1.5 Terminated lines

If all we did was launch signals down semi-infinite transmission lines, we would not get very much useful work done. We reallyneed to have a finite length line, and put something at the end...like a termination . So let's take a look at a terminated transmission line .

What happens when the signal gets to the load? We can assume some voltage $V_{\mathrm{L}}$ will appear across the load resistor, and hence a current $I_{\mathrm{L}}$ will flow through it. The most logical thing to assume would be that $V_{\mathrm{L}}=V_1^{+}$ . But , we quickly run into a contradiction. If $V_{\mathrm{L}}=V_1^{+}$ , then

What are we to do? We have an obvious contradiction. The telegrapher's equations permit two solutions to the transmission line problem: a signal going in the $x$ direction, $V_1^{+}$ , and a signal going in the $-x$ direction, $V_1^{-}$ . The only way out of our problem is to assume that when the $V_1^{+}$ signal gets to the load, a new signal , going in the $-x$ direction, is created, which then heads back towards the load. So, let's put in a $V_1^{-}$ and $I_1^{-}$ .

Going by that convention then, we equate voltage on either side of the termination

We could also have solved for $I_1^{+}$ in terms of $I_1^{-}$ and we would have found:

Let's take a break from equation manipulating, and think about what we have here. First of all, although theresult we have obtained is very important, the method we used to get there was even more so. What did we do? We postulated avoltage and current on the line, and then took a look to see if that solution resulted in a reasonable result. In this case itdid not. We were in gross violation of Kirchoff Laws! We rescued ourselves by taking the only possible escape route: we added anadditional voltage and current to the solution. Since $V^{-}$ and $I^{-}$ are related to each other in a different manner than $V^{+}$ and $I^{+}$ (by $-Z_0$ rather than $Z_0$ ), this gave us an additional degree of freedom so that we could simultaneously satisfy both the transmission line I-V relationships as well as the load I-Vrelationship.

Let's take a look at ${}_{\mathrm{vL}}$ for a minute. Over what range can it vary? A glance at shows that it depends on the range of $R_L$ . If we exclude the possibility of negative resistance (a reasonable exclusion) then $0\le R_L$∞ . As $R_L$ varies over this range, the voltage reflection coefficient goes from $-1\le {}_{\mathrm{vL}}\le 1$ . When $R_L$ is less than $Z_0$ , the reflection coefficient is negative. When $R_L$ is greater than $Z_0$ , the reflection coefficient is positive. When $R_L=Z_0$ the reflection coefficient is zero, and the line is said to be matched . In a matched transmission line, a signal traveling down it is completely absorbed by the load,and nothing is reflected from it. In this case we can have the incident voltage and current signals just equal load voltage andcurrents without the need to add reflected waves. For an unmatched transmission line, a signal incident on the load is(partially) reflected, and a new signal starts moving back down the line towards the source. An example of a transmission lineis a buss in a computer. What would be the implications of unmatched terminations on various connections between the bussand the computer circuitry?

Two special cases of terminated transmission lines that are of interest are $R_L=0$ (a shorted line) and $R_L$∞ (an open line). For $R_L=0$ , ${}_{\mathrm{vL}}=-1$ and so a signal with the same amplitude but opposite sign as the incident wave, is reflected back down the line. For $R_L$∞ , a signal with the same amplitude and sign gets reflected back down the line. It is easy toremember which is which, and also to make sure you have the right order in the equation for the reflection coefficient, whenyou keep in mind that the voltage across the load is the sum of $V_1^{+}$ and $V_1^{-}$ . If the line is terminated with a short circuit, $V_L$ must equal 0. Since

Now, let's go back to our terminated transmission line. What is going to happen to $V_1^{-}$ after it leaves the load? It obviously travels back down the line towards the generator. What happens when it getsback there? Time to look at .

We know

If you do not believe using superposition is valid, you can try doing the problem over again, but it shouldbe pretty obvious that we can write