# What is the signal of a receptor antenna irradiated with a plane wave?

Lets say there is a reception antenna connected to a detection system with a coax, and a transmitter sends a plane wave. What happens is that the antenna gets a certain power, following the [Schelkunoff and Friis. Antenna theory and practice 1952] or [Papas. Theory of electromagnetic wave propagation], the absorbed power will be $$$$P_{abs} = \frac{\lambda^2} {4\pi} g(\theta_0, \phi_0) S^{inc}(\theta_0, \phi_0) |\vec{p}_{rec}\cdot \vec{p}_{inc}|^2$$$$

Where $$\theta_0$$ and $$\phi_0$$ are angles from where the plane wave was emitted. $$S^{inc}$$ is incident pointing vector and $$\vec{p}$$ - are polarization vectors of incident wave and a reciever, and g - is a gain of a reciever

So the point is what is the amplitude of a received signal (=travelling wave) at a coax connected to the antenna (current or voltage)? Is it as following:

$$\frac{I_0^2Z} {2} = P_{abs}$$

Where Z is 50 Ohms (if the coax has 50 Ohms impedance), and the current is $$I =I_0 sin(\omega t)$$

• Taking a sinusoidal current, one have the power equal to $I^2R/2$ where I is the amplitude of the current. This value is average. basically it comes from the integration of square of current. Oct 22 '21 at 15:10
• @petr it's an old source of confusion, between Amplitude (peak value) and magnitude (RMS). You need to define terms carefully. But if you know $P_{abs}$ then you can work out RMS or peak current as normal. Your equation is correct if you mean $I = I{sin}(\omega t)$ Engineers more commonly use RMS as it's the standard and a bit simpler for calculating power, voltage drop, etc. Oct 22 '21 at 18:07
• Yup! Peak = 1.414 * RMS; RMS = Peak * .707. But I know that you knew that. :-) Oct 22 '21 at 18:59
• Thanks for the precisions. I come from the interface between engineering and physics. So, that's why the choice was $I =I_0 sin(\omega t)$ Oct 23 '21 at 8:10
• I have edited a post a bit in order to better reflect the problem Oct 23 '21 at 8:12