There's a 9 V battery with output resistance of 50 Ω, at time t = 0 it is connected to two lossless transmission lines connected in parallel: LTx1 and LTx2.

Both lines are identical, with characteristic impedance of 50 Ω, except for the load connected at the end of each one: an open circuit in LTx1 and a short circuit in LTx2.

The time it takes a wave to travel from end to end of either line is T, find the voltage distribution of both lines at time T = 5T/2.


Here's my solution so far:

  1. At time t = 0 the source sees two 50 Ω impedances in parallel, so the initial wave that propagates through both lines is 9 * (25/75) = 3 V.
  2. At time t = T in Ltx1 the 3 V arrives at the load and 3 V are reflected back, meanwhile in LTx2 -3 V are reflected back.
  3. At time t = 2T, there should be 6 V everywhere in LTx1 and 0 V in LTx2, the +3 V wave and -3 V wave arrive at the same time to the joint where both lines and the source meet.
  4. I've no idea how to handle the reflections at this joint.
  • \$\begingroup\$ First part (1). How can battery see two lines in parallel? \$\endgroup\$
    – johnnymopo
    Apr 9, 2021 at 14:14
  • \$\begingroup\$ what i mean is that the input impedance 'seen' from the battery into the transmision lines is the two characteristic impedances of the lines in parallel. \$\endgroup\$
    – Natt
    Apr 9, 2021 at 16:08
  • \$\begingroup\$ Perhaps I take the drawing at face value. Since the 1st stretch shows 50 Ohms, it only makes sense it has an electrical length. But I think that the 50 Ohm is the internal resistance. I am incorrect \$\endgroup\$
    – johnnymopo
    Apr 9, 2021 at 23:19
  • 1
    \$\begingroup\$ the 50 Ω rigth next to the 9 V is the series resistance of the battery, the other two 50 Ω that appear next to LTx1 and LTx2 are the characteristic impedances of each transmission line. \$\endgroup\$
    – Natt
    Apr 9, 2021 at 23:26

1 Answer 1


That homework really needs one to keep his head calm. But it can be solved without being a genius by using a brute force method: Build an operator equation. It fortunately can be simplified during the job because the R = Z =50 Ohm and the reflection factors at the ends of the lines are +1 and -1.

Normally homeworks do not get very detailed guidance, but I guess this case earns more because I present a not so common method.

One must build an equation for the total voltage Ux at the midpoint. It can be said in three ways. Elementary circuit theory says that Ux = E - IR where R is the resistance of the 50 Ohm resistor, I is its current and E = 9V. At the same time the Ux must fulfill what's expected from the cables - both cables at the same time.

The upper cable forces Ux = U1 + D(U1) where U1 is the voltage of the wave which travels towards the open end, D is a linear delay operator, delay is the travelling time forth and back =2T.

In the same way the lower cable forces Ux = U2 - D(U2) where U2 is the wave voltage going towards the shorted end. Minus comes from the fact that the shorted end returns the coming wave as inverted.

For currents you get I = ((U1) - D(U1))/Z + ((U2) + D(U2))/Z. Note that the returning current components must be subtracted from the components travelling towards the distant ends.

These equations give after some shuffling the wave voltage U1 as a quite simple formula of E, delayed E and delayed U1. This can be considered as recursive construction formula for U1. Being an operator formula it's not bound to reflection moments, it's valid continuously when time is greater or equal to zero, but it can be used as recursive calculation, too. The U1 contains at the midpoint everything which propagates in the upper cable towards the open end, also anything which has reflected from the midpoint and the shorted end of the lower cable.

The formula: U1 = (E-DE-(D^2)U1)/3 . Note that E-DE becomes zero after 2T. At the start DE = 0 and (D^2)U1 = 0.

You can calculate U1 with 2T timesteps. After having U1 calculated for long enough time calculate Ux = (1 + D)U1. Let T = 1us. Ux will jump at the start to 3V as you have already found. That voltage stays until t = 4us. Then Ux drops to -1V. That situation stays until t = 8us.

To get the recursion formula for U1 you get also U2 (=the wave voltage going towards the shorted end) presented with U1. Now you have all what's needed. You can calculate total voltages at both cables as U1+D(U1) and U2-D(U2).

One thing needs special attention. The formulas give wave voltages at the midpoint of the circuit at times 0, 2T, 4T, 6T...For some intermediate point you must interpolate by checking to where the parts which cause the next jump at the midpoint have propagated. Formally for total voltage in an intermediate point of the upper cable you need to calculate Da(U1)+Db(U1) where U1 is at the midpoint of the circuit, Da is means propagation delay Ti from the midpoint to the intermediate point and Db is 2T-Ti. For the shorted cable you calculate Da(U2)-Db(U2).

BTW. Simulation with Micro-Cap can give a good check how you succeeded. The circuit:

enter image description here

The 1MOhm resistors are inserted to prevent the forbidden "No DC path" -condition. The next circuit would do the same as the previous because in Micro-Cap the ground is a single point with zero width and length:

enter image description here.

In the first version the open end voltage must be calculated as v(4)-v(5), in the second version it's v(3)


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