During this experiment, I connected a 10m coaxial cable to pulse generator and at the load end I connected capacitor and inductor one by one, I got these waveforms. Can someone explain me the the reflection?
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1\$\begingroup\$ What do you want explained? The fact you know it is a reflection seems to indicate you already know enough to figure out why it exists. Although, in my opinion that doesn't look like a normal reflection or ringing. \$\endgroup\$– DKNguyenCommented Dec 10, 2020 at 17:11
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1\$\begingroup\$ If you have a nice pulse generator and a long coax, start with an open and short. Understand these first. \$\endgroup\$– Mattman944Commented Dec 10, 2020 at 17:12
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1\$\begingroup\$ I mean I know about open and short circuit reflections and everything. I am just trying to understand why there are ups and down in the reflection pulse. I am assuming that in capacitor, the initial decay is because the capacitor is charging for initial time and then discharging which gives it the gain in pulse reflection. I am not sure about what gives inductor that reflection at all. \$\endgroup\$– Jaimin PARMARCommented Dec 10, 2020 at 17:16
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1\$\begingroup\$ The initial negative fast decay when you connected the capacitor is because it is initially acting as a short circuit. Compare that with a totally shorted line. For the inductor, because initially it acts as an open circuit you get a positive pulse initially. Compare that with an open-circuit line. Explaining how those shapes evolve is pointless and complex. \$\endgroup\$– Andy akaCommented Dec 10, 2020 at 17:34
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\$\begingroup\$ OK, I believe that you understand how an open and short would behave. But, have you tried it with your setup? It may reveal weaknesses in your setup. I don't have access to a nice pulse generator now, but when I did, I could create some nice waveforms that I used to demonstrate transmission line reflections to young engineers. \$\endgroup\$– Mattman944Commented Dec 10, 2020 at 17:44
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1 Answer
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I'm only going to show what happens when the load is a capacitor. There are enough clues in this answer (and previous comments) to relate to the scenario when an inductor is the load.
To answer this properly and make it useful to others you have understand what happens at the sending end when the reflected pulse returns. For the capacitor load it is useful to consider (as a reminder) what happens when the t-line load is a true short circuit: -
If we replace the shorted end termination with a 500 pF capacitor we see this: -
So the capacitor-reflected pulse initially has exactly the same form as when the line was terminated in a short (at t = 200 ns). This is because when the pulse arrives at the load, the load (a 500 pF capacitor) initially behaves like a short circuit.
However, pretty immediately the capacitor starts converting the received pulse from the sending end into the blue voltage waveform seen in the image below. This waveform is reflected back to the source but it's initial value has to be -2.5 volts and so, if I mess around with this picture using "paint" I can start to build up what happens at the sending end when the reflection comes back from the capacitor: -
Of course, when the initial pulse ceases (after 50 ns) the capacitor at the end of the t-line starts to discharge but there is also a reflection "recoil" and a sharp rise of 2.5 volts is seen in the reflected voltage. The "recoil" happens because the 50 ns pulse is "spent" and this produces a near-instant rise in the reflected voltage. From that point onwards in time, the reflected waveform follows the capacitor discharge curve.
Micro-cap simulation file available on request. Transmission-line modeled as 20 metres of loss-less line with C = 100 pF/meter and L as 250 nH/metres. Source-end driven with a 5 volt pulse of 50 ns duration via 50Ω.
answered Dec 11, 2020 at 10:36
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\$\begingroup\$ @JaiminPARMAR do you know what to do next? Do you know about site protocol? \$\endgroup\$– Andy akaCommented Dec 11, 2020 at 16:13
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\$\begingroup\$ no what do I have to do? \$\endgroup\$ Commented Dec 11, 2020 at 16:17
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\$\begingroup\$ @JaiminPARMAR - spend a couple of minutes taking the site tour and note, that you have enough reputation (I've added to it by upvoting your question just now) to upvote useful answers and, once you have decided on the best answer to your question you can formally accept it. Any other related queries, please raise a comment. \$\endgroup\$– Andy akaCommented Dec 11, 2020 at 16:22