Let's derive the **transmission-line equations** (Telegrapher’s equations) for a **lossless line** from the equivalent circuit in the limit as \(\Delta z \rightarrow 0\). --- ### **Step 1: Model the Segment** Consider the segment of length \(\Delta z\) in **Figure (a)**: - Series inductance: \(L = l \Delta z\) - Shunt capacitance: \(C = c \Delta z\) Let: - \(v(z, t)\): Voltage at position \(z\) - \(i(z, t)\): Current at position \(z\) - \(v(z+\Delta z, t)\): Voltage at \(z+\Delta z\) - \(i(z+\Delta z, t)\): Current at \(z+\Delta z\) --- ### **Step 2: Apply KVL (Kirchhoff’s Voltage Law)** For the series inductor: \[ v(z, t) = v(z+\Delta z, t) + L \frac{\partial i}{\partial t} \] So, \[ v(z, t) - v(z+\Delta z, t) = l \Delta z \frac{\partial i(z, t)}{\partial t} \] Divide both sides by \(\Delta z\): \[ \frac{v(z, t) - v(z+\Delta z, t)}{\Delta z} = l \frac{\partial i(z, t)}{\partial t} \] Take the limit \(\Delta z \to 0\): \[ -\frac{\partial v}{\partial z} = l \frac{\partial i}{\partial t} \] Or, \[ \boxed{\frac{\partial v}{\partial z} = -l \frac{\partial i}{\partial t}} \] --- ### **Step 3: Apply KCL (Kirchhoff’s Current Law)** For the shunt capacitor: \[ i(z, t) = i(z+\Delta z, t) + C \frac{\partial v(z+\Delta z, t)}{\partial t} \] So, \[ i(z, t) - i(z+\Delta z, t) = c \Delta z \frac{\partial v(z+\Delta z, t)}{\partial t} \] Divide both sides by \(\Delta z\): \[ \frac{i(z, t) - i(z+\Delta z, t)}{\Delta z} = c \frac{\partial v(z+\Delta z, t)}{\partial t} \] Take the limit \(\Delta z \to 0\): \[ -\frac{\partial i}{\partial z} = c \frac{\partial v}{\partial t} \] Or, \[ \boxed{\frac{\partial i}{\partial z} = -c \frac{\partial v}{\partial t}} \] --- ### **Step 4: Final Transmission-Line Equations** \[ \boxed{ \begin{align*} \frac{\partial v}{\partial z} &= -l \frac{\partial i}{\partial t} \\ \frac{\partial i}{\partial z} &= -c \frac{\partial v}{\partial t} \end{align*} } \] These are the **lossless transmission-line (Telegrapher's) equations**. The specific arrangement of the lumped elements does **not** affect the result as \(\Delta z \to 0\), as all three circuits (a), (b), and (c) will yield the same equations in this limit. --- ## **Summary** - **Step 1:** Write KVL and KCL for the infinitesimal segment. - **Step 2:** Divide by \(\Delta z\), take the limit as \(\Delta z \to 0\). - **Step 3:** Obtain the transmission-line equations: \[ \boxed{ \frac{\partial v}{\partial z} = -l \frac{\partial i}{\partial t} \\ \frac{\partial i}{\partial z} = -c \frac{\partial v}{\partial t} } \]