$$f=\frac{1}{2\pi\sqrt{LC}}$$

where f is the frequency of oscillation