# Series inductor as discontinuity in transmission line

Dual to this question is the following circuit:

simulate this circuit – Schematic created using CircuitLab

An infinite transmission line (with characteristic impedance $Z_0$) ends upon a series inductor $L$, then another infinite transmission line (with the same characteristic impedance $Z_0$) begins.

A step signal of amplitude $V^+$ is going from left to right: it will come across the inductor and the current will "charge" it.

The following schematic is the equivalent circuit:

simulate this circuit

I followed a procedure similar to the previous one, and wrote the following equation for the charge process of the inductor:

$$I_L (t) = \frac{V^+}{2Z_0}(1 - \exp{(-t/\tau_L)})$$

where $\tau_L = L/(2Z_0)$. But now I would like to obtain the following result:

$$V^{++} (t) = V^+(1 - \exp{(-t/\tau_L)})$$

(exactly the dual of this) where $V^{++}$ is the voltage travelling from the inductor to the right infinite line.

I am supposing that $V^{++}$ is the voltage across the right impedance $Z_0$. So,

$$I_L(t) = \frac{V^{++}}{Z_0}$$

but anyway

$$\frac{V^{++}}{Z_0} = \frac{V^+}{2Z_0}(1 - \exp{(-t/\tau_L)})$$

$$V^{++} = \frac{V^+}{2}(1 - \exp{(-t/\tau_L)})$$

and there is an undesirable $2$ factor. I would like that $V^{++} \to V^+$ for $t \to \infty$, but when $I_L(t) \to V_0 / (2Z_0)$ there is an unavoidable voltage divider, maybe due to the circuit.

Is it possible to cancel this $2$ factor (and obtain exactly $V^{++} (t) = V^+(1 - \exp{(-t/\tau_L)})$ like in the capacitor)?

• Similar comments/answers to the previous question, then!
– Chu
Commented Jul 2, 2015 at 15:14
• @Chu is still there something not clear? Commented Jul 2, 2015 at 15:25
• Your equivalent circuit is wrong as it is missing both transmission lines ! Maybe you assume that you can omit them since you terminate them with Z0 at both ends ? BUT at the inductor you did not terminate with Z0. Commented Jul 2, 2015 at 16:01
• @IC_designer_Rimpelbekkie Yes, I omit them because they are infinite, that is, matched (terminated with $Z_0$). What do you mean by "at the inductor you did not terminate with $Z_0$"? The right line is represented by a $Z_0$ impedance; the left line is represented by the same impedance, with the generator which corresponds to the signal. Commented Jul 2, 2015 at 18:05
• You cannot simply cut a transmission line, insert a coil and continue the transmission line ! At the point where you cut it, the transmission line stops. A signal would reflect at that point since it is not terminated by Z0. You do not have an infinate transmission line, you have 2 infinate transmission lines which are both badly terminated at one side (at the coil). Commented Jul 2, 2015 at 21:21

... undesirable 2 factor. I would like that $V^{++} \to V^+$ for $t \to \infty$, but when $I_L(t) \to V_0 / (2Z_0)$ there is an unavoidable voltage divider, maybe due to the circuit.

The factor of 2 has nothing to do with the inductive discontinuity. To see that this is true, take the limit as $L\to{}0$, and you'll still have the factor of 2.

The factor of two is fundamental to using matched sources and loads with transmission lines. If you want to have a matched source generate a signal on a transmission line with amplitude $V$, you need the amplitude of that source to be $2V$.

Is it possible to cancel this $2$ factor (and obtain exactly $V^{++} (t) = V^+(1 - \exp{(-t/\tau_L)})$ like in the [capacitor][1])?

No. The factor of 2 was also there with the capacitive discontinuity, if you did your math right.

• Yes, it was also in the capacitor, but there the charged capacitor did replicate the $V+$ voltage at the beginning of the right line. Here instead the charged inductor is a short-circuit, as you rightly suggested. Thank you again for your useful answer. Commented Jul 2, 2015 at 16:06
• @BowPark, no the capacitor didn't replicate $V_+$. If you do the math right, in the capacitor case you should also get $V_{++}\to V_+/2$ as $t\to\infty$. Commented Jul 2, 2015 at 16:08
• oh, you are right!!! Commented Jul 2, 2015 at 18:03
• Sorry for bothering you again, but the conclusion is: a parallel (like the capacitor) or series (like the inductor) component as a discontinuity between two transmission lines introduces not only a delay for the signal, but it also splits in half the value of the voltage wave travelling through the discontinuity (in the best case, assuming that the lines are matched). Is it so? Commented Jul 2, 2015 at 18:25
• No, even if there's no discontinuity at all, the signal is halved. Even if there's no transmission line at all, actually. You have a source with 50-ohm output impedance, and a 50-ohm load. The voltage divider between them cuts the voltage in half. Commented Jul 2, 2015 at 20:29