be done? Well, have faith-there are two good rea- sons. First, ideal sampling leads to an understanding of the very important sampling theorem. Second, ideal sampling will help you easily grasp and understand the two practical sampling methods that follow. The Sampling Ideal sampling is simple, and it's shown in Figure 4.1. Here, an analog input X (t)-say, Carl's speech-enters the sampler. The sampler does one thing: it multiplies Carl's speech by the signal Chapter Four xs(t)=x(t).p(t) OTs 2Tg3Ts4T55Ts Figure 4.1 Ideal sampling (4.1) called an impulse train, shown in Figure 4.1. The output of the sampler is then (4.2) The multiplication of x(t) by the impulse train p(t) leads to the output shown on the output side of Figure 4.1. Here, we see that the output is made up of impulses at times kT of height x(kTs) ; that is, mathematically, x,(1)= XXXkT,)8(t-k7,) (4.3) In essence, this is everything there is to know about ideal sampling. Well, almost everything. The Information in the Samples After building the sampler, one telecommunication engineer began to wonder: "How much of the information in Carl's speech signal x(t) is lost when it's sampled to create X's (t) ?" We'll spend this section answering his question. The easiest way to answer it is to turn to the frequency domain for help. Let's first take a look at Carl's incoming speech signal, x (t). When described in the frequency domain, as X(f), we'll assume for this section that it looks like Figure 4.2(a). Next, we'll consider the impulse train p(t). The Fourier transform of p(t) is (4.4) where fs = 1/T and f is called the sampling rate. You'll find a picture of this signal in Figure 4.2(b). Source Coding and Decoding: Making it Digital Let's now evaluate Xs (f), the Fourier transform of the output signal X's (t). . We'll then use this to figure out how much of the information in Carl's speech signal x(t) is lost when it's sampled. Xs(f) corresponds to Xs(f)=F{xs(t)}=F{x(t).p(t)}. (4.5) To simplify Xs(f), we'll use the following property: Multiplication in th