The cut off frequency of the following fourth order filter is
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For a simple RC low-pass filter, cutoff frequency \(f_c = 1/(2\pi RC)\).
For a Sallen-Key low-pass filter with equal R and equal C (in the RC network part), its characteristic frequency is also \(f_0 = 1/(2\pi RC)\). The -3dB cutoff frequency depends on the Q-factor, which is set by the op-amp gain.
When cascading filter stages, the overall order adds up. The overall cutoff frequency calculation can be complex depending on the filter type (Butterworth, Chebyshev, etc.). If stages are identical, often the characteristic frequency of one stage is asked.
The circuit shows two cascaded identical second-order Sallen-Key low-pass filter stages. The overall filter is fourth-order.
For a Sallen-Key low-pass filter with equal resistors (R) and equal capacitors (C) in the RC network part (as shown in each stage: R = 1k\(\Omega\), C = 0.16\(\mu\)F for the frequency determining part), the natural frequency or characteristic frequency (\(f_c\) or \(f_0\)) of each second-order stage is given by:
\[ f_c = \frac{1}{2\pi RC} \]
Given:
R = 1 k\(\Omega\) = \(1 \times 10^3 \Omega\)
C = 0.16 \(\mu\)F = \(0.16 \times 10^{-6}\) F
Calculate \(f_c\) for one stage:
\[ f_c = \frac{1}{2\pi (1 \times 10^3 \Omega)(0.16 \times 10^{-6} \text{ F})} \]
\[ f_c = \frac{1}{2\pi (0.16 \times 10^{-3})} = \frac{1}{2\pi \times 0.00016} \]
\[ f_c = \frac{1}{0.0010053} \approx \frac{1}{0.001} \text{ Hz} = 1000 \text{ Hz} = 1 \text{ kHz} \]
Let's calculate more precisely:
\(RC = 10^3 \times 0.16 \times 10^{-6} = 0.16 \times 10^{-3} = 1.6 \times 10^{-4}\)
\(f_c = \frac{1}{2\pi \times 1.6 \times 10^{-4}} = \frac{10^4}{3.2\pi}\)
\(3.2\pi \approx 3.2 \times 3.14159 \approx 10.053\)
\(f_c \approx \frac{10000}{10.053} \approx 994.7 \text{ Hz}\).
This is very close to 1 kHz.
The gain of each non-inverting amplifier stage is \(K = 1 + R_2/R_1\). Here \(R_1 = 10\text{k}\Omega\) and \(R_2 = 12.35\text{k}\Omega\).
\(K = 1 + \frac{12.35 \text{k}\Omega}{10 \text{k}\Omega} = 1 + 1.235 = 2.235\). This gain affects the Q-factor and damping of the filter stage (e.g., for a Butterworth response, K is specific, usually 1.586 for Q=0.707).
However, the "cut off frequency" usually refers to the -3dB frequency, which for many standard filter designs (like Butterworth) is very close to the characteristic frequency \(f_c = 1/(2\pi RC)\) for each stage if the stages are designed to have their -3dB points at this frequency.
If it's a cascade of two identical Butterworth stages each with cutoff \(f_c\), the overall -3dB cutoff frequency of the 4th order filter will be lower than \(f_c\). However, often \(f_c\) is what is quoted as "the cutoff frequency" in simple terms.
Given the options, 1 kHz is the most direct calculation from \(1/(2\pi RC)\).
For a 4th order Butterworth filter made by cascading two 2nd order stages, if the overall -3dB cutoff is \(f_{c,overall}\), then each stage is typically designed with a specific \(f_0\) and Q that are slightly different from \(f_{c,overall}\).
But if the question implies each stage has its characteristic frequency at \(1/(2\pi RC)\) and asks for *this* frequency as "the cut off frequency" (perhaps of the individual sections contributing to the overall response):
The calculation \(f_c \approx 994.7 \text{ Hz}\) is approximately 1 kHz.
The "cut off frequency" of the overall fourth-order filter will be this value (1kHz) if the design is such that each identical stage has its -3dB point at 1kHz and they are directly cascaded (assuming no loading effects or that the stages are buffered, which they are by op-amps). In such a cascade of identical low-pass stages, the overall -3dB cutoff frequency becomes lower.
However, it is common in such questions to ask for the characteristic frequency \(f_c = 1/(2\pi RC)\) of the individual identical stages.
Given the options, 1 kHz is the result of this calculation.
\[ \boxed{1 \text{ kHz}} \]
