Question 1

Definition check: Band-pass filter passband and stopband  Does a band-pass filter reject all frequencies between its lower and upper critical frequencies and pass all others, or is it the reverse?

Accepted Answer

Introduction / Context: Identifying whether a filter passes or rejects a frequency band is fundamental when designing communications channels, audio crossovers, or sensor front-ends. The statement given swaps the meaning of band-pass and band-stop. Given Data / Assumptions: Lower critical frequency fL, upper critical frequency fH define the 3 dB points in many specifications. Band-pass filters intentionally pass frequencies between fL and fH. Outside this band, attenuation increases (stopbands). Concept / Approach: By definition, band-pass filters have a central passband and two stopbands (below fL and above fH). The magnitude response is highest within the band and rolls off outside it, often with slopes defined by filter order and type (Butterworth, Chebyshev, etc.). Step-by-Step Solution: Identify the passband: fL ≤ f ≤ fH for band-pass.Identify the stopbands: f < fL and f > fH for band-pass.Therefore, the claim that the band-pass “rejects within the band and passes outside” is incorrect; that is band-stop behavior.This holds irrespective of implementation (active, passive, digital IIR/FIR analogs). Verification / Alternative check: Examine a series RLC between source and load: around resonance f0, the impedance is low and the circuit passes the band near f0 (between fL and fH), attenuating frequencies far below or above—classic band-pass behavior. Why Other Options Are Wrong: “True” variants contradict the fundamental definition; Q or filter type does not change definitions. The “geometric mean only” statement confuses the center frequency with the width of the passband. Common Pitfalls: Assuming that because both band-pass and band-stop have two critical frequencies, their functional descriptions are interchangeable. Always tie the label to what happens inside the band. Final Answer: False

Question 2

Determining bandwidth of a resonant filter  Is the bandwidth of a resonant (RLC) filter governed by both the quality factor Q and the resonant frequency f0 (for example, BW = f0 / Q)?

Accepted Answer

Introduction / Context: Bandwidth quantifies the width of the passband (or stopband) around resonance for band-pass or band-stop filters. The quality factor Q reflects how underdamped the resonance is. Their relationship guides component selection for target selectivity. Given Data / Assumptions: Resonant frequency f0 for an RLC circuit. Quality factor Q defined as 2 * pi * (energy stored / energy dissipated per cycle) or, equivalently, f0 / BW for many resonant responses. Small-signal linear behavior. Concept / Approach: For classical second-order band-pass responses, BW = fH − fL and Q = f0 / BW, with f0 commonly near sqrt(fL * fH). This relation holds for both series and parallel RLC configurations when defined with standard half-power (−3 dB) points. Step-by-Step Solution: Identify fL and fH as the half-power frequencies.Compute BW = fH − fL.Relate to Q using Q = f0 / BW, so BW = f0 / Q.Thus, bandwidth depends on both f0 and Q; increasing Q narrows BW for fixed f0. Verification / Alternative check: Component-level: for a series RLC with R, L, C, Q ≈ (1/R) * sqrt(L/C) and f0 = 1 / (2 * pi * sqrt(L * C)). Substituting shows that BW depends on R through Q and on f0 via L and C, reinforcing the stated dependency. Why Other Options Are Wrong: It is not limited to series or active filters; the relationship is generic for second-order resonances. No minimum Q such as 10 is required for the formula to hold; it is definitional. Common Pitfalls: Confusing absolute bandwidth with fractional bandwidth (BW / f0). Also, mixing peak voltage bandwidth with power bandwidth; standard practice uses half-power points for BW. Final Answer: True

Question 3

Magnitude at the critical (cutoff) frequency of first-order filters  Is the output voltage equal to 63.3% of the maximum at the critical frequencies, or is a different percentage correct for the −3 dB point?

