Finding VSWR of dipole antenna

Status
Not open for further replies.
A

Alan0354

Full Member level 4
Joined
Sep 6, 2011
Messages
214
Helped
30
Reputation
60
Reaction score
29
Trophy points
1,308
Visit site
Activity points
2,709
This is a homework problem. The question is to find VSWR of lossless dipole antenna connected to a 50Ω lossless tx line. The length of antenna is l=λ/4, λ/2,3λ/4, λ. Assuming the antenna resonates are the given length.
For lossless dipole,\[Z_{in}=R_{in}+jX_{in}\] where \[X_{in}=-jZ_{ant}\cot(\frac{kl}{2})\] and \[\Gamma =\frac{Z_{in}-Z_0}{Z_{in}+Z_0}\]

Obviously, \[Z_{in} \]is complex except length l=nΠ/2 where \[X_{in}=0\]. But the answer from the solution manual claimed \[\Gamma =\frac{R_{in}-Z_0}{R_{in}+Z_0}\] for all cases.

I can agree with l=λ/2 and l=λ where \[X_{in}=0\]. But not for λ/4 and 3λ/4. Please help.
 

You are correct, your formula for gamma is only valid for Im(Zin) = 0 (that is Xin=0). You can see that when plotting the impedance on a Smith Chart. When there is an Im(Zin) part, the impedance moves towards the edge of the Smith Chart.

Can it be that the length of the antenna is specified as the length of one half of the dipole? So when they state 0.25lambda, total dipole length = 0.5lambda.
You probably know that Rin changes significantly for feedpoint in a current maximum or in a voltage maximum. A thin wire give higher Rin when feedpoint is in voltage maximum (for example a full-wave dipole).
 
Thanks for the reply. It clearly said the length l for the dipole. So λ/4 is λ/8 on each side. Even for the λ/2 dipole, it is given \[Z_{in}=73+j42.5\]. Obviously one cannot use \[R_{in}\] only for calculating the Γ.

Thanks
 

The question stated ASSUMING the dipole antenna is at resonant. What is resonant length of a dipole antenna? Dipole antenna is a series resonance circuit where at resonance, the impedance is at it's lowest and equal to \[R_{in}\]. But in λ/4 and 3λ/4, that's not resonance!!! Also according to

\[Z_{in}=-jZ_{ant}\cot(\frac{kl}{2})\]

This mean at l=λ, input impedance is \[\infty\]. That is not the definition of series resonance where impedance is at lowest and is pure resistance!!!
 

An antenna is said to be resonant when its input impedance is real, so Im(Zin) = 0. In real world this means a half wave dipole is somewhat shorter than a half wave. The thicker the dipole, the more length reduction is required to get it into resonance.

When you only use transmission line theory without correcting for radiation, a full wave dipole shows infinite impedance. But in real world the impedance is high (can be in the kOhm range), but not infinite. You can see the radiation as a kind of transmission line loss (but the loss does not result in dissipation).

Note that a full wave dipole is in reality somewhat shorter to get Im(Zin)=0.

- - - Updated - - -

Dipoles not equal to a nice resonance length can be brought into resonance, but then additional inductance or capacitance is required.
 

thanks for the reply. Yes I understand that. I have been doing a lot of search, sounds like dipole is really at resonance at length approx l=nλ/2 where n=1,3,5,7.... As this will give series resonance where the reactance is close to zero.

I think the key of the question in my first post is the book tell us to assume the dipole antenna is at resonance. This, is not true, but the book just use this as an example to make the problem simpler.
 

You can also use n= 2, 4, etc, but then you get a high impedance. In such case it behaves like a parallel resonant circuit instead of a series circuit.

Dipoles < 0.5lambda are brought into resonance with capacitive end plates, series inductance or meandering. When doing this, Re(Zin) drops due to less radiated power given certain input current.
 

But that will be harder to drive significant current and extra component is need to bring the impedance down.

From what I read, dipole is consider series resonance circuit, but I see your point of it's like parallel resonance at n= 2,4.....
 

You are right, driving a full-wave dipole from a coaxial cable needs an impedance match to get the power into it. In many situation such step can be made together with a balun function (for example in a 50 Ohms to 200 Ohms balun).

If you make a full-wave dipole from very fat conductors (or wide strips), and reduce the length to get it into resonance, then the impedance reduces significantly. You can get it below 100 Ohms. The useful bandwidth is about 33% more over a half-wave dipole. They are also usefull in certain arrays, as high impedance array elements can be combined easily to get (for example) a 50 Ohms feed point.
 
Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…