Simple Transmission Line Question

[HasuChObe]

Say you had a lossless transmission line that consists of an ideal source, a 50 ohm impedance, a 100 ohm impedance, and a 100 ohm terminator. The material is the same. You send a pulse down the line. When the pulse hits the impedance mismatch, you get a smaller reflection back, and a pulse bigger than the original pulse going toward the terminator. Since the material is the same, the pulse should have the same width (in space and time) as it did originally going down the line (prior to reflection) except now you have them where one is actually bigger than the initial pulse. How is energy reconciled in this case?

 

[uart]

The short answer is that the voltage is more but the current is less (in the transmitted pulse, compared to what it was in the incident pulse). More importantly, the product of voltage times current is also less.

 

[HasuChObe]

The short answer is that the voltage is more but the current is less (in the transmitted pulse, compared to what it was in the incident pulse). More importantly, the product of voltage times current is also less.

Hah; Of course I try to calculate field energy with only the E field :/ Good answer. Should be further along now :P Any inputs on the geometry changes from a smaller to larger impedance? I've concluded that the cross section of the dielectric part has to get bigger, but I could be wrong again -.-

 

[Bob S]

Are you assuming your ideal source has a 50 ohm output impedance? To solve the mismatch equation at the transmission line impedance step, you have to consider both the voltage V and the current I at the mismatch. For a forward (F) and reflected (R) signal you have 3 equations for the voltages and currents at the mismatch (Z₁=50 ohms, and Z₂=100 ohms). This is easily simulated in SPICE.

V_F1 = +Z₁·I_F1

V_F2 = +Z₂·I_F2,

V_R1 = -Z₁·I_R1

where F = forward, R = reflected.

[added] The minus sign comes from the direction of the Poynting vector.

Bob S

 

[Bob S]

Here in thumbnail is a simulation of the mismatch using LTSpice. A back-terminated voltage source drives a 50-ohm transmission line, which is connected to a 100-ohm transmission line, terminated at the end. For 1 volt in,

V_F1 = 1 volt
V_R1 = 0.333 volts
V_F2 = 1.333 volts
I_F1 = +20 mA
I_R1 = - 6.667 mA (note minus sign)
I_F2 = +13.333 mA

So V_F1 + V_R1 = V_F2
and I_F1 + I_R1 = I_F2

Bob S