Question

The cut-off frequency of an RC low-pass filter is:

• A. \( \frac{1}{(2\pi RC)} \)
• B. \( 2\pi RC \)
• C. \( \frac{R}{(2\pi C)} \)
• D. \( \frac{C}{(2\pi R)} \)

Hint:

The product \(RC\) represents the time constant \(\tau\) of the filter. The reciprocal of the time constant gives the cut-off frequency in radians per second (\(\omega_c = \frac{1}{\tau}\)). To find it in Hz, always divide by \(2\pi\), yielding \(f_c = \frac{1}{2\pi\tau} = \frac{1}{2\pi RC}\).

Answer 1

The Correct Option is A

Concept: An RC low-pass filter is constructed by placing a resistor \(R\) in series with the signal path and a capacitor \(C\) in parallel across the load terminals. The complex voltage transfer function \(H(j\omega)\) of this system is derived via a standard voltage divider relation: \[ H(j\omega) = \frac{V_{\text{out}}(j\omega)}{V_{\text{in}}(j\omega)} = \frac{\frac{1}{j\omega C}}{R + \frac{1}{j\omega C}} = \frac{1}{1 + j\omega RC} \] The cut-off frequency (also known as the corner frequency or half-power frequency) is defined as the frequency at which the magnitude of the transfer function drops to \(\frac{1}{\sqrt{2}}\) (or \(-3\text{ dB}\)) of its maximum passband value: \[ |H(j\omega_c)| = \frac{1}{\sqrt{1 + (\omega_c RC)^2}} = \frac{1}{\sqrt{2}} \]

Step 1: Solving for the angular cut-off frequency \(\omega_c\).
Equating the terms inside the square roots from our magnitude formulation: \[ 1 + (\omega_c RC)^2 = 2 \quad \implies \quad (\omega_c RC)^2 = 1 \quad \implies \quad \omega_c RC = 1 \] Therefore, the critical angular frequency in radians per second is: \[ \omega_c = \frac{1}{RC} \]

Step 2: Converting angular frequency to cyclic frequency \(f_c\).
We use the mathematical identity relating angular speed to linear frequency, \(\omega_c = 2\pi f_c\): \[ 2\pi f_c = \frac{1}{RC} \quad \implies \quad f_c = \frac{1}{2\pi RC} \] This expression matches Option (A).