Questions on tuned circuits and the Q factor

Question 1

State how the reactance of an inductor and of a capacitor each change as the frequency rises, and state what happens at resonance in a series LC circuit.

Accepted Answer

An inductor's reactance 2πfL rises with frequency; a capacitor's reactance 1/2πfC falls (1). At resonance the two are equal in size and opposite in sense, so they cancel and only the circuit's resistance is left to limit the current, which is therefore a maximum (1). The reactances do not become zero: each is still large, and the two cancel.

Question 2

Define the Q factor of a resonant circuit, and state the unit in which it is quoted.

Accepted Answer

Q = f₀/f_B, the resonant frequency divided by the bandwidth, the bandwidth being the width of the peak between the two frequencies at which the current has fallen to 1/√2 of its peak value (1). It is a pure number with no unit (1).

Question 3

A series LC circuit with resistance is driven by a variable-frequency source. State where the output must be taken to give a band-pass response and where to give a band-stop response.

Accepted Answer

Band-pass: take the output across the resistor, because the current, and so the pd across R, is largest at resonance (1). Band-stop: take the output across the series inductor and capacitor, whose combined reactance vanishes at resonance, so the output there falls to zero while everything far from resonance gets through (1).

Question 4

A resistor and a capacitor are connected in series across a signal source. State which component the output is taken across for a low-pass filter and which for a high-pass filter.

Accepted Answer

Output across the capacitor gives a low-pass filter, since the capacitor's reactance is large at low frequency and small at high, so high frequencies are shorted away (1). Output across the resistor gives a high-pass filter (1).

Question 5

Define the cut-off frequency of an RC filter, and state the output there as a fraction of the full output.

Accepted Answer

It is the frequency at which the capacitor's reactance equals the resistance, f_c = 1/2πRC (1), and there the output has fallen to 1/√2, about 0.71, of its full value, which is the half-power point (1). The output does not stop at the cut-off: the response falls away over a decade or more, with no sharp edge.

Question 6

State the effect of adding resistance to a tuned circuit on its resonant frequency, on the height of its resonance peak and on its bandwidth.

Accepted Answer

The resonant frequency does not move at all, since f₀ depends on L and C alone (1). The peak is lower and broader, because Q has fallen and the bandwidth f₀/Q has widened (1).

Question 7

An audio line carrying music from 20 Hz to 15 kHz is spoilt by a steady 1.0 kHz whistle from a faulty machine. Name the type of filter needed, and explain why neither a low-pass nor a high-pass filter can be used here.

Accepted Answer

A band-stop filter, a notch, centred on 1.0 kHz, with the output taken across the series inductor and capacitor (1). The wanted music lies both below and above the interference, so a filter with a single cut-off has to discard one side of it: a low-pass set below 1.0 kHz takes the whole treble away with the whistle, and a high-pass set above it takes the whole bass (1).

Question 8

A tuned circuit is built from an inductor of 1.0 mH and a capacitor of 100 nF. Calculate its resonant frequency.

Accepted Answer

LC = 1.0 × 10⁻³ × 100 × 10⁻⁹ = 1.0 × 10⁻¹⁰ (1), whose square root is 1.0 × 10⁻⁵ s (1). f₀ = 1/(2π × 1.0 × 10⁻⁵) = 1.59 × 10⁴ Hz, that is 15.9 kHz (1).

Question 9

A tuned circuit resonates at 15.9 kHz and has a Q factor of 10. Calculate its bandwidth and give the range of frequencies it passes.

Accepted Answer

f_B = f₀/Q = 15.9/10 = 1.59 kHz (1). The passband runs half a bandwidth either side of resonance (1), from 15.1 kHz to 16.7 kHz (1).

Question 10

A series circuit of 1.0 mH and 100 nF, resonating at 15.9 kHz, has a total resistance of 25 Ω. Using Q = 2πf₀L/R, calculate the Q factor and the bandwidth.

Accepted Answer

Q = 2π × 15915 × 1.0 × 10⁻³/25 (1) = 4.0 (1). f_B = f₀/Q = 15.9/4.0 = 4.0 kHz (1). Compare with 10 Ω, which gives Q = 10 and a bandwidth of 1.59 kHz: more resistance means less Q and a broader peak, at exactly the same centre frequency. R belongs in the denominator: written with R on top, more resistance would raise Q.

Question 11

A filter is made from a 1.6 kΩ resistor in series with a 0.10 μF capacitor. Calculate its cut-off frequency.

