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In Digital Communication can We use PAM in place of ASK
The Diagram is Given as


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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.

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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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  • \$\begingroup\$ Thanks for the Explanation.... I wonder in digital input,we have both amplitude and time as discrete...so Is the Output from the PAM Continuous in both Amplitude And Time??I have Seen Books describing PAM as Discrete-CW Modulation \$\endgroup\$ – MaMba Mar 8 '15 at 19:21
  • \$\begingroup\$ @SEXCrow: The final output is always analog even with digital communications, because you usually have to transmit the signal over a real (i.e. analog) channel. \$\endgroup\$ – Matt L. Mar 8 '15 at 19:40
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Amplitude Shift Keying (ASK) is a modulation scheme in which the modulating signal is digital, have only 2 voltage levels (for 2-ASK). It is Amplitude Modulation with digital modulating signal.

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Where as PAM is an modulation scheme in which the modulating signal is analog (continuous signals). PAM is natural sampling (not flat top) of modulating signal. So the amplitude of output signal depends on the modulating signal

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  • \$\begingroup\$ I believe the OP refers to digital PAM, whereas you correctly explained analog PAM. \$\endgroup\$ – Matt L. Mar 8 '15 at 16:49
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    \$\begingroup\$ @MattL. Ooops. The analog tag made me think that it is analog PAM. \$\endgroup\$ – nidhin Mar 8 '15 at 17:00

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