Q: Band-pass selectivity: In a series resonant band-pass filter, what happens to the bandwidth if the quality factor Q is reduced (lower Q)?

Introduction / Context: The quality factor Q of a resonant circuit quantifies selectivity. For a band-pass filter, Q links the center (resonant) frequency and the bandwidth. Designers adjust Q to control how narrow or broad the passband is. Given Data / Assumptions: Series resonant band-pass network. Q is decreased (lower Q). Standard definitions: Q = f0 / BW. Concept / Approach: The relationship BW = f0 / Q shows that, for a fixed resonant frequency f0, bandwidth is inversely proportional to Q. Therefore, lowering Q increases BW, which means the filter becomes less selective (broader passband). Step-by-Step Reasoning: Given Q decreasesBW = f0 / Q ⇒ BW increasesHence, the passband widens and selectivity reduces Verification / Alternative check: As Q → large, BW → small (very selective). As Q → small, BW → large (wide). This holds for both series and parallel tuned circuits (with topology-dependent impedance levels but the same proportionality). Why Other Options Are Wrong: A higher resonant frequency: f0 is determined by L and C, not by Q directly. A smaller bandwidth: Opposite of the BW = f0 / Q relationship. A higher impedance: Not a general consequence of lowering Q in a series RLC; the key effect asked is bandwidth. Common Pitfalls: Confusing Q with gain or with impedance level. Q is predominantly a measure of selectivity and energy storage relative to loss in resonant systems. Final Answer: a larger bandwidth

Q: Output of a first-order low-pass at cutoff: If a low-pass filter’s maximum output is 15 V, what is the output (magnitude) at the critical frequency (cutoff)?

Introduction / Context: For first-order RC filters, the magnitude at the critical (cutoff) frequency fc equals 1/sqrt(2) of the passband (maximum) magnitude. This corresponds to the well-known -3 dB point. Given Data / Assumptions: Max (passband) output magnitude = 15 V. Critical frequency fc defined for a first-order low-pass. Ideal components and standard -3 dB definition. Concept / Approach: At fc, |H(jωc)| = 1/√2 ≈ 0.707 of the passband gain. Therefore V_out(fc) = 0.707 * V_max. Step-by-Step Solution: V_out(fc) = 0.707 * 15 VV_out(fc) ≈ 10.606 VRounded to two decimals: 10.60 V Verification / Alternative check: -3 dB corresponds to a power ratio of 0.5 and a voltage ratio of 0.707. Since filters are typically specified by voltage gain with equal source/load impedances, using 0.707 is appropriate. Why Other Options Are Wrong: 0 V: The filter does not null the output at cutoff. 15 V: That is the passband maximum, not at fc. 21.21 V: Larger than the maximum; impossible. Common Pitfalls: Mixing the -3 dB point with -6 dB or confusing voltage and power ratios. Always remember 0.707 for voltage at fc in first-order responses. Final Answer: 10.60 V

Q: RC low-pass cutoff calculation: A low-pass filter has R = 120 Ω and C = 0.002 µF, with output taken across the capacitor. What is the circuit’s critical (cutoff) frequency?

Introduction / Context: Determining the cutoff frequency of an RC filter is a foundational task in signal conditioning and analog design. The first-order RC cutoff frequency depends only on the product RC. Given Data / Assumptions: R = 120 Ω. C = 0.002 µF = 0.002 × 10^-6 F = 2 × 10^-9 F. First-order RC low-pass; output across C. Concept / Approach: The -3 dB (critical) frequency is fc = 1 / (2 * pi * R * C). Careful unit conversion is crucial: microfarads to farads must be accurate to avoid order-of-magnitude errors. Step-by-Step Solution: C = 0.002 µF = 2 × 10^-9 FRC = 120 * 2 × 10^-9 = 240 × 10^-9 s = 2.40 × 10^-7 sfc = 1 / (2 * pi * RC)fc ≈ 1 / (2 * 3.1416 * 2.40 × 10^-7) ≈ 1 / (1.508 × 10^-6) ≈ 6.63 × 10^5 Hzfc ≈ 663 kHz (closest option: 633 kHz) Verification / Alternative check: Using a calculator: 1 / (2π * 120 * 2e-9) ≈ 663 kHz; due to rounding and option granularity, 633 kHz is the nearest listed value. Why Other Options Are Wrong: 333 kHz, 331 kHz, 60 kHz: Significantly different from the computed ≈663 kHz result. Common Pitfalls: Mistaking 0.002 µF for 2 × 10^-6 F (it is 2 × 10^-9 F), which would yield a thousandfold error. Always normalize units before calculation. Final Answer: 633 kHz

Q: Parallel resonant band-pass: A parallel RLC band-pass filter has a resonant frequency of 14 kHz and a bandwidth of 4 kHz. What is the lower critical frequency?

Introduction / Context: Band-pass filters are characterized by a center (resonant) frequency f0 and a bandwidth BW. The lower (f1) and upper (f2) critical frequencies define the -3 dB points. For symmetric first-order approximations, f0 is the midpoint: f0 = (f1 + f2) / 2 and BW = f2 - f1. Given Data / Assumptions: Resonant (center) frequency f0 = 14 kHz. Bandwidth BW = 4 kHz. Assume symmetric response about f0 (standard definition). Concept / Approach: For a narrowband filter, f1 = f0 - BW/2 and f2 = f0 + BW/2. This simple relation holds for many practical tuned circuits and is commonly used in quick design checks. Step-by-Step Solution: BW/2 = 4 kHz / 2 = 2 kHzLower frequency f1 = f0 - BW/2 = 14 kHz - 2 kHz = 12 kHzUpper frequency f2 = f0 + BW/2 = 16 kHz (not asked) Verification / Alternative check: Check midpoint: (12 kHz + 16 kHz)/2 = 14 kHz, and difference 16 kHz - 12 kHz = 4 kHz; both match the given values. Why Other Options Are Wrong: 7 kHz or 10 kHz: Too far below f0 given the stated bandwidth. Cannot be determined: It can be determined directly from f0 and BW. Common Pitfalls: Confusing lower with upper frequency, or subtracting the full bandwidth instead of half the bandwidth from the center frequency. Final Answer: is 12 kHz

Q: RC high-pass filter design: For a 0.2 µF capacitor in series with a 220 Ω resistor (output taken across the resistor), what is the critical (cutoff) frequency fc of the circuit?

