Awesome. Thanks for the comment. The reflection of pulses was always one of my favorite demos. Really helps to illuminate what's really going on when one visualizes a short pulse moving down the line and reflecting from the opposite end !
@@MegawattKS Exactly. I've studied this material for my Amateur Extra Class license, but seeing the reflections on the scope makes all the difference. In this case seeing really is believing.
Thanks! Glad it was helpful. It was a little tricky to do without the nice lab equipment in the university lab, but these new instruments work amazingly well. (It's just my really old scope that was a bit of an issue. Had to buy a really long coax so it could show long pulses needed due to its 60 MHz bandwidth ;-)
Just found your channel. It's amazing that, even with what I thought was a fair understanding of the topics, I end your videos with a MUCH greater understanding. Really appreciate your effort!
The remarkable thing is that the coaxial line looks locally like a 50Ω resistor, but does not convert the energy into heat like a resistor, but rather transports this energy away. And if the end of the line is not properly matched, the energy can flood back like a water wave on the beach. 🌊
Yes ! It's pretty cool. And the energy storage is not frequency selective like an LC circuit. The energy throughout the spectrum is just stored in the guided EM wave traveling down (and possibly back depending on the termination as you said) on the line ! :-)
Glad it was helpful ! Thanks for the comment. Sorry for the delay in replying. This was always one of my favorite experiment/demos when doing them in class.
I'm a bit late to the party but thank you for such a splendid series of lectures. If you are ever stuck for a lecture topic maybe you could consider an explanation of common mode currents on coax; esp the generation, detection/measurement and elimination. Thanks again. M0XYM
Thanks for the suggestion! I will write that down and try to do one on that in the future. When investigating the dipoles we made in class that were fed with coax and no balun, I did some EM sims. Was surprised to see how the RF currents traveled dozens of wavelengths along the outside surface, back toward the transmitter (in the free-space case). I think the literature calls this "surface wave" propagation. I will try to dig up some of that material and see if I can assemble a vid as you said. Thanks ! 73's
@@MegawattKS I'd really look forward to that! I've done a bit of experimentation - I made a 'probe' out of a split ferrite bead with about 10 turns of enamalled wire wrapped around one half of the ferrite and detected the current with a 'Tiny SA' spectrum analyser. I got CM currents when the dipole wasn't resonant - however I didn't see any CM if the feed-line was terminated with an open or high pure resistance - but I thought I would! My theory is a bit thin here and it isn't well documented in texts - over to you!! Thanks, 73 M0XYM
@@M0XYM Cool. Here's a theory: The main thing that sets up the surface-wave (which results in CM currents), is that the element of the dipole that connects to the coax center lead creates electric field lines that terminate on the other element of the dipole (which is connected to the coax shield). But it _also_ creates electric field lines that terminate on the exterior of the coax shield, which is at the same potential as that other element. These E field lines induce current on the outer-surface of the coax shield (which is unmatched by any other currrent - leading to CM current). Assuming I understood your test setups, I'm guessing you're not seeing anything when terminating in open or otherwise because there's no dipole rod sticking out at the end (using the shield as a counterpoise and setting up the CM current situaton). Here's an article that might help. They show the currents, but unfortunately not the E field's I was trying to describe. They also show a quarter-wave "bazooka" balun that can be used to address the problem: incompliancemag.com/article/balanced-unbalanced-antenna-structures-and-baluns/ K
@@MegawattKS Many thanks for this - and yes you are correct - I didn't have a dipole inserted - back to the bench! Please keep posting your lectures, they are truely excellent.
Yes - its a BNC-Tee connector. 1M to 2F. (plug to two sockets). I just have the tip of the 10x scope probe plugged (gently) into one of the two sockets. The ground clip lead of the scope probe is clipped to the ground contact of the signal generator. Ordinarily I don't like using the ground clip leads on probes at RF due to parasitic inductance, but for this demo with my 60 MHz scope, slow pulses, and long cables, everything is at a fairly low frequency (long wavelength) and we can get away with it 🙂
Great video. P.S. at 4:10 of the video you said ideally the nanovna puts out sine waves (you also mention this at 8:29 of the video). In reality the nanovna output is a square wave, and this is one reason it's able to operate at very high frequencies (uses the high harmonic content of the square wave to extend its operating range). Just FYI, Don
Good point ! Above 300 MHz, it's using the third harmonic on the version I have (fifth above 900 MHz). I think it singles out the needed harmonic in the receiver using a specific IF, so in a sense I suppose both are correct. It's using the specific sinewave components of the Fourier series as needed and rejecting the others in the receive channel. It's an amazing instrument !
