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On Sun, Oct 13, 2019 at 11:03 AM, W5DXP wrote:
If we replace the balun with a 50 ohm dummy load, disconnect the antenna, and
measure the impedance looking back down the feedline, what do we measure?
TLDetails says very close to 189+j1000 ohms, actually considering losses,
195+j994 ohms, i.e. a near-conjugate match.
I agree. This is true of any lossless (or essentially lossless) matching network (be it a tuned feedline or a lumped-element matching network) that has been tuned to match an impedance attached to the network's output to, say, 50 ohms, If you then disconnect the load and attach 50 ohms to the input of the network, and then measure the impedance presented by the tuner at its output (using your NanoVNA, for example), the measured impedance should be the complex conjugate of the impedance that had been attached at the network's output. See attached drawing. I've used this technique to determine the "match-space" of various antenna tuners (that is, the range of impedances a tuner can match to an SWR of 1:1). If I attach 50 ohms to a tuner's INPUT and then measure the impedance at its output terminals as I vary its tuning controls, I am measuring the complex-conjugates of the load impedances that this tuner can match to 50 ohms. For an example, see: Of course, if a network is lossy, the measured impedance presented by the tuner at its output (when its input is terminated with 50 ohms) will diverge from the actual complex-conjugate of the load impedance. - Jeff, k6jca
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From: QRP RX: On Sun, Oct 13, 2019 at 04:29 PM, W5DXP wrote: For maximum power transfer, the impedance looking back down the transmission line from the EDZ feedpoint needs to be 175+j1000 ohms, the conjugate of the feedpoint impedance.
That's not true. I was just quoting the following from: Maximum Power Transfer Theorem for AC Circuits: It can be stated as in an active network, the maximum power is transferred to the load when the load impedance is equal to the complex conjugate of an equivalent impedance of a given network as viewed from the load terminals. ("equivalent impedance" means the impedance transformed by the transmission line's SWR not being equal to 1:1.) Here's a simple example based on TLDetails: 100w XMTR---50 ohm coax---1:1 balun==========59.5 deg 600 ohm feedline======== EDZ Zfp=189 - j1000 ohms There is an SWR of ~12:1 on the 600 ohm feedline. The forward power is ~350w and the reflected power is ~250w. The SWR on the coax is 1:1, i.e. zero reflected energy is reaching the source. The power delivered to the antenna is 97.9 watts, i.e. 97.9% efficiency in power transfer. Where did the 250 watts of reflected power go? When it encountered a 50 ohm Z0-match at the balun, all the reflected power was redistributed back toward the antenna load through the processes of re-reflection and total destructive interference toward the XMTR. There's only 2.1 total watts lost in the 600 ohm feedline, i.e. 0.094 dB, mostly I^2*R losses. If we replace the balun with a 50 ohm dummy load, disconnect the antenna, and measure the impedance looking back down the feedline, what do we measure? TLDetails says very close to 189+j1000 ohms, actually considering losses, 195+j994 ohms, i.e. a near-conjugate match. The conservation of energy principle tells us that if zero reflected energy is reaching the source (XMTR) in a low loss system, that most of the reflected power is being redistributed back toward the load (antenna). That's what Z0-matches do. More info at:
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On 2019-10-12 8:18 p.m., W5DXP wrote: From: QRP RX: The loss in the cable due standing wave happens with poor impedance match between receiver/cable for RX and cable/antenna for TX. A poor impedance match is usually a bad idea for coax but a conjugate match using low loss parallel feedline is often acceptable. For instance, according to TLDetails, the matched line loss for 65 ft. of RG8x at 7 MHz is 0.5 dB. The loss in 65 ft. of 600 ohm parallel line at 7 MHz feeding a 50 ohm load is 0.2 dB even though the SWR on the 600 ohm feedline is 12:1. The loss in the impedance matched RG8x with an SWR of 1:1 is 2.5 times the loss in the impedance mismatched (but conjugately matched) 600 ohm feedline with an SWR of 12:1.
