# Step Response In a Hybrid Lumped-Distributed Network

### Introduction

Transmission line theory has been around for more than a century. However, the distributed nature of it has kept it interesting for all these years. One interesting subject is the step response in a hybrid lumped-distributed network. While for a purely lumped network it is quite easy to analyze and derive the response equation, for a distributed or a hybrid network, it may not be. This short post covers this topic briefly . It is assumed the reader has the fundamental knowledge of transmission lines.

### A Review of Transmission Line

Transmission lines are mainly defined with the following factors:

• Characteristic impedance $Z_{o}=\sqrt{\frac{R+j\omega L}{G+j\omega C}}$ where $R, L,G, C$ are the resistance, inductance, conductance and capacitance per unit length. In a lossless transmission line $Z_{o}=\sqrt{\frac{L}{C}}$.
• Propagation constant $\gamma=\alpha+j\beta$ where $\alpha$ is the loss (this post considers a lossless transmission line in which $\alpha$=0)  per unit length and $\beta$ is the wave number.
• The wave velocity (phase velocity) $\vartheta=\frac{{c}}{\sqrt{\epsilon_{r}}}$ where $c$ is the speed of the light and $\epsilon_{r}$ is he permitivity of the medium. For an IC designers $\epsilon_{r}$=4.2 as the signal travels in silicon-doixide.
• The signal wavelength $\lambda=\frac{\vartheta}{f}$ where $\vartheta$ is the wave velocity and $f$ is the frequency of the signal.

In a circuit where the signal wavelength is comparable to the size of the elements, the lumped analysis of the voltage and current (Kirchhoff circuit laws) is not valid anymore and should be treated with wave equation. In
this case, the signal travels like a wave on a transmission line and could have reflection if not terminated with a matched impedance load. If a transmission line with characteristic impedance of $Z_{o}$ is terminated with a load impedance of $Z_{L}$, then the reflection coefficient $\Gamma_{L}$ is defined $\frac{Z_{L}-Z_{o}}{Z_{L}+Z_{o}}$.

### Step Response Equation

We now can use the terminologies we defined earlier to derive the step response of a distributed network. Consider the circuit illustrated in the following figure.

Figure 1. Exciting a transmission line with a length of $l$. The signal travels with a velocity of $\vartheta$ through the line. The reflection coefficients are $\Gamma_{S}$ and $\Gamma_{L}$ at the begining and end of the line respectively.

The source has an impedance of $Z_{S}$ driving a transmission line with a length of $l$ and characteristic impedance of $Z_{o}$. The transmission line is then terminated with $Z_{L}$$\Gamma_{S}$ and $\Gamma_{L}$ are the reflection coefficients at the source and load respectively. Assume at $t=0$ the source voltage $V_{S}$ is launched. Then there is a voltage divide ratio between the source impedance, $Z_{S}$ , and $Z_{o}$ of the line. This defines the forward voltage travelling from source to the load at $t=0$. We call this $V_{1}=V_{S}\frac{Z_{o}}{Z_{o}+Z_{S}}$ . It takes $t_{0}=\frac{l}{\vartheta}$ for $V_{1}$ to get to the other side of the line where it sees $Z_{L}$. The reflected wave is now $V_{2}=\Gamma_{L}V_{1}$. The same analysis is applied for the travelling signal at each end point of the transmission

To derive the equation of the voltage at each node one need to usethe superposition of each reflected waves. Let's start with $V_{I}$.Looking at Figure 1 it is shown this node will have transition at time $0,2t_{0},4t_{0},...,2nt_{0}$. So:

\begin{align}
V_{I}(t)&=V_{1}u(t)+(V_{2}+V_{3})u(t-2t_{0})+(V_{4}+V_{5})u(t-4t_{0})\\ \nonumber
&=+...+(V_{2n}+V_{2n+1})u(t-2nt_{0})
\label{eq:VI}
\end{align}

Similarly, $V_L$ is derived as:

\begin{align}
V_{L}(t)&=(V_{1}+V_{2})u(t-t_{0})+(V_{3}+V_{4})u(t-3t_{0})\\ \nonumber
&=+...+(V_{2n-1}+V_{2n})u(t-(2n-1)t_{0})
\label{eq:VL}
\end{align}

To complete the discussion in this section the dual circuit of Figure 1 is also analyzed. Figure 2 shows the Norton equivalent of $V_S$ and $Z_S$. Since the reflection coefficient for current always has a negative sign, the sign of each reflected current is toggled every other time. This has been shown in Figure 2.

Figure 2. Dual circuit of Figure 1. The reflection coefficient for current always keeps a negative sign.

Similar to equation 2 one can derive $I_L(t)$ with the same exact format. For a purely resistive network, both $\Gamma_S$ and $\Gamma_L$ are real numbers. This makes ploting the step response quite easy. For a complex source and load impedance such as RC, RL and RLC networks, the reflection coefficients are not real anymore and further consideration is required to derive the step response.

Consider a $Z_S = 650 \Omega$, a 2mm transmission line with $Z_o = 90 \Omega$ that is terminated with a $Z_L = 130 \Omega$ in silicon-dioxide medium. Plot $I_I(t)$ and $I_L(t)$.

• $\Gamma_S = \frac{650-90}{650+90} = 0.756$
• $\Gamma_L = \frac{130-90}{130+90} = 0.181$
• $I_1 = I_S\frac{650}{650+90} = 0.878I_S$
• $t_0 = \frac{2x10^{-3}}{1.5x10^8} = 13.33ps$

By plugging the above numbers into the equations in Figure 2 one can derive $I_I(t) = 0.878u(t)-0.039u(t-26.6ps)-0.005u(t-53.33ps)$ and $I_L(t) = 0.719u(t-13.33ps)+0.099u(t-39.99ps)+0.014u(t-66.66ps)$. Figure 3 shows the waveforms of $I_I(t)$ and $I_L(t)$.

Figure 3. Step responses of $I_I$ and $I_L$.

Reference:
[1]. Brian C. Wadell, Transmission Line Design Handbook, Artech House, 1991.
[2]. Stuart M. Wentworth, Applied Electromagnetics : Early Transmission Lines Approach, Wiley, 2007.