# Calculating reflections

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

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

Line
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: 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 ?
• 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? Aug 8, 2013 at 16:38
• 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? Aug 8, 2013 at 16:40
• @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. Aug 8, 2013 at 18:35