Questions: Passive Filter Transfer Function Analysis

Question 1

An RC low-pass filter has cutoff frequency ω_c = 1/RC. At exactly ω = ω_c, what is the magnitude of the transfer function |H(jω)|?

• A. 1 — the signal passes through at full amplitude at the cutoff
• B. 1/√2 ≈ 0.707, corresponding to −3 dB attenuation
• C. 0 — the signal is completely blocked at the cutoff frequency
• D. 1/2 — the signal amplitude is exactly halved

Answer: (B)

At the cutoff frequency, |H(jω_c)| = 1/√(1 + (ω_c·RC)²) = 1/√(1+1) = 1/√2 ≈ 0.707. This is the standard definition of the cutoff: power is halved (−3 dB), not amplitude. The signal is neither fully passed nor fully blocked — that only happens asymptotically at ω = 0 and ω → ∞. Option D is a common confusion; amplitude of 1/2 would correspond to −6 dB.

Question 2

A designer needs a passive filter that rolls off at −40 dB/decade. What determines whether this is achievable and how?

• A. By choosing a small enough RC time constant in a first-order filter
• B. By using a second-order RLC filter — rolloff rate is set by circuit order, not component values
• C. By cascading two resistors without any reactive components
• D. By lowering the supply voltage, which steepens the attenuation slope

Answer: (B)

Rolloff rate is determined by the order of the filter. Each pole in the transfer function contributes −20 dB/decade beyond the cutoff. A first-order RC or RL filter always rolls off at −20 dB/decade regardless of R and C values — changing those shifts the cutoff frequency, not the slope. A −40 dB/decade rolloff requires a second-order filter (e.g., an RLC circuit). Component values set *where* the rolloff begins; circuit order sets *how steeply* it falls.

Question 3

For a first-order RC low-pass filter, the phase shift ∠H(jω) approaches −90° as ω → ∞.

• T. True
• F. False

Answer: True

The phase is ∠H(jω) = −arctan(ωRC). As ω → ∞, arctan(ωRC) → π/2, so the phase approaches −90°. At ω = 0 it is 0°; at the cutoff frequency it is −45°. This progressive phase shift represents increasing time delay at higher frequencies — a critical practical concern in control systems, where accumulated phase shift reduces phase margin and can cause instability.

Question 4

The transfer function H(jω) should be recalculated for each new input frequency, just as phasor analysis requires knowing the frequency in advance.

• T. True
• F. False

Answer: False

This conflates phasor analysis with transfer function analysis. Phasor analysis computes a single voltage ratio at one chosen frequency. The transfer function H(jω) is an analytic function of ω that characterizes the circuit's response across ALL frequencies simultaneously — you evaluate it at any ω of interest without redoing the circuit analysis. This is the conceptual shift: from reasoning about 'what happens at this frequency' to 'what does this circuit do to every frequency at once.'

Question 5

What is the key conceptual difference between using the transfer function H(jω) to analyze a filter versus performing a single-frequency phasor analysis?

Model answer: Phasor analysis yields a single gain and phase value at one specific frequency. The transfer function treats frequency as a variable, producing an analytic expression that describes the circuit's gain and phase shift at every frequency simultaneously — revealing the passband, stopband, rolloff rate, and cutoff in one expression.

The power of the transfer function is that it turns the question 'what happens at this frequency?' into 'what does this circuit do to the entire spectrum?' The magnitude |H(jω)| gives the gain at each frequency; the phase ∠H(jω) gives the time delay. Both are read directly from the same expression derived once from the voltage divider ratio. This is what allows filter design to specify performance across a bandwidth rather than at a single point.