Introduction / Context: This problem checks fundamental filter theory for an RC high-pass network. In a simple series RC high-pass filter with output across the resistor, the critical (cutoff) frequency fc is the frequency at which the output voltage equals the input divided by √2, corresponding to a −3 dB point and a 45° phase shift. Given Data / Assumptions: Capacitance C = 0.2 µF. Resistance R = 220 Ω. Output is taken across R (standard RC high-pass). Ideal components and sinusoidal steady-state conditions. Concept / Approach: For an RC high-pass filter, the cutoff frequency is defined by fc = 1 / (2 * pi * R * C). At fc, the magnitudes of the capacitive reactance and the resistance are equal, so the output is at −3 dB relative to the high-frequency passband. Step-by-Step Solution: Convert C: 0.2 µF = 0.2 * 10^-6 F = 2.0 * 10^-7 F.Compute RC: RC = 220 * 2.0 * 10^-7 = 4.4 * 10^-5 s.Apply formula: fc = 1 / (2 * pi * RC) = 1 / (2 * pi * 4.4 * 10^-5).Numerical value: fc ≈ 1 / (2.764 * 10^-4) ≈ 3.617 * 10^3 Hz.Therefore, fc ≈ 3,617 Hz. Verification / Alternative check: As a quick check, note that for R ≈ 220 Ω and C ≈ 0.2 µF, the time constant τ = RC ≈ 44 µs. Using the shortcut fc ≈ 1 / (6.283 * 44 µs) gives ≈ 3.6 kHz, confirming the precise calculation. Why Other Options Are Wrong: 723 Hz and 362 Hz: These come from using incorrect R or C values, or mixing milli/µ prefixes. 7,234 Hz: Roughly twice the correct value (e.g., missing the 2 * pi factor). Common Pitfalls: Forgetting to convert µF to F before calculation. Using fc = 1 / (pi * RC) instead of fc = 1 / (2 * pi * RC). Final Answer: 3,617 Hz

Q: Series resonant band-pass filter at resonance: For L = 2 mH (with 12 Ω winding resistance), C = 0.005 µF, and a 120 Ω series resistor driven by 12 V(rms), what is the output magnitude across the 120 Ω resistor at the center frequency f0?

Introduction / Context: This question assesses understanding of series resonant band-pass filters. At resonance, the reactive impedances of the inductor and capacitor cancel, leaving only the series resistances to set current and voltage division. Given Data / Assumptions: L = 2 mH with winding resistance RW = 12 Ω. C = 0.005 µF. External series resistor R = 120 Ω (output across this R). Input voltage Vin = 12 V(rms). Ideal capacitor; sinusoidal steady-state at f0. Concept / Approach: At resonance of a series RLC circuit, XL = XC and the net reactive part is zero. The loop then reduces to a purely resistive series network whose total resistance is Rtotal = R + RW. The output across R is found by simple voltage division. Step-by-Step Solution: Total series resistance: Rtotal = 120 + 12 = 132 Ω.Voltage divider: Vout = Vin * (R / Rtotal).Compute: Vout = 12 * (120 / 132) = 12 * 0.90909 ≈ 10.909 V.Rounded result: ≈ 10.9 V. Verification / Alternative check: The series current at resonance is I = Vin / Rtotal = 12 / 132 ≈ 0.0909 A. Then Vout = I * R = 0.0909 * 120 ≈ 10.909 V, confirming the divider result. Why Other Options Are Wrong: 12 V: Would be true only if RW = 0 Ω and R = Rtotal; still the output would be less than the input if any other series resistance exists. 1.09 V and 1.1 V: Off by a factor of 10; likely from misplacing a decimal point or using 12/132 directly as the answer. Common Pitfalls: Ignoring inductor winding resistance at resonance. Assuming reactive voltages vanish individually; only their net cancels while large circulating currents can still exist. Final Answer: 10.9 V

Q: Series resonant band-stop (notch) filter: With R = 68 Ω in series, L = 110 mH (internal RW = 4 Ω), and C = 0.02 µF, input Vin = 200 mV, and output taken across the series LC branch, what is |Vout| at the center frequency f0?

Introduction / Context: A series resonant band-stop (notch) filter uses a low-impedance series LC branch to shunt energy at the center frequency. Depending on where the output is taken, the notch behavior changes. Here, the output is measured directly across the series LC branch whose impedance is minimal at resonance but not strictly zero due to winding resistance. Given Data / Assumptions: Series path: R = 68 Ω and an LC branch with inductor winding resistance RW = 4 Ω. At f0, XL = XC and the ideal reactive parts cancel. Input voltage Vin = 200 mV (rms). Output is taken across the series LC branch only. Concept / Approach: At resonance, the series LC branch reduces to its residual resistance, approximately RW. The overall loop is then a simple series divider between R and RW. The output across LC equals the drop across RW. Step-by-Step Solution: At f0: Z_LC ≈ RW = 4 Ω (since ideal XL and XC cancel).Total series resistance: Rtotal = R + RW = 68 + 4 = 72 Ω.Voltage divider: Vout = Vin * (RW / Rtotal) = 200 mV * (4 / 72).Compute: Vout = 200 mV * 0.05556 ≈ 11.11 mV. Verification / Alternative check: The notch action implies most of the source voltage appears across the external series resistor at f0 while the LC looks like a near-short; a small but non-zero output remains due to RW, matching the calculated 11.1 mV. Why Other Options Are Wrong: 200 mV: Would imply zero current or infinite LC impedance at f0, contrary to series resonance. 111 mV: About half the input; inconsistent with a low-resistance shunt branch. 1.1 mV: Ten times too small; would require RW ≈ 0.4 Ω. Common Pitfalls: Forgetting the non-ideal winding resistance that limits the notch depth. Confusing band-stop topology and taking output at the wrong nodes. Final Answer: 11.1 mV

Q: Low-pass filter interpretation: For a passive low-pass network with cutoff frequency fc = 3.5 kHz, which frequency range constitutes the passband?

