Numerical related to Impedance Scaling

Perform Impedance scaling to the following network for C_(new) = 10 μF.

fig: LPF

\[
\begin{aligned}
R_{\text{old}} &= 1 \ \Omega \\
C_{\text{old}} &= 1 \ \text{F}
\end{aligned}
\]

\[
\text{Now, we have:} \quad C_{\text{new}} = 10 \ \mu\text{F}
\]

\[
\text{We know:} \quad C_{\text{new}} = \frac{C_{\text{old}}}{K_m}
\]

\[
10 \ \mu\text{F} = \frac{1 \ \text{F}}{K_m}
\]

\[
K_m = \frac{1 \ \text{F}}{10 \ \mu\text{F}} = \frac{1 \ \text{F}}{1 \times 10^{-5} \ \text{F}} = 10^5
\]

\[
K_m = 10^5
\]

\[
\text{We have:} \quad R_{\text{new}} = K_m \times R_{\text{old}} = 10^5 \times 1 \ \Omega
\]

\[
R_{\text{new}} = 100 \text{K} \ \Omega
\]

Then the new circuit becomes

fig: After Impedance Scaling

The transfer function for 1st Figure is:

\[
T_{\text{old}}(s) = \frac{1}{s + 1}
\]

\[
T_{\text{new}}(s) = \frac{\frac{1}{R_{\text{new}} C_{\text{new}}}}{s + \frac{1}{R_{\text{new}} C_{\text{new}}}} = \frac{1}{s + 1}
\]

Thus, we can say that there is no change in the following transfer function while performing magnitude Scaling.