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I'm in the process of getting to grips with RF design and am having some trouble understanding why we should always match antennas to \$50\Omega\$.

I understand from the maximum power transfer theorem that if you have a fixed source (e.g. \$50\Omega\$) that you cannot change as a designer/user, then the maximum power transfer occurs if you set your load impedance to be the same as your source impedance. Therefore, if you are using an antenna as a transmitter, then you'd want to make the antenna (load) impedance equal to the impedance of the source for maximum power transfer into the antenna - so you match your antenna to \$50\Omega\$.

However, what if you are only using your antenna to receive? In that case the fixed variable is the load impedance (i.e. your receiver) which will be fixed at \$50\Omega\$ as below (I've also drawn the transmission line for completeness):

schematic

simulate this circuit – Schematic created using CircuitLab

If \$Z_L\$ is fixed (\$50\Omega\$), then the maximum received power will be when the voltage across its terminals is maximized. The voltage across \$Z_L\$ is (potential divider): $$ V_{Z_L} = V_s \frac{Z_L}{Z_L+Z_S} $$ \$Z_L\$ is fixed but we can "control" \$Z_S\$. Maximum voltage across \$Z_L\$ (and hence maximum power to the load) occurs when \$Z_S=0\Omega\$

Therefore, following this logic, you'd ideally want your receiving antenna to have a very low impedance (as close to \$0\Omega\$ as possible).

I'm probably missing quite a few fundamental concepts here but at the moment, the only reason I can think to match a receiving antenna to \$50\Omega\$ is in order to avoid re-reflections back to the load. However, if the load itself is already matched well to its \$50\Omega\$ transmission line, there should be minimal reflections already.

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  • \$\begingroup\$ Well, in general, as you say, maximum power transfer rules applies when you have a given source impedance and you can only modify your load impedance. In that case you cannot do better than matching the impedances. But when you design the output impedance of a 'source', you try to choose the best impedance that give you the maximum voltage (zero output resistance) or current (infinite output resistance). This will also increase the power transferred to the load with respect to a matched impedance case. Avoiding reflections and complying to standards is IMO what makes you choose a matched Z. \$\endgroup\$ – Sredni Vashtar May 8 '20 at 15:48
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However, what if you are only using your antenna to receive?

An antenna (e.g. monopole) has a complex impedance characteristic versus frequency: -

enter image description here

Picture from my answer here.

So, if you are tuned into a frequency that makes the antenna exactly a quarter wavelength (0.25 on the X axis above), it presents an impedance that is roughly 37 ohms resistive plus about 21 ohms inductive.

This means that you can maximize receive power transfer with a little bit of series capacitance to cancel the inductive reactance and, "produce" a receiving source having pure 37 ohms. Feed this into a matching 37 ohms to extract the most receive power.

But there's really nothing much else you can do.

A quarter wave monopole transforms the impedance of free space (roughly 377 ohms) down to 37 ohms when used at the "right frequency". Use it off-centre and the reactance can go positive (inductive) or negative (capacitive).

Therefore, following this logic, you'd ideally want your receiving antenna to have a very low impedance (as close to 0Ω as possible).

Well, you can do this - you can make the antenna intentionally "short" and it presents quite a bit of capacitive reactance and the radiation resistance will drop down to a fraction of 37 ohms (maybe 5 ohms). However, your received signal is also much smaller because the transformation ratio has dropped even lower. But, you can tune out the capacitance with series inductance and get a very highly electrically tuned antenna. Nice in some circumstances but bad in others.

Old crystal radios never had anything like a quarter wave monopole because if they did it would be nearly the length of 4 football fields (for the good old BBC long wave service at circa 200 kHz) BUT, the length was still fairly long and it's the length of the antenna that gives you the signal intensity (up to a certain point). It was still regarded as a "short" antenna though (electrically).

if you are using an antenna as a transmitter, then you'd want to make the antenna (load) impedance equal to the impedance of the source for maximum power transfer into the antenna - so you match your antenna to 50 Ω

You've got this a bit backwards - for a lot of set-ups, the antenna is somewhat distant to the transceiver and are connected via coaxial cable. Coax has a transmission impedance and, at that precise transmission impedance, signals are not reflected from a matching load. This is quite important. And so an antenna is usually matched to the coax but, an antenna (standard ones like monopoles and dipoles) don't look like 50 Ω so they are matched via a balun or resistor network at the antenna (load) end. This can mean that the transmitter end needn't match to anything like 50 Ω and, can drive its full power to the load.

