Compact discs (CDs) have become a very popular medium for recording and playing music. CDs store information in a digital manner; that is, the music is sampled at a very high rate, and the samples are recorded on the disc. The trick is to sample so quickly that the reproduction sounds continuous.

Question

Compact discs (CDs) have become a very popular medium for recording and playing music. CDs store information in a digital manner; that is, the music is sampled at a very high rate, and the samples are recorded on the disc. The trick is to sample so quickly that the reproduction sounds continuous. The industry standard sampling rate is 44.1 kHz- one sample every 22.7 μs.

One interesting aspect regarding the analog-to-digital conversion that takes place inside the unit recording a CD is called the Nyquist criterion. This criterion states that in the analog conversion, any signal components at frequencies above half the sampling rate (22.05 kHz in this case) cannot be faithfully reproduced. Therefore, recording technicians filter these frequencies out before any sampling occurs, yielding higher fidelity to the listener.
Let us design a series of low-pass filters to perform this task.

Accepted Answer

Suppose, for example, that our specification for the filter is unity gain at dc and 20 dB of attenuation at 22.05 kHz. Let us consider first the simple RC filter in Fig. 12.79.

The transfer function is easily found to be

[latex]\mathbf{G}_{v 1}(s)=\frac{\mathbf{V}_{o 1}}{\mathbf{V}_{ ext {in }}}=\frac{1}{1+s R C}[/latex]

Since a single-pole transfer function attenuates at 20 dB/decade, we should place the pole frequency one decade before the -20 dB point of 22.05 kHz.

Thus,

[latex]f_{P}=\frac{1}{2 \pi R C}=2.205 \mathrm{kHz}[/latex]

If we arbitrarily choose C = 1 nF, the resulting value for R is 72.18 kΩ, which is reasonable. A Bode plot of the magnitude of [latex]\mathbf{G}_{v 1}(s)[/latex] is shown in Fig. 12.80. All specifications are met but at the cost of severe attenuation in the audible frequency range. This is undesirable.

An improved filter is shown in Fig. 12.81. It is a two-stage low-pass filter with identical filter stages separated by a unity-gain buffer.

The presence of the op-amp permits us to consider the stages independently. Thus, the transfer function becomes

[latex]\mathbf{G}_{v 2}(s)=\frac{\mathbf{V}_{o 2}}{\mathbf{V}_{ ext {in }}}=\frac{1}{[1+s R C]^{2}}[/latex]

To find the required pole frequencies, let us employ the equation for [latex]\mathbf{G}_{v 2}(s)[/latex] at 22.05 kHz, since we know that the gain must be 0.1 (attenuated 20 dB ) at that frequency. Using the substitution s = jω, we can express the magnitude of [latex]\mathbf{G}_{v 2}(s)[/latex] as

[latex]\left|\mathbf{G}_{v 2} ight|=\left\{\frac{1}{1+\left(22,050 / f_{p} ight)^{2}} ight\}=0.1[/latex]

and the pole frequency is found to be 7.35 kHz. The corresponding resistor value is 21.65 kΩ. Bode plots for [latex]\mathbf{G}_{v 1}[/latex] and [latex]\mathbf{G}_{v 2}[/latex] are shown in Fig. 12.82. Note that the two-stage filter has a wider bandwidth, which improves the fidelity of the recording.

Let us try one more improvement-expanding the two-stage filter to a four-stage filter. Again, the gain magnitude is 0.1 at 22.05 kHz and can be written

[latex]\left|\mathbf{G}_{v 3} ight|=\left\{\frac{1}{\left[1+\left(22,050 / f_{p} ight)^{2} ight]^{2}} ight\}=0.1[/latex]

The resulting pole frequencies are at 15 kHz, and the required resistor value is 10.61 kΩ. Figure 12.83 shows all three Bode plots. Obviously, the four-stage filter, having the widest bandwidth, is the best option (discounting any extra cost associated with the additional active and passive circuit elements).

Question 12.38: Compact discs (CDs) have become a very popular medium for re...

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