What roll-off is. Beyond its cut-off frequency, a filter's gain falls at a fixed rate on a log-log (Bode) plot. That rate is the roll-off, and it is set entirely by the number of poles — i.e. by the filter order n.

The governing rule. Every pole contributes 20 dB/decade (equivalently 6 dB/octave) of attenuation:

\(\text{Roll-off} = 20n\ \text{dB/decade} = 6n\ \text{dB/octave}\)

Apply it to the two orders.

First order (n = 1): \(20 \times 1 = 20\ \text{dB/decade}\) (6 dB/octave).

Second order (n = 2): \(20 \times 2 = 40\ \text{dB/decade}\) (12 dB/octave).

The change asked for.

\(\Delta = 40 - 20 = 20\ \text{dB/decade}\)

Read the question carefully — this is where marks are usually lost. The question asks for the change in roll-off, not the roll-off of the second-order filter. 40 dB/decade (option 4) is the second-order slope itself, and it is the tempting wrong answer; the increase over the first-order filter is 20 dB/decade.

Why each pole gives 20 dB/decade. A single-pole low-pass transfer function is

\(H(j\omega)=\dfrac{1}{1+j\omega/\omega_0}\)

For ω ≫ ω₀ the magnitude tends to ω₀/ω, so a ten-fold rise in frequency divides the gain by 10, i.e. −20 dB; n cascaded poles multiply that effect, giving 20n dB/decade. Each order also adds 90° of phase lag, which matters for stability in feedback loops.

Hence, the roll-off increases by 20 dB/decade.