Calculate the cutoff frequency using the formula cutoff

Solved Step by Step With Explanation- Calculate High-Pass Filter Cutoff

Questions

b. Calculate the cutoff frequency if R₁ = 2.1 kN,

C10.05mF, RG =10 kQ, and RF = 50 kn.

A non-inverting amplifier is a type of electronic circuit that amplifies an input voltage while maintaining its polarity (i.e., the output voltage is in the same phase as the input voltage). It's called "non-inverting" because the output voltage doesn't undergo a 180-degree phase inversion as it does in an inverting amplifier.

Given:

V+ = Vi * (R1 / (R1 + 1/SC1))

V+ = Vi * (2.1 * 10^3 / (2.1 * 10^3 + 2 * 10^4 * S * C1))

Vo = (1 + 50/10) * Vi * (2.1 * 10^3 / (2.1 * 10^3 + 20 * 10^4 * π * f * C1))

Vo = 6 * Vi * (2.1 * 10^3 / (2.1 * 10^3 + 20 * 10^4 * π * f * C1))

H(s=jω) = (6jω) / (j^2 * 9.524 + 9.524 * jω)

Now, let's solve for the cutoff frequency:

|H(s=jω)| = |(6jω) / (-9.524 * (1 + jω))|

At the cutoff frequency, the magnitude will be -3 dB, which corresponds to a gain of 0.7071:

6ωcutoff / |-9.524 - 9.524jωcutoff| = 0.7071

6ωcutoff / √(9.524^2 + (9.524ωcutoff)^2) = 0.7071