1. Whether the load impedance is great or less than the characteristic impedance, there is a reflection. 2. In either case, there is no signal lost. Because of these fundamental flaws alone, I can't recommend this video.
What i don't understand is the explanation of the signal relfexion when there is an open circuit, why would a signal travel through a place where there is no end , in circcuits no current flows through a dead end cable.
His explanations are junk is why you don't understand, not your fault. Think of a coaxial cable ('transmission line') as an inifinite series of inductors and capacitors, then apply an ac signal at one end. The energy in the signal heads down the cable charging up all the L's and C's as in goes - when it gets to the end, the only place that energy (volts and amps) can go is back down the cable, when it is out of phase with the signal going up the line. The core of this is that it is an alternating current and ac behaves very differently to dc. In the dc case you would be correct, but for ac, you have a very complex situation. I've probably not explained that well enough, but you can get hold of Walt Maxwells books 'Reflections' as free pdf downloads, and he explains much better than I how this stuff works.
In 1:37 it is stated that NO REFLECTION happens when ZL < ZC.......Guess that's wrong, in every mismatch there is reflection to my knowledge!? (i.g. a short -in extreme- also will produce a 100% reflection)
Yes. You are absolutely correct. And optimum diameter ratio for voltage breakdown is 2.7 corresponding to 60-ohms impedance. So 50 ohm is a good compromise. And funny enough if you take a 77 ohm transmission line with air insulation and change out with a PE isolation you get a 51 ohm transmission line.
Dear could you plz explain how can electrical signal reflection occur against high potential at the part of transmission. for optical fire it is obvious, but electrical signal reflection is bit confusing, as electricity can not propagate against high potential.
Hi Shersada. Reflection only when happens when the load impedance is higher than the characteristic impedance of the cable under test. If the cable is connected to any potential, whether it is high or low, the load impedance is essentially zero. Therefore, if the cable has any potential reflection does not happen.
Thank you for your question. An electrical signal reflects back to its source when it encounters a load impedance higher than the impedance of the source. If by high potential, you mean connecting the cable to a high voltage, that will be equivalent to shorting the cable. In this case, no reflection occurs.
This is very good as an instructive video, but I think there are two errors that you make and need to correct. First is the comment that 50 ohm cable is for low loss applications. That is not true. Most 50 ohm coax such as RG8 or RG213 is rather high loss cable compared to 75 ohm coax, such as RG6, and particularly compared to higher impedance non-coax cable such as 450 ohm window line. This is because the for a given power level, the current levels will be higher in 50 ohm coax vs. higher impedance transmission line, and thus I^2R losses are usually higher, especially if there are standing waves present due to load impedance mismatch. The loss data of coaxial cables and parallel conductor cables are well documented, and are readily found online from coax manufacturers or in texts such as the ARRL Antenna Book or ARRL Handbook. Second, you say that there is no reflection if the load impedance is less than the cable impedance. That is not true either. Any impedance mismatch causes reflected power in the form of reflected voltage or current waves. For smaller load impedance, part of the voltage wave is reflected out of phase relative to the incident wave to partially or fully cancel the voltage wave in order to adjust the voltage/current ratio in the lower impedance load to a smaller value (thus smaller impedance). The ultimate being a short circuit termination, where 100% of the voltage wave is reflected 180 degrees out of phase in order to cancel the voltage to zero at the termination. The reverse is true for an open circuit termination. Mismatch of any kind causes signal loss due to reflected power being consumed by lossy cables. These points can be confirmed by looking up any text on the subject. Also, just a suggestion: it is always more pleasant to hear a real human voice and not an AI created robot voice in these videos.
2.47 min i think its totally wrong. To convert 50 ohms to 75 ohms you can't use 25 ohms resistor as you mentioned. then source impedance of 50 ohms not satisfied . Any body can correct me if I wrong www.amateurtele.com/images/Amateurtelecom-impedance_matching_schematic.png
The exact purpose of adding the 25 Ohms resistor is to change the source impedance to 75 Ohms. This is to optimize for loss and reflection in the system. Clearly, in this case, the source impedance is 75 Ohms, and no longer 50 Ohms. And of course, it does not have to 50.
