In Digital Communication can We use PAM in place of ASK
The Diagram is Given as
The Book Later Talks about Symbols.
I Don't Understand, in PAM we don't have specific signal as input like in ASK we provide the keying as the digital input, so we can either have 2-ASK high/low(1 or 0) but in PAM the O/P value of the PAM circuit depends on the modulating signal.
Someone please Explain.
Answer
In ASK, a carrier is multiplied by a set of discrete amplitudes, depending on the information bits. In practice, binary ASP (BASK) is often used, where one of the amplitudes is zero, i.e. for a digital \$0\$, the modulated signal is zero, and for a digital \$1\$ the modulated signal is the carrier multiplied with some fixed amplitude. This important special case of ASK is called on-off keying (OOK).
Digital PAM is a digital modulation format where a pulse is multiplied by the current data symbol. In this sense, ASK can be seen as a special case of digital PAM, where the pulse is a sinusoid with the carrier frequency over one symbol interval. In general, the digital PAM signal can be written as
$$s(t)=\sum_{k=0}^{\infty}A_kp(t-kT)\tag{1}$$
where \$A_k\$ are the discrete symbols, \$p(t)\$ is the pulse function, and \$T\$ is the symbol interval. Note that \$A_k\$ can be any set of discrete symbols. Usually the number of symbols is a power of \$2\$. E.g., if the number of symbols is \$2^m\$, each symbol carries \$m\$ bits. Note that the signal \$s(t)\$ in (1) can be modulated by a carrier. In the general case, the symbols \$A_k\$ can be complex-valued, and a passband signal is generated by
$$\tilde{s}(t)=\Re\{s(t)e^{j\omega_c t}\}\tag{2}$$
where \$\omega_c\$ is the carrier frequency in radians. Equations (1) and (2) are the general representation of digital passband PAM. Quadrature amplitude modulation (QAM) and phase shift keying (PSK) are special cases.
Note that the type of PAM explained in nidhin's answer is analog PAM.
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