Look at its output using your scope and you will see it’s output is a 50 percent duty cycle square wave regardless of frequency. Yes, really an amazing device. I have the NanoVNA-F with push buttons and could not be happier.
Good discussion. I agree. Squarewaves are presented physically to the device under test. But 'effectively' only the one desired sinewave in the spectrum of the squarewave is in play in the data collection / measurement result. As long as the DUT is linear, this works :-) To elaborate: Based on what I see with my spectrum analyzer, below 300 MHz, they offset the LO frequency by 5 kHz in the receive mixing scheme (which implies the DSP must use a 5 kHz IF). The third and fifth harmonics etc within the squarewaves being mixed will convert to frequencies outside this 5 kHz IF (e.g. 15 kHz, 25 kHz, ...) and be ignored, so effectively only the fundamental sinewave in the Fourier series of the port 1 squarewave signal is being used for the reflected (S11) or transmitted (S21) measurements taken below 300 MHz. But, it gets even more interesting in the higher frequencies where harmonics are used. My SpecAn shows that if I set the VNA to 600 MHz CW, it's outputting 200.000 MHz plus expected harmonics of the square wave from port 1 as expected to produce the 600 MHz third harmonic - but port 2 shows a 120.001 MHz fundamental (and squarewave harmonics). [I can't see this LO bleedthrough well enough on port 1 in the midst of the strong forward wave signal so I switched to port 2]. This implies they are using harmonic mixing (as expected) above 300 MHz, but with the internally produced squarewave LO feeding the RX channel mixer set to inject the _5th_ subharmonic (120 MHz) - with a 0.001 MHz offset needed to produce the needed 5 kHz IF. I.e. the 5th harmonic sinewave in the 120.001 MHz square wave's Fourier spectrum at 600.005 MHz mixes with the third harmonic sinewave in the 200.000 MHz port 1 forward wave squarewave's Fourier series to produce the 5 kHz IF that is processed. So only the 600 MHz sinewave is being received and processed and all the other harmonics and mixing products are ignored. Amazing this all works - but to be fair, they didn't totally invent it. The original 8410 VNA system developed by HP used rich harmonic waveforms and very-low IF in the receiver section to work into GHz regions, back in the 60's :-) 73
How can the phase of the open reflection be zero degrees? It is not "in phase" with the stimulus, so phase must mean something else. Does the VNA measure the voltage and current phase? How does a capacitor or inductor "change phase" for the VNA? I understand, for example, that current leads voltage in a cap. Or is the VNA actually looking at the timing of the wave form? So for a given length of coax, the reflection peak, positive or negative, can vary in timing from being in sync with the stimulus, or can get farther from the stimulus peak, until it reaches a position of then getting closer to the next peak...If you have a vid on this, I'd be happy to watch it.
Its zero-degrees in the sense that the forward and reverse traveling waves are in-phase with each other, but _only_ at the open-circuit load location. They are in general not in-phase with the stimulus since there's a time (and hence phase) delay. Unless the line is an integer number of half-wavelengths long. Here's the more detailed explanation: Define "in-phase" to mean that at a given time t the waveform value is at the same point on the sinewave cycle (e.g. the positive-going zero crossing). The forward traveling wave (voltage and current - which are in-phase with each other due to the characteristic impedance of the line) hits the open after traveling towards it at high speed (e.g. 0.7 times the speed of light). When it arrives there, the current has no-where to go since its an open circuit. So for charge conservation reasons (KCL), there must be a new current wave "launched" back into the line in the opposite direction. That current sinusoid sees the line as a resistance of value Zo and creates an associated reflected voltage wave Vref = Iref * Zo. Since the reflected current wave is the same magnitude as the forward one, Vref has the same magnitude as the forward one (Vfwd). Vref is also in-phase with Vfwd since the current reversed direction instantaneously at the open-circuit - so there is no time delay and hence no phase delay (at that point in space). Of course the VNA is at the other end and sees things differently. Hence the need for the calibration step. Hope that helps.
The simplest way I can think of is to setup to read S11 in dB readout mode and then calibrate the VNA to an open (nothing on it's SMA connector). Then put on a long piece of cable (with suitable connector/adapters), read the negative S11 value in dB, divide by 2, and then divide by the cable length to get dB/unit-length.