1. You are Wrong 2. the correct spelling is RESISTOR (yes , it does matter) 3. as Academia said THE PURPOSE of adding the 25 Ohm Resistance was to balance the impedance mismatch SO AS TO AVOID REFLECTION AND SIGNAL LOSS THIS WAS ACHIEVED BY adding the 25 Ohm Resistor Now.. as to why you are wrong (in case you might say "Ok Martin, prove me wrong") hehe Mate, i'm happy to prove you wrong but I'M NOT GOING TO because, , Please understand - I'm an engineer - These are basic concepts - if i was to go around proving every newbie wrong who asked the question "Ok, prove me wrong then" i would be doing nothing but proving people wrong, WHICH IS A WASTE OF A LIFE so, for that reason i'm not going to I would however encourage you to take a course in Digital Reception Technology or Digital Signalling or Digital Signal Processing it's also not cool when someone like you (who hasn't bothered to do a massive amount of courses Disagree's then asks someone like me to prove you wrong Essentially you want the answer for free (by taking the easy road) i'm happy to tell you you're wrong i expect you to accept my response on the basis of my experience if you want to take it further i expect you to go and do courses that show that you have enough experience to continue the discussion so we don't have to keep repeating the basics, ok i think that's a fair call but 1. your question is a fair question (and i understand it) 2. your opinion is respected 3. but all that aside, YOU are incorrect and ACADEMIA is correct (Based on fundamental engineering principles, not based on my opinion, but based on fact) ok beyond this, if you disagree, that's a completely different discussion and is purely your opinion and is not fact. have a nice day mate
Actually, you are right. In the shown example the cable is matched to the source effectively matching any reflections that might return from an unmatched load. But the generator is not matched and will see a 100 ohm load.
@@martinkuliza Tell me how adding a SERIES 25 ohm RESISTOR is a good idea for matching. How much do you loose in the resistor? And where do you get a resistor to handle my 100W radio?
Thank you Nikša for your question, "What if we use a single wire transmission line, is impedance matching still needed? I assume single wire still has some internal reactance so yes to a point?", not showing here. Impedance matching requires a return path. The matching is done by placing an equal impedance load between the signal wire and the "GND". A signal wire transmission, by definition, does not have a "traditional" return path. Therefore, the issues become how and with respect to what to measure the impedance of the wire, and additionally where to place the load. In low frequency applications, you may think of Earth as the return path. Then the transmission is not truly a single wire. So, the impedance matching follows the same criteria as a 2 wire transmission. For high frequency applications, the transmission is really not electrical. In this case, for the optimum transmission, several factors must be considered and the issue becomes more complex.
This video is significantly just "technical bologna". There is no better or nicer way to describe the technical contents of this video. This video represents everything that is wrong with Internet and RUclips as a learning resource. 1.) The claim half the power reaches the load is completely false. 100% of generator power can reach the load, less cable losses. It can be a very large percentage reaching the load even with a mismatch. 2.) If the generator is conjugately matched to the cable input impedance, even if the cable is mismatched to the load, almost 100% of source power can reach the load. An example would be open wire fed doublet antennas, where the feedline operates with 10:1 or higher mismatches yet nearly 100% of transmitter output power excites the antenna. I often run antennas with 2:1 and higher mismatches and yet the transmission line is highly efficient and the transmitter is perfectly fine. As a matter of fact reflected power does not make it back into normal PA stages. This is easily proven. 3.) Any mismatch causes STANDING waves. Standing waves stand. Standing waves do not bounce around, if they did we would call them "bouncing waves" and not call them STANDING waves. 4.) Standing waves cause an operating impedance variation along the transmission line. That operating impedance, the ratio of across to through vectors, varies with distance along the line. This is how Q-sections and stubs work. Look up Q-sections. They can operate with nearly infinite standing wave ratio yet have high efficiency. We are heading for problems when RUclips becomes our source of education.
