Homework Eight

How does a CD player work with no oversampling, but digital filtering (1x oversampling)?

Nick Christman

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Recall that when the audio for a music CD is produced it has an infinite amount of data points which can be expressed as $\scriptstyle x(t)$ in the time domain [Figure 1] and $\scriptstyle X(f)$ in the frequency domain [Figure 2].

Figure 1: Audio data as a function of time, x(t)

Figure 2: Audio data as a function of frequency, X(f)

When one wants to store the data of $\scriptstyle x(t)$ (i.e. reading the data onto a CD in this case) an infinite number of data points is not ideal -- in fact, it is impossible. Therefore, we must sample the data at the (typical) rate of $\scriptstyle f_{s}={\frac {1}{T}}$. This will give us a periodic function which we will call $\scriptstyle x(nT){\mbox{, where }}n\in \mathbb {Z} {\mbox{ and }}T$ is the sampling period. It is common to allow a sampling rate of twice the frequency at which a human can hear (i.e. 2 x 22 kHz) -- that means, $\scriptstyle f_{s}=44$ kHz.

Recall from class that in order to sample the original function, $\scriptstyle x(nT)$, we need to use a delta function. In other words, the continuous function of $\scriptstyle x(nT)$ can be written as $\sum _{n=-\infty }^{\infty }x(nT)\delta (t-nT_{0})$ [Figure 3]. As one may expect, in the frequency domain we should get the same plot as in Figure 2, but it should repeat. In the frequency domain, this can be expressed as ${\frac {1}{T}}\sum _{n=-\infty }^{\infty }X(f-{\frac {n}{T}})$ [Figure 4].

Figure 3: x(nT) written as impulses for sampling purposes

Figure 4: x(nT) written as impulses for sampling purposes in the frequency domain

As one may suspect, in order to get a more accurate function you must take more data points -- in other words, a higher sampling rate (i.e. 2x, 8x, etc. oversampling) leads to a more accurate collection of data. For this case, we are asked to looked a digital sampling, or 1x oversampling. In order to accomplish this, we will convolve our function from above with a Finite Impulse Response filter (FIR filter). Let's call it g(t and define it as $\sum _{-M}^{M}g({\frac {mT}{p}})\delta (t-{\frac {mT}{p}})$ where $\scriptstyle p$ is the oversampling rate. In our case $\scriptstyle p=1$. Because we are dealing with digital sampling, the convolution will result in the original function [Figure 5].

Figure 5: x(nT) after the convolution mentioned above