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I'm confused why sometimes when we look at the termination effect on a transmission line, we don't consider the overall input impedance looking into the transmission line including the load? For example, I have this: transmission line with series termination:

We say, the voltage at node A before the wave propagates down the transmission line is only 1/2 of Vin because we treat it as voltage divider of Rs and Zo (voltage at node A = Vi * Zo/(Rs + Zo)). Why in this case we are not considering the total Zin (blue, the impedance looking into the transmission line) for the voltage divider? Why do we only use Zo to form voltage divider with Rs? But, sometimes, why do we try to calculate for Zin? Thanks.

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    \$\begingroup\$ I think that's the transmission line peculiarity: when \$ R_L \$ and \$ Z_0 \$ are matched, \$ Z_{in} = Z_0 \$ . And impedances have to be matched to avoid reflections, which is indicated by a Standing Wave Ratio greater than 1. \$\endgroup\$ – user59864 Sep 18 '17 at 7:34
  • \$\begingroup\$ I can't quite tell what you are asking, but perhaps you are wondering why you need to match the characteristic impedance of a particular transmission line? Like RG-6 and RG-11 and RG-59 at 72 Ohms, but RG-8 and RG-58 at 50 ohms? If that is what you are asking then there's your answer - remember that a transmission line is like running inductors with capacitors across them all the way - which determine the transmission line characteristic impedance. Is that what you mean? \$\endgroup\$ – SDsolar Sep 18 '17 at 7:56
  • \$\begingroup\$ Let's say Zo = 50 Ohm = series termination resistor, Rs, but my RL = 1k. Why is node A voltage calculated from Vin * Zo/(Rs+Zo) instead of Vin * Zin/(Rs+Zin) where Zin is the input impedance looking into the transmission line from the blue line I drew that will include the load resistance RL? Why do we use Zo instead of Zin to find voltage at node A before the wave propagates down the transmission line? \$\endgroup\$ – user3618703 Sep 18 '17 at 8:12
  • \$\begingroup\$ If the impedances are matched then Zin=Z0. In your case Z0=50 and ZL=1k, so not matched, then the equation is much different, because the equivalent impedance Zin becomestottaly different. \$\endgroup\$ – Marko Buršič Sep 18 '17 at 12:04
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The characteristic impedance of the transmission line can be thought of an equivalent impedance seen into a long chain of series LC networks. The impedance which you are talking about is the impedance which the input voltage signal sees when the at the time signal is applied (t=0, at the time of input step).
Since the signal has not propagated down the transmission line yet, it has no idea that there is a load RL sitting after the transmission line. All it sees is an infinite series of differential capacitors and inductors. Thus we see only the characteristic impedance of the transmission line (T-Line) and the net impedance seen at the input will be iven by your equation.
After the propagation delay through transmission line it would see a discontinuity of the impedance and some of the signal will be reflected back from the load back into T-Line. The impedance seen now will not be given by your equation.
After the transients have settled down, the T-Line is just a wire with no impedance (assuming lossless T-Line) and the impedance seen now is just RL.
Note that the transmission line model, is for modelling the reflections and all different transient behavior due to finite transmission delay through a piece of wire. After the transients have settled down, there is no difference between T-line and normal wire. Thus, we should see no impedance from T-Line in steady state (for step inputs).

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The input impedance of a transmission line will be its characteristic impedance if the end terminator equals Zo. So, if Zo = RL then the input impedance to the line will be Zo irrespective of length.

If RL does not equal Zo then you get problems with line mismatches and reflections and these vary with operating frequency to cause a significant headache for digital transmission systems.

So, the assumption is that the terminating resistor RL equals the characteristic impedance Zo. This means that if RS = Zo then you have a simple potential divider.

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  • \$\begingroup\$ Series termination is used at the source and it doesn't require RL to be equal to Zo. RL is not a termination. It's just a load. That means the input impedance at blue, Zin, is not equal to Zo. Therefore, that's why I asked why we use Zo to calculate for the voltage at node A instead of Zin. \$\endgroup\$ – user3618703 Sep 18 '17 at 21:08
  • \$\begingroup\$ At the time of applying a pulse (for instance) the impedance instantly presented by the t-line is the characteristic impedance irrespective of RL. When the signal has reached the far end and meets an impedance other than Zo, there will be a reflection that travels back up the t-line. That reflection is line length dependent and will disrupt the original signal at point A but, if RS=Zo then there will be no counter reflection. \$\endgroup\$ – Andy aka Sep 19 '17 at 12:02
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here we are dealing with transient analysis of transmission lines.But the formula which you said is for steady state condition.Intially the current sees only characteristic impedance.Finally in steady state condition the current will be according to your formula.It is just like in open circuited transmission line, intially in transient state current flows because the signal can't see the open load it sees only characteristic impedance,but finally current becomes zero which is same as current from formula.

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If you're calculating for steady state- where the voltage source is a steady sine wave then you consider the impedance looking into the transmission line calculated from Z0 and RL. On the other hand, if you apply a voltage step to the line, initially you only consider Z0 because the source doesn't 'see' RL till the wave has had time to propagate.

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