fully reproduced the original analog message from the samples alone. If we increase the frequency of an analog message x(t) from 1 to 2 kHz we will have the lowest component of the sampled signal at 2 kHz, the solid spectral line at 1 kHz is moved to the right, the next spectral component at 8 kHz - 2 kHz = 6 kHz, and the solid line at 7 kHz is moved to the left. Low- pass filtering will still give the original 2-kHz message. Now if we increase the frequency beyond 4 kHz to, say, 5 kHz, we will get components at 5 kHz and 8 kHz - 5 kHz = 3 kHz, and lowpass filtering will give a 3-kHz signal instead of the original 5-kHz signal. Reproduction will not work anymore because the frequency of the analog signal has exceeded half of the sampling frequency. We have seen that the sampling frequency must be more than twice the highest frequency component of the original signal to be encoded; otherwise, the message spectra around zero frequency and sampling frequency will over- lap. This can be seen from the spectrum Y(f) in Figure 3.8 if we imagine what happens if W > f12. From the spectrum of the sampled signal Y(f) in Figure 3.8, we also see that the message can be completely reconstructed from a PAM signal with a 4-kHz lowpass filter if W < f12. This requirement is fundamental for all digital signal processing. Introduction to Telecommunications Network Engineering The highest frequency of voice that will be transmitted is chosen to be 3,400 Hz and the sampling frequency is standardized at 8,000 Hz, leaving enough guard band for filtering. Samples are then taken at intervals of Ts = 125 us. In the sampling process a PAM signal y(t) is created. The amplitudes of PAM pulses follow the original analog signal. Note that the samples are still analog, having any analog value between the minimum and maximum values of the original signal. 3.6.2 Quantizing In the previous section we utilized sampling that produces a PAM signal that represents discrete but still analog values of the original analog message at the sampling instants