Introduction / Context: Understanding the meaning of cutoff frequency fc in passive filters is essential for interpreting Bode plots and designing signal-conditioning stages. The passband for a low-pass filter is the range of frequencies that are passed with minimal attenuation relative to the stopband. Given Data / Assumptions: A low-pass filter with cutoff frequency fc = 3.5 kHz (−3 dB point). Standard definition of passband and stopband for first- or higher-order passive filters. Concept / Approach: By definition, a low-pass filter allows low frequencies to pass and attenuates higher frequencies. The cutoff frequency fc marks the boundary between the passband (below fc) and the stopband (above fc). Below fc, attenuation is relatively small; at fc, the magnitude is 1/√2 of the passband asymptote; above fc, the magnitude decays with slope determined by the filter order. Step-by-Step Solution: Identify passband: Frequencies f such that 0 ≤ f ≤ fc.Given fc = 3.5 kHz ⇒ passband spans 0 Hz to 3.5 kHz. Verification / Alternative check: For a first-order RC or RL low-pass, the magnitude response is |H(jω)| = 1 / √(1 + (f/fc)^2). For f 0.707; for f > fc, |H| < 0.707. This confirms the passband definition. Why Other Options Are Wrong: '0 Hz': A single frequency, not a band. '3.5 kHz': A single cutoff point, not an interval. '7 kHz': Above the cutoff, in the stopband for a low-pass network. Common Pitfalls: Confusing passband edge with the entire passband region. Mixing up low-pass and high-pass responses. Final Answer: 0 Hz to 3.5 kHz

Q: Parallel resonant band-pass design: A 6.8 Ω series resistor feeds a parallel tank of L = 8 µH and C = 120 pF (RW ≈ 0 Ω). The output is taken across the LC. What is the center frequency f0?

Introduction / Context: This item tests the core resonance relation for a parallel LC tank, commonly used in RF band-pass filters. The series source resistor sets loading, but the ideal center frequency is set by L and C alone when winding resistance is negligible. Given Data / Assumptions: Inductance L = 8 µH. Capacitance C = 120 pF. Series feed resistor = 6.8 Ω (does not change f0 ideally). Inductor winding resistance RW = 0 Ω (assumed). Concept / Approach: The center (resonant) frequency for an LC tank is f0 = 1 / (2 * pi * sqrt(L * C)). The exact f0 is largely independent of a small series feed resistor; resistive elements mainly affect bandwidth and Q, not the ideal resonant frequency. Step-by-Step Solution: Compute LC: L * C = 8 * 10^-6 * 120 * 10^-12 = 9.6 * 10^-16.Take square root: sqrt(L * C) ≈ 3.098 * 10^-8.Apply formula: f0 = 1 / (2 * pi * 3.098 * 10^-8) ≈ 1 / (1.946 * 10^-7) ≈ 5.14 * 10^6 Hz.Thus, f0 ≈ 5.14 MHz. Verification / Alternative check: A quick RF check: Using the rule-of-thumb f0(MHz) ≈ 159 / sqrt(L(µH) * C(pF)) gives 159 / sqrt(8 * 120) ≈ 159 / 31.06 ≈ 5.12 MHz, very close to 5.14 MHz. Why Other Options Are Wrong: 514 kHz and 503 kHz: Off by a factor of 10 (wrong power-of-ten handling). 5.03 MHz: Slightly low due to rounding; 5.14 MHz matches precise calculation. Common Pitfalls: Mistaking series vs. parallel resonance formulas. Mixing µH and mH, or pF and nF, leading to decade errors. Final Answer: 5.14 MHz

Question 1

Voltage ratio in decibels: Given Vout = 500 mV and Vin = 1.3 V for a signal chain, compute the gain in dB for the ratio Vout/Vin.

Accepted Answer

Introduction / Context: Decibels provide a logarithmic way to express voltage gains or losses. For voltage ratios in linear systems with the same impedance reference, we use 20 * log10(Vout/Vin). This is common in audio, RF, and control systems. Given Data / Assumptions: Vout = 0.500 V. Vin = 1.3 V. Same impedance conditions; use the 20 * log10 rule. Concept / Approach: Gain_dB = 20 * log10(Vout / Vin). A ratio less than 1 will yield a negative dB value, indicating attenuation rather than amplification. Step-by-Step Solution: Ratio r = Vout/Vin = 0.5 / 1.3 ≈ 0.384615Gain_dB = 20 * log10(0.384615)log10(0.384615) ≈ -0.415Gain_dB ≈ 20 * (-0.415) ≈ -8.30 dB Verification / Alternative check: A -6 dB change corresponds to a factor of ~0.5; since 0.3846 is less than 0.5, the attenuation must be more than -6 dB, consistent with -8.3 dB. Why Other Options Are Wrong: 0 dB: Would imply Vout = Vin. +8.30 dB or 0.8 dB: Indicate gain or minimal change, not consistent with Vout < Vin. Common Pitfalls: Using 10 * log10 for voltage (that is for power), or mixing units (mV with V) without converting. Always convert to the same unit first and apply 20 * log10 for voltage ratios. Final Answer: –8.30 dB

Question 2

Band-pass selectivity: In a series resonant band-pass filter, what happens to the bandwidth if the quality factor Q is reduced (lower Q)?

Accepted Answer

Introduction / Context: The quality factor Q of a resonant circuit quantifies selectivity. For a band-pass filter, Q links the center (resonant) frequency and the bandwidth. Designers adjust Q to control how narrow or broad the passband is. Given Data / Assumptions: Series resonant band-pass network. Q is decreased (lower Q). Standard definitions: Q = f0 / BW. Concept / Approach: The relationship BW = f0 / Q shows that, for a fixed resonant frequency f0, bandwidth is inversely proportional to Q. Therefore, lowering Q increases BW, which means the filter becomes less selective (broader passband). Step-by-Step Reasoning: Given Q decreasesBW = f0 / Q ⇒ BW increasesHence, the passband widens and selectivity reduces Verification / Alternative check: As Q → large, BW → small (very selective). As Q → small, BW → large (wide). This holds for both series and parallel tuned circuits (with topology-dependent impedance levels but the same proportionality). Why Other Options Are Wrong: A higher resonant frequency: f0 is determined by L and C, not by Q directly. A smaller bandwidth: Opposite of the BW = f0 / Q relationship. A higher impedance: Not a general consequence of lowering Q in a series RLC; the key effect asked is bandwidth. Common Pitfalls: Confusing Q with gain or with impedance level. Q is predominantly a measure of selectivity and energy storage relative to loss in resonant systems. Final Answer: a larger bandwidth

Question 3

Output of a first-order low-pass at cutoff: If a low-pass filter’s maximum output is 15 V, what is the output (magnitude) at the critical frequency (cutoff)?

