Question

For an active low-pass filter of first order, the roll-off is:

• A. \( -10\text{ dB/dec} \)
• B. \( -40\text{ dB/dec} \)
• C. \( -20\text{ dB/dec} \)
• D. \( -60\text{ dB/dec} \)

Hint:

The roll-off rate for an \(n\)-th order filter is given by \(n \times (-20\text{ dB/decade})\). - 1st-order: \(-20\text{ dB/dec}\) - 2nd-order: \(-40\text{ dB/dec}\) - 3rd-order: \(-60\text{ dB/dec}\)

Answer 1

The Correct Option is C

Concept: The stopband attenuation rate of an analog active filter is called its roll-off rate, which describes how quickly the filter attenuates signals past its cut-off frequency. The transfer function magnitude for a standard first-order low-pass filter operating at high frequencies (\(\omega \gg \omega_c\)) simplifies asymptotically to: \[ |H(j\omega)| \approx \frac{\omega_c}{\omega} \] When we map this frequency response onto a logarithmic decibel scale for a Bode plot representation, the gain expression is formulated as: \[ A_{\text{dB}} = 20 \log_{10} \left| H(j\omega) \right| \approx 20 \log_{10}\left(\frac{\omega_c}{\omega}\right) = 20 \log_{10}(\omega_c) - 20 \log_{10}(\omega) \] The order of the filter (\(n\)) scales this slope line linearly by a factor of \(-20\text{ dB/decade}\) per order.

Step 1: Quantifying attenuation across a single decade frequency shift.
A decade change means the angular frequency increases tenfold from an initial value \(\omega_1\) to a new value \(\omega_2 = 10\omega_1\). Let us compute the change in gain (\(\Delta A_{\text{dB}}\)) across this frequency interval: \[ \Delta A_{\text{dB}} = A_{\text{dB}}(\omega_2) - A_{\text{dB}}(\omega_1) \] \[ \Delta A_{\text{dB}} = \left[ 20 \log_{10}(\omega_c) - 20 \log_{10}(10\omega_1) \right] - \left[ 20 \log_{10}(\omega_c) - 20 \log_{10}(\omega_1) \right] \] The constant term involving \(\omega_c\) cancels out directly: \[ \Delta A_{\text{dB}} = -20 \log_{10}(10\omega_1) + 20 \log_{10}(\omega_1) \] Applying logarithmic subtraction properties: \[ \Delta A_{\text{dB}} = -20 \left[ \log_{10}(10\omega_1) - \log_{10}(\omega_1) \right] = -20 \log_{10}\left(\frac{10\omega_1}{\omega_1}\right) \] \[ \Delta A_{\text{dB}} = -20 \log_{10}(10) \] Since \(\log_{10}(10) = 1\): \[ \Delta A_{\text{dB}} = -20 \times 1 = -20\text{ dB} \]

Step 2: Conclusion.
This result shows that for every tenfold increase in signal frequency in the stopband region, the gain drops by exactly \(20\text{ dB}\). Therefore, a first-order filter provides a roll-off rate of \(-20\text{ dB/decade}\) (or \(-6\text{ dB/octave}\)), matching Option (C).