Another way is to do an insertion-loss measurement using S21. This video might help. I've keyed it up at the point where that's done on a 7 meter cable so we can see the cable loss... ruclips.net/video/LfGzf1_g2_c/видео.html
First, note that it is the return-loss or SWR that matters, not really inductance. Fortunately, as noted in the video, the NanoVNA is fundamentally measuring return-loss and can report it in dB when in the S11 trace mode (if properly calibrated). In general, most hams and experienced engineers will consider a 10 dB return loss (S11 = -10 dB or better) as generally good enough. For a transmitter, that means 90% of the power radiates (assuming no losses in the cable). Getting a better percentage radiated won't actually increase range much (though it can be a little nicer on the electronics). This is the same as the classic goal of < 2:1 SWR. Some people try to tune for 20 dB return loss (S11 = -20 dB, or VSWR = 1.2:1). But that may be hard to do with a fan-dipole which I assume is being used to work over a broad bandwidth. Hope that helps. For more, please see this video: ruclips.net/video/hLPNOp0RXD8/видео.html
Yes - I think so. It's been a while, but from looking at the pics and trying to remember, I think I had the other sig-gen output outputting the same waveform and used it to trigger the scope.
I've watched many videos about the NanoVNA and Smith charts. This video is among the best. Thanks so much for your straight forward presentation!
Awesome. Thanks for the comment. The reflection of pulses was always one of my favorite demos. Really helps to illuminate what's really going on when one visualizes a short pulse moving down the line and reflecting from the opposite end !
@@MegawattKS Exactly. I've studied this material for my Amateur Extra Class license, but seeing the reflections on the scope makes all the difference. In this case seeing really is believing.
Best presentation of Smith charts that I've seen. Nice work. Thank you.
Thank you sir. I worked on M.R.I. equipment for years. You brought back fond memories.
You're very welcome. Sounds like interesting work !
Wonderful illustration and great to see both time domain and polar representations.
Thanks! Glad it was helpful. It was a little tricky to do without the nice lab equipment in the university lab, but these new instruments work amazingly well. (It's just my really old scope that was a bit of an issue. Had to buy a really long coax so it could show long pulses needed due to its 60 MHz bandwidth ;-)
I love your series. I am learning so much! Thank you for doing these demonstrations.
You are so welcome! Thanks for leaving the comment - and sorry for the delay in replying.
Just found your channel. It's amazing that, even with what I thought was a fair understanding of the topics, I end your videos with a MUCH greater understanding. Really appreciate your effort!
Awesome, thank you!
excellent presentation.. Thank you for showing this.
Fantastic description of the basics for newcomers and makers, NO MATHS!
Thanks. Glad to hear it made sense! This was one of my favorite demos in RF and microwaves courses I used to teach :-)
The remarkable thing is that the coaxial line looks locally like a 50Ω resistor, but does not convert the energy into heat like a resistor, but rather transports this energy away. And if the end of the line is not properly matched, the energy can flood back like a water wave on the beach. 🌊
Yes ! It's pretty cool. And the energy storage is not frequency selective like an LC circuit. The energy throughout the spectrum is just stored in the guided EM wave traveling down (and possibly back depending on the termination as you said) on the line ! :-)
Excellent explanation and demo. Outstanding! Thank you.
Glad it was helpful ! Thanks for the comment. Sorry for the delay in replying. This was always one of my favorite experiment/demos when doing them in class.
Like this demonstration thoroughly.
Thanks, very good explanation!