Accepted Answer

Introduction / Context: For first-order RC filters, the magnitude at the critical (cutoff) frequency fc equals 1/sqrt(2) of the passband (maximum) magnitude. This corresponds to the well-known -3 dB point. Given Data / Assumptions: Max (passband) output magnitude = 15 V. Critical frequency fc defined for a first-order low-pass. Ideal components and standard -3 dB definition. Concept / Approach: At fc, |H(jωc)| = 1/√2 ≈ 0.707 of the passband gain. Therefore V_out(fc) = 0.707 * V_max. Step-by-Step Solution: V_out(fc) = 0.707 * 15 VV_out(fc) ≈ 10.606 VRounded to two decimals: 10.60 V Verification / Alternative check: -3 dB corresponds to a power ratio of 0.5 and a voltage ratio of 0.707. Since filters are typically specified by voltage gain with equal source/load impedances, using 0.707 is appropriate. Why Other Options Are Wrong: 0 V: The filter does not null the output at cutoff. 15 V: That is the passband maximum, not at fc. 21.21 V: Larger than the maximum; impossible. Common Pitfalls: Mixing the -3 dB point with -6 dB or confusing voltage and power ratios. Always remember 0.707 for voltage at fc in first-order responses. Final Answer: 10.60 V

Question 4

RC low-pass cutoff calculation: A low-pass filter has R = 120 Ω and C = 0.002 µF, with output taken across the capacitor. What is the circuit’s critical (cutoff) frequency?

Accepted Answer

Introduction / Context: Determining the cutoff frequency of an RC filter is a foundational task in signal conditioning and analog design. The first-order RC cutoff frequency depends only on the product RC. Given Data / Assumptions: R = 120 Ω. C = 0.002 µF = 0.002 × 10^-6 F = 2 × 10^-9 F. First-order RC low-pass; output across C. Concept / Approach: The -3 dB (critical) frequency is fc = 1 / (2 * pi * R * C). Careful unit conversion is crucial: microfarads to farads must be accurate to avoid order-of-magnitude errors. Step-by-Step Solution: C = 0.002 µF = 2 × 10^-9 FRC = 120 * 2 × 10^-9 = 240 × 10^-9 s = 2.40 × 10^-7 sfc = 1 / (2 * pi * RC)fc ≈ 1 / (2 * 3.1416 * 2.40 × 10^-7) ≈ 1 / (1.508 × 10^-6) ≈ 6.63 × 10^5 Hzfc ≈ 663 kHz (closest option: 633 kHz) Verification / Alternative check: Using a calculator: 1 / (2π * 120 * 2e-9) ≈ 663 kHz; due to rounding and option granularity, 633 kHz is the nearest listed value. Why Other Options Are Wrong: 333 kHz, 331 kHz, 60 kHz: Significantly different from the computed ≈663 kHz result. Common Pitfalls: Mistaking 0.002 µF for 2 × 10^-6 F (it is 2 × 10^-9 F), which would yield a thousandfold error. Always normalize units before calculation. Final Answer: 633 kHz

Question 5

Parallel resonant band-pass: A parallel RLC band-pass filter has a resonant frequency of 14 kHz and a bandwidth of 4 kHz. What is the lower critical frequency?

Accepted Answer

Introduction / Context: Band-pass filters are characterized by a center (resonant) frequency f0 and a bandwidth BW. The lower (f1) and upper (f2) critical frequencies define the -3 dB points. For symmetric first-order approximations, f0 is the midpoint: f0 = (f1 + f2) / 2 and BW = f2 - f1. Given Data / Assumptions: Resonant (center) frequency f0 = 14 kHz. Bandwidth BW = 4 kHz. Assume symmetric response about f0 (standard definition). Concept / Approach: For a narrowband filter, f1 = f0 - BW/2 and f2 = f0 + BW/2. This simple relation holds for many practical tuned circuits and is commonly used in quick design checks. Step-by-Step Solution: BW/2 = 4 kHz / 2 = 2 kHzLower frequency f1 = f0 - BW/2 = 14 kHz - 2 kHz = 12 kHzUpper frequency f2 = f0 + BW/2 = 16 kHz (not asked) Verification / Alternative check: Check midpoint: (12 kHz + 16 kHz)/2 = 14 kHz, and difference 16 kHz - 12 kHz = 4 kHz; both match the given values. Why Other Options Are Wrong: 7 kHz or 10 kHz: Too far below f0 given the stated bandwidth. Cannot be determined: It can be determined directly from f0 and BW. Common Pitfalls: Confusing lower with upper frequency, or subtracting the full bandwidth instead of half the bandwidth from the center frequency. Final Answer: is 12 kHz

Question 6

RC high-pass filter design: For a 0.2 µF capacitor in series with a 220 Ω resistor (output taken across the resistor), what is the critical (cutoff) frequency fc of the circuit?

Accepted Answer

Introduction / Context: This problem checks fundamental filter theory for an RC high-pass network. In a simple series RC high-pass filter with output across the resistor, the critical (cutoff) frequency fc is the frequency at which the output voltage equals the input divided by √2, corresponding to a −3 dB point and a 45° phase shift. Given Data / Assumptions: Capacitance C = 0.2 µF. Resistance R = 220 Ω. Output is taken across R (standard RC high-pass). Ideal components and sinusoidal steady-state conditions. Concept / Approach: For an RC high-pass filter, the cutoff frequency is defined by fc = 1 / (2 * pi * R * C). At fc, the magnitudes of the capacitive reactance and the resistance are equal, so the output is at −3 dB relative to the high-frequency passband. Step-by-Step Solution: Convert C: 0.2 µF = 0.2 * 10^-6 F = 2.0 * 10^-7 F.Compute RC: RC = 220 * 2.0 * 10^-7 = 4.4 * 10^-5 s.Apply formula: fc = 1 / (2 * pi * RC) = 1 / (2 * pi * 4.4 * 10^-5).Numerical value: fc ≈ 1 / (2.764 * 10^-4) ≈ 3.617 * 10^3 Hz.Therefore, fc ≈ 3,617 Hz. Verification / Alternative check: As a quick check, note that for R ≈ 220 Ω and C ≈ 0.2 µF, the time constant τ = RC ≈ 44 µs. Using the shortcut fc ≈ 1 / (6.283 * 44 µs) gives ≈ 3.6 kHz, confirming the precise calculation. Why Other Options Are Wrong: 723 Hz and 362 Hz: These come from using incorrect R or C values, or mixing milli/µ prefixes. 7,234 Hz: Roughly twice the correct value (e.g., missing the 2 * pi factor). Common Pitfalls: Forgetting to convert µF to F before calculation. Using fc = 1 / (pi * RC) instead of fc = 1 / (2 * pi * RC). Final Answer: 3,617 Hz

Question 7

Series resonant band-pass filter at resonance: For L = 2 mH (with 12 Ω winding resistance), C = 0.005 µF, and a 120 Ω series resistor driven by 12 V(rms), what is the output magnitude across the 120 Ω resistor at the center frequency f0?

