I've always understood characteristic impedance as the impedance seen through the dielectric, by a sharp edge signal. This video helped me to "grok" it stronger. Thanks for these videos, Robert! I learn so much from you, Eric Bogatin, Rick Hartley, and others.
I've been teaching electronics for over 40 years and THIS is why people have such a difficult time with very simple subject matter. You will never read 50 ohms using a DC meter... period. The "characteristic" impedance only applies to signals that are within the design parameters of the transmission line. How do you explain the characteristic impedance of a waveguide to students, then? In coaxial lines, the characteristic impedance (Z0) is determined by the dielectric constant of the chosen insulator (you did not even broach this) as the signal is wholly contained within the cable (assuming no ingress or egress). If the inner and outer conductors' radii are maintained, Z0 is a function of the dielectric constant , alone, and the choice of dielectric (PE, PTFE, air, FE, etc.) will also affect the propagation velocity. Impedance is the complex summation of inductive reactance, capacitive reactance, and resistance. There is NO WAY to ever see the impedance of the circuit without the reactances, which are frequency dependent. A DC incident signal with a steep leading edge will contain far too much HF to be useful. Most TDR's, for this reason, use a sinusoidal waveform (half wave) for the incident signal. More accurate and reliable TDR's use complex waveforms (spread spectrum reflectometry). Q: "Why don't you measure 50 Ohms on a 50 Ohm cable?" A: The simple answer is that impedance is complex (multiple formulas and parameters) and Ohm Meters are not designed to measure anything but resistance.
Exactly, Retired Engineer! Time domain refectometry (TDR) names the speed of an electric signal through a characteristic coaxial impedance at about sixty percent of light-speed in vacuum.
This. It is unbelievable how difficult it is to find any material nor any lecturer who could teach anything both right and understandably about transmission line impedances, reflections, and most of the RF phenomenons. I am a skilled professional on electronics and electricals, but even I still struggle to understand those subjects. After long-term struggle though, I understand the most basics, including what this video is trying to say and why it is wrong. At the beginning of this struggle, I was very frustrated about the prevailing teaching concepts to suddenly start telling and calculating about mystical waves which reflect within cable and need specifically designed cable, without ever explaining what they are in electrical terms. And not mentioning about basic things applying always, even though negligibly in low frequencies, such as any transmission line having charasteristic capacitance and inductance, and reflection occuring and magnetic flux and electric field existing even with slow dc pulses in regular cables. Mentioning those would have given consistent perspective, instead of making confusing thinkings that cranking up the frequency suddenly brings us into another mystery world where exists things not existing in the LF world, where ohms mean totaly diffent thing, and where we on the other hand don't even think about basic voltage and current calculations anymore... The , even after getting over that conceptual stumblestone, I still find that most materials, videos, and lectures have either way too much omissions, oversimplifications, even outright errors, OR they are way too hard to understand for unfamilar about those subjects, even if skilled in electricals otherwise.
Oh man, you could not be more wrong. A transmission line at the instant a signal is applied to it looks like an impedance equal to its characteristic impedance. An ideal 50 ohm line will look like a 50 ohm resistor when a wave is launched into it. The characteristic impedance is the ratio between voltage and current of a propagating wave inside a minuscule amount of T-line. When you connect an ohmmeter to it (instantly), it will launch a wave into it, and the wave impedance will look like 50 ohms. Q: "Why don't you measure 50 Ohms on a 50 Ohm cable?" A: you will, until the reflection comes back and changes the reading. If you had a fast enough "ohmmeter", you'd see 50 ohms until the first reflection comes back.
Many informed individuals don't seem to understand that these inductances and capacitances do NOT return the energy they absorb to the transmitter; instead they pass it along to their neighbors. Thus, the energy is transmitted down the transmission line. The transmitter ONLY sees that the energy it delivers doesn't come immediately back (+/- 90 deg.) as it would from C or L (or LC). Instead, the energy seems to be lost as from a pure resistance. IF the cable were lossless and light-seconds long, the Ohm meter WOULD measure it as 50 Ohm... at least for a few seconds. As far as the transmitter is concerned, it does not matter whether the energy is converted to heat, excites a cavity, or radiates into space -- since it is lost to the transmitter, it looks like a resistance.
Taking the paradigm of DC resistance with a meter people understand and transfer it to RF is kinda waste of time. As you said most coax cables do not support the speed of light, due dielectric absorbtion, which is minimized in better coax with sparse placement of plastic spacers. Both speed and power is lost. RF Impedance is better explained otherwise.