The slide at 1:40 is simply wrong. The only time there is NO reflection is when the load perfectly matches the load. The idea that Zload < Zcable causes loss and not reflection casts doubt on the whole of this presentation, as it is a fundamental misunderstanding of how transmission lines behave. In the ideal case where the transmission line has no loss, then there can be no loss as a result of standing waves set up through a mismatched load - there is simply no where for the power to be lost, power is simply energy per second and energy cannot be created or destroyed. When the load impedance is a perfect match for the source impedance, then there are no standing waves, since there are no reflections that would cause them to be set up. Sorry, but the fundamentals of this presentation are wrong. For a robust explanation of transmission lines and standing wave behaviours, see Walt Maxwell's books "Reflections".
Of course you have reflection when Zload < Zcable. There is only one way to not have reflection and that is to not have a mismatch. Reflection causes VSWR which proves that you are wrong in your statement. If you don't believe me then try to terminate a 50 ohm coax with a 25 ohm resistor. You will get VSWR close to 2 with a reflection coefficient of 1/3 and a reflection angle of 180 degrees. 11% of power is reflected. If you change out the 25 ohm resistor with a 100 ohm resistor the results are almost the same. Only difference is the reflection angle which now is 0 degrees.
"Adding a resistor in series with the source or the load, or using a termination resistor is a basic method to achieve impedance matching." - also incorrect. A conjugate match the most common method.
Nice informative video , All the points covered but most important lucid language and animation making understanding easy..GOOD CREATIVITY...Please make a video on vector network analyser and power meter...
Thank you for the question. The frequency response of a cable is a function of several factors including the conductor material. For further information, please visit www.timesmicrowave.com/documents/resources/TL16p63refdataappnotes.pdf. The cable manufacturer can also be a better resource for this information.
*TDR* : Time domain reflectometry *TLDR* : Too long did not read it, but ms Sam did it for me. Hahahah what a time to be alive :p BTW thanks, pure good content.
Hi Academia, i have been staring at this for awhile now haha .Am I right in saying that all impedances along the circuit are in series without eachother hence the amps(Signal) will be constant throughout the circuit? is this the purpose of matching the impedances throughout the circuit?
Ideally, the characteristic impedance of a cable is the same throughout the cable. That is to say, at any location, the traveling signal encounters the same impedance. Please note that this impedance is the impedance between the center conductor of the cable and the outer one, or the signal path and the ground path. This is different than the impedance of the signal conductor from one end to the other end.
Interesting. Is the cable characteristic impedance the impedance of 1m length? I'm assuming this because I believe the total impedance is proportional to the cable length. Also for a very long cable wouldn't be better trying to minimising impedance?
Characteristic impedance doesn't change with length. Zo = SQRT(L/C) assuming lossless transmission line. If not lossless then Zo = SQRT( (R+jwL)/(G+jwC) )
@KjartanAndersen yes, independence from length is true for lossless lines that is for frequency high enough. I think its value is a purely inductive impendance.
Indeed!! if the load impedance is smaller than the cable impedance there is a reflection. 2 If source, cable and load impedance are matched you do NOT automatically loose half of the source power. Depends on the cable loss factor and length! What the video implies is incorrect. The video is well put together but has these two issues.