In hopes of making things a little clearer to the newly arrived in
the world of r.f. energy transfer... Electrical energy, including r.f., one can say can be moved in the
form of current or in the form of e.m.f., the "push" on the energy, the voltage. (Of course, there is always some of each in any such transfer.) The reason that 600 Ohm open wire line, above, is so attractive is that the energy transfer is being made by voltage. Increasing voltage means decreasing current and so decreasing I2R losses. High powered short wave stations used open wire technique for that very reason, r.f. energy had to be "transported" a long way to get out to the vast antenna farms such stations used. 600 Ohm and 1,200 Ohm spacings were used. It's the same reason that hydroelectric installations, distant from
the consumer city, "transport" energy in the form of voltage rather than current. Very high voltage brings it's own set of losses from leakage across thousands of insulators, corona and other effects. It is the business of engineers to figure the best balance between the savings from low current and the losses from high voltage in the design of a particular long distance power line. In a radio station where a random length antenna is in use, the
impedance at the antenna feed point can vary hugely. A tuner at the transmitter can be adjusted to match whatever impedance is seen at the transmitter end of the open wire line. The standing waves on the open wire line involve little current and so the losses to I2R, as noted in the post above, are inconsiderable. John
at radio station VE7AOV
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On Sun, Oct 13, 2019 at 04:29 PM, W5DXP wrote:
For maximum power transfer, the impedance looking back down the transmission
line from the EDZ feedpoint needs to be 175+j1000 ohms, the conjugate of the
feedpoint impedance.
That's not true. If you have X not equals to zero, it means that load returns back RF energy. And it means that you have higher loss in the cable. The thing that you're talking about is that coax cable can be used as impedance transformer. But coax transformer works due to standing wave. And standing wave make higher loss in the coax cable. The maximum power transfer is achieved when X=0 and complex Z is equals for all components: Ztransmitter = Zcable = Zantenna. If some impedance is different, you will have higher loss.
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On 2019-10-12 11:32 a.m., David KD4E wrote: Does not a mismatch create problems with loss and noise?
If I connect a 50 ohm coax to a 300 ohm TV antenna, without a
transformer, should I not anticipate a poor quality received signal?
David, just to deal with that exact example, commercial digital
television,...no. Television stations, like other broadcasters, saturate their market
areas in signal so about anything will serve as an adequate antenna. Just look at those rubbish "magic" all channel antennas that are peddled by hard sell artists and that appear in peg board bubble packs! Their sole technical merit is a complicated name yet they work fine within stations' prime markets. A wire coat hanger would work just as well. The second reason for "no" is that these days we are dealing with
digital technique. The margin in signal strength between a perfect picture and no picture is very slim. Rather than noise, it involves pixelation as the receiver struggles to assemble a proper display. John
at radio station VE7AOV
Or am I misunderstanding?
David, KD4E
You know, after all the talk about matching impedances, I'm really
wondering .... we talk about POWER in watts but that's pretty much for
transmitters. But we talk about receivers in?? VOLTS. Well, okay,
microvolts,
usually. Somehow, I don't see a good reason to CARE about matching
receiver input impedance. Maybe it lowers the antenna bandwidth or Q?
Impedances have PHASE, but voltages do not. Care to comment?
Bill
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From: Starsekr: Bill, you get maximum power transfer when the impedance is matched. Jim, according to the AC (RF) maximum power transfer theorem, you get maximum power transfer when the load impedance is *conjugately* matched. Luckily, coax with a Z0 of 50+j0 ohms is a trivial conjugate match to a 50-j0 ohm load. For a non-trivial example, the feedpoint impedance of an Extended Double Zepp antenna may be 175-j1000 ohms. For maximum power transfer, the impedance looking back down the transmission line from the EDZ feedpoint needs to be 175+j1000 ohms, the conjugate of the feedpoint impedance.
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On Sun, Oct 13, 2019 at 07:48 AM, Starsekr wrote:
I like your answer, but for maximum energy transfer, the systems (transmitter
output, cable, antenna) and (antenna, cable, receiver input) should all be
matched. Each time you have a mismatch, there is a reflection, and signal
loss.
yes, exactly. But for cheap receiver there is no need high sensitivity, because it works with very strong signals and some mismatch didn't affect reception quality much, because the nearby transmitter signal is strong enough even with these losses.
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On Sat, Oct 12, 2019 at 06:51 PM, QRP RX wrote:
When you use TX, standing wave in the cable depends on cable load impedance
(antenna). This is why it is very critical to match your cable with antenna.
Note, I wrote cable (not TX output), this is not mistake, you're needs to
match your cable impedance with antenna.
When you use RX, standing wave in the cable depends on cable load impedance
(receiver input impedance). This is why antenna match is not critical for RX.
But for RX you're needs to match your RX input with cable impedance, otherwise
you will have the same issue as for TX.
I like your answer, but for maximum energy transfer, the systems (transmitter output, cable, antenna) and (antenna, cable, receiver input) should all be matched. Each time you have a mismatch, there is a reflection, and signal loss. Either energy lost in the final tank coil or energy loss at the antenna for your two examples. Dr. Ben Tongue did some work on crystal radio optimizing and tried to make each part of his crystal sets see the same impedance. If you lose signal in the antenna-cable part of the receiving system, you lose atmospheric noise by the same ratio, but the receiver noise is still there and your total signal/ noise ratio goes down.
