x(t). The more rapidly x(t) changes, the higher frequency the components it contains. This explains why the sampling frequency is related to the highest frequency of the analog signal to be sampled. From the time-domain figure we under- stand that the sampling frequency must be much higher than the highest Operation principle Time domain: of a sampling Original analog message Sampled signal circuit: Relay Sampling instants Frequency domain: Spectrum of an analog message Spectrum of sampled signal B = 4 kHz 3.4 kHz fm = 1 kHz fm = 1 kHz 8 kHz Figure 3.8 Sampling. Signals Carried over the Network frequency of the analog message. Otherwise, rapid changes of signal x(t) between sampling instants could not be described by sample values. The accurate answer to how much higher it should be can be understood more easily via the frequency domain. The frequency-domain descriptions in Figure 3.8 show the spectrum of x(t) and the sampled signal y(t). Before sampling, the spectrum X(f) of x(t) contains speech frequencies up to 3.4 kHz, shown as a dashed line in the fig- ure. As an example of the frequency components of speech we drew the spec- trum of a 1-kHz cosine wave as a solid spectral line at the 1-kHz point on the frequency axis. After sampling, the spectrum of the message also appears around the sampling frequency. If the message contains a single 1-kHz frequency com- ponent, after sampling we will have components at 1 kHz, 8 kHz - 1 kHz = 7 kHz, and at 8 kHz + 1 kHz = 9 kHz, as seen in the figure. In addition to these components, sampling also generates components around double sam- pling frequency, three times sampling frequency, and SO forth. The reproduction of an original signal from a sampled signal is per- formed by a lowpass filter and in the case of voice the bandwidth B = 4 kHz, that is, half the sampling frequency. We see from Figure 3.8 that this filter would let through only a 1-kHz component of the spectrum, that is, the actual original analog signal. With the help of the lowpass filter we have suc- cess