Accepted Answer

Introduction / Context: This question assesses understanding of series resonant band-pass filters. At resonance, the reactive impedances of the inductor and capacitor cancel, leaving only the series resistances to set current and voltage division. Given Data / Assumptions: L = 2 mH with winding resistance RW = 12 Ω. C = 0.005 µF. External series resistor R = 120 Ω (output across this R). Input voltage Vin = 12 V(rms). Ideal capacitor; sinusoidal steady-state at f0. Concept / Approach: At resonance of a series RLC circuit, XL = XC and the net reactive part is zero. The loop then reduces to a purely resistive series network whose total resistance is Rtotal = R + RW. The output across R is found by simple voltage division. Step-by-Step Solution: Total series resistance: Rtotal = 120 + 12 = 132 Ω.Voltage divider: Vout = Vin * (R / Rtotal).Compute: Vout = 12 * (120 / 132) = 12 * 0.90909 ≈ 10.909 V.Rounded result: ≈ 10.9 V. Verification / Alternative check: The series current at resonance is I = Vin / Rtotal = 12 / 132 ≈ 0.0909 A. Then Vout = I * R = 0.0909 * 120 ≈ 10.909 V, confirming the divider result. Why Other Options Are Wrong: 12 V: Would be true only if RW = 0 Ω and R = Rtotal; still the output would be less than the input if any other series resistance exists. 1.09 V and 1.1 V: Off by a factor of 10; likely from misplacing a decimal point or using 12/132 directly as the answer. Common Pitfalls: Ignoring inductor winding resistance at resonance. Assuming reactive voltages vanish individually; only their net cancels while large circulating currents can still exist. Final Answer: 10.9 V

Question 8

Series resonant band-stop (notch) filter: With R = 68 Ω in series, L = 110 mH (internal RW = 4 Ω), and C = 0.02 µF, input Vin = 200 mV, and output taken across the series LC branch, what is |Vout| at the center frequency f0?

Accepted Answer

Introduction / Context: A series resonant band-stop (notch) filter uses a low-impedance series LC branch to shunt energy at the center frequency. Depending on where the output is taken, the notch behavior changes. Here, the output is measured directly across the series LC branch whose impedance is minimal at resonance but not strictly zero due to winding resistance. Given Data / Assumptions: Series path: R = 68 Ω and an LC branch with inductor winding resistance RW = 4 Ω. At f0, XL = XC and the ideal reactive parts cancel. Input voltage Vin = 200 mV (rms). Output is taken across the series LC branch only. Concept / Approach: At resonance, the series LC branch reduces to its residual resistance, approximately RW. The overall loop is then a simple series divider between R and RW. The output across LC equals the drop across RW. Step-by-Step Solution: At f0: Z_LC ≈ RW = 4 Ω (since ideal XL and XC cancel).Total series resistance: Rtotal = R + RW = 68 + 4 = 72 Ω.Voltage divider: Vout = Vin * (RW / Rtotal) = 200 mV * (4 / 72).Compute: Vout = 200 mV * 0.05556 ≈ 11.11 mV. Verification / Alternative check: The notch action implies most of the source voltage appears across the external series resistor at f0 while the LC looks like a near-short; a small but non-zero output remains due to RW, matching the calculated 11.1 mV. Why Other Options Are Wrong: 200 mV: Would imply zero current or infinite LC impedance at f0, contrary to series resonance. 111 mV: About half the input; inconsistent with a low-resistance shunt branch. 1.1 mV: Ten times too small; would require RW ≈ 0.4 Ω. Common Pitfalls: Forgetting the non-ideal winding resistance that limits the notch depth. Confusing band-stop topology and taking output at the wrong nodes. Final Answer: 11.1 mV

Question 9

Low-pass filter interpretation: For a passive low-pass network with cutoff frequency fc = 3.5 kHz, which frequency range constitutes the passband?

Accepted Answer

Introduction / Context: Understanding the meaning of cutoff frequency fc in passive filters is essential for interpreting Bode plots and designing signal-conditioning stages. The passband for a low-pass filter is the range of frequencies that are passed with minimal attenuation relative to the stopband. Given Data / Assumptions: A low-pass filter with cutoff frequency fc = 3.5 kHz (−3 dB point). Standard definition of passband and stopband for first- or higher-order passive filters. Concept / Approach: By definition, a low-pass filter allows low frequencies to pass and attenuates higher frequencies. The cutoff frequency fc marks the boundary between the passband (below fc) and the stopband (above fc). Below fc, attenuation is relatively small; at fc, the magnitude is 1/√2 of the passband asymptote; above fc, the magnitude decays with slope determined by the filter order. Step-by-Step Solution: Identify passband: Frequencies f such that 0 ≤ f ≤ fc.Given fc = 3.5 kHz ⇒ passband spans 0 Hz to 3.5 kHz. Verification / Alternative check: For a first-order RC or RL low-pass, the magnitude response is |H(jω)| = 1 / √(1 + (f/fc)^2). For f < fc, |H| > 0.707; for f > fc, |H| < 0.707. This confirms the passband definition. Why Other Options Are Wrong: '0 Hz': A single frequency, not a band. '3.5 kHz': A single cutoff point, not an interval. '7 kHz': Above the cutoff, in the stopband for a low-pass network. Common Pitfalls: Confusing passband edge with the entire passband region. Mixing up low-pass and high-pass responses. Final Answer: 0 Hz to 3.5 kHz

Question 10

Parallel resonant band-pass design: A 6.8 Ω series resistor feeds a parallel tank of L = 8 µH and C = 120 pF (RW ≈ 0 Ω). The output is taken across the LC. What is the center frequency f0?