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For active load with no reactance (such as cal-kit load, terminator or just some resistor) software correction may works ok. But when you try to measure reactive loads and resonant circuits, you will get errors. Because hardware 50 ohm ports will affect your circuit parameters.
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The problem here is that NanoVNA has hardware 50 ohm input and output. You cannot fix it with software settings. Even if you apply error-adapter correction for 75 ohm, your NanoVNA ports are still 50 ohm and it will affect Q-factor of measured device. So, you will get shifted resonance frequency, shifted bandwidth and other errors.
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Mini Circuits and several other companies sell 50-75 Ohm transformersjust for situations like this.??? There is also a 7 dB 'Min-Loss Pad'.? An
attenuator with resistance values such that one side is 50 Ohms and theother side 75 Ohms.? For S11 you would need to cal on the otherside of the 7 dB.pad.For S21/S12 you would need two pads and cal out the 14 dB of loss.
Kent
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On Saturday, October 12, 2019, 10:55:44 PM CDT, Starsekr via Groups.Io <Starsekr@...> wrote: On Sat, Oct 12, 2019 at 11:06 AM, Bill Hemmings wrote:
You know, after all the talk about matching impedances, I'm really
wondering .... we talk about POWER in watts but that's pretty much for
transmitters. But we talk about receivers in? VOLTS. Well, okay, microvolts,
usually. Somehow, I don't see a good reason to CARE about matching
receiver input impedance. Maybe it lowers the antenna bandwidth or Q?
Impedances have PHASE, but voltages do not. Care to comment?
Bill, you get maximum power transfer when the impedance is matched.? Wiki "Maximum Power Transfer Theorem" for details.? This applies to transmitter to coax to antenna to air (the ether).? Or the ether to antenna to coax to receiver.? Some energy is lost in the resistance of the coax wire, and in the inductance and capacitance that we measure as Xl and Xc.? When there is a change in impedance, a discontinuity, some of that energy is absorbed at the discontinuity and some is reflected, causing a standing wave.? The reflection moving back down the coax also loses the same proportion of energy as the first wave.? Where it hits the transmitter matching unit or final, is reflecfted forward to the antenna and loses energy again in the coax .? Etc.? Signal strength (in the ether) is measured in micro volts (uV) per meter because it is a field, not just a current in a wire.? If you have a mismatched receiving antenna system, you lose some of the signal, (and atmospheric noise, too) but retain the noise inside the receiver, so your signal to noise ratio goes down. also see: Jim, KA6TPR
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On Sat, Oct 12, 2019 at 11:06 AM, Bill Hemmings wrote:
You know, after all the talk about matching impedances, I'm really
wondering .... we talk about POWER in watts but that's pretty much for
transmitters. But we talk about receivers in? VOLTS. Well, okay, microvolts,
usually. Somehow, I don't see a good reason to CARE about matching
receiver input impedance. Maybe it lowers the antenna bandwidth or Q?
Impedances have PHASE, but voltages do not. Care to comment?
Bill, you get maximum power transfer when the impedance is matched. Wiki "Maximum Power Transfer Theorem" for details. This applies to transmitter to coax to antenna to air (the ether). Or the ether to antenna to coax to receiver. Some energy is lost in the resistance of the coax wire, and in the inductance and capacitance that we measure as Xl and Xc. When there is a change in impedance, a discontinuity, some of that energy is absorbed at the discontinuity and some is reflected, causing a standing wave. The reflection moving back down the coax also loses the same proportion of energy as the first wave. Where it hits the transmitter matching unit or final, is reflecfted forward to the antenna and loses energy again in the coax . Etc. Signal strength (in the ether) is measured in micro volts (uV) per meter because it is a field, not just a current in a wire. If you have a mismatched receiving antenna system, you lose some of the signal, (and atmospheric noise, too) but retain the noise inside the receiver, so your signal to noise ratio goes down. also see: Jim, KA6TPR
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From: QRP RX: The loss in the cable due standing wave happens with poor impedance match between receiver/cable for RX and cable/antenna for TX. A poor impedance match is usually a bad idea for coax but a conjugate match using low loss parallel feedline is often acceptable. For instance, according to TLDetails, the matched line loss for 65 ft. of RG8x at 7 MHz is 0.5 dB. The loss in 65 ft. of 600 ohm parallel line at 7 MHz feeding a 50 ohm load is 0.2 dB even though the SWR on the 600 ohm feedline is 12:1. The loss in the impedance matched RG8x with an SWR of 1:1 is 2.5 times the loss in the impedance mismatched (but conjugately matched) 600 ohm feedline with an SWR of 12:1.