Accepted Answer

Introduction / Context: Critical (cutoff) frequencies for first-order low-pass or high-pass filters are commonly defined at the −3 dB point. Remembering the exact amplitude ratio prevents errors when reading Bode plots or specifying bandwidths. Given Data / Assumptions: First-order filter (RC or RL), linear, sinusoidal steady state. Critical frequency fc = 1 / (2 * pi * τ) where τ is the time constant. At fc, the magnitude response is down by 3 dB relative to the passband. Concept / Approach: At −3 dB, the ratio of output to maximum magnitude is 1 / sqrt(2) ≈ 0.707 (70.7%). The number 63.3% is associated with time-domain exponential settling (1 − e^(−1)) at one time constant, not with frequency-domain cutoff. Step-by-Step Solution: Compute |H(jωc)| = 1 / sqrt(1 + (ωc * τ)^2) for a low-pass with ωc * τ = 1.Then |H(jωc)| = 1 / sqrt(2) ≈ 0.707.Convert to decibels: 20 * log10(0.707) ≈ −3 dB.Hence, the correct percentage is about 70.7%, not 63.3%. Verification / Alternative check: Time-domain vs frequency-domain: at t = τ after a step input to a first-order low-pass, the output reaches 63.2% (often rounded 63.3%) of its final value. This figure is unrelated to the −3 dB amplitude in the steady-state sinusoidal frequency response. Why Other Options Are Wrong: “True” variants conflate time-constant settling with −3 dB amplitude. The RC vs RL distinction is irrelevant; both first-order filters share the same normalized magnitude at cutoff. Current vs voltage does not change the 1 / sqrt(2) relationship for the transfer variable. Common Pitfalls: Memorizing 63% without context. Always separate time-constant step responses (63.2% at t = τ) from frequency response cutoffs (70.7% at −3 dB). Final Answer: False

Question 4

RC high-pass topology and phase behavior  For a standard RC high-pass filter (series capacitor, shunt resistor), is the output taken across the resistor and does the output lead the input in phase for much of the spectrum?

Accepted Answer

Introduction / Context: Knowing where to measure the output in first-order filters is the key to predicting both magnitude and phase behavior. The RC high-pass places the capacitor in series and the resistor to ground, with the output across the resistor. Given Data / Assumptions: Series capacitor C, shunt resistor R to ground. Output node is the junction between C and R (across R). Transfer function: H(jω) = jωRC / (1 + jωRC). Concept / Approach: The magnitude of H rises from near 0 at very low frequency to ~1 at high frequency (20 dB/decade slope). The phase is positive (lead) at low frequency, approaching +90°, and tends to 0° at high frequency, meaning the output generally leads the input except in the asymptotic high-frequency limit. Step-by-Step Solution: Write H(jω) = jωRC / (1 + jωRC).Magnitude: |H| = (ωRC) / sqrt(1 + (ωRC)^2).Phase: ∠H = 90° − arctan(ωRC), which is a positive angle (lead) for finite ω.Therefore, output is across R and exhibits phase lead relative to input over most of the band. Verification / Alternative check: At ω = 1 / RC, |H| = 1 / sqrt(2) ≈ 0.707 and phase = +45°, a clear lead. At ω → ∞, phase → 0°, consistent with diminishing lead as the passband flattens. Why Other Options Are Wrong: “False” contradicts the textbook topology and phase result. “True only above cutoff” misses the fact that the phase lead is strongest below cutoff. Component ideality and value ratios do not alter the qualitative behavior. Common Pitfalls: Swapping output node to the capacitor (which would create a low-pass) or assuming phase lag because many filters lag; here, the high-pass leads. Final Answer: True

Question 5

Terminology: Why critical frequencies are often called the −3 dB points  Are the critical (cutoff) frequencies of first-order filters commonly referred to as the −3 dB frequencies?

Accepted Answer

Introduction / Context: “Critical frequency,” “cutoff frequency,” and “−3 dB frequency” are terms frequently used interchangeably for first-order filters and many normalized second-order prototypes. Understanding why helps in reading datasheets and designing networks. Given Data / Assumptions: First-order RC or RL filters in sinusoidal steady state. Cutoff defined where the magnitude falls to 1 / sqrt(2) of the passband value. Decibel conversion for voltage ratios: 20 * log10(Vout / Vin). Concept / Approach: At the cutoff frequency fc, |H(j2 * pi * fc)| = 1 / sqrt(2) ≈ 0.707 for first-order filters. Converting to decibels, 20 * log10(0.707) ≈ −3.01 dB, hence the shorthand “−3 dB frequency.” This convention extends to many higher-order responses when defining half-power points for bandwidth measurements. Step-by-Step Solution: Compute the normalized magnitude at cutoff: |H| = 1 / sqrt(2).Convert to dB: 20 * log10(1 / sqrt(2)) ≈ −3.01 dB.Associate fc with the half-power point because power ∝ V^2, and (0.707)^2 = 0.5 → −3 dB.Therefore, fc is commonly called the −3 dB frequency. Verification / Alternative check: Measure a first-order low-pass Bode plot; the intersection of the 0 dB low-frequency asymptote and the −20 dB/decade high-frequency asymptote occurs at fc and is labeled −3 dB, matching the analytic result. Why Other Options Are Wrong: It is not limited to Butterworth or RC; RL and many normalized second-order filters use the same naming for their half-power points. Load reactance does not alter the definition of the filter’s own cutoff when referenced to the designed transfer function. Common Pitfalls: Confusing −3 dB (voltage) with −3 dB (power) without recognizing that −3 dB in voltage implies half power only when the load impedance is constant across frequency. Nevertheless, the naming convention for cutoff remains −3 dB for first-order sections. Final Answer: True

Question 6

First-order filters: A basic single-pole RC or RL filter exhibits a magnitude slope (roll-off rate) of approximately 20 dB per decade beyond its cutoff frequency. State whether this is true or false.