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  • \$\begingroup\$ Dear Andy, Thank you for your reply. It does help a little bit. I think what I am missing is: what is the relationship between the power received and antenna impedance? Is there a better electrical model that might describe how the antenna functions (instead of the simplistic one in my question)? \$\endgroup\$ – Qbort May 8 '20 at 16:36
  • \$\begingroup\$ @Qbort antenna theory is quite complex but if you model an antenna on the graph in my answer then, for a monople you won't be far wrong. But there are other antenna types that have different impedances so, it's horses for courses. \$\endgroup\$ – Andy aka May 8 '20 at 16:48
  • \$\begingroup\$ @Qbort there's no straightforward relationship, given freedom to vary the antenna design. All kinds of techniques can raise or lower the impedance of an antenna without having a substantial impact on effective aperture, which is what really dictates receive power. 50 and 75 ohms do happen to be pretty good matches to half-wave dipoles, but they also happen to be good values for minimizing loss in practical (not ideal) coax. It's all engineering compromises. \$\endgroup\$ – hobbs May 9 '20 at 3:48
  • \$\begingroup\$ Ta v much @Neil_UK \$\endgroup\$ – Andy aka May 9 '20 at 9:33
  • \$\begingroup\$ @hobbs and Andy aka, thank you very much both for your help with this. I think I have a clearer idea. So just to summarise (is this correct?): Transmitting antenna: you want to match it to the 50 ohm coaxial cable, but not just for maximising power transfer, also to ensure you get no reflections back from the antenna. Receiving antenna: As I stated in my question, you do want to lower the impedance of your receiving antenna as far as possible (instead of matching to 50 ohms). However you only have the reactive (capacitive/inductive) part of the impedance to play with. \$\endgroup\$ – Qbort May 11 '20 at 9:08
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You want the impedance of the antenna to match that of whatever is driving it (transmit), or whatever the antenna is feeding in to (receive).

There is nothing magical about 50 ohms. In theory, it could be 75 ohms (and is for CATV and TV applications), 60 ohms, 90 ohms, or anything else. The 50 ohm standard for RF transmission lines and interfaces came about a long time ago, and almost all RF & microwave test gear was built to accommodate that standard. Hence 50 ohms became a de-facto standard, and so to maximize power transfer source and load impedances need to both be 50 ohms.

EDIT 1 - added tutorial reference

Here's a nice tutorial that explains it mathematically:

https://www.tutorialspoint.com/network_theory/network_theory_maximum_power_transfer_theorem.htm

And it doesn't matter what direction the power is flowing. The need for the load to match the source for maximum power transfer is the same in both cases.

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  • \$\begingroup\$ Dear @SteveSh, thank you for your reply but I still do not understand the logic. I can totally get the reasoning for a transmitting antenna to have to be matched to 50ohms (assuming the source impedance is also 50 ohms). However, I feel the story is very different if we are talking about a receiving antenna. In this case the antenna becomes the "source" and the load is fixed (50 ohm). As explained in my question, the way I see it is that the maximum transfer will occur when the antenna impedance is 0 ohms. \$\endgroup\$ – Qbort May 8 '20 at 14:51
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Good question. After reading the free "RF Design" magazine for many years, that after repeatedly/thoroughly reading "The Amateur Radio Handbook" and admiring those EIMAC 4CX-1000A triodes shown in the back, AND examining the primarily narrow-band behavior of RF circuits (antennas are NARROW BAND, maximum-power-transfer matching PI networks are NARROW BAND), I sat back and considered the impedance of 10pF of Cin (Cbe + Cob with Miller Effect) at 1,000 MHz.

That impedances is 16 ohms. Whereas dipoles are 73 ohms. And vertical whips are 37 ohms.

So we design for lowest added noise in the active circuits (the transistors) because that mindset provides the best possible Signal Noise Ratio, and SNR is the key to enjoyable music and minimal data-packet re-transmission requests.

And we achieve best SNR (to minimize "loss of signal" on your phone) by matching.

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    \$\begingroup\$ Hi there, thank you for taking the time to answer. I must say I don't understand what you are trying to say - would you be able to re-phrase or elaborate? \$\endgroup\$ – Qbort May 8 '20 at 16:30
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Fatal error in the question. One cannot control Zs without changing Vs at the same time, those quantities are coupled together. They both depend on the used antenna construction if the RF field is given and we assume the antenna is always directed for maximum available Vs at that place.

Of course the questioner were right if Zs and the caught Vs from a given RF field were independent, but they aren't except in case there's a series reactive part in antenna's Zs. Like another answer has already said, by inserting a negative external series reactance the total Zs becomes smaller and the receiver will get more power like the questioner expects.

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    \$\begingroup\$ Like I said, I'm still trying to learn RF design. The bits I have learned so far are from reading on the internet but still don't have a good grasp. Could you expand on "Fatal errors in the question" and what are the reasons why Zs and power received are tightly coupled? \$\endgroup\$ – Qbort May 8 '20 at 14:20

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