2
\$\begingroup\$

Compute the voltage on the end of coaxial line (\$Z_0=75~\Omega\$, \$l=100~m\$,\$\tau=5~ns/m\$) at time \$t=1.6~\mu s\$ after connecting the line to voltage source (\$U_0=6~V\$,\$R_i=0~\Omega\$), if line is ended by resistor \$150~\Omega\$.

Found this example in older test. I have no idea how the schema should look like, nor what \$\tau\$ is and how does it affect the computation. Does anybody understand what is needed to do? I don't know result.

\$\endgroup\$
  • \$\begingroup\$ This is a classic transmission line problem. en.wikipedia.org/wiki/Transmission_line \$\endgroup\$ – Matt Young Jan 16 '15 at 16:03
  • 2
    \$\begingroup\$ \$\tau\$ is a measure of the speed at which an energy wave propagates down the cable. With a 100 m cable, and a \$\tau\$ of 5 ns/m (a velocity of 0.2 m/ns), it takes 500 ns for anything that happens at one end of the cable to be "seen" at the other. This is the basic interval between reflections. 6V is applied to the line at time 0, the first reflection occurs at the terminator at time 500 ns, the second reflection occurs at the source at time 1.0 us, etc. Time 1.6 us is shortly after the third reflection, at the terminator, has occurred (at 1.5 us). \$\endgroup\$ – Dave Tweed Jan 16 '15 at 16:51
  • \$\begingroup\$ Nope, I don't get it. I have spend about 3 hours searching through web, watching YT education videos about transmission lines, I didn't find any solved example, and I can't see how to use information I have to compute the voltage in that time on that load. \$\endgroup\$ – user50222 Jan 16 '15 at 19:55
  • \$\begingroup\$ Is the telegrapher's equations involved somehow? What form of it? The source seem to be DC... it use to be AC in other materials. \$\endgroup\$ – user50222 Jan 16 '15 at 19:57
2
\$\begingroup\$

First, read this answer for a general description of what's going on.

In your specific example, at t = 0, 6V is applied to the transmission line, which initially has 0V everywhere along its length. Since the source has zero resistance, the full 6V appears across the 75Ω impedance of the line, and a current of 6V/75Ω = 80 mA flows into it. This wave propagates toward the terminator.

At t = 500 ns, this wave reaches the 150Ω termination. However, at 6V, only 40 mA would ordinarily flow through this termination, so there's an excess of 40 mA in the line. The voltage rises above 6V by enough to cancel this excess current in the line, keeping in mind that as the voltage rises, more current flows through the termination, too. For every volt above 6V that the voltage rises, 1/3 of the excess current flows through the 150Ω terminator, and 2/3 of it flows back into the 75Ω impedance of the line. Since we have an excess of 40 mA, the voltage rises until another 13.33 mA flows into the terminator and 26.67 mA flows back into the line. Since there's now a total of 53.33 mA flowing through the terminator, the voltage rises to 53.33 mA × 150 Ω = 8 volts. This wave (a step from 6V to 8V) propagates back toward the source. In general, you can say that this ratio of line impedance to termination impedance always reflects 1/3 of the incident wave back, with the same polarity.

At t = 1000 ns = 1.0 µs, this wave reaches the source. However, since the source has zero impedance, the voltage there must remain at 6V. In effect, this creates a -2.0 V step that directly cancels the +2.0 V step created by the terminator. However, the current in the line was reduced by the terminator from 80 mA initially to 53.33 mA, and this is the value of the current throughout the line at this moment in time. But in order to create this new negative-going step, it is necessary for the current in the line to drop by another 2.0 V / 75 Ω = 26.67 mA. This step propagates toward the terminator, but now the current in the line behind it is reduced to a total of 26.67 mA. In other words, a zero-ohm termination (even at the source) reflects the incident wave at 100% amplitude, but the voltage is inverted and the incremental current is doubled.

At t = 1.5 µs, this new step (ΔV = -2.0 V) reaches the terminator. But once again, the current and voltage do not match what the terminator requires (40 mA @ 6 V), and so the terminator once again reflects 1/3 of this wave back toward the source, making the line voltage 6.0V - 2.0V/3 = 5.333 V. This, by the way, is the answer to the original question, since this is the voltage that exists at the terminator from 1.5 µs until the next reflection arrives from the source at 2.5 µs.

These reflections keep going on, with the amplitude being reduced by 1/3 at each reflection at the terminator, until everything eventually settles down with the line voltage at 6V and 40 mA flowing through the line and the terminator.

The voltage at the termination goes like this: $$ \begin{matrix} time (\mu s) & & voltage \\ \hline 0.0 - 0.5 && 0.000 V\\ 0.5 - 1.5 && 6.000 V + 1/3 \cdot 6.000 V = 8.000 V\\ 1.5 - 2.5 && 6.000 V - 1/9 \cdot 6.000 V =5.333 V\\ 2.5 - 3.5 && 6.000 V + 1/27 \cdot 6.000 V = 6.222 V\\ 3.5 - 4.5 && 6.000 V - 1/81 \cdot 6.000 V = 5.926 V\\ 4.5 - 5.5 && 6.000 V + 1/243 \cdot 6.000 V = 6.025 V\\ 5.5 - 6.5 && 6.000 V - 1/729 \cdot 6.000 V = 5.992 V\\ \end{matrix} $$ ... and so on.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.