I just solved a problem in my homework. I had to calculate the reflection waves and I was given the following data:

R0 = 120 Ω line impedance
Ri = 90 Ω resistance at input
Rb = 1 kΩ termination resistance

l = 0.5m length of the line
δ = 6 ns/m

Source voltage
U0 = 0.4 V logical 0 - low voltage
U1 = 4.8 V logical 1 - high voltage

I was also given the following graph:
enter image description here

Showing I have to calculate reflections for when signal goes from 0>1 / low to high voltage basically when we have a pulse.

First I calculated Tau, T = l * δ = 3 ns, and then I calculated the reflection coefficients at R0 and Rb using formula ρx = \$ \frac{Rx - R0}{Rx = R0} \$

With that I was able to calculate for the part before the pulse:

ui(0-) = ub(T-) = U0 \$ \cdot \$ \$ \frac{Rb}{Rb + Ri} \$

Followed by the rise:

ui(0+) = ui(0-) + ΔU \$ \cdot \$ \$ \frac{R0}{R0 + Ri} \$

And then I was simply able to calculate reflection voltages at certain time for instance first traveling voltage from the change:

u0(1) = ΔU \$ \cdot \$ \$ \frac{R0}{R0 + Ri} \$

And then reflection voltage at T on the end of the line:

ub(T+) = ub(T-) + u0(1) + u0(2)

After that I can just calculate the reflections until 5 Tau to see if the signal stabilizes anyhow.

  • My question is how do my calculations and formulas change when the graph provided would be inverse showing a drop in the pulse from 4.8V to 0.4V ?
  • \$\begingroup\$ I'm not sure I want to figure out what all of your different \$u(t)\$ variables are supposed to represent, but why do you think the formulas would change? \$\endgroup\$
    – The Photon
    Commented Aug 8, 2013 at 16:38
  • 1
    \$\begingroup\$ What happens if you plug in 4.8 V for \$u_i(0^-)\$ and 0.4 for \$u_i(0^+)\$, and why don't you think that's the right solution? \$\endgroup\$
    – The Photon
    Commented Aug 8, 2013 at 16:40
  • \$\begingroup\$ @ThePhoton Turns out I just messed up with variables if graph is inverted then U<sub>1</sub> = 0.4 and U<sub>0</sub> = 4.8 and then we you calculate the reflection at 5 Tau it comes out to 0.38 or rounded 0.4 which would equal the U<sub>1</sub> = 0.4 which means signal has stabilized and we have no more reflections. \$\endgroup\$ Commented Aug 8, 2013 at 18:35

2 Answers 2


First of all, you wait for 5 tau for the transient to extinguish, provided that tau is the time constant of an exponential signal. When transmission lines are involved, the decay at each reflection is dependent on the reflection coefficient at the load and the generator. In general, you do not wait 5 line delays! (Consider the limiting case of a negligible impedance source driving a line terminated in an open or short circuit, where you have theoretically infinite ringing).

Then, as Ignacio explains, you can use superposition. Keep in mind that the solution for a constant signal is what you would obtain from classic circuit theory: the input and output ports collapse into lumped nodes. You can get the solution as a voltage divider. If you are able to compute vo(t) as the response to a unit step u(t), the response to A·u(t) is A·vo(t). And the response to a sum of inputs is the sum of their individual responses...


You can use linearity of the wave equation. A pulse going from 4.8V to .4V is like the original pulse inverted, plus a constant 5.2V. So take your previous solution, invert it and add 5.2V .


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