A general filter circuit is shown in the figure: If R 1 = R 2 =

SOLUTION

For the given network, we analyze the filter characteristics in the following steps:

Step 1:

At low frequency (ω → 0), the capacitance impedance will be:

\({X_c} = \frac{1}{{j\omega C}} = \frac{1}{{j\left( 0 \right)C}} = \infty \)

i.e. the capacitor acts as an open-circuit and the equivalent circuit for low frequency will be:

With V⁺ = V^- = V_A, applying nodal at the inverting terminal, we get:

\(\frac{{{V_A} - {V_{in}}}}{{{R_1}}} + \frac{{{V_A} - {V_0}}}{{{R_2}}} = 0\)

With R₁ = R₂ = R_A, the above expression becomes:

2V_A = V_i + V₀    ---(1)

Similarly, applying nodal at the non-inverting terminal, we get:

\(\frac{{{V_A} - {V_{in}}}}{{{R_3}}} + \frac{{{V_A} - 0}}{{{R_4}}} = 0\)

With R₃ = R₄ = R_B, the above expression becomes:

2V_A = V_in  ---(2)

Using Equation (2), equation (1) can be written as:

2V_A = 2V_A + V₀

V₀ = 0 V

Since the above analysis was done for low frequencies, we conclude that the network stops or does not allow low-frequency signals to pass.

Step 2:

At high frequency (ω → ∞), the capacitive impedance becomes:

\({X_C} = \frac{1}{{j\omega C}} = 0\) 

The equivalent circuit is redrawn as:

The voltage at the non-inverting terminal will be:

\({V^ + } = \frac{{{V_i} \times {R_4}}}{{{R_3} + {R_4}}}\)

With R₃ = R₄ = R_B, we get:

\({V^ + } = \frac{{{V_i}}}{2}\)

\({V^ + } = {V^ - } = {V_0} = \frac{{{V_i}}}{2}\)

From these results, we conclude that the given network is a high pass filter.