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it doesn't means that impedance match cable/antenna is not needed for RX and impedance match transmitter/cable is not needed for TX. Such type of mismatch leads to signal loss. But this loss is not so high as loss due to standing wave in the cable.
The loss in the cable due standing wave happens with poor impedance match between receiver/cable for RX and cable/antenna for TX.
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On Sat, Oct 12, 2019 at 09:06 PM, Bill Hemmings wrote:
Somehow, I don't see a good reason to CARE about matching
receiver input impedance.
it doesn't related with volts or watts, impedance mismatch has the same poor effect for any volts and watts. Impedance mismatch just reduce signal level. But when you use cable, there is another issue - standing wave in the cable. Standing wave multiplies all losses in the cable. This is why standing wave in the cable leads to dramatic signal loss. When you use TX, standing wave in the cable depends on cable load impedance (antenna). This is why it is very critical to match your cable with antenna. Note, I wrote cable (not TX output), this is not mistake, you're needs to match your cable impedance with antenna. When you use RX, standing wave in the cable depends on cable load impedance (receiver input impedance). This is why antenna match is not critical for RX. But for RX you're needs to match your RX input with cable impedance, otherwise you will have the same issue as for TX. If you use RX with proper impedance match between RX input and cable impedance (both impedances should be equals, for example 50 ohm), there is no standing wave in the cable. If your antenna has bad impedance match with cable, it just will leads to low signal level, but there is no multiple losses and distortions in the cable due to standing wave, because there is no standing wave. Your receiver can use high gain amplifier in order to compensate low signal level from antenna. This is why antenna match is not critical for receiver. The same output impedance of transmitter is not critical, it just leads to lower efficiency, but there is no multiple losses and distortions due standing wave. Because standing wave is missing when you transmit to antenna which has good match with cable.
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Right on! There are many questions that don't seem to be asked, and answers don't
always much sense without some hands-on tests and data.
I got started with this round watching comparisons of assorted Baofeng-type antennas
for those really cheap Chinese walkie-talkies. (on youtube)
Then I got to wondering what difference does the impedance make.
One thing that makes some experimenting easy is telescoping antennas like the Nagy -704.
You can move the ''resonance'' all over the place!? For test signals there's things like garage door
openers and tire-pressure sensors.
Bill
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On 10/12/19 1:32 PM, David KD4E wrote: Does not a mismatch create problems with loss and noise?
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Somehow, I don't see a good reason to CARE about matching
receiver input impedance. Maybe it lowers the antenna bandwidth or Q? Certainly Q, which is good or bad, depending on application. Mismatch also plays with imbalance, so that e.g. coax from antenna to receiver can effectively become part of antenna, picking up noise between outdoor antennas and indoor receivers.
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Does not a mismatch create problems with loss and noise?
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Does not a mismatch create problems with loss and noise?
If I connect a 50 ohm coax to a 300 ohm TV antenna, without a
transformer, should I not anticipate a poor quality received signal?
Or am I misunderstanding?
David, KD4E
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Show quoted text
You know, after all the talk about matching impedances, I'm really
wondering .... we talk about POWER in watts but that's pretty much for
transmitters. But we talk about receivers in? VOLTS. Well, okay, microvolts,
usually. Somehow, I don't see a good reason to CARE about matching
receiver input impedance. Maybe it lowers the antenna bandwidth or Q?
Impedances have PHASE, but voltages do not. Care to comment?
Bill
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You know, after all the talk about matching impedances, I'm really
wondering .... we talk about POWER in watts but that's pretty much for
transmitters. But we talk about receivers in? VOLTS. Well, okay, microvolts,
usually. Somehow, I don't see a good reason to CARE about matching
receiver input impedance. Maybe it lowers the antenna bandwidth or Q?
Impedances have PHASE, but voltages do not. Care to comment?
Bill
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On 10/12/19 6:50 AM, Dr. David Kirkby from Kirkby Microwave Ltd wrote: On Sat, 12 Oct 2019 at 03:37, Starsekr via Groups.Io <Starsekr=
[email protected]> wrote:
David,
Could you give the group a short form on how this is done, and what a
NanoVNA user would do and see to measure something like a piece of RG6 or a
75-300 ohm transformer? Not to take away from your sales, but what kind of
error would a pad made from 1% carbon resistors introduce, compared to a
commercial pad? I don¡¯t sell 75 ohm kits, neither do I sell matching pads - only 50 ohm
kits and consultancy services.
I heard a report somewhere on here that the receiver port on the NanoVNA
is not particularly close to 50 ohms and varies on the batch a given device
is from. That¡¯s something I would like to fix myself, which could be done
with a few 1% resistors, after measuring the impedance.
Dave
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Starsekr
Oristo
