# Length of coaxial cable, based on signal reflection

1. Feb 2, 2014

### xWaffle

1. The problem statement, all variables and given/known data
We measured the time between a signal source, and it's reflection coming back through our probe after going through an open-ended coaxial cable.

My teacher told us this: the cable has a polyethylene insulator between central wire and the grounding web, which has a dielectric constant of 2.3, which slows the speed of the signal in the wire. Find the length of the wire. You may need to use the web to look up cable properties.

2. Relevant equations
The resistor that 'cancelled' the reflected signal was 51 ohms, so I'm assuming that is the characteristic resistance of the cable. I found that RG-9/U cables have this characteristic resistance.

I have also found that in a transmission line, signal velocity is the reciprocal of the square root of the capacitance-inductance product, where inductance and capacitance are typically expressed as per-unit length.

v = 1/ sqrt(LC)

But I can't find the characteristic L or C of this wire anywhere.

3. The attempt at a solution
I'm not sure. Is this analog to waves in classical mechanics? What does the dielectric constant have to do with the speed of the signal if it's surrounding the wire that the signal has to travel through?

I'm not really sure how I am supposed to do this, and that is literally all the information provided to us. This is for a lab writeup. The delay in the reflected signal I measured as just about 300 nanoseconds.

Last edited: Feb 2, 2014
2. Feb 2, 2014

3. Feb 2, 2014

### xWaffle

Thank you, I think I got what I needed from there. It just seemed like too simple of a problem.
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Now, would anyone happen to be able to tell me why a signal inverts iitself when coming back down the coaxial cable if we've shorted the other end?

I know the math that explains it, I just can't wrap my head around what is physically happening with the signal.

The reflection coefficient

$\Gamma = \frac{Z_{load} - Z_{0}}{Z_{load} + Z_{0}}$

where Zload is the impedance of the load, and Z0 is the impedance of the cable.

When Zload = Z0, Γ is zero, and there is no reflection.

When the cable is open-ended, Zload = ∞, and you can rearrange it so that Γ = 1, in other words the signal is reflected back identically.

When the cable is shorted, Zload = 0, and you that Γ = -1, in other words the signal is reflected back inverted.

What is physically happening to the signals here? I'm trying to find an analog to waves on a string (attached to a wall at one end), but I really can't wrap my head around it.

4. Feb 3, 2014

### CWatters

Consider an organ pipe with a closed (=shorted) end. The air is not able to move at that end so there must be a node there (At a node the amplitude is zero). So you have two waves moving in the pipe, the forward wave and the reflected wave. How would these two waves need to be arranged at the end so that they interfere with each other to produce a node? For that to happen the reflected wave has to have the opposite amplitude to the forward wave, eg it's inverted.

Ultimately it comes down to conservation of energy. There is no energy loss due to resistive heating (P=IV) in either a short (V=0) or open circuit (I=0).

5. Feb 3, 2014

### CWatters

In the case of a string attached to a wall...

The amplitude must be zero so again there must be a node there. For that to occur the amplitude of the reflected wave must be opposite to the forward wave.

Actually nothing is totally rigid so the fixed end can move very slightly storing energy in the molecular bonds in the attachment point. Perhaps think of those bonds as very tiny ideal stiff springs that briefly store the energy arriving in the forward wave and then returning that energy to the string to form the reflected wave.