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Why would signal get reflected without proper ohmic termination?

Taking the example of a signal generator connected to an oscilloscope via a BNC cable, I understand the "virtual" voltage divider between the source 50 Ohms and the cabling 50 Ohms, which would divide a 2 V signal to output a 1 V signal.

But I don't understand why, if at the scope end, when NOT using a 50 Ohm terminator, there is signal reflection?

Or, why is there no signal reflection, when terminated with 50 Ohms?

Thanks
 
Thanks. It was a very nice video. I enjoy watching these old time videos, not just for the nostalgia but many of them take the time to explain fundamentals even better than many modern videos.

HOWEVER this video still did not answer what is perplexing me.

I've actually watched several videos before this one, and read stuff, but am having trouble grasping what happens right at the terminus of the cabling.

I understand all about the distributed inductance and capacitance, but what is the "magic" that happens at termination to cause reflection?

As far as I can figure, for an open ended cable, as the signal reaches the end, there is no further conductor to go, and (perhaps?) this causes a voltage spike due to creating a potential difference between the two conductors. And this potential difference causes a current propagation to be reflected back along the cable. Not sure if this is correct.

Now, with the shorted end, I don't understand why it would get reflected, and not dissipated as heat even though there is no resistor at the end. It could be dissipated as heat and even visible radiation right at the end. (I know it must be dissipated along the cable, but why not peak at the shorted end?)

So for any resistance quantity that is less or over the optimal (50 ohms) I figure it is a gradation of the above two extreme scenarios.

Appreciate if you guys can help me get my head around the above part. Why is there a magical non-reflective number for impedance matching?

(Related question: Is the 50 ohms at the end in parallel with a further load? i.e., can you put in a 50 ohm resistor across the conductors at the end, and then put another load in parallel with that 50 ohm resistor? Or is the 50 ohm in series with another load, which doesn't make sense anyway because they would add up).
 

davenn

Moderator
but what is the "magic" that happens at termination to cause reflection?

As far as I can figure, for an open ended cable, as the signal reaches the end, there is no further conductor to go, and (perhaps?) this causes a voltage spike due to creating a potential difference between the two conductors. And this potential difference causes a current propagation to be reflected back along the cable. Not sure if this is correct.

Now, with the shorted end, I don't understand why it would get reflected, and not dissipated as heat even though there is no resistor at the end. It could be dissipated as heat and even visible radiation right at the end. (I know it must be dissipated along the cable, but why not peak at the shorted end?)

So for any resistance quantity that is less or over the optimal (50 ohms) I figure it is a gradation of the above two extreme scenarios.

it showed you all that in the video from around the 11 minute mark

maybe you didn't watch the whole video


Why is there a magical non-reflective number for impedance matching?

nothing magical about it ...... when the impedance of the line = the impedance of the load, all available energy is passed to the load

Any mismatch will cause a reflection of energy from the point of the mismatch
with the reflection being proportional to the size of the mismatch



Dave
 
Thanks for the excellent AT&T video.

I did watch the other video all the way; the AT&T one was a good sequel.
 
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