Assuming your coax is RG058 (specified impedance of 50 ohms), the velocity factor is 65, *not* 100, so your hypothetical measurement time should be a bit over 3 seconds.
At 5:00, the first reflection will actually be 2 V. If the meter's output impedance is 50 Ohm, it will absorb this transient via the voltage divider to yield 1 V; otherwise, the power will be reflected again and again until the transient is absorbed by system losses..
I understand imprdance as a value that had to be measured from a load indirectly or with a TDR. One indirect method being with an ammeter and a voltmeter and the other being a voltmeter comparing the voltage across a load and one across a variable resistor. The voltages across the two being equal reflecting the resistors current value being the impedance. What he seems to be saying here is that current for a time induces current in a return wire creating a temporary circuit in which messurements could be made.
The impedance of a coaxial cable refers to the capacitance reactance of the wire and the shield and is dependent on the dielectric and separation distance of the conductors among other factors
The transmission line is a series inductor and a parallel capacitance over and over as you go down the line, so it takes that time for your little one gold signal to charge all of those reactions and they build up until you will eventually read an open circuit. But the characteristic impedance of that line did not change
I think is easy to test, just need a osciloscope, a long coax cable and a single shot step signal and a resistor, you can measure on channel 2 the voltaje across the resistor and so you know the current, and you should see a constant current for the first nanosecond and then slowly go to 0, ( a few nanoseconds).
"Why don’t you measure 50 OHM on a 50 OHM cable? " Mostly because I don't want to; which is not the same as wanting not to, but merely not seeing much need for it. If you make it long enough it will indeed measure 50 ohms end-to-end with a DC ohmmeter.
I think the simplest explanation is to say that your meter can't measure the far-end impedance until the signal has a chance to get there and back; the 50 ohm characteristic impedance marks what your meter would see while awaiting the "real" measurement.
just for my understanding, the virtual coax cable to the moon has to be very thick in order to have 50ohm on this long distance right? and what about the capacity, you didnt say anything about that. i mean if the cable is so long than the capacity should be big too. and a big cap looks like 0 ohm at the beginning of loading, how does that fit together with what you said?
If you put any sort of cable even to low earth orbit you'd have millions of volts until whatever you made got destroyed. It's a thought exercise and erroneous at that.
@Google user And as far as where the power comes from. It's about 100 volts per meter above surface, you do the math. Ever seen the ionosphere, those lovely pictures.
When we have a 50ohm cable, it is always 50ohm and it is not relevant how long it is. It is the same for PCB traces - when they are 50OHM, they are always 50OHM and it doesnt't matter how long they are. Why is that? Because when we say 50OHM cable, sometimes people imagine a cable with 50OHM resistance, but it is not. We are talking about 50OHM impedance - and simply to say, impedance is combination of resistance + capacitance + inductance. This means, the cable can have 50OHMs even if resistance is not 50OHM (simply to say, resistance is what you would see on a standard DVM) because it will be build also from capacitance and inductance of the cable. However, yes, the real signal in real cable would have losses - it means, after some time the signal would get smaller - that is what the length of the cable will influence. But loss was not taken into consideration in this example - just to make the explanation more simple. PS: in reality, DVM doesn't measure resistance, but it calculates it from a known voltage and measuring the current - that is very important.
It is also the propagation of the EM field. Both the discreet components transmission line model and Maxwell's Equations approach work. That's hits on a pretty profound philosophical thing... We have no clue what's "really" happening. We only have models which do a good job (or not) of predicting measurements we make. In a lot of cases there is more than one model/explanation which work equally well.
The way is to put impulse generator and oscilloscope on one side of the cable and tune variable resistor on the other side. Once you see no reflections thus the value of resistor is matched to cable impedance. Some oscilloscopes have function generator built-in, thus you can do this measurement using one device.
I'm surprised he didn't say anything about displacement current. That's the only thing (from my understanding) that would ever give a non-infinite resistance reading on an open circuit.
Am I right if I say changing the transmission line impédance is all about changing conductor distance (one from each other, and considering same conductor and diaelectric material)?
Fogot to say it's and impedance and not a resistance. DC just wont measure it. it's just like 1000's of little transmission lines holding hands all matched to 50ohm. so load one end with 50 ohm and you see 50 ohm's at the other. that's how i remember it. that's it if it's with coax, switch to waveguide and they all let go of their hands and the rest is done with magic. ;-)
It will measure 59 ohms while the wave travels out. Probably for such a short time it will not register on the meter. An infinite line would always measure 50 ohms.
