It is important to have your definitions straight. The source voltage is measured **after** the source's output impedance, as shown in below image:

^{simulate this circuit – Schematic created using CircuitLab}

You are right that V1 in the image is double amplitude with respect to "Source Voltage", but that voltage is not part of the equation.

So with an ideal cable and matched in- and output impedances, the source voltage is equal to the voltage at the load end and with that reflection coefficient is 0:

# Reflection coefficient

\$\varGamma = \dfrac{Z_L - Z_S}{Z_L + Z_S} = \dfrac{50 - 50 }{50 +50} = 0\$

# VSWR

VSWR for a properly terminated line is defined as:

\$VSWR = \dfrac{V_{MAX}}{V_{MIN}} = \dfrac{1+|\varGamma|}{1+|\varGamma|} = \dfrac{1+0}{1-0} = 1\$

**VSWR for an shorted line**

\$\varGamma = \dfrac{Z_L - Z_S}{Z_L + Z_S} = \dfrac{0 - 50}{0 + 50} = -1\$

\$VSWR = \dfrac{1+|\varGamma|}{1+|\varGamma|} = \dfrac{1+1}{1-1} = \dfrac{2}{0} = \infty\$

Which basically means the voltage along the line varies between 2 and 0 times the source voltage.

Voltage standing wave ratio is the ratio between the largest amplitude and the lowest amplitude found along the line. If the lowest amplitude nears zero, then VSWR will go to infinity.

**VSWR for an open line**

\$\varGamma = \dfrac{Z_L - Z_S}{Z_L + Z_S} = \dfrac{\infty - 50}{\infty + 50} = 1\$

\$VSWR = \dfrac{1+|\varGamma|}{1+|\varGamma|} = \dfrac{1+1}{1-1} = \dfrac{2}{0} = \infty\$

Which again basically means the voltage along the line varies between 2 and 0 times the source voltage.

Coaxial-cable in the above illustration is a standard 50 Ohm intrinsic impedance BNC cable. I know its length and lets say it is 20 meters long. Since we use lumped element model we will not use 50 Ohm right? And capacitance and the inductance will vary with length? In other words, how can I model this cable in LTspice?

First, when we talk about transmission lines, we talk about *characteristic impedance*. "Intrinsic impedance" is not a term that has any specific meaning in the area of transmission lines.

A lumped element model of a transmission line with 50 ohms characteristic impedance does not involve a 50 ohm resistive element in series. Characteristic impedance describes the ratio between voltage and current in the travelling wave that can propagate along the line. It doesn't cause any power loss like a series resistance would.

It might involve a series of capacitive and inductive elements in a pi or T section arrangement. A pi-section model of a lossless unbalanced line would look like this:

^{simulate this circuit – Schematic created using CircuitLab}

C1 and C6 would have half the value of C3, C4, and C5, because the intermediate capacitors actually each represent the shunt legs of two pi sections in parallel. The total capacitance should add up to the line's capacitance per unit length times its length. The total inducance should add up to the line's inductance per unit length times its length.

Obviously this model will fail when the frequency gets too high, because the first and last capacitance elements will effectively short out signals approaching in the forward and reverse directions. By increasing the number of sections you can reduce the capacitance per section in the model, and so increase the frequency where this issue occurs.

So now we have a transducer output impedance which is Rout=Zout, coaxial cable impedance which is 50 Ohm, and we have a load Rload=Zload which is lets say 100M.

The load impedance is not particularly realistic. Typical scope inputs are 1 or 10 Megohms. Scopes designed for measuring reasonably high frequencies will usually have an option to program an input impedance of 50 ohms.

So in this case to achieve impedance matching throughout the line I need a 50 Ohm resistor in parallel with Rload just before Rload to make the Rload 50 Ohm.

Yes, if your scope doesn't have a 50 ohm input impedance option, and reflections become an issue, you can add a 50-ohm parallel resistance at the input to reduce these reflections. It will also reduce the signal seen by the scope.

And I also need to know the Rout and I need to add a series or parallel resistor to it such that its equivalent or Thevenin would be 50 Ohm.

You don't strictly need to match both ends of the transmission line. If you match one end very well it will eliminate reflections that reach that end, so you won't see ringing from multiple reflections.

The 50 Ohm parallel resistor will load the transducer and I will have more error right? It seems to me for sending data this might not be problem but in this case the signal's voltage level is important. What would you suggest in this case?

You could either provide a buffer amplifier at the transducer to produce a signal with low output impedance.

You could move the scope closer to the transducer so that the line can be shorter and impedance matching not needed.

You could provide an RC filter at the transducer output to reduce the edge transition speed so that less high frequency signal is present and impedance matching is not needed.

## Best Answer

Think about the signal on the transmission line as being made up of a forward-travelling wave and a reverse-travelling wave.

At any given point on the line, the ratio of voltage to current in each of these two independent waves is determined by the characteristic impedance \$Z_0\$ of the line.

The forward travelling wave was initially created by the power source driving the line.

The reverse-travelling wave only exists because something caused reflections of the forward wave to return back up the line. Reflections only occur when there's some change in the line geometry, a circuit attached to the line, or a mismatched termination.

Your formula for \$Z_{in}\$ applies to one specific situation: There is a mismatched termination on an otherwise ideal line and you want to know the ratio of voltage and current (due to the combination of forward and reverse waves) at some location up the line from the termination. It accounts for the fact that the phase of the forward wave is different at this point on the line then where it encounters the load, and that the phase of the reverse wave has also changed as it traveled back down the line to where you're measuring.

It doesn't mean that the characteristic impedance of the line is changed in any way: The voltage/current ratio in the forward wave is still determined by \$Z_0\$ and the voltage/current ratio in the reverse wave is still determined by \$Z_0\$. And if there isn't a discontinuity or geometry change at this point in the line, there won't be any additional reflection created because \$Z_{in} \ne Z_0\$ at this point on the line.

The apparent input impedance will only go to infinity if the load has reflected 100% of the energy in the signal (\$|\Gamma_L|=1\$). In this case, in fact there can't be any energy delivered to the load, because all of the energy is reflected back in the reverse wave.