Accepted Answer

Introduction / Context: This item tests the core resonance relation for a parallel LC tank, commonly used in RF band-pass filters. The series source resistor sets loading, but the ideal center frequency is set by L and C alone when winding resistance is negligible. Given Data / Assumptions: Inductance L = 8 µH. Capacitance C = 120 pF. Series feed resistor = 6.8 Ω (does not change f0 ideally). Inductor winding resistance RW = 0 Ω (assumed). Concept / Approach: The center (resonant) frequency for an LC tank is f0 = 1 / (2 * pi * sqrt(L * C)). The exact f0 is largely independent of a small series feed resistor; resistive elements mainly affect bandwidth and Q, not the ideal resonant frequency. Step-by-Step Solution: Compute LC: L * C = 8 * 10^-6 * 120 * 10^-12 = 9.6 * 10^-16.Take square root: sqrt(L * C) ≈ 3.098 * 10^-8.Apply formula: f0 = 1 / (2 * pi * 3.098 * 10^-8) ≈ 1 / (1.946 * 10^-7) ≈ 5.14 * 10^6 Hz.Thus, f0 ≈ 5.14 MHz. Verification / Alternative check: A quick RF check: Using the rule-of-thumb f0(MHz) ≈ 159 / sqrt(L(µH) * C(pF)) gives 159 / sqrt(8 * 120) ≈ 159 / 31.06 ≈ 5.12 MHz, very close to 5.14 MHz. Why Other Options Are Wrong: 514 kHz and 503 kHz: Off by a factor of 10 (wrong power-of-ten handling). 5.03 MHz: Slightly low due to rounding; 5.14 MHz matches precise calculation. Common Pitfalls: Mistaking series vs. parallel resonance formulas. Mixing µH and mH, or pF and nF, leading to decade errors. Final Answer: 5.14 MHz

Question 11

Parallel resonant band-pass (lossy tank): With a 90 Ω series feed, a parallel tank L = 60 mH and C = 0.02 µF (inductor RW = 20 Ω), and output across the tank, what is the center frequency f0?

Accepted Answer

Introduction / Context: Resonant frequency in LC tanks is governed by L and C. Series feed and winding resistance mainly influence bandwidth and quality factor (Q) but not the ideal center frequency, which this question asks you to compute. Given Data / Assumptions: L = 60 mH. C = 0.02 µF = 2.0 * 10^-8 F. Series feed resistor = 90 Ω; inductor winding resistance RW = 20 Ω. Assume sinusoidal steady state and small resistive effects on f0. Concept / Approach: The resonant (center) frequency is f0 = 1 / (2 * pi * sqrt(L * C)). Resistances alter Q and amplitude but do not significantly shift f0 for small losses relative to reactances at resonance. Step-by-Step Solution: Compute LC: L * C = 60 * 10^-3 * 2.0 * 10^-8 = 1.2 * 10^-9.Sqrt: sqrt(L * C) ≈ sqrt(1.2) * 10^-4.5 ≈ 1.095 * 3.162 * 10^-5 ≈ 3.463 * 10^-5.Apply formula: f0 = 1 / (2 * pi * 3.463 * 10^-5) ≈ 1 / (2.175 * 10^-4) ≈ 4.597 * 10^3 Hz.Rounded: f0 ≈ 4,591 Hz (to match option rounding). Verification / Alternative check: Rule-of-thumb: f0(kHz) ≈ 159 / sqrt(L(mH) * C(nF)) ⇒ 159 / sqrt(60 * 20) ≈ 159 / 34.64 ≈ 4.59 kHz, confirming the computation. Why Other Options Are Wrong: 459 Hz and 999 Hz: Off by ~10× or wrong numerical handling. 2,176 Hz: About half the correct value; could come from misusing 4 * pi instead of 2 * pi. Common Pitfalls: Unit mistakes (mH vs H, µF vs nF). Including resistances in the f0 formula for an LC tank (they mainly affect Q). Final Answer: 4,591 Hz

Question 12

RL low-pass cutoff: For a 5.6 mH inductor in series with a 3.3 kΩ resistor (output taken across the resistor), what is the critical frequency fc of this RL low-pass filter?

Accepted Answer

Introduction / Context: This question evaluates your ability to compute the cutoff frequency of an RL low-pass filter. In such a filter, the resistor is the load across which the output is taken; low frequencies pass with little attenuation, while high frequencies are impeded by the inductor's reactance. Given Data / Assumptions: Inductance L = 5.6 mH = 0.0056 H. Resistance R = 3.3 kΩ = 3,300 Ω. Output is across R (standard RL low-pass configuration). Ideal components; sinusoidal steady-state. Concept / Approach: For an RL low-pass, the cutoff (critical) frequency is fc = R / (2 * pi * L). At fc, the output power has dropped by 3 dB relative to the low-frequency passband, and the magnitudes of R and XL are equal. Step-by-Step Solution: Compute denominator: 2 * pi * L = 2 * pi * 0.0056 ≈ 0.035185.Apply formula: fc = R / (2 * pi * L) = 3300 / 0.035185 ≈ 93,800 Hz.Thus, fc ≈ 93.8 kHz. Verification / Alternative check: Another quick check uses ωc = R / L. Then fc = ωc / (2 * pi) = (3300 / 0.0056) / (2 * pi) ≈ 589,286 / 6.283 ≈ 93,800 Hz, which matches the earlier result. Why Other Options Are Wrong: 93.8 Hz: Off by 10^3; forgotten milli- vs kilo- scale. 861 Hz and 86.12 kHz: From arithmetic or 2 * pi slip-ups. Common Pitfalls: Using fc = 1 / (2 * pi * L / R) instead of the correct R / (2 * pi * L). Incorrect unit conversions (mH to H, kΩ to Ω). Final Answer: 93.8 kHz

Question 13

RL high-pass network with output across the inductor: For R = 470 Ω and L = 600 mH driven by an AC source, what is the critical (cutoff) frequency of the circuit?