Suppose a 1 volt signal is applied to a very long , unterminated transmission line. Suppose one can measure both the phase and amplitude of the current amplitude in 1 usec Will the measurement reveal 1v/50ohm = 0.02 Amp current until the open circuit transmission line reflects BACK ? Does the 1v x 0.02 Amp current produce a 0.02watt dissipation during this interval ?. Will the so callled 50ohm be heated ? NO!
the initial "charging" pulse going down the cable DOES have its current & voltage in phase so it DOES INITIALLY APPEAR PURELY RESISTIVE (not REACTIVE) so it DOES appear for a FEW nS to apparently be a straight 50 ohm resistor (with no reactance).... BUT this energy goes into creating an electromagnetic wave charging down the line NOT HEAT (even though it is in phase it is NOT lost as heat).... Think of this analogy... I'm trying to charge a rechargeable battery... I put 12 volts into the battery at 4 amps... DC so by definition in phase & apparently a 3 ohm "resistor" ... but again that energy is NOT dissipated as 48W of resistive heat (here it goes into chemical energy storage)..... Although RESISTIVE is in-phase V&I it does not always mean HEAT. The interesting thing is that the driving signal generator will have its own output impedance... if the signal generator only for a very short instant puts out power but then quickly switches back to zero volts out the line will have appeared to resistively "absorb" the initial power but 2 transit times of the line later it will spit the reflection back out and dissipate the energy back into the output stage of the sig gen (the problems of driving unmatched/badly terminated lines with a high SWR). As the line can return this energy it shows it hasn't been wasted as heat.
agreed, but the thing that is really mind-blowing is that for"normal" flat plate capacitors the capacitance is dictated by the separation distance of the plates but for co-axial cables (and capacitors) the distance between the inner & outer cable isn't the determining factor. Here the capacitance is set by the ratio of the inner to outer cable diameter (when you have seen the cables a few times you can actually "see" the impedance from their geometry simply looking at the end of it, be that 1/4" cable, 1/2" or 7/8" 50ohm or 75ohm).
This is convoluted. Just stop about the DC Ohmmeter sampling every 1/2 seconds. The 50 ohm number works when you are modeling impedance due to transmission line effects.
I can't believe there are people who don't know this! I learned this problem in basic electromagnetics class in my undergrad. And on top of that, there's tons of people that are confidently wrong about this. I'll use it as an interview question.
You are messing up 2 different things. The 50 Ohm is not like a reqular reistor R but impedamce Z. In the right frequency range it is 50 OHm. You cant simply measure 50 Ohm by using an Ohm meter. Also you have to deal with terminated or not terminated coax ends.
Did you even watch the video? He’s talking about a real multimeter that takes a “DC” resistance reading every 0.5s. Of course the first reading will be 50 ohms when connected to a long coax. He also said multiple times that the other end does not have to be terminated.
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I've always understood characteristic impedance as the impedance seen through the dielectric, by a sharp edge signal. This video helped me to "grok" it stronger.
Thanks for these videos, Robert! I learn so much from you, Eric Bogatin, Rick Hartley, and others.
I've been teaching electronics for over 40 years and THIS is why people have such a difficult time with very simple subject matter.
You will never read 50 ohms using a DC meter... period. The "characteristic" impedance only applies to signals that are within the design parameters of the transmission line. How do you explain the characteristic impedance of a waveguide to students, then?
In coaxial lines, the characteristic impedance (Z0) is determined by the dielectric constant of the chosen insulator (you did not even broach this) as the signal is wholly contained within the cable (assuming no ingress or egress). If the inner and outer conductors' radii are maintained, Z0 is a function of the dielectric constant , alone, and the choice of dielectric (PE, PTFE, air, FE, etc.) will also affect the propagation velocity.
Impedance is the complex summation of inductive reactance, capacitive reactance, and resistance. There is NO WAY to ever see the impedance of the circuit without the reactances, which are frequency dependent. A DC incident signal with a steep leading edge will contain far too much HF to be useful. Most TDR's, for this reason, use a sinusoidal waveform (half wave) for the incident signal. More accurate and reliable TDR's use complex waveforms (spread spectrum reflectometry).
Q: "Why don't you measure 50 Ohms on a 50 Ohm cable?"
A: The simple answer is that impedance is complex (multiple formulas and parameters) and Ohm Meters are not designed to measure anything but resistance.
