So when talking to one board, the other 9 are high impedance;
This isn't quite right. The input impedance of the receiver doesn't change when it is listening or not listening. So all 10 loads will be high impedance (or capacitive).
If this is the case, should I series terminate each line according to the transmission line impedance (ribbon cable, so ~100 Ohms)?
This won't do any good. 100 ohms in series with a high impedance is still a high impedance. If you are going to terminate these lines at the receiving end, you would need the termination to be in parallel with the load. But be careful before you do that and make sure your driver can actually drive a 100 ohm load.
Series terminations are more often seen at the source, since the driver tends to be low impedance, and, say, 95 ohms in series with the driver might match a 100-ohm line reasonably well.
If you have no series termination and high impedance, you get up to 2*Vin at the source on the back reflection. That was just one wire; I just scaled that up for 9 more reflections. Is my logic flawed?
Yes, your logic is flawed. Because if you split up the signal to lead to 9 (or 10) loads, only a fraction of the energy would travel down each line. If all the lines were the same length, you'd end up with a total reflection of 2*Vin (or probably a little less because some of the signal would have been reflected back at the fanout point, and returned out of phase with the other reflections).
So what should you actually do?
Depending on your design constraints, you could try
- Connecting the loads in daisy-chain configuration and provide a parallel termination only at the end.
or
- Use a fan-out buffer to drive the signal to each load separately
It is actually valid, the way this question is worded, that the left and right side's reflection coefficient don't know about each other. This is because the time of interest expires after 120 ns. As you computed, one leg is 40ns long, the other is 25ns. If you consider that the for one load to affect the reflection of the other the step voltage would have to travel round trip on one leg, (e.g. 2 × 40 ns) and round trip on another to get t point A twice (e.g. 2 × 25 ns), it would take 130 ns. Therefore, an assumption of not considering the loads knowledge of each other is somewhat valid here as this effect happens 10 ns after the observation ends.
That is to answer about the multiple loads. I'll leave the rest of your homework to you.
Best Answer
That homework really needs one to keep his head calm. But it can be solved without being a genius by using a brute force method: Build an operator equation. It fortunately can be simplified during the job because the R = Z =50 Ohm and the reflection factors at the ends of the lines are +1 and -1.
Normally homeworks do not get very detailed guidance, but I guess this case earns more because I present a not so common method.
One must build an equation for the total voltage Ux at the midpoint. It can be said in three ways. Elementary circuit theory says that Ux = E - IR where R is the resistance of the 50 Ohm resistor, I is its current and E = 9V. At the same time the Ux must fulfill what's expected from the cables - both cables at the same time.
The upper cable forces Ux = U1 + D(U1) where U1 is the voltage of the wave which travels towards the open end, D is a linear delay operator, delay is the travelling time forth and back =2T.
In the same way the lower cable forces Ux = U2 - D(U2) where U2 is the wave voltage going towards the shorted end. Minus comes from the fact that the shorted end returns the coming wave as inverted.
For currents you get I = ((U1) - D(U1))/Z + ((U2) + D(U2))/Z. Note that the returning current components must be subtracted from the components travelling towards the distant ends.
These equations give after some shuffling the wave voltage U1 as a quite simple formula of E, delayed E and delayed U1. This can be considered as recursive construction formula for U1. Being an operator formula it's not bound to reflection moments, it's valid continuously when time is greater or equal to zero, but it can be used as recursive calculation, too. The U1 contains at the midpoint everything which propagates in the upper cable towards the open end, also anything which has reflected from the midpoint and the shorted end of the lower cable.
The formula: U1 = (E-DE-(D^2)U1)/3 . Note that E-DE becomes zero after 2T. At the start DE = 0 and (D^2)U1 = 0.
You can calculate U1 with 2T timesteps. After having U1 calculated for long enough time calculate Ux = (1 + D)U1. Let T = 1us. Ux will jump at the start to 3V as you have already found. That voltage stays until t = 4us. Then Ux drops to -1V. That situation stays until t = 8us.
To get the recursion formula for U1 you get also U2 (=the wave voltage going towards the shorted end) presented with U1. Now you have all what's needed. You can calculate total voltages at both cables as U1+D(U1) and U2-D(U2).
One thing needs special attention. The formulas give wave voltages at the midpoint of the circuit at times 0, 2T, 4T, 6T...For some intermediate point you must interpolate by checking to where the parts which cause the next jump at the midpoint have propagated. Formally for total voltage in an intermediate point of the upper cable you need to calculate Da(U1)+Db(U1) where U1 is at the midpoint of the circuit, Da is means propagation delay Ti from the midpoint to the intermediate point and Db is 2T-Ti. For the shorted cable you calculate Da(U2)-Db(U2).
BTW. Simulation with Micro-Cap can give a good check how you succeeded. The circuit:
The 1MOhm resistors are inserted to prevent the forbidden "No DC path" -condition. The next circuit would do the same as the previous because in Micro-Cap the ground is a single point with zero width and length:
In the first version the open end voltage must be calculated as v(4)-v(5), in the second version it's v(3)