A first-order low-pass RC filter has R = 1 kΩ and C = 1 μF. Approximately what is the gain in dB at a frequency ten times the corner frequency?
AApproximately 0 dB — the signal passes without attenuation
BApproximately −3 dB — the corner frequency marks only a slight rolloff
CApproximately −20 dB — one decade above the corner produces −20 dB of attenuation
DApproximately −40 dB — the rolloff doubles past the corner frequency
The corner frequency is ω_c = 1/RC = 1/(1000 × 10⁻⁶) = 1000 rad/s. At ten times the corner frequency (10ω_c), the Bode approximation gives −20 dB/decade × 1 decade = −20 dB of attenuation. The −20 dB/decade rolloff is the signature of a single-pole (first-order) filter: every factor-of-10 increase in frequency above the corner attenuates by another 20 dB. The −40 dB/decade rolloff belongs to a two-pole (second-order) filter.
Question 2 Multiple Choice
Why does a first-order filter produce exactly −20 dB/decade rolloff rather than −10 or −40 dB/decade?
A−20 dB/decade is set by the IEC standard for passive filter design
BThe transfer function magnitude falls as 1/ω for large ω; each decade of frequency multiplies ω by 10, and 20·log₁₀(10) = 20 dB
CEnergy stored in the capacitor decreases exponentially with frequency
D−20 dB/decade is the rolloff of any passive filter regardless of order
For a first-order low-pass filter, |H(jω)| ≈ ω_c/ω for ω >> ω_c. When frequency increases by a decade (factor of 10), magnitude decreases by a factor of 10, which is 20·log₁₀(10) = 20 dB. Each additional pole adds another factor of 1/ω to the magnitude rolloff, adding another 20 dB/decade — so a second-order filter gives −40 dB/decade, third-order −60 dB/decade, and so on.
Question 3 True / False
In a first-order RC circuit, which element you take the output voltage across determines whether you get a low-pass or high-pass response.
TTrue
FFalse
Answer: True
The capacitor's impedance Z_C = 1/(jωC) is large at low frequencies and small at high frequencies. Taking V_out across the capacitor: at low frequencies the capacitor has high impedance and captures most of the voltage (low-pass). Taking V_out across the resistor: at high frequencies the capacitor has low impedance, leaving most of the voltage drop on the resistor (high-pass). Same circuit, different measurement point — different filter type.
Question 4 True / False
At the corner frequency of a first-order low-pass RC filter, the output voltage is exactly half the input voltage.
TTrue
FFalse
Answer: False
At the corner frequency, the gain is 1/√2 ≈ 0.707 (not 0.5), corresponding to −3 dB. The factor of 1/√2 comes from |H(jω_c)| = 1/√(1 + (ω_c/ω_c)²) = 1/√2. A gain of 0.5 would be −6 dB. The phase shift at the corner frequency is −45° for a low-pass filter. Both the −3 dB magnitude and −45° phase are signatures that uniquely identify the corner frequency.
Question 5 Short Answer
Why is the −3 dB corner frequency also called the 'half-power frequency,' and what is its relationship to the time-domain time constant τ?
Think about your answer, then reveal below.
Model answer: Power is proportional to voltage squared. At the corner frequency the voltage gain is 1/√2, so delivered power is (1/√2)² = 1/2 of the passband power — hence 'half-power frequency.' The corner frequency is ω_c = 1/τ, where τ = RC (or L/R for RL filters). A larger time constant means the capacitor charges and discharges more slowly in the time domain, so it cannot follow rapid (high-frequency) changes — equivalently, the corner frequency is lower and the passband is narrower. Smaller τ means faster transient response and higher corner frequency (wider bandwidth). Time constant and bandwidth are reciprocals.
This relationship is not coincidental — it is the same physics in two domains. A capacitor with large τ acts as a better integrator (averages fast fluctuations), which corresponds directly to more aggressive high-frequency attenuation. Understanding both perspectives simultaneously is essential for filter design: the time-domain spec (settling time) and the frequency-domain spec (bandwidth) are linked by τ = 1/ω_c.