Accepted Answer

f_c = 1/(2πRC) = 1/(2π × 1600 × 0.10 × 10⁻⁶) (1) = 1/(2π × 1.6 × 10⁻⁴) (1) = 995 Hz, just under 1.0 kHz (1). Reading 0.10 μF as 0.10 × 10⁻⁹ F puts the cut-off a thousand times too high.

Question 12

A low-pass filter is to have a cut-off frequency of 2.5 kHz using a 4.7 kΩ resistor. Calculate the capacitance required, and state which component the output is taken across.

Accepted Answer

C = 1/(2πf_cR) (1) = 1/(2π × 2500 × 4700) = 1.35 × 10⁻⁸ F, that is 13.5 nF (1). The output is taken across the capacitor (1). Rearranging to C = 2πf_cR inverts the relationship and gives 7.4 × 10⁷ F; the answer should be a small fraction of a microfarad.

Question 13

A radio's tuned circuit uses a fixed 1.0 mH inductor with a variable capacitor adjustable from 40 pF to 360 pF. Calculate the resonant frequency at each end of the range, and state the ratio of highest to lowest frequency.

Accepted Answer

At 360 pF: f₀ = 1/(2π√(1.0 × 10⁻³ × 360 × 10⁻¹²)) = 2.65 × 10⁵ Hz, that is 265 kHz (1). At 40 pF: 7.96 × 10⁵ Hz, that is 796 kHz (1). The ratio is 3.0 to 1 (1). Note the square root at work: nine to one in capacitance is only three to one in frequency.

Question 14

A measured resonance curve peaks at 40.0 kHz, and the current has fallen to 1/√2 of its peak value at 38.4 kHz and at 41.6 kHz. Calculate the bandwidth and the Q factor.

Accepted Answer

f_B = 41.6 − 38.4 = 3.2 kHz (1). Q = f₀/f_B = 40.0/3.2 = 12.5 (1), a pure number. The bandwidth spans both half-power points: taking only the gap from the peak to one of them, 1.6 kHz, doubles the Q.

Question 15

A low-pass filter is built from a 3.3 kΩ resistor in series with a 22 nF capacitor. Calculate its cut-off frequency, and calculate the fraction of a 20 kHz signal that reaches the output, given that a low-pass filter's output fraction is 1/√(1 + (f/f_c)²).

Accepted Answer

f_c = 1/(2πRC) = 1/(2π × 3300 × 22 × 10⁻⁹) = 2192 Hz, about 2.19 kHz (1). At 20 kHz the ratio f/f_c = 20 000/2192 = 9.12 (1). Output fraction = 1/√(1 + 9.12²) = 0.109, so about 11% of the 20 kHz signal still gets through (1). Well above the cut-off the output falls roughly as f_c/f, so a signal 9.1 times the cut-off arrives at about a ninth of its size rather than at nothing at all.

Question 16

A receiver's tuned circuit is centred on 200 kHz and must pass a signal 8.0 kHz wide, so its bandwidth may not be less than 8.0 kHz. The only inductor available is 250 μH. Calculate the capacitance needed, the largest Q factor the circuit may have, and the smallest total circuit resistance that is acceptable, using Q = 2πf₀L/R. Comment on what would go wrong if the resistance were made very much smaller.

Accepted Answer

Rearranging f₀ = 1/(2π√(LC)) gives C = 1/(L(2πf₀)²) (1) = 1/(250 × 10⁻⁶ × (2π × 2.00 × 10⁵)²) = 2.53 × 10⁻⁹ F, that is 2.53 nF (1). Q = f₀/f_B = 200/8.0 = 25, the largest Q allowed (1). Rearranging the Q expression, R = 2πf₀L/Q (1) = 2π × 2.00 × 10⁵ × 250 × 10⁻⁶/25 = 12.6 Ω, the smallest resistance that is acceptable (1). Less resistance would push Q above 25 and the bandwidth below 8.0 kHz (1), so the circuit would start shaving the outer sidebands off the station it is tuned to and the received audio would lose its top frequencies (1).

Question 17

Speech occupying 300 Hz upwards shares a line with 50 Hz mains hum. A filter built from a 22 kΩ resistor is to have a cut-off frequency of 120 Hz. State which type of filter is needed, calculate the capacitance, and estimate the fraction of the hum that survives, given that a high-pass filter's output fraction is (f/f_c)/√(1 + (f/f_c)²).

Accepted Answer

A high-pass filter, since the wanted signal lies above the interference (1). C = 1/(2πf_cR) = 1/(2π × 120 × 22 × 10³) = 6.03 × 10⁻⁸ F, about 60 nF (1). At 50 Hz, f/f_c = 50/120 = 0.417, so the output fraction is 0.38 (1): about 38% of the hum still gets through, while speech at 300 Hz passes at 0.93 (1).

