# How can I convert from voltage TDR into impedance TDR

Based on the circuit below I have built the voltage based TDR diagram.

Then I went to the ADS TDR simulation and constructed by the manual the circuit below.

In the ADS circuit they say that IMPEDANCE=V/I=V/(delta(V)/R)=V*R/Voltage_drop_over_R

I can't see the logic of this formula.

Why does measuring impedance on the left side of the transmission line over time means its the impedance at some location in the transmission line?

• Try using mathjax i.e. $Z = \dfrac{V}{\frac{\delta V}{R}}$ = $Z = \dfrac{V}{\frac{\delta V}{R}}$ Commented Mar 11, 2021 at 12:50
• I fully appreciate how some people pronounce english words differently and my spelling is often poor, yet I recommend Grammarly for anyone. Commented Mar 11, 2021 at 12:50
• None of the symbols in your formula appear in either schematic, so I don’t know what they refer to. However, if V is the voltage drop across a 2-terminal component A in any branch, R is a resistor in the same branch, and I is (voltage drop across R)/R, then it is certainly true that V/I is the impedance at A (the voltage across A divided by the current through A). This does not tell us the impedance anywhere else in the circuit. Commented Mar 11, 2021 at 13:50
• Hello 10ppb,I have implemented an official keysight manual shown bellow. they say its impedance at every point(time varying location) youtube.com/watch?v=U3be5KkGmAc&t=121s Commented Mar 11, 2021 at 14:31

$$\Z(t)=\dfrac{V(t)}{I(t)} = \dfrac{δV(t)}{δI(t)}\$$
Similarily yet different for a Solar Panel's MPPT operating point. $$\Z_{mpt}=\dfrac{δV(t)}{δI(t)}\$$ where V*I is max and $$\Z_{mpt}\dfrac{V_{mpt}}{I_{mpt}}= \dfrac{V_{oc}}{I_{sc}}\$$