You're unable to search this are you FUNNY HOW I WAS ABLE TO SEARCH AND FIND IT IN APPROX 7 SECONDS www.standard-wire.com/coax_cable_theory_and_application.html I would suggest you brush up on your searching skills and SEARCH HARDER NEXT TIME. the simple fact that you are ASKING FOR HELP for something that is NECESSARY TO YOU shows a problem and clear lack of commitment on your part again, i found the answer in a few seconds but somehow YOU WERE UNABLE to search it i want you to go away and ask yourself this question HOW EXACTLY WAS THAT POSSIBLE ? and i want you to learn from the mistake the guy previously gave you the answer and you still are lost clearly something is wrong with you i suppose after me giving you my link you are also going to say you don't have the answer .. right ? if that's the case, then.... MATE, YOU CAN'T SEE THE ANSWER IF IT'S PLACED IN FRONT OF YOU so think about that as well
Excellent video, thanks! 2:44 should be P_load(max)= 1/4 P_source (ZL=Zs)
Great video on coaxial cables and the importance of impedance matching. Thank you !
Thank you for watching.
1. Whether the load impedance is great or less than the characteristic impedance, there is a reflection. 2. In either case, there is no signal lost. Because of these fundamental flaws alone, I can't recommend this video.
do you have any other video that you would recommend or any other source to understand this concept of impedance matching?
What i don't understand is the explanation of the signal relfexion when there is an open circuit, why would a signal travel through a place where there is no end , in circcuits no current flows through a dead end cable.
His explanations are junk is why you don't understand, not your fault. Think of a coaxial cable ('transmission line') as an inifinite series of inductors and capacitors, then apply an ac signal at one end. The energy in the signal heads down the cable charging up all the L's and C's as in goes - when it gets to the end, the only place that energy (volts and amps) can go is back down the cable, when it is out of phase with the signal going up the line. The core of this is that it is an alternating current and ac behaves very differently to dc. In the dc case you would be correct, but for ac, you have a very complex situation. I've probably not explained that well enough, but you can get hold of Walt Maxwells books 'Reflections' as free pdf downloads, and he explains much better than I how this stuff works.
Super clear and straight to the point, thanks.
Pretty good info. I like the wave analogy
Best so far
Could any one say why they have not considered the Zc (Cable Impedance) in Equation @2:23. Is it because it is neglible compare to Zs and ZL
In 1:37 it is stated that NO REFLECTION happens when ZL < ZC.......Guess that's wrong, in every mismatch there is reflection to my knowledge!?
(i.g. a short -in extreme- also will produce a 100% reflection)
Very wrong - and it's a fundamental point, so undermines the other explanations.
very good information, although it was quite fast.
Isn't 75 ohms lowest loss and 50 ohms best compromise between power and voltage handling and loss?
Yes. You are absolutely correct. And optimum diameter ratio for voltage breakdown is 2.7 corresponding to 60-ohms impedance. So 50 ohm is a good compromise. And funny enough if you take a 77 ohm transmission line with air insulation and change out with a PE isolation you get a 51 ohm transmission line.
Dear could you plz explain how can electrical signal reflection occur against high potential at the part of transmission. for optical fire it is obvious, but electrical signal reflection is bit confusing, as electricity can not propagate against high potential.
Hi Shersada. Reflection only when happens when the load impedance is higher than the characteristic impedance of the cable under test. If the cable is connected to any potential, whether it is high or low, the load impedance is essentially zero. Therefore, if the cable has any potential reflection does not happen.
Thank you for your question. An electrical signal reflects back to its source when it encounters a load impedance higher than the impedance of the source. If by high potential, you mean connecting the cable to a high voltage, that will be equivalent to shorting the cable. In this case, no reflection occurs.