Exactly, Retired Engineer! Time domain refectometry (TDR) names the speed of an electric signal through a characteristic coaxial impedance at about sixty percent of light-speed in vacuum.
This. It is unbelievable how difficult it is to find any material nor any lecturer who could teach anything both right and understandably about transmission line impedances, reflections, and most of the RF phenomenons. I am a skilled professional on electronics and electricals, but even I still struggle to understand those subjects.
After long-term struggle though, I understand the most basics, including what this video is trying to say and why it is wrong.
At the beginning of this struggle, I was very frustrated about the prevailing teaching concepts to suddenly start telling and calculating about mystical waves which reflect within cable and need specifically designed cable, without ever explaining what they are in electrical terms. And not mentioning about basic things applying always, even though negligibly in low frequencies, such as any transmission line having charasteristic capacitance and inductance, and reflection occuring and magnetic flux and electric field existing even with slow dc pulses in regular cables. Mentioning those would have given consistent perspective, instead of making confusing thinkings that cranking up the frequency suddenly brings us into another mystery world where exists things not existing in the LF world, where ohms mean totaly diffent thing, and where we on the other hand don't even think about basic voltage and current calculations anymore...
The , even after getting over that conceptual stumblestone, I still find that most materials, videos, and lectures have either way too much omissions, oversimplifications, even outright errors, OR they are way too hard to understand for unfamilar about those subjects, even if skilled in electricals otherwise.
Oh man, you could not be more wrong.
A transmission line at the instant a signal is applied to it looks like an impedance equal to its characteristic impedance. An ideal 50 ohm line will look like a 50 ohm resistor when a wave is launched into it. The characteristic impedance is the ratio between voltage and current of a propagating wave inside a minuscule amount of T-line. When you connect an ohmmeter to it (instantly), it will launch a wave into it, and the wave impedance will look like 50 ohms.
Q: "Why don't you measure 50 Ohms on a 50 Ohm cable?"
A: you will, until the reflection comes back and changes the reading. If you had a fast enough "ohmmeter", you'd see 50 ohms until the first reflection comes back.
Many informed individuals don't seem to understand that these inductances and capacitances do NOT return the energy they absorb to the transmitter; instead they pass it along to their neighbors. Thus, the energy is transmitted down the transmission line. The transmitter ONLY sees that the energy it delivers doesn't come immediately back (+/- 90 deg.) as it would from C or L (or LC). Instead, the energy seems to be lost as from a pure resistance. IF the cable were lossless and light-seconds long, the Ohm meter WOULD measure it as 50 Ohm... at least for a few seconds. As far as the transmitter is concerned, it does not matter whether the energy is converted to heat, excites a cavity, or radiates into space -- since it is lost to the transmitter, it looks like a resistance.
Taking the paradigm of DC resistance with a meter people understand and transfer it to RF is kinda waste of time.
As you said most coax cables do not support the speed of light, due dielectric absorbtion, which is minimized in better coax with sparse placement of plastic spacers. Both speed and power is lost.
RF Impedance is better explained otherwise.
Assuming your coax is RG058 (specified impedance of 50 ohms), the velocity factor is 65, *not* 100, so your hypothetical measurement time should be a bit over 3 seconds.
At 5:00, the first reflection will actually be 2 V. If the meter's output impedance is 50 Ohm, it will absorb this transient via the voltage divider to yield 1 V; otherwise, the power will be reflected again and again until the transient is absorbed by system losses..
This makes a lot of sense thanks. It also explains how the impedance is seen by something which is switching on the order of nanoseconds.
Thank you for your presentation .
I understand imprdance as a value that had to be measured from a load indirectly or with a TDR.
One indirect method being with an ammeter and a voltmeter and the other being a voltmeter comparing the voltage across a load and one across a variable resistor. The voltages across the two being equal reflecting the resistors current value being the impedance.
What he seems to be saying here is that current for a time induces current in a return wire creating a temporary circuit in which messurements could be made.
This is a really GOOD explanation - thank you!
The impedance of a coaxial cable refers to the capacitance reactance of the wire and the shield and is dependent on the dielectric and separation distance of the conductors among other factors
And the inductance.
The transmission line is a series inductor and a parallel capacitance over and over as you go down the line, so it takes that time for your little one gold signal to charge all of those reactions and they build up until you will eventually read an open circuit. But the characteristic impedance of that line did not change
I think is easy to test, just need a osciloscope, a long coax cable and a single shot step signal and a resistor, you can measure on channel 2 the voltaje across the resistor and so you know the current, and you should see a constant current for the first nanosecond and then slowly go to 0, ( a few nanoseconds).
we created also a video with measurements: th-cam.com/video/U60y4JC0Wxs/w-d-xo.html
IMHO Derek of Veritasium should have contacted Dr Eric as his technical consultant for his light bulb video.
