0
\$\begingroup\$

I see there are already a few questions asked wether it is better to physcally tune the antenna for said frequency and impedance or use a matching network. It seems the conscenus is often the physical way is prefered firstly. But id like to know exactly why. From my current understanding, the matching network will infact reduce any reflective power due to mismatching, logically then that "not-reflected-power" would lead to more power to the antenna and hence more power to the air? I suspect just because more power is going to the antenna, it is probably wasted in some form and not 100% propagated, so in essence matching a mismatch will always help partly, but just not as if the antenna is actually tuned for that frequency.

So id like to know if im on the right track here, and if so, in a case i had matching network to increase power to the antenna since im not TX at its natural frequency, AND hyperthetically there is no ohmic loss in the antenna and matching network, would the now matched system behave asif it was at its original resonante frequency?

Edit 2: I gather from existing answers in reality that non relfected power is still going to be wasted in parastici losses

\$\endgroup\$

2 Answers 2

Reset to default
2
\$\begingroup\$

From my current understanding, the matching network will infact reduce any reflective power due to mismatching, logically then that "not-reflected-power" would lead to more power to the antenna and hence more power to the air?

Generally true i.e. whether it's matched due to physical length tuning or, matched due to an impedance matching network then, the result is the same with the following caveats: -

  • An antenna that is a lot different in length compared to the physical optimum may have a very low radiation resistance and, that can make it trickier to match to.
  • An antenna with a lower-than-ideal radiation resistance will have a loss-resistance that is much more significant i.e. greater losses have to be incurred and hence less power transferred to free-space.
  • An antenna with a low radiation resistance will necessarily require a matching network that is high in Q-factor and, as a result, the usable bandwidth of the antenna will be significantly reduced.

A simple quarter-wave monopole might have a radiation resistance of 37 Ω but, it will also have a series loss of a couple of ohms and, when you reduce the antenna length, you disproportionately lower the radiation resistance and, quite soon, losses ramp up to near 50% or worse. There is nothing you can do about correcting these losses.

\$\endgroup\$
6
  • \$\begingroup\$ Thankyou for the straightforward responses. So i would assume in the 1/4 monopole case, the most optimum solution for a 50ohm receiver/emmiter is to match the 37 to to 50 at said frequency, and I would assume assuming ideal matching there would be no loss in doing so? Also what about if the antenna had 0 real resistance, but wasnt at a naturall frequency, where does that transmitted power go? Thanks \$\endgroup\$
    – George kirby
    Commented Jan 5 at 21:53
  • 1
    \$\begingroup\$ @Georgekirby I used 37 ohm as the value of radiation resistance because, an ideal monopole antenna actually and theoretically has this value. You then match the 37 ohm of the antenna to the feedline (probably 50 ohm). Ideal matching means minimal loss. For an antenna to be valid it must present a radiation resistance to the electrical terminals or, power doesn't get transmitted and will be wholly reflected. \$\endgroup\$
    – Andy aka
    Commented Jan 5 at 22:09
  • \$\begingroup\$ I see that makes sense, on the second note Apologies when i said 0 real resistance i mean the part that isnt its ‘radiation resistance’, as in, when for example the antenna is not at its resonant frequency, but there is a matching network in place to prevent reflections, where does the non propagated energy go? \$\endgroup\$
    – George kirby
    Commented Jan 5 at 22:20
  • \$\begingroup\$ @Georgekirby please rephrase your question because I'm getting confused by it (it's late in the evening round here so I may not reply till tomorrow) \$\endgroup\$
    – Andy aka
    Commented Jan 5 at 23:48
  • 1
    \$\begingroup\$ @Georgekirby the matching network will improve the power to the antenna but, as I said in my 3rd bullet point in my answer, the Q-factor of the matching network may restrict the usable bandwidth of the signals transmitted to the antenna. Also, because the antenna may not be an ideal quarter wave dipole (for instance) the radiation shape may be affected; the power will still be transmitted but, the lobe shapes (in extreme cases) may make the antenna not as one might want it. If you need more detail on how this might pan out, I suggest you ask a new question and get other contributors involved. \$\endgroup\$
    – Andy aka
    Commented Jan 9 at 14:01
0
\$\begingroup\$

The physically bigger a conductor is, the lower its resistance is. Ferromagnetic and dielectric materials used to increase inductance and capacitance of small components also absorb energy. Thus, physically larger inductors and capacitors tend to have lower loss. The limiting case is where the inductors and capacitors in the matching network are as big as possible: they are the antenna itself.

\$\endgroup\$
5
  • \$\begingroup\$ Hi, little confused on relating this to the question of radiated power on the antenna. I thought given such high frequencies the capacitor and inductors in matching networks are fairly low values? \$\endgroup\$
    – George kirby
    Commented Jan 5 at 21:58
  • 1
    \$\begingroup\$ Compare them to the inductance or capacitance of a wire of the same physical size as your component. \$\endgroup\$
    – John Doty
    Commented Jan 5 at 22:34
  • 1
    \$\begingroup\$ Power lost in resistance or hysteresis is not radiated. \$\endgroup\$
    – John Doty
    Commented Jan 5 at 22:34
  • \$\begingroup\$ I see so this is in terms of why the matching network might not always be a perfect thing. So say in an ideal world with no real resistance, would the now matched entire sytem radiate all that incoming power? \$\endgroup\$
    – George kirby
    Commented Jan 5 at 22:59
  • 1
    \$\begingroup\$ If you have a perfect lossless matching network and your antenna itself has no ohmic or other dissipation, then all of the power you feed to your antenna will be radiated, yes. Conservation of energy. \$\endgroup\$
    – John Doty
    Commented Jan 5 at 23:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.