Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

can anyone prove how this?!!

Status
Not open for further replies.

hitman1988

Member level 5
Joined
Nov 17, 2009
Messages
83
Helped
5
Reputation
10
Reaction score
4
Trophy points
1,288
Location
iraq
Activity points
1,769
hi
i want to prove that the impedence of a loop antenna is four times of the impedence of a dipole antenna ?
 

Hello,

The impedance of a folded dipole is about 4 times the value of a half wave dipole. This is not valid for a full wave loop.

Given 1A feed current, a half wave dipole generates a far field. The strength depends on the orientation of the observing point to the dipole (donut shaped radiation pattern).

Integrating the power intensity over the full sphere gives the radiated power. From the radiated power and the feed current follows the radiation resistance. (about 73 Ohms for a very thin half wave resonating dipole).

When you make a folded dipole, the total length is 1 lambda. At the bends you have a current minimum and the current direction inverses. This follows from the standing wave pattern at 0.25 lambda from a shorted transmission line. You have high voltage, but (virtually) no current. Left and right from the current node, current flow is opposite, but e same strength.

Because of the change in current direction, the current in the shorted section has same direction as the fed section. When the shorted section has same diameter as the fed section, the current in the shorted part is same as current in the fed part. So it is like having two dipoles in parallel close to each other.

As the distance between the shorted part and fed part of a folded dipole is << lambda, the field produces by the two parts reinforce each other. Given the 1A feed current, the H and E field doubles (with respect to the HW open dipole), When H, E field doubles, the intensity increases with factor 4. So total radiated power increases with factor 4. Hence the radiation resistance must be 4 times higher (given the 1A feed current).

Now your loop.
Where the currents have its maximum value, the distance in a loop is no longer small w.r.t. Wavelength. In a direction perpendicular to your loop, the fields of the two halves are in phase and therefore field intensity becomes 4 times higher, but in the plane of the loop, the fields are no longer in phase. This is because of the distance in traveled path length. So the field intensity does not becomes 4 times higher. Hence the total radiated field is not 4 times higher. The impedance of a full wave (square) loop is therefore less then that of folded dipole.

Knowing the basic principle enables you to do the full mathematics.
 

    hitman1988

    Points: 2
    Helpful Answer Positive Rating
Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top