Question

A monopole antenna extends vertically over a perfectly conducting plane, and has a linear current distribution. If the length of the antenna is \(0.01 \lambda\). what value of \(I_{0}\) is required to \((a)\) provide a radiation-field amplitude of \(100 \mathrm{mV} / \mathrm{m}\) at a distance of \(1 \mathrm{mi}\), at \(\theta=90^{\circ} ;(b)\) radiate a total power of 1 W? Assume free space above the plane.

Short answer

In summary, for the given monopole antenna with a linear current distribution, to provide a radiation-field amplitude of 100 mV/m at a distance of 1 mi and θ = 90°, the required peak current, \(I_0\), is approximately 0.0267 A/λ. To radiate a total power of 1 W, the required peak current, \(I_0\), is approximately 0.2827 A/λ.

Step-by-step solution

Case (a): Find \(I_0\) to provide radiation-field amplitude of 100 mV/m at a distance of 1 mi and \(\theta=90^\circ\)

The radiation-field amplitude formula for a monopole antenna with a linear current distribution is given by: \(E_r = \dfrac{60 k I_0 \sin\theta}{\pi r}\) where \(E_r\) is the radiation-field amplitude, \(k\) is the antenna length in terms of wavelength, \(I_0\) is the peak current, \(\theta\) is the angle, and \(r\) is the distance from the antenna. We need to find the value of \(I_0\) when \(E_r = 100 \, \text{mV/m}\), \(k = 0.01\lambda\), \(\theta = 90^\circ\), and \(r = 1 \, \text{mi}\). First, convert the distance from miles to meters: \(r = 1 \, \text{mi} \times 1609.34 = 1609.34 \, \text{m}\) Next, plug the given values into the formula and solve for \(I_0\): \(100 \times 10^{-3} \, \text{V/m} = \dfrac{60 \times 0.01\lambda \times I_0 \times \sin(90)}{\pi \times 1609.34}\) \(I_0 = \dfrac{100 \times 10^{-3} \pi \times 1609.34}{60 \times 0.01\lambda}\) Finally, calculate the value of \(I_0\): \(I_0 \approx 0.0267 \dfrac{\text{A}}{\lambda}\) So \(I_0 \approx 0.0267 \frac{\text{A}}{\lambda}\) is required to provide a radiation-field amplitude of 100 mV/m at a distance of 1 mi and \(\theta = 90^\circ\).

Case (b): Find \(I_0\) to radiate a total power of 1 W

The total power (\(P_{\text{total}}\)) radiated by a monopole antenna with a linear current distribution is given by: \(P_{\text{total}} = 40 \pi^2 k^2 I_0^2\) where \(k\) is the antenna length in terms of wavelength, and \(I_0\) is the peak current. We need to find the value of \(I_0\) when \(P_{\text{total}} = 1 \, \text{W}\) and \(k = 0.01\lambda\). Plug the given values into the formula and solve for \(I_0\): \(1 = 40 \pi^2 (0.01\lambda)^2 I_0^2\) \(I_0^2 = \dfrac{1}{40 \pi^2 (0.01\lambda)^2}\) Finally, calculate the value of \(I_0\): \(I_0 \approx 0.2827 \dfrac{\text{A}}{\lambda}\) So \(I_0 \approx 0.2827 \frac{\text{A}}{\lambda}\) is required to radiate a total power of 1 W.

Key concepts

Antenna Current Distribution

Antenna current distribution plays a crucial role in defining the radiation characteristics of an antenna, including the radiation-field amplitude and the total radiated power. In the case of a monopole antenna, which extends vertically over a perfectly conducting ground plane, the current is assumed to have a linear distribution.

This means that the current varies linearly with the height of the antenna. The current value is maximum at the base of the antenna (often denoted as \( I_0 \)) and decreases to zero at the tip. This variation in current creates an electromagnetic field around the antenna, which is responsible for the radiation properties of the antenna.

For a monopole antenna with a short length relative to the wavelength (\( k = 0.01\lambda \)), such as in our exercise scenario, the current distribution is approximated to be linear. This simplification is practical for understanding and solving problems related to monopole antennas with short lengths. With the base current \( I_0 \) known, engineers can determine many characteristics of the antenna's performance, including the fields it will produce at a given distance and the power it will radiate.

Radiation-Field Amplitude

The radiation-field amplitude refers to the strength of the electric field radiated by an antenna at a certain distance. For instance, in our problem, we are asked to find the peak current, \( I_0 \), necessary to achieve a radiation-field amplitude of 100 mV/m at a distance of 1 mile from the monopole antenna.

The formula presented for the monopole antenna takes into account the antenna's linear current distribution and the angle relative to the plane of the ground (denoted as \( \theta \)). The amplitude of the radiated electric field is crucial, as it directly correlates to how effectively the antenna can transmit signals over a distance. It's important for students to understand that the desired radiation-field amplitude at a specific location depends on the operating current, the antenna geometry, and the angle from the antenna (in our case, \( \theta = 90^\circ \)), among other factors.

To solve the problem, converting units to ensure consistency within the formula is essential. For example, converting the distance from miles to meters ensures that the answer for \( I_0 \) is in the correct unit of amperes per wavelength \( \frac{\text{A}}{\lambda} \) and facilitates the correct application of the formula.

Antenna Radiation Power

Antenna radiation power is the total power radiated into space by the antenna. It is a fundamental characteristic representing the antenna's ability to convert electrical power from a transmitter into electromagnetic waves. In our example, the task is to decide on the appropriate value of \( I_0 \) in order for the antenna to radiate a power of 1 Watt.

The formula for the total radiated power by a monopole antenna with a short length in terms of the wavelength involves the square of the peak current \( I_0 \) and the square of the antenna length as a fraction of the wavelength \( k \). The radiated power is proportional to these squares, which implies that small changes in current or antenna length can significantly impact the power output.

The antenna's efficiency and how effectively it can be matched to the transmitter are directly tied to the peak current and the antenna length. In our setup, the total power is 1 W and the antenna length is a small fraction of the wavelength (\( k = 0.01\lambda \)), leading to a straightforward calculation once the form of the equation is understood. This concept is paramount for students who are learning about antenna design and operation.