Next, we try doubling that to make an 8 second wide pulse. Although the response of the first lump is more step-like, this step smooths out in transmission further along the line, so that the 23rd lump is again not very flat.
Finally, lets try a 12 second wide pulse. Still not very satisfactory, but somewhat better in this time regime. However, further reflections continue to smooth out the initial step. What is causing this loss in clarity with time? A first thought would be dissipation of some type in the system - and in a real system that would be a good guess. However, this ideal capacitor/inductor system has no place for dissipation to occur. Thus, the decay in the response with time that we are seeing is artificial and numerical in nature. When working with numerical solutions you must always be watching for things like this that don't quite make sense: they are an indication of an error in your numerical analysis, time steps that are too large, lack of convergence, etc.
In spite of the numerical losses occuring in our solution, one thing we can see clearly with the above series of plots is that the width of the input pulse does not affect the propagation velocity. (The pulse travels from lump 3 to lump 23 in exactly the same time period.) In fact, if you try changing the amplitude of the pulse height, you will also find that does not affect propagation velocity.
We can see several things here: first indeed the single step pulse we input for 12 seconds travels down the line, reflects from the end, comes back, reflects from the start, travels down, reflects from the end, etc. What happens when the pulse reflects from the initial end (at about timestep 70)? Now it travels down the wire with a voltage of the opposite sign! Why is this? The boundary condition on the left end of the system is a zero voltage input (after the initial step pulse), so the inductor 1 is effectively attached to ground after the initial pulse is sent. Thus, when the current is coming back down the line towards the startpoint, capacitor 1 charges up, then discharges through inductor 1. When the capacitor is discharged (V1=0), the current in inductor 1 is still going and since the current in the inductor cannot stop or reverse instantaneously, the inductor continues to pull current through the capacitor such that capacitor 1 becomes negatively charged! Eventually the current in the inductor reaches zero and reverses to go back down the line, but now a negative voltage pulse will be sent from capacitor 1 down the line and back!
Not quite distinct enough, especially toward the end of the single down and back traverse. Let's put the pulses 4 seconds apart:
OK, so it seems that two four second pulses, four seconds apart can travel down and back the transmission line distinct. Since this transmission line can carry 2 pulses in 12 seconds, that makes the bandwidth 2/12=0.167 pulses/second.
We see that the voltage 3 rises to approximately unity and stays high for the duration, unlike the pulse input in which the voltage 3 was also a pulse. When the front end of the step input reaches the start point again, there is a doubling of the voltage in V3. For voltage 23, we see that the level rises to approximately unity, then also doubles upon reflection (about timestep 45) and stays elevated for a while! Let's carry this out a little longer, long enough for a pulse to make two complete down and back traversals of the line:
Now we see that the voltage 23 drops back to unity, then to zero. Let's keep going:
We can see that the reflections of the step input are very complex and similar in nature to the reflections with a pulse, with the important change that the boundary condition on the left end is now a constant voltage source, not ground. This accounts for the differences in behavior between the two responses. Finally, let's look at this in a mesh plot: