2
$\begingroup$

enter image description here

The transmission line is without loss of length λ / 4, the load $Z_L$ is a short circuit and $Z_G = Z_0$.

At $t = 0$, an incident wave of voltage $V_1$ and current $I_1$ such that $V_1 = Z_0 * I_1$ travel to the load, before $V_1$ reaches plane B the source $V_G$ delivers a power $P_G = V_G^2 / (Z_G + Z_0)$, then the waves $V_1$, $I_1$ reach the plane B and are fully reflected, generating reflected waves $V_2$, $I_2$ such that $V_2 = Z_0 * I_2$ that travel towards the source $V_G$, both form standing waves of total voltage $V = V_1 + V_2$ and total current $I = I_1 + I_2$ such that in plane A, $V / I = Z_{in} = ∞$ then the source sees an open circuit at A and cannot deliver any more energy.

If the above is correct, my questions are:

  1. once the $V_2$, $I_2$ waves reached the source, what happens to them?
  2. If the answer is that they dissipate in $Z_G$, where does that energy come from? since the source sees an open circuit ($Z_{in} = ∞$) and for this to happen I need $V_1$, $I_1$ and $V_2$, $I_2$ at all times.
$\endgroup$
2
$\begingroup$

The source resistor will dissipate the vector sum of what can be modeled as the forward and reverse currents. Thus, for a pure sinusoid in steady state, if the forward and reverse currents are equal in magnitude but 180 degrees out of phase, they will vector sum to zero, which leads to the resistor dissipating zero power.

Linear time invariant systems are nice because you can sum things (or even make up non-zero things that sum to zero.)

Yet another way to look at it is that, in steady state, the instantaneous generated voltage on the left side of the resistor will equal the instantaneous reflected voltage on the right side, at all times. Thus with no voltage difference across the resistor, there will be no current through the resistor, thus no power dissipation. (assuming ideal lumped component models, and etc.)

$\endgroup$
1
  • $\begingroup$ The answer is perfect from the point of view of the current, now I can see what is happening there. From the voltage point of view, only correct that on the right side of the resistance Zg the instantaneous voltage is the sum of the incident plus the reflected one, for example: left side VG Volts and right side (Vg / 2 + Vg / 2) Volts, with the length of the line and the gamma = -1 in the load all make sense. Thank you! $\endgroup$ May 22 at 3:42

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.