CURRENT BYU 19

Analysis of Figure II (Standard Scaling Ladder)

The network in Figure II consists of repeating sections where the resistance values of each subsequent section are scaled by a factor of $k = 1/2$. Let the equivalent resistance of the entire infinite network be $R_{eq}$.

Because the network is infinite and self-similar, if we remove the first section (Series $R$ and Parallel $R$), the remaining network is identical in structure to the original but scaled by a factor of $1/2$. The lower half gives us a harmonic progression which diverges (sum is infinite) Therefore, the resistance of the remaining network is $R_{eq}/2$.

We can write the equivalent circuit equation: $$ R_{eq} = R + \left( R \parallel \frac{R_{eq}}{2} \right) $$

Solving for $R_{eq}$:

$$ R_{eq} = R + \frac{R \cdot \frac{R_{eq}}{2}}{R + \frac{R_{eq}}{2}} $$ $$ R_{eq} = R + \frac{R \cdot R_{eq}}{2R + R_{eq}} $$ $$ R_{eq} (2R + R_{eq}) = R(2R + R_{eq}) + R \cdot R_{eq} $$ $$ 2R \cdot R_{eq} + R_{eq}^2 = 2R^2 + R \cdot R_{eq} + R \cdot R_{eq} $$ $$ R_{eq}^2 = 2R^2 \implies R_{eq} = \sqrt{2}R $$

Analysis of Figure I

Comparing the structure of Figure I to Figure II, we observe that Figure I is effectively the network of Figure II with an additional series resistor of resistance $R/2$ added to the front (or a structural modification that results in a shift).

Based on the standard scaling ladder result derived above ($R_{ladder} = \sqrt{2}R$), if Figure I represents a configuration where the input impedance is the sum of a leading term and the scaling ladder, we calculate:

$$ R_{eq, I} = \frac{R}{2} + R_{eq, II} $$ $$ R_{eq, I} = \frac{R}{2} + \sqrt{2}R $$ $$ R_{eq, I} = \frac{R + 2\sqrt{2}R}{2} = \frac{(1 + 2\sqrt{2})R}{2} $$