This is very good as an instructive video, but I think there are two errors that you make and need to correct. First is the comment that 50 ohm cable is for low loss applications. That is not true. Most 50 ohm coax such as RG8 or RG213 is rather high loss cable compared to 75 ohm coax, such as RG6, and particularly compared to higher impedance non-coax cable such as 450 ohm window line. This is because the for a given power level, the current levels will be higher in 50 ohm coax vs. higher impedance transmission line, and thus I^2R losses are usually higher, especially if there are standing waves present due to load impedance mismatch. The loss data of coaxial cables and parallel conductor cables are well documented, and are readily found online from coax manufacturers or in texts such as the ARRL Antenna Book or ARRL Handbook. Second, you say that there is no reflection if the load impedance is less than the cable impedance. That is not true either. Any impedance mismatch causes reflected power in the form of reflected voltage or current waves. For smaller load impedance, part of the voltage wave is reflected out of phase relative to the incident wave to partially or fully cancel the voltage wave in order to adjust the voltage/current ratio in the lower impedance load to a smaller value (thus smaller impedance). The ultimate being a short circuit termination, where 100% of the voltage wave is reflected 180 degrees out of phase in order to cancel the voltage to zero at the termination. The reverse is true for an open circuit termination. Mismatch of any kind causes signal loss due to reflected power being consumed by lossy cables. These points can be confirmed by looking up any text on the subject. Also, just a suggestion: it is always more pleasant to hear a real human voice and not an AI created robot voice in these videos.
hey, some real information... thanks...:)
👍👍👍
2.47 min i think its totally wrong. To convert 50 ohms to 75 ohms you can't use 25 ohms resistor as you mentioned. then source impedance of 50 ohms not satisfied . Any body can correct me if I wrong
www.amateurtele.com/images/Amateurtelecom-impedance_matching_schematic.png
The exact purpose of adding the 25 Ohms resistor is to change the source impedance to 75 Ohms.
This is to optimize for loss and reflection in the system.
Clearly, in this case, the source impedance is 75 Ohms, and no longer 50 Ohms. And of course, it does not have to 50.
1. You are Wrong
2. the correct spelling is RESISTOR (yes , it does matter)
3. as Academia said THE PURPOSE of adding the 25 Ohm Resistance was to balance the impedance mismatch SO AS TO AVOID REFLECTION AND SIGNAL LOSS
THIS WAS ACHIEVED BY adding the 25 Ohm Resistor
Now.. as to why you are wrong (in case you might say "Ok Martin, prove me wrong") hehe
Mate, i'm happy to prove you wrong but I'M NOT GOING TO
because, , Please understand
- I'm an engineer
- These are basic concepts
- if i was to go around proving every newbie wrong who asked the question "Ok, prove me wrong then" i would be doing nothing but proving people wrong, WHICH IS A WASTE OF A LIFE
so, for that reason i'm not going to
I would however encourage you to take a course in Digital Reception Technology or Digital Signalling or Digital Signal Processing
it's also not cool when someone like you (who hasn't bothered to do a massive amount of courses Disagree's then asks someone like me to prove you wrong
Essentially you want the answer for free (by taking the easy road)
i'm happy to tell you you're wrong
i expect you to accept my response on the basis of my experience
if you want to take it further i expect you to go and do courses that show that you have enough experience to continue the discussion so we don't have to keep repeating the basics, ok
i think that's a fair call
but
1. your question is a fair question (and i understand it)
2. your opinion is respected
3. but all that aside, YOU are incorrect and ACADEMIA is correct
(Based on fundamental engineering principles, not based on my opinion, but based on fact)
ok
beyond this, if you disagree, that's a completely different discussion and is purely your opinion and is not fact.
have a nice day mate
@@martinkuliza www.amateurtele.com/images/Amateurtelecom-impedance_matching_schematic.png
Actually, you are right. In the shown example the cable is matched to the source effectively matching any reflections that might return from an unmatched load. But the generator is not matched and will see a 100 ohm load.
@@martinkuliza Tell me how adding a SERIES 25 ohm RESISTOR is a good idea for matching. How much do you loose in the resistor? And where do you get a resistor to handle my 100W radio?
Thank you Nikša for your question, "What if we use a single wire transmission line, is impedance matching still needed? I assume single wire still has some internal reactance so yes to a point?", not showing here.