Ah, you got your $THANKS button at last! Have a coffee on me, appreciate your work and especially loved your 100 tips series. Cheers... Kiwi Clark
Thank you so much Clark! Yes, they finally enabled this for everyone.
"Why don’t you measure 50 OHM on a 50 OHM cable? "
Mostly because I don't want to; which is not the same as wanting not to, but merely not seeing much need for it. If you make it long enough it will indeed measure 50 ohms end-to-end with a DC ohmmeter.
I think the simplest explanation is to say that your meter can't measure the far-end impedance until the signal has a chance to get there and back; the 50 ohm characteristic impedance marks what your meter would see while awaiting the "real" measurement.
If you had an infinitely long piece of 50 ohm coax, you would measure a DC resistance of 50 ohms with an ohm meter.
just for my understanding, the virtual coax cable to the moon has to be very thick in order to have 50ohm on this long distance right? and what about the capacity, you didnt say anything about that. i mean if the cable is so long than the capacity should be big too. and a big cap looks like 0 ohm at the beginning of loading, how does that fit together with what you said?
If you put any sort of cable even to low earth orbit you'd have millions of volts until whatever you made got destroyed. It's a thought exercise and erroneous at that.
@Google user I could go on but aint nobody got time fo dat
@Google user And as far as where the power comes from. It's about 100 volts per meter above surface, you do the math. Ever seen the ionosphere, those lovely pictures.
@Google user Put a wire on a model rocket and watch lightning come down.
When we have a 50ohm cable, it is always 50ohm and it is not relevant how long it is. It is the same for PCB traces - when they are 50OHM, they are always 50OHM and it doesnt't matter how long they are. Why is that? Because when we say 50OHM cable, sometimes people imagine a cable with 50OHM resistance, but it is not. We are talking about 50OHM impedance - and simply to say, impedance is combination of resistance + capacitance + inductance. This means, the cable can have 50OHMs even if resistance is not 50OHM (simply to say, resistance is what you would see on a standard DVM) because it will be build also from capacitance and inductance of the cable. However, yes, the real signal in real cable would have losses - it means, after some time the signal would get smaller - that is what the length of the cable will influence. But loss was not taken into consideration in this example - just to make the explanation more simple. PS: in reality, DVM doesn't measure resistance, but it calculates it from a known voltage and measuring the current - that is very important.
The current with an open line is the charging of the distributed parasitic capacitance of the line.
It is also the propagation of the EM field. Both the discreet components transmission line model and Maxwell's Equations approach work.
That's hits on a pretty profound philosophical thing... We have no clue what's "really" happening. We only have models which do a good job (or not) of predicting measurements we make. In a lot of cases there is more than one model/explanation which work equally well.
@@travcollier Agreed.
signals do not propagate at c when going down a transmission line. They propagate at around 70% c because the dielectric isn't free space.
In coax the signal propagates not at the speed of light, but 2/3rds or less.
Is there any simple way to find out Impedance of a coaxial cable or Antenna wire which says 75 ohms & 300 ohms respectively?
The way is to put impulse generator and oscilloscope on one side of the cable and tune variable resistor on the other side. Once you see no reflections thus the value of resistor is matched to cable impedance.
Some oscilloscopes have function generator built-in, thus you can do this measurement using one device.
I'm surprised he didn't say anything about displacement current. That's the only thing (from my understanding) that would ever give a non-infinite resistance reading on an open circuit.
He did, in the full version of the video: th-cam.com/video/Lp_b8gQpxW8/w-d-xo.html
With such a long transmission line, should we see similar initial results if the end is shorted?
yep
Is this also why 10 Mbit coax cables had 50 Ohm terminal resistors?
Am I right if I say changing the transmission line impédance is all about changing conductor distance (one from each other, and considering same conductor and diaelectric material)?
yes, e.g. in PCB when you change distance between track and reference plane, the impedance will change
Fogot to say it's and impedance and not a resistance. DC just wont measure it. it's just like 1000's of little transmission lines holding hands all matched to 50ohm. so load one end with 50 ohm and you see 50 ohm's at the other. that's how i remember it. that's it if it's with coax, switch to waveguide and they all let go of their hands and the rest is done with magic. ;-)
Incredible!!!
