Let's consider the N number of baseband signals containing message information. Each signal is bandlimited to FM and represented by m1(t), m2(t), … mn(t). The communication channel does not require a bandwidth larger than NFM. The sampling of the baseband signal m1(t) is not greater than the time interval Ts = 1/2FM. It means that the successive signals are also sampled at the same time interval gap (1/2FMN). It is the gap between the two successive signals (for example, m1(t) and m2(t)). If the channel bandwidth is high, demultiplexing would be easier and can be achieved directly. But, if the channel bandwidth is restricted, the basebands signals may combine resulting in crosstalk, which may affect the quality of the signal.
A low pass filter is used to recover the message signal at the output in a similar way like other modulation processes. Suppose the first baseband signal is transmitted at t = 0. The response will be represented by:
S1 = I1ωc sin ωct/πωct
At t = 0, the response shows the peak proportional to the message signal m1(t). The second baseband signal needs to be transmitted at t= 1/2FC. The response IS GIVEN BY:
S2 = I2ωc sin ωc (t - 1/2FC)/πωc(t - 1/2FC)
The response of both the baseband signals is shown below:
The dashed line represents the response of the second baseband signal. The Sampling at time intervals prevents the crosstalk with the recovery of the original baseband signals m1(t) and m2(t).
We have discussed the instantaneous Sampling of baseband signals, which is the process of sending multiple signals using electronic switches. It was based on the revolution of rotary switches present at both the transmitting and receiving ends. But, sampling using such switches is not feasible. The energy of the instantaneous samples is sometimes infinite. It gives rise to small peak values that may introduce back noise in the signals. Hence, Natural Sampling is considered an efficient multiplexing method in Pulse Amplitude Modulation. Here, the sampling waveform S(t) consists of a train of pulses of duration τ. The sampled signal S(t)m(t) consists of a train of pulses following the waveform of the message signal m(t). The top of the signal S(t) also follows the pattern of the message waveform.
The waveforms of the message signal m(t), sampling signal S(t), and the sampled signal S(t)m(t) are shown below:
Like instantaneous sampling technique, it passes the sampled signals through the low pass filter with the cut-off frequency FM to recover the original message signal.
The sampling waveform is represented by:
S(t) = τ/TS + 2 τ/TS (C1cos2π t/ TS + C2cos4π t/ TS + … )
Where,
Cn = sin(nπτ/TS)/(nπτ/TS)
The sampled signal S(t)m(t) is given by:
S(t)m(t) = m(t) τ /TS + 2 τ /TS (m(t)C1cos2π t/ TS + m(t)C2cos4π t / TS + … )
In the natural sampling process, the sampled waveform follows the shape of the message waveform. It is not used frequently. Instead, the flat top sampling process is used. Though, the reconstruction of the original message signal is not easy. But, the design of flat top sampling electronic circuitry is simple as compared to the natural sampling process.
The baseband signal = m(t)
The Fourier transform of the baseband signal = M(jω)
The spectral density of flat-top sampling process is given by:
Spectral density = τ sin(ωτ/2) / (ωτ/2) M(jω)
The message signal has a flat spectral density over the range 0 to FM. The original signal can be recovered if the spectral density of the sampled signal and the original message signal are the same. The signal recovery can be performed using a low pass filter, as discussed above.
The transfer function of the network is given by:
H(jω) = τ /dT sin(ωτ/2) / (ωτ/2)
The transform using the product of H(jω) and M(jω) is given by:
Transform = τ /TS sin(ωτ/2) / (ωτ/2) M(jω)
We have discussed that the time interval between the different message signals is T/TS = 1/N. N is the total number of signals that need to be multiplexed. As N becomes larges, the time interval T/TS becomes smaller. Signal recovery through holing is an alternative signal recovery process that does not use amplifiers to raise the output level. The amplifiers are avoided because they may introduce noise in the circuit. Here, the flat top sampled is taken, as discussed above.
The output waveform is the same as the flat-top signal except for the holding part. The sampled pulses are extended and held until the next signal occurs. It means that there is no gap between the two adjacent pulses. The extension is represented with the dashed line in the sampled waveform, as shown below:
The circuit through which the holding operation is performed is shown below:
The switch is connected with the amplifier and a capacitor. The switch opens before the occurrence of the edge with no voltage. The switch closes after the occurrence of the edge of a sample pulse, allowing the capacitor to charge. It holds the voltage until the next operation is repeated. The amplifier selected for the process is a low impedance amplifier.
The baseband signal = m(t)
The Fourier transform of the baseband signal = M(jω)
We consider that the received samples are the flap top samples to encompass the interval between the instantaneous samples. The spectral density of sampled and the held signal is given by:
Spectral density = sin(ωTs/2) / (ωTs/2) M(jω)
It is the same as the density of the flap-top sampling process, except the T is replaced by the sampling time.
The magnitude of the sampled and held spectrum is shown below:
The unequal waveform is due to some distortion using the flat-top sampling process in the range 0 to FM. But, the distortion can be corrected by using an equalizer. It helps in removing distortion from the sampled signal. In audio instruments, it helps in removing unwanted sounds.
The magnitude of the instantaneous sampled signal is shown below:
The advantages of Pulse Amplitude Modulation are as follows:
The disadvantages of Pulse Amplitude Modulation are as follows:
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