Accepted Answer

Introduction / Context: Single-pole filters are the simplest building blocks in analog electronics. Understanding their asymptotic roll-off rate is essential for quick Bode plot sketches, noise estimation, and cascade design for higher-order responses. Given Data / Assumptions: Filter is first-order (one reactive element and one resistive element). Type is RC or RL (either low-pass or high-pass topology). Cutoff (break) frequency is defined where magnitude is 1/√2 of passband (−3 dB point). Concept / Approach: A first-order transfer function has the form |H(jω)| = 1 / √(1 + (ω/ωc)^2) for a low-pass (or its dual for high-pass). On a log-magnitude (dB) plot, the asymptotes differ by 20 dB per decade between successive decades of frequency beyond the break point, yielding the well-known slope of 20 dB/dec (≈6 dB/octave). Step-by-Step Solution: Write transfer magnitude in dB: 20log10|H(jω)|.For ω ≫ ωc in a low-pass, |H| ≈ ωc/ω ⇒ 20log10(ωc/ω) = −20*log10(ω/ωc).Increase ω by a factor of 10 (one decade): magnitude changes by −20 dB.For a high-pass, the same 20 dB/dec applies below cutoff (increasing as frequency rises). Verification / Alternative check: Construct Bode asymptotes: horizontal in the passband, then a straight line with slope ±20 dB/dec starting one decade away from the break, intersecting at the corner. SPICE simulations confirm within a small margin near the corner where the exact response deviates by up to 3 dB. Why Other Options Are Wrong: “False” would imply a different slope for first-order filters; slopes like 40 dB/dec or 60 dB/dec correspond to second- or third-order networks (additional poles). Common Pitfalls: Confusing per-octave and per-decade slopes (6 dB/oct vs 20 dB/dec); mixing passband ripple (active filters) with simple RC/RL responses; misidentifying the −3 dB point as the slope rather than the breakpoint. Final Answer: True

Question 7

RL high-pass behavior: In a series RL high-pass filter, the output is taken across the inductor (L), and the output voltage leads the input phase. True or false?

Accepted Answer

Introduction / Context: RL and RC first-order networks can form either low-pass or high-pass filters depending on where the output is taken. Knowing which node provides high-pass behavior and the phase relationship is crucial for signal-conditioning and phase-lead networks. Given Data / Assumptions: Series RL network with input applied across the series combination. High-pass output is required. Linear, time-invariant components; sinusoidal steady-state (phasor) analysis applies. Concept / Approach: In a series RL divider, taking output across L yields a high-pass response because |ZL| = ωL grows with frequency. The transfer function becomes H(jω) = jωL / (R + jωL), which approaches 1 at high frequency and 0 at low frequency. The phase of H is 90° − arctan(ωL/R), a positive angle for any ω > 0, meaning the output leads the input. Step-by-Step Solution: Define ZR = R and ZL = jωL.Compute H(jω) = Vout/Vin = ZL / (ZR + ZL) = jωL / (R + jωL).Magnitude: |H| = ωL / √(R^2 + (ωL)^2), rising with ω (high-pass).Phase: ∠H = 90° − arctan(ωL/R) ∈ (0°, 90°), so Vout leads Vin. Verification / Alternative check: At ω ≪ R/L, |H| ≈ ωL/R → 0 (blocked at low frequency). At ω ≫ R/L, |H| → 1 and phase → 0° to 90° lead limit; scope measurements confirm the output zero-crossing precedes the input for midband HP operation. Why Other Options Are Wrong: “False” would imply either output is not across L for RL high-pass or that phase does not lead, contradicting both the impedance rise of L with frequency and the phasor angle of jωL/(R + jωL). Common Pitfalls: Swapping node choice: output across R is a low-pass. Also, forgetting that phase lead approaches +90° only ideally; with finite R, lead is between 0° and 90°. Final Answer: True

Passive Filters Questions