Let's mix resistance and impedance and try to explain the 1-1 input response of a network to a step function😢
Absolutely!
Huh?
This was a source of confusion for me. It has nothing to do with resistance. Seems easier to understand if you think of conductor spacing.
It will measure 59 ohms while the wave travels out. Probably for such a short time it will not register on the meter. An infinite line would always measure 50 ohms.
informative video, very good!
impedance is not resistance
Mind blown
How can i make 50 ohm pcb antenna using track?
this may help: Designing a PCB patch antenna for WiFi and Bluetooth | KiCad | Philip Salmony th-cam.com/video/cOtv0ddR3aI/w-d-xo.html
I didn't understand a thing unfortunately. Using DC here confuses me a bit.
I think if the coaxial cable was infinitely long, and ideal in every sense, it would read 50 ohms.
Suppose a 1 volt signal is applied to a very long , unterminated transmission line.
Suppose one can measure both the phase and amplitude of the current amplitude in 1 usec
Will the measurement reveal 1v/50ohm = 0.02 Amp current until the open circuit transmission line reflects BACK ?
Does the 1v x 0.02 Amp current produce a 0.02watt dissipation during this interval ?.
Will the so callled 50ohm be heated ?
NO!
the initial "charging" pulse going down the cable DOES have its current & voltage in phase so it DOES INITIALLY APPEAR PURELY RESISTIVE (not REACTIVE) so it DOES appear for a FEW nS to apparently be a straight 50 ohm resistor (with no reactance).... BUT this energy goes into creating an electromagnetic wave charging down the line NOT HEAT (even though it is in phase it is NOT lost as heat).... Think of this analogy... I'm trying to charge a rechargeable battery... I put 12 volts into the battery at 4 amps... DC so by definition in phase & apparently a 3 ohm "resistor" ... but again that energy is NOT dissipated as 48W of resistive heat (here it goes into chemical energy storage)..... Although RESISTIVE is in-phase V&I it does not always mean HEAT.
The interesting thing is that the driving signal generator will have its own output impedance... if the signal generator only for a very short instant puts out power but then quickly switches back to zero volts out the line will have appeared to resistively "absorb" the initial power but 2 transit times of the line later it will spit the reflection back out and dissipate the energy back into the output stage of the sig gen (the problems of driving unmatched/badly terminated lines with a high SWR). As the line can return this energy it shows it hasn't been wasted as heat.
Some can teach....some can't
You will if it's infinity long
This was a total waste of good electrons..
Capacitance will come into this as well
agreed, but the thing that is really mind-blowing is that for"normal" flat plate capacitors the capacitance is dictated by the separation distance of the plates but for co-axial cables (and capacitors) the distance between the inner & outer cable isn't the determining factor. Here the capacitance is set by the ratio of the inner to outer cable diameter (when you have seen the cables a few times you can actually "see" the impedance from their geometry simply looking at the end of it, be that 1/4" cable, 1/2" or 7/8" 50ohm or 75ohm).
It is not in the cable.
This is convoluted. Just stop about the DC Ohmmeter sampling every 1/2 seconds. The 50 ohm number works when you are modeling impedance due to transmission line effects.
I can't believe there are people who don't know this! I learned this problem in basic electromagnetics class in my undergrad. And on top of that, there's tons of people that are confidently wrong about this.
I'll use it as an interview question.
It is school dependent.
This is a good video of why it's 50 ohms.
th-cam.com/video/I-OnQZJv35I/w-d-xo.html
Why does a coil measure as a short? The proper terminology is "50 OHM IMPEDANCE" cable. Calling a dogs tail a leg does not make it a leg.
complex impedance
Because impedance ain't DC simple stuff.
Yo TikTok generation learn NOTHING with these byte-sized clips. Disgustingly sad, what the world is turning into!
Perhaps read the description. Full video here th-cam.com/video/Lp_b8gQpxW8/w-d-xo.html
Ok boomer
@@NavinF And proud of it!!!
You are messing up 2 different things. The 50 Ohm is not like a reqular reistor R but impedamce Z. In the right frequency range it is 50 OHm. You cant simply measure 50 Ohm by using an Ohm meter. Also you have to deal with terminated or not terminated coax ends.
Did you even watch the video? He’s talking about a real multimeter that takes a “DC” resistance reading every 0.5s. Of course the first reading will be 50 ohms when connected to a long coax. He also said multiple times that the other end does not have to be terminated.
DC measurement on AC ... not gonna work.
If you don't know the answer to this question, you need more than a few utube videos