Question 18

An audio line is to have 50 Hz mains hum removed by a band-stop filter built from an inductor in series with a 1.0 μF capacitor. Calculate the inductance required, comment on whether the component is practical, and state where the output of the filter must be taken.

Accepted Answer

The notch sits at the resonant frequency, so L = 1/(C(2πf₀)²) (1) = 1/(1.0 × 10⁻⁶ × (2π × 50)²) = 10.1 H (1). That is an enormous inductor, physically large, heavy and resistive, so a real 50 Hz notch is built round an op-amp instead (1). The output is taken across the series L and C, whose combined reactance vanishes at 50 Hz (1). At mains frequency the inductance comes out in henries, not millihenries: an answer of 10 mH means a factor of 10³ has been dropped in the prefixes.

Question 19

Medium-wave stations are spaced 9 kHz apart. A receiver is tuned to a station on 909 kHz and its designer has wound the tuned circuit to a Q factor of 300. Calculate the bandwidth this gives, state the Q the receiver should have instead, and explain what the listener would hear with the Q of 300.

Accepted Answer

f_B = f₀/Q = 909/300 = 3.03 kHz (1). The bandwidth should be about the channel spacing, 9 kHz, which needs Q = 909/9 = 101 (1). At Q = 300 the passband is only 3.03 kHz wide, far narrower than the 9.0 kHz the station occupies (1), so the outer sidebands, which carry the highest audio frequencies, are cut away and the programme arrives dull and muffled (1).

Question 20

A parallel LC circuit is driven at its resonant frequency. Explain why the current drawn from the supply falls to a minimum there, and explain why the current circulating in the loop can nevertheless be far larger than the supply current.

Accepted Answer

Both branches share the same pd, and at resonance the two branch currents are equal in size and in antiphase: the capacitor's current leads the applied pd by a quarter of a cycle and the inductor's lags it by a quarter (1). The two therefore cancel where the branches meet, so the current drawn from the supply falls to a minimum and the pair presents its largest impedance (1). Inside the loop, energy goes on sloshing between the inductor's magnetic field and the capacitor's electric field, and the supply has only to make up what the loop's resistance takes each cycle, so the circulating current is Q times the supply current (1).

Question 21

A hobbyist wants a model-aircraft receiver that responds to the exact carrier frequency, 35.000 MHz, and rejects every other frequency completely, so winds its tuned circuit to the largest Q the coil will bear, 3500. (a) Explain why no tuned circuit, whatever Q it is wound to, can ever pass a single frequency and reject everything else. (b) The control signal needs a bandwidth of 20 kHz to carry all its channels without distortion. Calculate the bandwidth a Q of 3500 actually gives at 35.000 MHz, and state whether the receiver will reproduce the signal correctly.

Accepted Answer

Every real coil and capacitor has some resistance, so Q is always finite and the bandwidth f₀/Q it gives is never zero, however large Q is wound (1). Even an imaginable zero-resistance circuit passing one exact frequency would still be useless, because the transmitted signal is itself spread over a band of frequencies around the carrier rather than concentrated on a single one, so a filter narrower than that band would strip out the wanted information along with everything else (1). f_B = f₀/Q = 35.000 × 10⁶/3500 = 10000 Hz, that is 10 kHz (1). That is narrower than the 20 kHz the control signal needs, so the receiver will not reproduce it correctly: the outer parts of its spectrum are cut away (1). The signal needs a Q no higher than 1750.

Question 22

Speech occupying 300 Hz to 3.4 kHz on an audio line is spoilt by a steady 1.0 kHz tone from a faulty machine. A band-stop filter centred on 1.0 kHz with a Q factor of 2.0 is proposed. Calculate the width of the notch and the two frequencies it runs between, and explain why no notch filter, whatever Q it is built to, can remove the tone without touching the speech.

Accepted Answer

f_B = f₀/Q = 1000/2.0 = 500 Hz (1), and the notch runs half a bandwidth either side of 1.0 kHz, from 750 Hz to 1250 Hz (1). The whole of that band lies inside the 3.1 kHz the speech occupies, so every speech component between 750 Hz and 1250 Hz is attenuated along with the tone (1). A larger Q narrows the notch but never closes it, because every real inductor and capacitor has resistance and Q stays finite; and the speech itself contains components at 1.0 kHz, so even a notch of zero width would take wanted signal away with the interference (1).