Continue to Site

Welcome to

Welcome to our site! 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.

half wave dipole antenna

Not open for further replies.


Newbie level 5
Jun 26, 2015
Reaction score
Trophy points
Activity points
Why the current distribution of a half wave dipole antenna is sinusoid ?
I mean how the length of the antenna is related to the wave period ?
Is there a formula for get this result ?

YES FOR sure thereis a relation between antenna length and current period(or correspond frequency).
each antenna works under certain frequency range(named bandwidth)
you could convert this frequency to wave length by landa=c0/f formula
then since antenna is half wave it means antenna length is landa/2

Yeah. The length of each arm of a dipole is Lambda/4 at frequency obtain from formula f=c/lambda. So it is equal to the length of half sinosoid. if your decrease length more it shows a triangular current pattern at that frequency. It is simply because current is zero at open ends. decreasing size of dipole more result in lower gain and directivity at that particular frequency.
Hi thanks for response
I know that every antenna works at certain frequency
But my question is that why the length of the antenna is related to the wave frequency
I mean is that any relation that gives us sinusoid distribution along half wave dipole ?
Or how can we get how is the current distribution ?

to get current distribution along any antenna numerical method likes MoM, FEM, TLM, .. . can be useful.
I think the best efficient method is MoM for a linear dipole antenna. NEC is a non-commercial software based on MoM to do this.
Hi.. The answer can be stated from transmission line (tx) theory.

Two-wire tx line with source at one end and open at other, will have standing wave patterns of current and voltage which repeats lambda/2 apart..Current is zero at open end, max at lambda/4 from end, and zero again at lambda/2.. this repeats over line.. Now imagine the tx line itself is lambda/4 long.. Now flare the top conductor upwards and the bottom conductor downwards.. What you get is two lambda/4 segments with a source at centre, and hence a lambda/2 half-wave dipole.

It is obvious now the current is zero at top end, and goes to a maximum towards center when it reaches lambda/4.. similarly for the lower segment, current starts with zero at bottom end and goes to maximum at center when it reaches lambda/4..
Hence you get a half sinusoid.
  • Like
Reactions: amirsde and FvM


    Points: 2
    Helpful Answer Positive Rating


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

Part and Inventory Search

Welcome to