- #1
cjs94
- 16
- 0
Homework Statement
This is one part of a wider question, I'm only posting the part I'm having trouble with.
$$
\begin{align}
\text{Given an impedance network } B &= \frac{Z_1 \parallel Z_3}{Z_2 + Z_1 \parallel Z_3} \\
\text{show that: } \frac{1}{B} &= 1 + \frac{R_2}{R_1} + j\frac{\omega CR_2}{1 + \omega CR_3}
\end{align}
$$
Homework Equations
$$\begin{align}
Z_1 &= R_1 \\
Z_2 &= R_2 \\
Z_3 &= R_3 + C_1 \text{(in series)}
\end{align}$$
The Attempt at a Solution
$$\begin{align}
Z_3 &= R_3 - j\frac{1}{\omega C_1} \\
Z_1 \parallel Z_3 &= \frac{Z_1 Z_3}{Z_1 + Z_3} \\
&= \frac{R_1 \left( R_3 - j\frac{1}{\omega C_1} \right)}{R_1 + R_3 - j\frac{1}{\omega C_1}} \\
\frac{1}{B} &= \frac{Z_1 \parallel Z_3}{Z_1 \parallel Z_3} + \frac{Z_2}{Z_1 \parallel Z_3} \\
&= 1 + \frac{R_2}{\frac{R_1 \left( R_3 - j\frac{1}{\omega C_1} \right)}{R_1 + R_3 - j\frac{1}{\omega C_1}}} \\
&= 1 + \frac{R_2 \left( R_1 + R_3 - j\frac{1}{\omega C_1} \right)}{R_1 \left( R_3 - j\frac{1}{\omega C_1} \right)}\\
&= 1 + \frac{R_2 \left( R_3 - j\frac{1}{\omega C_1} \right)}{R_1 \left( R_3 - j\frac{1}{\omega C_1} \right)} + \frac{R_2 R_1}{R_1 \left( R_3 - j\frac{1}{\omega C_1} \right)}\\
&= 1 + \frac{R_2}{R_1} + \frac{R_2}{R_3 - j\frac{1}{\omega C_1}} \\
&= 1 + \frac{R_2}{R_1} + \frac{j\omega C_1 R_2}{1+ j\omega C_1 R_3}
\end{align}$$
I just can't get the right hand element to match, I've got an extra ##j##. What am I doing wrong?