# Finding Jso for Antenna Array in Vacuum

• robert25pl
In summary, the conversation discusses an antenna array consisting of two parallel sheets with current densities, a propagating wave in vacuum, and the requirement for a specific time average of the Poynting vector. The question is how to calculate the needed surface current Jso. One suggestion is to find the total electric field and magnetic field, and then use the given Poynting vector to determine Jso.

#### robert25pl

Antenna array consists of two infinite plane parallel sheets in the xy plane spaced half a wavelength apart and having current densities. Wave is propagating in vacuum.

$$J_{s1} = J_{so}cos(\omega t) \vec{i}$$ z = 0

$$J_{s2} = J_{so}sin(\omega t) \vec{i}$$ $$z = \frac{\lambda}{2}$$
The time average of the Poynting vector is required to e 500 W/m^2 to the right of second antenna ($$z > \frac{\lambda}{2}$$ )
I have to find surface current Jso is needed?
I think that I have to find total E = E1 + E2 and then H. Poynting vector is given but I don't know how to surface current Jso from that. Any suggestions. Thanks

Last edited:

To find the required surface current Jso for the antenna array in vacuum, we can use the time-averaged Poynting vector equation:

<S> = 0.5Re(E x H*)

First, we can calculate the electric field E by adding the individual electric fields from the two infinite plane parallel sheets:

E = E1 + E2

Since the electric fields are perpendicular to the sheets, we can use the superposition principle to add them. E1 is in the positive x direction and E2 is in the negative x direction, so the total electric field E is simply:

E = (E1 - E2)

Substituting the given expressions for E1 and E2:

E = Jso(cos(\omega t) - sin(\omega t))

Next, we can calculate the magnetic field H by using the relationship H = (1/\eta) x E, where \eta is the impedance of vacuum (377 ohms).

H = \frac{1}{377}(E1 - E2)

Substituting the expression for E:

H = \frac{1}{377}Jso(cos(\omega t) - sin(\omega t))

Finally, we can substitute the expressions for E and H into the Poynting vector equation and equate it to the given value of 500 W/m^2:

<S> = 0.5Re(E x H*) = 500 W/m^2

0.5Re(Jso(cos(\omega t) - sin(\omega t)) x \frac{1}{377}Jso(cos(\omega t) - sin(\omega t))) = 500 W/m^2

Simplifying and solving for Jso:

Jso = \sqrt{\frac{1000}{377}} = 5.31 A/m

Therefore, a surface current of Jso = 5.31 A/m is needed for the antenna array in vacuum to have a time-averaged Poynting vector of 500 W/m^2 at a distance greater than half a wavelength from the second antenna.

## 1. What is "Jso" in relation to antenna arrays?

Jso stands for "Johnson noise" and refers to the thermal noise generated by the random movement of electrons in a conductor, such as the metal components of an antenna array.

## 2. Why is the search for Jso important in a vacuum environment?

In a vacuum, there is no air or other matter to absorb or disperse the Johnson noise. This means that the noise generated by the antenna array will be more pronounced and can potentially interfere with the reception of signals. Therefore, it is crucial to find the Jso and reduce it as much as possible in vacuum environments.

## 3. How is Jso typically measured in antenna arrays?

Jso is usually measured using a spectrum analyzer or a noise figure meter. These tools can detect and quantify the level of Johnson noise in the antenna array.

## 4. What factors can affect the level of Jso in an antenna array?

The level of Johnson noise can be affected by several factors, such as the temperature of the antenna array, the resistance of the conductors, and the bandwidth of the signals being received. Higher temperatures, higher resistance, and wider bandwidths can all increase the level of Jso.

## 5. How can the level of Jso be reduced in an antenna array?

To reduce the level of Johnson noise in an antenna array, the temperature can be lowered, high-quality conductors with lower resistance can be used, and the bandwidth can be narrowed. Additionally, shielding and filtering techniques can also be employed to minimize the effects of Jso on signal reception.

• Introductory Physics Homework Help
Replies
17
Views
241
• Introductory Physics Homework Help
Replies
29
Views
1K
• Introductory Physics Homework Help
Replies
64
Views
1K
• Introductory Physics Homework Help
Replies
13
Views
594
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
9
Views
568
• Electrical Engineering
Replies
15
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
793