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Chapter 12 Practice Problems

Practice Problems

Problem 1 — RC Low-Pass Filter Design

Design an RC low-pass filter with a cutoff frequency of 3.4 kHz (telephone voice band upper limit).

(a) Select C = 47 nF and calculate the required R.

(b) Calculate the actual cutoff frequency with a standard 1.0 kΩ resistor (the nearest standard value, if R comes out near 1 kΩ — check your calculation).

(c) What is the filter's attenuation at 34 kHz (one decade above cutoff)?

(d) What is the attenuation at 68 kHz (two decades above)?

Solution

(a) Required R:

\[R = \frac{1}{2\pi f_c C} = \frac{1}{2\pi \times 3400 \times 47\times10^{-9}} = \frac{1}{1.003\times10^{-3}} = 997\ \Omega \approx 1.0\text{ kΩ}\]

(b) Actual cutoff with R = 1.0 kΩ:

\[f_c = \frac{1}{2\pi \times 1000 \times 47\times10^{-9}} = \frac{1}{295.3\times10^{-6}} = 3{,}386\text{ Hz}\]

Very close to the target — within 0.4%.

(c) Attenuation at 34 kHz (10× above \(f_c\)): for a first-order filter, one decade above the cutoff attenuates by approximately −20 dB.

More precisely: \(|H| = 1/\sqrt{1+(10)^2} = 1/\sqrt{101} = 0.0995\), \(|H|_{dB} = -20.04\text{ dB} \approx -20\text{ dB}\)

(d) At 68 kHz (20× above \(f_c\)): \(|H| = 1/\sqrt{1+(20)^2} = 1/\sqrt{401} = 0.0499\), \(|H|_{dB} = -26.0\text{ dB}\). Note: two decades above cutoff does NOT give −40 dB for a first-order filter — it gives approximately −26 dB (the −20 dB/decade rule is only asymptotic).

Problem 2 — RC High-Pass Filter for Hum Removal

A 60 Hz power line hum contaminates an audio signal. Design an RC high-pass filter with \(f_c = 80\) Hz to attenuate the hum.

(a) With C = 220 nF, calculate R.

(b) Calculate the attenuation of the 60 Hz hum component.

(c) Calculate the attenuation at 1 kHz (in the signal passband) — should be less than −3 dB.

Solution

(a) Required R:

\[R = \frac{1}{2\pi f_c C} = \frac{1}{2\pi \times 80 \times 220\times10^{-9}} = \frac{1}{110.6\times10^{-6}} = 9{,}042\ \Omega \approx 9.1\text{ kΩ}\]

Use 10 kΩ (nearest E24 value; actual \(f_c = 72.3\) Hz — use this for parts b, c).

(b) At 60 Hz with \(f_c = 72.3\) Hz: \(\omega RC = f/f_c = 60/72.3 = 0.830\)

High-pass: \(|H| = \omega RC / \sqrt{1+(\omega RC)^2} = 0.830/\sqrt{1+0.689} = 0.830/1.300 = 0.638\) [|H|{dB} = 20\log]}(0.638) = -3.90\text{ dB

(c) At 1 kHz: \(f/f_c = 1000/72.3 = 13.83\)

[|H| = 13.83/\sqrt{1+13.83^2} = 13.83/\sqrt{192.3} = 13.83/13.87 = 0.997] [|H|{dB} = 20\log]}(0.997) = -0.026\text{ dB

Virtually no attenuation at 1 kHz — the signal is unaffected.

Problem 3 — RLC Band-Pass Filter

Design a series RLC band-pass filter centered at 10 kHz with a quality factor Q = 25. Use L = 1 mH.

(a) Calculate the required capacitance C.

(b) Calculate the resistance R.

(c) Calculate the 3 dB bandwidth.

(d) What are the lower and upper −3 dB frequencies?

Solution

(a) From \(f_0 = 1/(2\pi\sqrt{LC})\):

\[C = \frac{1}{(2\pi f_0)^2 L} = \frac{1}{(2\pi\times10{,}000)^2 \times 0.001} = \frac{1}{3.948\times10^9 \times 0.001} = \frac{1}{3.948\times10^6} = 253.3\text{ pF}\]

(b) From \(Q = \omega_0 L / R\):

\[R = \frac{\omega_0 L}{Q} = \frac{2\pi \times 10{,}000 \times 0.001}{25} = \frac{62.83}{25} = 2.51\ \Omega\]

(c) Bandwidth:

\[\text{BW} = \frac{f_0}{Q} = \frac{10{,}000}{25} = 400\text{ Hz}\]

(d) The −3 dB frequencies straddle \(f_0\):

[f_L = f_0 - \frac{\text{BW}}{2} = 10{,}000 - 200 = 9{,}800\text{ Hz}] [f_H = f_0 + \frac{\text{BW}}{2} = 10{,}000 + 200 = 10{,}200\text{ Hz}]

Problem 4 — Decibels and Audio Levels

A microphone produces −50 dBV. A preamplifier with 40 dB gain amplifies it to line level.

(a) Calculate the output voltage level in dBV.

(b) Calculate the output voltage in volts (RMS).

(c) Convert the output level to dBu.

(d) A second stage amplifier has 12 dB gain. What is its output in dBV?

Solution

(a) Output level in dBV:

\[L_{out} = L_{in} + G = -50 + 40 = -10\text{ dBV}\]

(b) Convert −10 dBV to volts:

\[V = 10^{-10/20} = 10^{-0.5} = 0.316\text{ V RMS}\]

(c) Converting dBV to dBu: \(0\text{ dBV} = 1\text{ V} = 2.218\text{ dBu}\) (since \(1\text{ V}/0.7746\text{ V} = 1.291\) and \(20\log(1.291) = 2.218\text{ dB}\))

\[L_{dBu} = L_{dBV} + 2.218 = -10 + 2.218 = -7.78\text{ dBu}\]

(d) After second stage: \(-10 + 12 = +2\text{ dBV}\), which is \(10^{2/20} = 1.26\text{ V RMS}\).

Problem 5 — Filter Dynamic Range

A high-quality audio recording system has a noise floor of −90 dBV and a maximum output before clipping of +10 dBV.

(a) Calculate the dynamic range in dB.

(b) What is the noise floor voltage in microvolts?

(c) A 60 dB-gain preamplifier adds its own noise of −110 dBV referred to input. What is the noise floor referred to output?

(d) What headroom does the system have above a nominal operating level of 0 dBV?

Solution

(a) Dynamic range:

\[\text{DR} = +10 - (-90) = 100\text{ dB}\]

(b) Noise floor voltage:

\[V_{noise} = 10^{-90/20} = 10^{-4.5} = 31.6\text{ μV RMS}\]

(c) Preamplifier input noise: −110 dBV. After 60 dB gain:

\[L_{noise,out} = -110 + 60 = -50\text{ dBV}\]

(d) Headroom = Maximum level − Nominal level = \(+10 - 0 = 10\text{ dB}\). This 10 dB headroom accommodates transient peaks 3.16× the nominal level without clipping.