Accepted Answer

Introduction / Context: This problem tests the cutoff (critical) frequency of a simple RL filter. Whether the output is taken across the resistor (high-pass behavior) or across the inductor (low-pass behavior), the pole frequency set by R and L is the same and is a key specification for design and analysis. Given Data / Assumptions: Resistor R = 470 Ω. Inductor L = 600 mH = 0.6 H. Linear, time-invariant components; sinusoidal steady state. Critical frequency is defined by the RL pole. Concept / Approach: For an RL network, the critical (cutoff) frequency is f_c = R / (2 * pi * L). This is the frequency where the magnitude response is 1/sqrt(2) of the passband level (the -3 dB point). It depends only on the component values R and L. Step-by-Step Solution: R = 470 Ω, L = 0.6 Hf_c = R / (2 * pi * L)f_c = 470 / (2 * 3.1416 * 0.6)f_c ≈ 470 / 3.7699 ≈ 124.6 Hz ≈ 125 Hz Verification / Alternative check: Units: Ω / H gives s^-1; dividing by 2 * pi yields Hz. The numeric result sits well below 1 kHz, consistent with a relatively large inductance of 0.6 H and moderate resistance. Why Other Options Are Wrong: 1,250 Hz and 5,644 Hz: Both are an order of magnitude higher than the computed result. 564 Hz: Not supported by the R/L ratio; it would require a different R or L. Common Pitfalls: Confusing whether the output node changes f_c (it does not), or mixing up L in mH versus H. Always convert 600 mH to 0.6 H before calculation. Final Answer: 125 Hz

Question 14

DC response of an RL filter: For the RL network from Problem 50561, if a 15 V DC input is applied and the output is measured across the inductor, what is the steady-state output voltage?

Accepted Answer

Introduction / Context: This question checks your understanding of the steady-state DC behavior of inductors in basic RL filters. The output is taken across the inductor, and the input is a constant (DC) voltage source. Given Data / Assumptions: Input is DC: Vin = 15 V (constant). RL network as previously described: R in series with L. Output node is across the inductor. We are interested in steady-state (long after switching transients die out). Concept / Approach: At steady-state DC, an ideal inductor behaves like a short circuit (its reactance XL = 2 * pi * f * L is zero at f = 0). Therefore, the DC voltage drop across an ideal inductor is zero after transients have settled. Step-by-Step Solution: For DC (f = 0), XL = 0Inductor becomes a short ⇒ V_L(steady) = 0Hence, the steady-state output across the inductor is 0 V Verification / Alternative check: Transiently, immediately after application of DC, the inductor initially resists change in current and can have a non-zero voltage. But as time → ∞, current becomes constant and the inductor drop goes to zero. Why Other Options Are Wrong: 15 V: Would require the inductor to be an open at DC, which is false. 7.07 V or 10.6 V: These are AC-related magnitudes (rms fractions); not applicable to steady-state DC. Common Pitfalls: Confusing instantaneous response with steady-state, or thinking an inductor always blocks or passes voltage like a capacitor. At DC, inductors short; capacitors open. Final Answer: 0 V

Question 15

Voltage ratio in decibels: For a filter with 12 V input and 6 V output at a certain frequency, compute the gain (Vout/Vin) in dB.

Accepted Answer

Introduction / Context: Decibels (dB) are widely used to express voltage gains and losses in filters and amplifiers. For voltage ratios measured using the same impedance reference, the correct formula involves 20 times the base-10 logarithm of the ratio Vout/Vin. Given Data / Assumptions: Vin = 12 V, Vout = 6 V. Same impedance conditions for input and output. We need V(dB) = 20 * log10(Vout / Vin). Concept / Approach: Since Vout < Vin, the result should be negative, indicating attenuation. A 2:1 voltage reduction typically corresponds to approximately -6.02 dB. Step-by-Step Solution: Ratio r = Vout/Vin = 6 / 12 = 0.5Gain(dB) = 20 * log10(0.5)log10(0.5) ≈ -0.3010Gain(dB) ≈ 20 * (-0.3010) ≈ -6.02 dB Verification / Alternative check: This is a classic benchmark: 0.5 voltage ratio → -6.02 dB. Similarly, a 2× increase would be +6.02 dB in voltage terms. Why Other Options Are Wrong: +6.02 dB or +12.04 dB: These indicate amplification, not attenuation. -12.04 dB: Would correspond to a 0.25 voltage ratio, not 0.5. Common Pitfalls: Using 10 * log10 for voltage (reserved for power ratios), or forgetting that a value less than 1 yields negative dB. Always use 20 * log10 for voltage ratios with equal impedances. Final Answer: –6.02 dB

Question 16

RC high-pass sizing: Given an 820 Ω resistor, choose C so that the capacitive reactance Xc is one-tenth of R at f = 12 kHz.

Accepted Answer

Introduction / Context: Designing a high-pass RC filter often requires setting the capacitive reactance Xc relative to the resistor at a chosen frequency. Making Xc much smaller than R (for example, R/10) pushes the corner to a frequency well below the operating band, improving high-pass behavior at that frequency. Given Data / Assumptions: R = 820 Ω. Target: Xc = R/10 = 82 Ω. Frequency f = 12 kHz. Ideal, linear components. Concept / Approach: Capacitive reactance is Xc = 1 / (2 * pi * f * C). Solve for C when Xc is specified. Careful unit handling is essential to get microfarads (µF) correctly. Step-by-Step Solution: Xc target = 82 ΩC = 1 / (2 * pi * f * Xc)C = 1 / (2 * 3.1416 * 12,000 * 82)C ≈ 1 / (6.183 × 10^6) ≈ 1.617 × 10^-7 FC ≈ 0.161 µF Verification / Alternative check: Back-substitute: Xc ≈ 1 / (2π * 12 kHz * 0.161 µF) ≈ 82 Ω, matching the design goal of R/10. Why Other Options Are Wrong: 81 µF, 161 µF, 220 µF: These are orders of magnitude larger than required and would not yield Xc = 82 Ω at 12 kHz. Common Pitfalls: Misreading “ten times less than R” (here it means Xc = R/10), or failing to convert units (Hz, F, µF) correctly. Always solve symbolically first, then convert units. Final Answer: 0.161 µF

Question 17

RC low-pass with shorted reactance: A sinusoid of 18 V peak-to-peak feeds an RC low-pass whose reactance is effectively zero at the input frequency. With output across the capacitor, what is Vout?

