Question

Consider a lossless half-wave dipole in free space, with radiation resistance, \(R_{\mathrm{rad}}=73\) ohms, and maximum directivity \(D_{\max }=1.64\). If the antenna carries a 1-A current amplitude, \((a)\) how much total power (in watts) is radiated? \((b)\) How much power is intercepted by a \(1-\mathrm{m}^{2}\) aperture situated at distance \(r=1 \mathrm{~km}\) away? The aperture is on the equatorial plane and squarely faces the antenna. Assume uniform power density over the aperture.

Short answer

= (1 * 73) / 2 = 73 / 2 = 36.5 W So, the total radiated power (Prad) is 36.5 watts. #tag_title# Step 2: Calculate the power density (S) at 1 km distance #tag_content# To find the power intercepted by a 1-square meter aperture situated 1 km away from the antenna, we first need to calculate the power density (S) at that distance. We can use the following formula: S = (Prad * G) / (4 * π * d²), where Prad is the total radiated power, G is the antenna's directivity, and d is the distance from the antenna. In this problem, Prad = 36.5 W, G = 1.64 (given), and d = 1000 m (1 km). Using these values, we can find the power density at 1 km away from the antenna: S = (36.5 W * 1.64) / (4 * π * (1000 m)²) = (59.78 W) / (4 * π * 1000000 m²) = 59.78 W / 12566370.61 m² ≈ 4.76 x 10⁻⁶ W/m² So, the power density (S) at 1 km distance is approximately 4.76 x 10⁻⁶ W/m². #tag_title# Step 3: Calculate the intercepted power by the 1-square meter aperture #tag_content# Lastly, to find the power intercepted by a 1-square meter aperture, we simply multiply the power density (S) by the aperture's area (A): P_intercepted = S * A Since A = 1 m², the intercepted power by the 1-square meter aperture is: P_intercepted = 4.76 x 10⁻⁶ W/m² * 1 m² = 4.76 x 10⁻⁶ W Therefore, the power intercepted by a 1-square meter aperture situated 1 km away from the antenna is approximately 4.76 x 10⁻⁶ watts.

Step-by-step solution

Calculate the total radiated power (Prad)

To calculate the total radiated power (Prad), we can use the following formula: Prad = (I² * Rrad) / 2, where I is the current amplitude and Rrad is the radiation resistance. For this problem, I = 1 A and Rrad = 73 ohms. Using these values, we can find the total radiated power in watts: Prad = ((1 A)² * 73 ohms) / 2

Key concepts

Radiation Resistance

When studying antennas, one of the fundamental concepts is the radiation resistance, which represents the portion of an antenna's resistance that is directly responsible for the radiation of electromagnetic waves. It's a theoretical resistance that, if it were present in a resistor, would dissipate power in a similar fashion to how the antenna radiates power into space.

For instance, a lossless half-wave dipole has a theoretical radiation resistance, often denoted as \(R_{\mathrm{rad}}\), that varies depending on the antenna's design and operation frequency. In our textbook example, the dipole's radiation resistance is given as 73 ohms. The power radiated by the antenna, denoted as \(P_{\mathrm{rad}}\), can be calculated using the formula \(P_{\mathrm{rad}} = \frac{I^2 \cdot R_{\mathrm{rad}}}{2}\), where \(I\) is the current amplitude passing through the dipole. The divisor '2' accounts for the fact that the current has a sinusoidal nature; therefore, the power calculation involves the root mean square (RMS) value of the current.

Dipole Antenna

The dipole antenna is one of the most straightforward and widely used antenna types in radios and wireless communication. It consists of two metallic rods or wires that are aligned collinearly and fed at the center. The length of the dipole is crucial as it determines the frequency at which the antenna operates most efficiently.

In the provided exercise, we look at a half-wave dipole, which means that the total length of the antenna is about half the wavelength of the frequency it's designed to transmit or receive. This length makes the dipole resonant, allowing it to radiate energy efficiently. For the half-wave dipole, the maximum radiation occurs in the perpendicular direction to the axis of the dipole and its minimum radiation, which is virtually zero, occurs along the axis. Despite its simplicity, the dipole antenna is fundamental for understanding other more complex antenna designs.

Directivity

The concept of directivity in the realm of antennas refers to the measure of how 'focused' the radiated power is in a certain direction. An antenna with high directivity sends out energy in a narrow beam, which is more intense and can reach longer distances in that direction, compared to an antenna that spreads the energy in all directions.

In our textbook case, we evaluated an antenna with a maximum directivity of \(D_{\max }=1.64\). This indicates that the antenna radiates power more efficiently in its preferred direction versus an isotropic radiator, which is a theoretical point source that radiates equally in all directions. Antennas with higher directivity are ideal for applications like satellite communications, where the signal needs to reach a specific location without unnecessary dissipation of power.

Free Space

In antenna theory, the term free space refers to a perfect vacuum through which electromagnetic waves can propagate without any attenuation due to medium. However, in a practical sense, when we talk about free space in regards to radio wave propagation, we often mean the condition where waves travel without significant obstruction or reflection, such as in the atmosphere of Earth above the ground level or outer space.

The exercise mentioned assumes that the dipole antenna is in free space, implying no loss of power due to the transmission medium. This idealization simplifies calculations and serves as a baseline when comparing with real-world scenarios where the medium (like air or buildings) can absorb or reflect some of the energy. The power intercepted by an aperture at a distance within free space can be analyzed without accounting for medium-based losses, which is crucial for understanding actual antenna behaviors in a typical operating environment.