I'm a bit late to the party but thank you for such a splendid series of lectures. If you are ever stuck for a lecture topic maybe you could consider an explanation of common mode currents on coax; esp the generation, detection/measurement and elimination. Thanks again. M0XYM
Thanks for the suggestion! I will write that down and try to do one on that in the future. When investigating the dipoles we made in class that were fed with coax and no balun, I did some EM sims. Was surprised to see how the RF currents traveled dozens of wavelengths along the outside surface, back toward the transmitter (in the free-space case). I think the literature calls this "surface wave" propagation. I will try to dig up some of that material and see if I can assemble a vid as you said. Thanks ! 73's
@@MegawattKS I'd really look forward to that! I've done a bit of experimentation - I made a 'probe' out of a split ferrite bead with about 10 turns of enamalled wire wrapped around one half of the ferrite and detected the current with a 'Tiny SA' spectrum analyser. I got CM currents when the dipole wasn't resonant - however I didn't see any CM if the feed-line was terminated with an open or high pure resistance - but I thought I would! My theory is a bit thin here and it isn't well documented in texts - over to you!! Thanks, 73 M0XYM
@@M0XYM Cool. Here's a theory: The main thing that sets up the surface-wave (which results in CM currents), is that the element of the dipole that connects to the coax center lead creates electric field lines that terminate on the other element of the dipole (which is connected to the coax shield). But it _also_ creates electric field lines that terminate on the exterior of the coax shield, which is at the same potential as that other element. These E field lines induce current on the outer-surface of the coax shield (which is unmatched by any other currrent - leading to CM current). Assuming I understood your test setups, I'm guessing you're not seeing anything when terminating in open or otherwise because there's no dipole rod sticking out at the end (using the shield as a counterpoise and setting up the CM current situaton). Here's an article that might help. They show the currents, but unfortunately not the E field's I was trying to describe. They also show a quarter-wave "bazooka" balun that can be used to address the problem: incompliancemag.com/article/balanced-unbalanced-antenna-structures-and-baluns/ K
@@MegawattKS Many thanks for this - and yes you are correct - I didn't have a dipole inserted - back to the bench! Please keep posting your lectures, they are truely excellent.
Great information. What connector are you using at the signal generator? is it just a double ended BNC?
Yes - its a BNC-Tee connector. 1M to 2F. (plug to two sockets). I just have the tip of the 10x scope probe plugged (gently) into one of the two sockets. The ground clip lead of the scope probe is clipped to the ground contact of the signal generator. Ordinarily I don't like using the ground clip leads on probes at RF due to parasitic inductance, but for this demo with my 60 MHz scope, slow pulses, and long cables, everything is at a fairly low frequency (long wavelength) and we can get away with it 🙂
great content, very informative and didactic. Please keep it going :D
Thanks for the encouragement. Will try to make more. (Just starting on a new "Radio Design 101" series.)
Great video. P.S. at 4:10 of the video you said ideally the nanovna puts out sine waves (you also mention this at 8:29 of the video). In reality the nanovna output is a square wave, and this is one reason it's able to operate at very high frequencies (uses the high harmonic content of the square wave to extend its operating range). Just FYI, Don
Good point ! Above 300 MHz, it's using the third harmonic on the version I have (fifth above 900 MHz). I think it singles out the needed harmonic in the receiver using a specific IF, so in a sense I suppose both are correct. It's using the specific sinewave components of the Fourier series as needed and rejecting the others in the receive channel. It's an amazing instrument !
Look at its output using your scope and you will see it’s output is a 50 percent duty cycle square wave regardless of frequency. Yes, really an amazing device. I have the NanoVNA-F with push buttons and could not be happier.
Good discussion. I agree. Squarewaves are presented physically to the device under test. But 'effectively' only the one desired sinewave in the spectrum of the squarewave is in play in the data collection / measurement result. As long as the DUT is linear, this works :-) To elaborate:
Based on what I see with my spectrum analyzer, below 300 MHz, they offset the LO frequency by 5 kHz in the receive mixing scheme (which implies the DSP must use a 5 kHz IF). The third and fifth harmonics etc within the squarewaves being mixed will convert to frequencies outside this 5 kHz IF (e.g. 15 kHz, 25 kHz, ...) and be ignored, so effectively only the fundamental sinewave in the Fourier series of the port 1 squarewave signal is being used for the reflected (S11) or transmitted (S21) measurements taken below 300 MHz.
But, it gets even more interesting in the higher frequencies where harmonics are used. My SpecAn shows that if I set the VNA to 600 MHz CW, it's outputting 200.000 MHz plus expected harmonics of the square wave from port 1 as expected to produce the 600 MHz third harmonic - but port 2 shows a 120.001 MHz fundamental (and squarewave harmonics). [I can't see this LO bleedthrough well enough on port 1 in the midst of the strong forward wave signal so I switched to port 2]. This implies they are using harmonic mixing (as expected) above 300 MHz, but with the internally produced squarewave LO feeding the RX channel mixer set to inject the _5th_ subharmonic (120 MHz) - with a 0.001 MHz offset needed to produce the needed 5 kHz IF. I.e. the 5th harmonic sinewave in the 120.001 MHz square wave's Fourier spectrum at 600.005 MHz mixes with the third harmonic sinewave in the 200.000 MHz port 1 forward wave squarewave's Fourier series to produce the 5 kHz IF that is processed. So only the 600 MHz sinewave is being received and processed and all the other harmonics and mixing products are ignored.
