Circuit to Transfer Function

Derive the $A=V_{out}/V_{in}$ for the following circuit:

We can see that:

• Stage 1 is a frequency-dependent voltage divider
• Stage 2 is a non-inverting amplifier
• Since the two stages are independent, we have $A=A_1\cdot A_2$.

For stage 1, we have:

$$\begin{array}{r}{A}_1=\frac{V_{o_1}}{V_{in}}\end{array}$$

Since it’s a frequency-dependent voltage divider, we have:

$$\begin{array}{rl}{A}_1& =\frac{V_{o_1}}{V_{in}}\\ & =\frac{Z_L}{R_1+Z_L},Z_L=sL\\ & =\frac{sL}{R_1+sL}=\frac{\frac{sL}{R_1}}{\frac{sL}{R_1}+1}\\ & =\frac{s/{\omega }_c}{s/{\omega }_c+1}\text{(Standard form for 1st-order HPF)}\end{array}$$

Note that $\tau =\frac{L}{R_1}$ and ${\omega }_c=\frac{1}{\tau }=\frac{R_1}{L}$

Qualitatively:

• At low frequency, $L_1$ is essentially a closed switch, so $V_{o_1}=0$
• At high frequency, $L_1$ is essentially an open switch, so $V_{o_1}=V_{in}$
• Therefore, we know that this is a high pass filter qualitatively.

Since we know that stage 2 is just a non-inverting amplifier, we have:

$$A=A_1{A}_2=\left(\frac{s/{\omega }_c}{s/{\omega }_c+1}\right)\left(1+\frac{R_f}{R_i}\right)$$