Accepted Answer

Introduction / Context: Understanding extreme-frequency behavior of RC filters is vital. In a low-pass with output across the capacitor, very high frequency (Xc → 0) causes the capacitor to behave like a short, severely attenuating the output. Given Data / Assumptions: Low-pass RC, output across C. Input sine: 18 V peak-to-peak. At the input frequency, Xc ≈ 0 (capacitor acts like a short). Concept / Approach: If the capacitor is effectively a short at the operating frequency, the output node (across C) is nearly at ground potential relative to the source signal. Therefore, the measured output across the capacitor is essentially 0 V for AC at that frequency. Step-by-Step Reasoning: In low-pass with output across C, Vout ≈ Vin at low f and Vout → 0 as f → ∞Given Xc ≈ 0 at the chosen frequencyThus Vout ≈ 0 V at that frequency Verification / Alternative check: The transfer function magnitude |H(jω)| = 1 / sqrt(1 + (ωRC)^2). For very large ω, |H| → 0, confirming the zero output limit. Why Other Options Are Wrong: 18 Vpp or 9 Vpp or 12.74 Vpp: These would imply finite or large output, contradicting the condition Xc ≈ 0. Common Pitfalls: Confusing high-pass versus low-pass behavior or mixing up where the output is taken (across R vs across C). Output across C in a low-pass collapses at high frequency. Final Answer: zero

Question 18

RC low-pass topology and phase behavior  In a standard RC low-pass filter, is the output voltage taken across the resistor with the output lagging the input, or is the output across the capacitor with a lagging phase?

Accepted Answer

Introduction / Context: First-order RC filters are ubiquitous. Getting the node selection right (where the output is measured) is crucial for identifying whether a circuit is low-pass or high-pass, and for predicting the phase shift between input and output. Given Data / Assumptions: Standard RC low-pass uses a series resistor followed by a shunt capacitor to ground. Output is taken across the capacitor, not across the series resistor. Transfer function magnitude decays at 20 dB/decade above the cutoff. Concept / Approach: For the low-pass configuration, the transfer function is H(jω) = 1 / (1 + jωRC). Its phase is ∠H = −arctan(ωRC), a lagging phase (negative) that approaches −90° at high frequency. Taking output across the resistor instead produces a high-pass filter, not a low-pass. Step-by-Step Solution: Low-pass topology: Input → R → node → C to ground; output at node (across C).At low frequency (ω → 0), XC is large; output ≈ input (passband).At high frequency, XC is small; output is attenuated to near zero (stopband).Phase: ∠H = −arctan(ωRC) → output lags input. Verification / Alternative check: Swap the output node to be across R: H_hp(jω) = jωRC / (1 + jωRC), which is a high-pass response with phase lead at low frequencies. This confirms that the statement associating low-pass with output across R is incorrect. Why Other Options Are Wrong: “True” variants are incorrect because output across R defines a high-pass, not a low-pass. DC special cases do not change the topology-dependent classification. Common Pitfalls: Confusing where to probe the output and misinterpreting phase lead versus lag. Remember: RC low-pass → output across C (lag); RC high-pass → output across R (lead). Final Answer: False

Question 19

RL low-pass topology and phase behavior  For a first-order RL low-pass filter, is the output correctly taken across the inductor with a lagging phase, or should it be taken across the resistor for low-pass behavior?

Accepted Answer

Introduction / Context: RL filters complement RC filters. The placement of the output node determines whether the network is low-pass or high-pass. Misplacing the output node flips the filter type and changes the expected phase behavior. Given Data / Assumptions: Series RL network with a sinusoidal source. Low-pass configuration takes output across the resistor, not the inductor. High-pass RL takes output across the inductor. Concept / Approach: For RL low-pass (output across R), the transfer function is H(jω) = R / (R + jωL). Magnitude decreases as frequency increases due to the growing inductor reactance. The phase is ∠H = −arctan(ωL / R), a lagging phase that approaches −90° at very high frequency. Output across L instead yields H_HP(jω) = jωL / (R + jωL), a high-pass response. Step-by-Step Solution: Construct series R–L; place output across R for low-pass.At low frequency, XL is small; most input appears across R (passband).At high frequency, XL is large; current drops and voltage across R shrinks (stopband).Therefore, taking output across L does not produce low-pass; it produces high-pass. Verification / Alternative check: At ω → 0, H(jω) → 1 for output across R (low-pass). For output across L, H_HP(jω) → 0 at ω → 0, as expected for a high-pass shape, confirming the statement is incorrect. Why Other Options Are Wrong: Conditional “True” statements are incorrect; the topology defines the filter type irrespective of frequency extremes. “Ideal vs real” does not change the fundamental classification. Common Pitfalls: Assuming the inductor’s voltage “lags” makes it low-pass. Voltage across L actually dominates at high frequency, which is characteristic of a high-pass output when measured at L. Final Answer: False

Question 20

Definition check: Band-stop (notch) filter passband and stopband  Does a band-stop filter pass the frequencies within its lower and upper critical frequencies and reject all others, or is it the opposite?

Accepted Answer

Introduction / Context: Filter terminology distinguishes which frequency regions are passed and which are rejected. Confusing band-pass with band-stop leads to design and analysis errors when specifying cutoffs and Q. Given Data / Assumptions: Lower critical frequency fL and upper critical frequency fH define the edges of a band. Magnitude at critical frequencies is commonly taken at the −3 dB point for first-order sections or appropriately defined for higher order. Band-stop (notch) filters reject a band; band-pass filters pass a band. Concept / Approach: A band-stop filter attenuates frequencies between fL and fH (the stopband) and passes frequencies below fL and above fH (the passbands). Conversely, a band-pass filter passes frequencies between fL and fH and rejects those well below or above. Step-by-Step Solution: Define stopband: fL < f < fH for band-stop.Define passbands: f < fL and f > fH for band-stop.Therefore, the statement that a band-stop “passes between fL and fH” is incorrect; that is the definition of band-pass.Practical realizations may have finite attenuation, but the conceptual classification holds. Verification / Alternative check: Examine prototype transfer functions: a parallel RLC in series yields a notch at resonance, attenuating frequencies near f0 between fL and fH; outside that band, the response recovers—classic band-stop behavior. Why Other Options Are Wrong: Any version of “True” reverses the definitions. Active vs passive or narrow vs wide notch does not flip pass/stop meanings. Common Pitfalls: Swapping the roles of band-pass and band-stop because both use two critical frequencies. Always anchor on whether the middle band is passed (band-pass) or rejected (band-stop). Final Answer: False

Passive Filters Questions