Antenna resistance

  • #1
Hello community

Sorry if this is a repost, but I don't find what i am looking for.

I am reading the book "anntenas", by Kraus and Marhefka

In page 30, says the antenna impedances are complex. Z{[tex]\tau[/tex]} = +
R{[tex]\tau[/tex]} + jX{[tex]\tau[/tex]} ..

Could anyone please tell me what does the complex part imply in real life, and why is it regarded as complex.
Thank you

edit: the tau is supposed to be subscripted, but it is appearing as superscript :s

Answers and Replies

  • #2
Impedance is used with complex notation for oscillation current. Suppose you have current oscillation at frequency ω and amplitude I0. Then the current is given by:

[tex]I = Re(I_0 e^{i \omega t}) = I_0 cos(\omega t)[/tex]

The complex voltage is related to complex current using analog of Ohm's law.

[tex]V = Re(Z I_0 e^{i \omega t})[/tex]

If Z is real, this is no different from Ohm's Law. But let's substitute your complex form in.

[tex]V = Re((R + iX) I_0 e^{i \omega t})[/tex]

[tex]V = I_0 Re(R e^{i \omega t} + X e^{i (\omega t + \frac{\pi}{2})})[/tex]

[tex]V = I_0 R cos(\omega t) - I_0 X sin(\omega t)[/tex]

The first term is still just IR, but second term is phase-shifted by 90°. This expression can also be re-written using a single cosine function.

[tex]V = I_0 \sqrt{R^2 + X^2} cos(\omega t + \tan^{-1}(\frac{X}{R}))[/tex]

In other words, the norm of Z gives you the ratio of peak voltage to peak current, and so plays a role of effective resistance, while the angle of Z in complex plane gives you the phase shift between voltage and current.
  • #3
hi K2

Thanks you for the reply

then let me ask the next obvious question.
why don't we then just use any 2 dimensional plane for the Z vector?

etheta is still sin(theta) + cos(theta), albeit the sin and cos part is summed, and not distinguishable
  • #4
Because then you can't write V = ZI. Multiplication is not defined in vector spaces, but is defined in complex space.

It's just a convenience, though. You can solve any circuit using ordinary voltage and current by breaking up each into sin and cos components. It just gets a little messy. Once you know a few tricks, working with impedances is very fast.
  • #5
hi K2

thank you very much.

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