As you will recall from Chapter 3, the bandwidth of a signal is the total range of frequencies occupied by that signal. When we decompose an ASK signal, we get a spectrum of simple frequencies. The signal at the center of this spectrum is the carrier fc. At either side are signals with frequencies fc-Nbaud/2, fc + Nbaud/2, fc-3 Nbaud/2, fc + 3 Nbaud/2, and so on. For practical purposes, however, only the carrier frequency and the two closest side frequencies are needed (see Figure 4.3-4).
Bandwidth requirements for ASK encoding are calculated using the formula
BW=(1+d)x Nbaud where
BW is the bandwidth Nbaud is the baud rate d is a factor related to the condition of the line (with a minimum value of 0)
As you can see, the minimum bandwidth required for transmission is equal to the baud rate.
Example 4.3-1
Find the bandwidth for an ASK signal transmitting at 2000 bit/s. Transmission is in half-duplex mode.
In ASK the baud rate and bit rate are the same. The baud rate is therefore 2000. An ASK signal requires a bandwidth equal to its baud rate. Therefore, the bandwidth is 2000 Hz.
Example 4.3-2
Given a bandwidth of 5000 Hz for an ASK signal, what are the baud rate and bit rate? Solution
In ASK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But because the baud rate and the bit rate are also the same for ASK, the bit rate is 5000 bit/s.
Given a bandwidth of 10,000 Hz (l000 to 11,000 Hz), draw the full-duplex ASK diagram of the system. Find the carriers and the bandwidths in each direction. Assume there is no gap between the bands in two directions. Solution
For full-duplex ASK the bandwidth for each direction is
BW=10,000/2=5000 Hz The carrier frequencies can be chosen at the middle of each band (see Figure 4.3-5).
fc(forward) =1000+5000/2=3500 Hz fc(backward) =11000-5000/2=8500 Hz
Fig. 4.3-5 Solution to Example 4.3-3
Fig. 4.3-5 Solution to Example 4.3-3
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