Signal reflection on long cable

Thread Starter

bug13

Joined Feb 13, 2012
1,885
Hi

Can someone explain to me why we have signal reflection on long cable? What is the equivalent circuit of a long cable?

Thanks guys!
 

t_n_k

Joined Mar 6, 2009
5,455
Hi

Can someone explain to me why we have signal reflection on long cable? What is the equivalent circuit of a long cable?

Thanks guys!
The issue isn't about the cable length. One can have reflections in short cables. The issue is about impedance mismatch at the ends of the cable. Reflections don't commence until the travelling wave propagates to the point of discontinuity - such as the mismatched load impedance.
 

Papabravo

Joined Feb 24, 2006
14,672
Hi

Can someone explain to me why we have signal reflection on long cable? What is the equivalent circuit of a long cable?

Thanks guys!
Transmission lines are modeled as a distributed series inductance (microhenries per foot), and distributed shunt capacitance (nanofarads per foot). This ideal transmission line leads to a partial differential equation, which given certain boundary conditions, can be solved for the explicit waveforms.
 
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crutschow

Joined Mar 14, 2008
25,647
Reflections in a cable are related to the maximum energy transfer theorem which states that the maximum energy is transferred when the source (cable) impedance matches the load impedance. If the two impedances are not matched then the energy not transferred to the load has nowhere else to go and is reflected back down the cable.
 

Brownout

Joined Jan 10, 2012
2,390
Hi

Can someone explain to me why we have signal reflection on long cable? What is the equivalent circuit of a long cable?

Thanks guys!
The principle of conservation of energy. Unless the energy propagating along the cable is converted or absorbed, it must continue to propagate. As the propagation medium doesn't extend past the end, the energy must return towards the source, along the cable (less whatever propagated into free space)
 

anhnha

Joined Apr 19, 2012
881
Reflections in a cable are related to the maximum energy transfer theorem which states that the maximum energy is transferred when the source (cable) impedance matches the load impedance. If the two impedances are not matched then the energy not transferred to the load has nowhere else to go and is reflected back down the cable.
Can I say that maximum power (W) is transferred the source (cable) impedance matches the load impedance?

I read some materials and saw that they proved about power.
Are they equivalent?
If there is a mismatch, then energy is reflected but if it is transferred in a very long time, then most energy will also be transferred to load?
 

Papabravo

Joined Feb 24, 2006
14,672
In regions of constant impedance, a wave will propagate in both directions. At each impedance discontinuity a portion of the signal is reflected and a portion is transmitted. These portions are often measured by a "reflection" coefficient and a "transmission" coefficient.
 

w2aew

Joined Jan 3, 2012
219

Thread Starter

bug13

Joined Feb 13, 2012
1,885
I agree that the old AT&T video showing a mechanical representation of wave propagation and reflection is simply fantastic.

You might also find one of my videos on the same topic useful to understand reflections.

http://www.youtube.com/watch?v=Il_eju4D_TM
A video worth a thousand word, thanks a lot.

Can I use a short cable to recreate the experiment? I have a cheap function generator from eBay, can do up to 20M, and a digital scope. The longest wire I can find at home is just normal house appliance cord, will that work?
 

nsaspook

Joined Aug 27, 2009
7,733
You might also find one of my videos on the same topic useful to understand reflections.

http://www.youtube.com/watch?v=Il_eju4D_TM
Cool. We can see the effects of a discontinuity in impedance in fiber optic cables also.
I built a optical cable tester for work using the PIC18 CTMU module as the timing element. When looking at the optical receiver signal at the top trace (the bottom trace is the transmitter signal) of the scope you can the reflected energy nodes change and grow as the fiber cable is slightly pulled from the photo-diode in the receiver connector. This also has the effect of changing the signal pulse timing (it creates the optical equivalent of a RC time delay) and is what caused the FTL neutrino 'loose cable' mess a few years ago.

