Return to Question

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)?

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)?

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.

1. In the equivalent circuit, is \$ V^{++} \$ the voltage across the right impedance \$ Z_0 \$? If yes, one could write

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

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.

2. Is it possible to cancel this 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.

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)?