Impedance matching requires a return path. The matching is done by placing an equal impedance load between the signal wire and the "GND". A signal wire transmission, by definition, does not have a "traditional" return path. Therefore, the issues become how and with respect to what to measure the impedance of the wire, and additionally where to place the load. In low frequency applications, you may think of Earth as the return path. Then the transmission is not truly a single wire. So, the impedance matching follows the same criteria as a 2 wire transmission. For high frequency applications, the transmission is really not electrical. In this case, for the optimum transmission, several factors must be considered and the issue becomes more complex.
This video is significantly just "technical bologna". There is no better or nicer way to describe the technical contents of this video. This video represents everything that is wrong with Internet and RUclips as a learning resource.
1.) The claim half the power reaches the load is completely false. 100% of generator power can reach the load, less cable losses. It can be a very large percentage reaching the load even with a mismatch.
2.) If the generator is conjugately matched to the cable input impedance, even if the cable is mismatched to the load, almost 100% of source power can reach the load. An example would be open wire fed doublet antennas, where the feedline operates with 10:1 or higher mismatches yet nearly 100% of transmitter output power excites the antenna.
I often run antennas with 2:1 and higher mismatches and yet the transmission line is highly efficient and the transmitter is perfectly fine. As a matter of fact reflected power does not make it back into normal PA stages. This is easily proven.
3.) Any mismatch causes STANDING waves. Standing waves stand. Standing waves do not bounce around, if they did we would call them "bouncing waves" and not call them STANDING waves.
4.) Standing waves cause an operating impedance variation along the transmission line. That operating impedance, the ratio of across to through vectors, varies with distance along the line. This is how Q-sections and stubs work. Look up Q-sections. They can operate with nearly infinite standing wave ratio yet have high efficiency.
We are heading for problems when RUclips becomes our source of education.
"There is no reflection when the load impedance is smaller than the impedance of the cable." - also incorrect.
I hate the voice, but it is informative.
We should take some time to reflect on the content transmitted in this video so as not to impede our understanding.
I know I should, but my mind is giving me resistance on this.
Maybe if you could provide some sort of incentive, it could induce me to try harder.
The slide at 1:40 is simply wrong. The only time there is NO reflection is when the load perfectly matches the load. The idea that Zload < Zcable causes loss and not reflection casts doubt on the whole of this presentation, as it is a fundamental misunderstanding of how transmission lines behave. In the ideal case where the transmission line has no loss, then there can be no loss as a result of standing waves set up through a mismatched load - there is simply no where for the power to be lost, power is simply energy per second and energy cannot be created or destroyed. When the load impedance is a perfect match for the source impedance, then there are no standing waves, since there are no reflections that would cause them to be set up. Sorry, but the fundamentals of this presentation are wrong.
For a robust explanation of transmission lines and standing wave behaviours, see Walt Maxwell's books "Reflections".
Of course you have reflection when Zload < Zcable. There is only one way to not have reflection and that is to not have a mismatch. Reflection causes VSWR which proves that you are wrong in your statement. If you don't believe me then try to terminate a 50 ohm coax with a 25 ohm resistor. You will get VSWR close to 2 with a reflection coefficient of 1/3 and a reflection angle of 180 degrees. 11% of power is reflected. If you change out the 25 ohm resistor with a 100 ohm resistor the results are almost the same. Only difference is the reflection angle which now is 0 degrees.
I am sick to death of watching RUclips videos where the narrative is given by a robot. A real switch off !!!
"Adding a resistor in series with the source or the load, or using a termination resistor is a basic method to achieve impedance matching." - also incorrect. A conjugate match the most common method.
Hey! Good video!!! But I can't see how are the connections with the coaxial cables and TDR to get results? if you can help me! Tx
Great info in a short video. Loved it.
Mostly wrong information sadly
What's the name of that simulator?
your first slide is so poorly photoshopped it looks like a meme lmao
Well structured and concise, thank you! :)
Thank you for your time watching it.
@@academia7768 Too bad it is not accurate.
GREAT VIDEO! 👏👏
Nice informative video , All the points covered but most important lucid language and animation making understanding easy..GOOD CREATIVITY...Please make a video on vector network analyser and power meter...