Amazing this all works - but to be fair, they didn't totally invent it. The original 8410 VNA system developed by HP used rich harmonic waveforms and very-low IF in the receiver section to work into GHz regions, back in the 60's :-)
73
@@MegawattKS Awesome explanation!
Just beautiful.
This is gold.
Thanks !
Thanks
How can the phase of the open reflection be zero degrees? It is not "in phase" with the stimulus, so phase must mean something else. Does the VNA measure the voltage and current phase? How does a capacitor or inductor "change phase" for the VNA? I understand, for example, that current leads voltage in a cap. Or is the VNA actually looking at the timing of the wave form? So for a given length of coax, the reflection peak, positive or negative, can vary in timing from being in sync with the stimulus, or can get farther from the stimulus peak, until it reaches a position of then getting closer to the next peak...If you have a vid on this, I'd be happy to watch it.
Its zero-degrees in the sense that the forward and reverse traveling waves are in-phase with each other, but _only_ at the open-circuit load location. They are in general not in-phase with the stimulus since there's a time (and hence phase) delay. Unless the line is an integer number of half-wavelengths long. Here's the more detailed explanation: Define "in-phase" to mean that at a given time t the waveform value is at the same point on the sinewave cycle (e.g. the positive-going zero crossing). The forward traveling wave (voltage and current - which are in-phase with each other due to the characteristic impedance of the line) hits the open after traveling towards it at high speed (e.g. 0.7 times the speed of light). When it arrives there, the current has no-where to go since its an open circuit. So for charge conservation reasons (KCL), there must be a new current wave "launched" back into the line in the opposite direction. That current sinusoid sees the line as a resistance of value Zo and creates an associated reflected voltage wave Vref = Iref * Zo. Since the reflected current wave is the same magnitude as the forward one, Vref has the same magnitude as the forward one (Vfwd). Vref is also in-phase with Vfwd since the current reversed direction instantaneously at the open-circuit - so there is no time delay and hence no phase delay (at that point in space). Of course the VNA is at the other end and sees things differently. Hence the need for the calibration step. Hope that helps.
Wonderful
Thank you!
Thanks!
Great video I am trying to use the nano Vna to see coax loss at UHF > 450mhz any ideas?
The simplest way I can think of is to setup to read S11 in dB readout mode and then calibrate the VNA to an open (nothing on it's SMA connector). Then put on a long piece of cable (with suitable connector/adapters), read the negative S11 value in dB, divide by 2, and then divide by the cable length to get dB/unit-length.
Another way is to do an insertion-loss measurement using S21. This video might help. I've keyed it up at the point where that's done on a 7 meter cable so we can see the cable loss... ruclips.net/video/LfGzf1_g2_c/видео.html
@@MegawattKS Much appreciated will give it a go.
thanks, how would i check my fan dipole? i mean what is a good number coming back? to yaesu 757 for indtance?
First, note that it is the return-loss or SWR that matters, not really inductance. Fortunately, as noted in the video, the NanoVNA is fundamentally measuring return-loss and can report it in dB when in the S11 trace mode (if properly calibrated). In general, most hams and experienced engineers will consider a 10 dB return loss (S11 = -10 dB or better) as generally good enough. For a transmitter, that means 90% of the power radiates (assuming no losses in the cable). Getting a better percentage radiated won't actually increase range much (though it can be a little nicer on the electronics). This is the same as the classic goal of < 2:1 SWR. Some people try to tune for 20 dB return loss (S11 = -20 dB, or VSWR = 1.2:1). But that may be hard to do with a fan-dipole which I assume is being used to work over a broad bandwidth. Hope that helps. For more, please see this video: ruclips.net/video/hLPNOp0RXD8/видео.html
Hi what is ch2 of scope and trigger on scope connected to? Sig gen?
Yes - I think so. It's been a while, but from looking at the pics and trying to remember, I think I had the other sig-gen output outputting the same waveform and used it to trigger the scope.
@@MegawattKS thanks, great videos by the way!
I would like to give you more than onle like 😊.
Thanks. You just did :-)