About 0:26 in the video you can see the size of a reflection node and the pulse delay change when the gap increases.
https://flic.kr/p/bmmGau

http://en.wikipedia.org/wiki/Faster-than-light_neutrino_anomaly#Measurement_errors
 
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MrChips

Joined Oct 2, 2009
22,089
Yes, you can try this but pay attention to the numbers.

Speed of light is 3 x 10^8 m/s.
The inverse of that is about 3ns/m.
If we use a velocity factor of about 66%, we end up with 5ns/m.

Thus 10m of cable would give you a round-trip reflection of about 100ns.

So you need:

1) a signal generator with fast rise-time that is less than 10ns
2) an oscilloscope with 100MHz bandwidth or higher
3) about 10m of cable
 

Thread Starter

bug13

Joined Feb 13, 2012
1,885
Yes, you can try this but pay attention to the numbers.

Speed of light is 3 x 10^8 m/s.
The inverse of that is about 3ns/m.
If we use a velocity factor of about 66%, we end up with 5ns/m.

Thus 10m of cable would give you a round-trip reflection of about 100ns.

So you need:

1) a signal generator with fast rise-time that is less than 10ns
2) an oscilloscope with 100MHz bandwidth or higher
3) about 10m of cable
Thanks for the number, can I understand the number like this: if the cable is say 1m, the refection won't be significance.

And how about a 1m cable, but the cable is highly inductive and/or capacitive, will I see it?
 

MrChips

Joined Oct 2, 2009
22,089
That's not correct thinking.

The round-trip with a 1-metre cable will be 10ns. This produces "ringing" at the edges of the signal. The reason you may not see it is a combination of the rise-time of the signal may not be fast enough and the bandwidth of your oscilloscope is not high enough.
 

Thread Starter

bug13

Joined Feb 13, 2012
1,885
That's not correct thinking.

The round-trip with a 1-metre cable will be 10ns. This produces "ringing" at the edges of the signal. The reason you may not see it is a combination of the rise-time of the signal may not be fast enough and the bandwidth of your oscilloscope is not high enough.
Hi MrChip

Thanks for correcting me, that's actually clear thing up a lot. So if I understand it correctly, inductive and capacitive of a cable has no/little effect of the refection of a signal transmitting on the cable?
 

MrChips

Joined Oct 2, 2009
22,089
Did I say that? I don't think so.

The inductive and capacitive properties will determine the characteristic impedance of the cable which in turn will affect transmission line reflections.
 

crutschow

Joined Mar 14, 2008
25,647
So if I understand it correctly, inductive and capacitive of a cable has no/little effect of the refection of a signal transmitting on the cable?
Actually it's the opposite. The characteristic impedance of transmission cable is determined mainly by it's distributed capacitance and inductance. It's interesting that the value of this is given in ohms, the same as a resistance, and is theoretically independent of frequency and cable length. The distributed resistance of a real cable wire is normally much less than its characteristic impedance.
 

nsaspook

Joined Aug 27, 2009
7,733
It's interesting that the value of this is given in ohms, the same as a resistance, and is theoretically independent of frequency and cable length. The distributed resistance of a real cable wire is normally much less than its characteristic impedance.
Yes it is, the distributed LC elements model the impedance of the dielectric between the conductors where the energy flows. Real resistance (in conductors and possible dielectric loss) is seen as transmission losses and the dielectric impedance is seen as the confined space that the EM EB fields travel in-phase when impedance matched so it's seen as resistive but not dissipative (most of the energy is constrained between the conductors). When we have reflections and standing waves from an impedance mismatch the phase of the EM fields changes making part of the cable impedance reactive along it's length.

 

Thread Starter

bug13

Joined Feb 13, 2012
1,885
Did I say that? I don't think so.

The inductive and capacitive properties will determine the characteristic impedance of the cable which in turn will affect transmission line reflections.
No you did not say that, it's just me try to understand what you said, and repeat it in a different way, if you confirm it, I know I get it right, otherwise I know there is something I still don't understand.
 
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