material effect on resonant frequency of cylindrical wire antenna

Hi.
I know that we use from copper for cylindrical wire antenna.
But I want to know that what is material effect on antenna performance?
for example,if conductivity of a wire antenna reduced,antenna parameters,such as resonant frequency,input impedance and gain what would change?
thanks

cylindrical wire antenna ? what ? like a Helix ( Helical) ?
well the 2 most common wire types would be Aluminium or Copper
I would doubt that any of the parameters you mentioned would be noticably or measureably affected when comparing one made of each type

Dave

Hi Dave
I know that aluminium and copper are good condcutor and results are same.
But you suppose you have a finite conductivity that is depende on frequency.now,what is your answer?

I agree with davenn. You probably can expect an effect in two cases
- using very thin wire with a DC wire resistance above e.g. 1 ohm
- using metals with low conductivity, e.g. NiCr or stainless steel

Calculating the antenna wire resistance (considering skin effect) and relating it to antenna radiation resistance should give a first estimation. For exact effect determination, use an EM solver.

Hi FvM.thanks
how relate skin depth to radiation resistance?while ohmic resistance is related to skin depth by:
ohmic resistance = 1/(2πa)*√(wμ/(2*σ))
skin depth = √2/(w*μ*σ)

I want to extract a analytical formula.where can I found a relation between radiation resistance and skin depth?

suppose,we have two monopole antenna,first with conductivity = 1e3 and second with conductivity = 1e8
I found that resonant frequency in second case increased.
why?
do you have any idea?

Some things to consider if you want to develop some (approximate) formula incuding everything:

The current distribution is dominated by a standing wave pattern. This is the case when in a current maximum Rrad + Rloss << characteristic impedance of antenna wire . This is the easiest case as radiation resistance (the part of the total resistance that is caused by radiation) is independent of resistance due to ohmic loss.

The current distribution is no long dominated by a standing wave (so the traveling wave component becomes important). This happens when Rrad+Rloss is no longer very small with respect to the characteristic impedance of the antenna wire. A sign of this is low Q factor. This one is already harder to handle as additional loss does change the radiation pattern, hence the total radiated power given certain input current.

The transmission line model is no longer valid. This is the case when X(L') is no longer large with respect to Rloss'.
L' is inductance/m of the antenna wire, Rloss' is the AC resistance/m of antenna wire. If this happens, the characteristic impedance becomes a complex number.

Given your conductivity range, you may run into situations where all three aspects are present and then answering your question is impossible without deep theoretical investigation.

Hi wim.
actually,in first case my antenna is not a conductor only.suppose we have a monopole antenna that is sorrounded by a lossy dielectric coating.
Like as previous my topic ('dielectric coated antenna'),when imaginary part of coating relative permitivity increase,electrical length of antenna decrease and so resonant frequency increase.

I can not understand why electrical length decrease.
I know that in this case propagation velocity decrease.But how electrical length decrease?

What values did you use for er' ( Re(er) ) and er" ( Im(er) )?
In your case you have a series circuit of a dielectric layer and air dielectric (the major part). Therefore the dielectric displacement current density (or just capacitive current) is mostly determined by the air dielectric.

Increasing er"/er' ratio in your thin dielectric layer reduces the capacitive current and increases the ohmic current (loss current) in the tin dielectric layer. In the extreme case (that is large er"/er'), all displacement current is no longer supported by er', but by er". So increasing the loss, reduces the capacitive effect of the thin dielectric. A reduced capacitive effect increases the propagation velocity, so the resonant frequency becomes higher. So you are right, electrical length reduces.

why I have a series circuit of a dielectric layer and air dielectric (the major part)?

At short distance from the conductor surface, the E-field is perpendicular to the surface (assuming rod with high conductivity, so no E-field component parallel to the metal-dielectric interface. The E-field lines first "see" your thin dielectric layer, and than the ambient air.

For the displacement current same situation. The current goes through the dielectric and than into air. Displacement current density in dielectric is same as in the air, of course all very close to the media interface.

If you would look to E*ds, just a fraction of E*ds is inside your thin dielectric. With "thin" I mean thickness << radius of wire. So rest (major part) of E*ds is present in air.

thanks a lot.
now can you exactly explain why when propagation velocity decrease,reosnant frequency increase?what is formula between them?

No, I can't explain what you want. When propagation velocity increases (due to reduced overall er'), resonant frequency increases also.

I know propagation velocity increase.it's ok.
my problem is that how relate resonant frequency to velocity.I know that:
resonant-frequency = propagation-velocity / wavelength.

Now,when propagation velocity increase,both resonant frequency and wavelength change.How you can say from this formula that resonant frequency increase,while wavelength change also.