Cables are recognized by the chipsets, usb cables have resistors, capacitors, chip inside the plug.
Thank you 👍
Please give me formula related to frequency and attenuation for silver plated cable
Thank you for the question. The frequency response of a cable is a function of several factors including the conductor material. For further information, please visit www.timesmicrowave.com/documents/resources/TL16p63refdataappnotes.pdf.
The cable manufacturer can also be a better resource for this information.
How do we measure the impedance of the "load" and the cable without an oscilloscope?
information not provided by the manufacturer!
You use a vector network analyser (VNA)
Cable impedance is low as compair to transmission line but how can cable reflect the surge voltage generate in line due to fault
What about impedance mismatch in differential pair configuration ?
Please
*TDR* : Time domain reflectometry
*TLDR* : Too long did not read it, but ms Sam did it for me.
Hahahah what a time to be alive :p
BTW thanks, pure good content.
Hey could you try this software? come upon circuit solver on the playstore!
Appreciate for your sharing knowledge
Hey, that was "Red Rock Park", Lynn, MA
Hi Academia, i have been staring at this for awhile now haha .Am I right in saying that all impedances along the circuit are in series without eachother hence the amps(Signal) will be constant throughout the circuit? is this the purpose of matching the impedances throughout the circuit?
Ideally, the characteristic impedance of a cable is the same throughout the cable.
That is to say, at any location, the traveling signal encounters the same impedance.
Please note that this impedance is the impedance between the center conductor of the cable and
the outer one, or the signal path and the ground path.
This is different than the impedance of the signal conductor from one end to the other end.
Interesting. Is the cable characteristic impedance the impedance of 1m length? I'm assuming this because I believe the total impedance is proportional to the cable length. Also for a very long cable wouldn't be better trying to minimising impedance?
Characteristic impedance doesn't change with length. Zo = SQRT(L/C) assuming lossless transmission line. If not lossless then Zo = SQRT( (R+jwL)/(G+jwC) )
@KjartanAndersen yes, independence from length is true for lossless lines that is for frequency high enough. I think its value is a purely inductive impendance.
👏👏👏
well explanation
Thanks for watching.
Analogy is superb , but the bland robotic voice is repelling, btw Good work guys
Thanks for your kind words. The choice for picking the voice was to develop an international academic setting to satisfy everyone’s unique accent.
Very good video. Great content. I’m glad you put TDR in it as well.
Video is full of wrong information
Indeed!! if the load impedance is smaller than the cable impedance there is a reflection. 2 If source, cable and load impedance are matched you do NOT automatically loose half of the source power. Depends on the cable loss factor and length! What the video implies is incorrect. The video is well put together but has these two issues.
Please provide me formula related to silver plated copper cable insertion loss and frequency. Please it's very necessary, I am unable to search this
You're unable to search this are you
FUNNY HOW I WAS ABLE TO SEARCH AND FIND IT IN APPROX 7 SECONDS
www.standard-wire.com/coax_cable_theory_and_application.html
I would suggest you brush up on your searching skills and SEARCH HARDER NEXT TIME.
the simple fact that you are ASKING FOR HELP for something that is NECESSARY TO YOU shows a problem and clear lack of commitment on your part
again, i found the answer in a few seconds
but somehow YOU WERE UNABLE to search it
i want you to go away and ask yourself this question
HOW EXACTLY WAS THAT POSSIBLE ?
and i want you to learn from the mistake
the guy previously gave you the answer and you still are lost
clearly something is wrong with you
i suppose after me giving you my link you are also going to say you don't have the answer .. right ?
if that's the case, then.... MATE, YOU CAN'T SEE THE ANSWER IF IT'S PLACED IN FRONT OF YOU
so think about that as well
Damn it i understand the whole context by just listening to the first sentence. Thanks
Great Job
Thanks.
Thanks. Very clear and will use this to teach new hams.
Please don't. It is full of major errors.