# 0.7 Bits to symbols to signals  (Page 3/8)

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A Gray code has the property that the binaryrepresentation for each symbol differs from its neighbors by exactly one bit.A Gray code for the translation of binary into 4-PAM is

$\begin{array}{ccc}01\hfill & \to \hfill & +3\hfill \\ 11\hfill & \to \hfill & +1\hfill \\ 10\hfill & \to \hfill & -1\hfill \\ 00\hfill & \to \hfill & -3\hfill \end{array}$

Mimic the code in naivecode.m to implement this alternative and plot the number of errors as a function ofthe noise variance v . Compare your answer with [link] . Which code is better?

## Symbols to signals

Even though the original message is translated into the desired alphabet, it is not yet ready for transmission:it must be turned into an analog waveform. In the binary case, a simple method is to use a rectangular pulseof duration $T$ seconds to represent $+1$ , and the same rectangular pulse inverted (i.e., multiplied by $-1$ ) to represent the element $-1$ . This is called a polar non-return-to-zero line code.The problem with such simple codes is that they use bandwidth inefficiently.Recall that the Fourier transform of the rectangular pulse in time is the $\text{sinc}\left(f\right)$ function in frequency [link] , which dies away slowly as $f$ increases. Thus, simple codes like the non-return-to-zeroare compact in time, but wide in frequency, limiting the number of simultaneous nonoverlapping users ina given spectral band.

More generally, consider the four-level signal of [link] . This can be turned into an analog signal for transmission by choosinga pulse shape $p\left(t\right)$ (that is not necessarily rectangular and not necessarily of duration $T$ ) and then transmitting

$\begin{array}{cc}\hfill p\left(t-kT\right)& \text{if}\phantom{\rule{4.pt}{0ex}}\text{the}\phantom{\rule{4.pt}{0ex}}k\text{th}\phantom{\rule{4.pt}{0ex}}\text{symbol}\phantom{\rule{4.pt}{0ex}}\text{is}\phantom{\rule{4.pt}{0ex}}1\hfill \\ \hfill -p\left(t-kT\right)& \text{if}\phantom{\rule{4.pt}{0ex}}\text{the}\phantom{\rule{4.pt}{0ex}}k\text{th}\phantom{\rule{4.pt}{0ex}}\text{symbol}\phantom{\rule{4.pt}{0ex}}\text{is}\phantom{\rule{4.pt}{0ex}}-1\hfill \\ \hfill 3p\left(t-kT\right)& \text{if}\phantom{\rule{4.pt}{0ex}}\text{the}\phantom{\rule{4.pt}{0ex}}k\text{th}\phantom{\rule{4.pt}{0ex}}\text{symbol}\phantom{\rule{4.pt}{0ex}}\text{is}\phantom{\rule{4.pt}{0ex}}3\hfill \\ \hfill -3p\left(t-kT\right)& \text{if}\phantom{\rule{4.pt}{0ex}}\text{the}\phantom{\rule{4.pt}{0ex}}k\text{th}\phantom{\rule{4.pt}{0ex}}\text{symbol}\phantom{\rule{4.pt}{0ex}}\text{is}\phantom{\rule{4.pt}{0ex}}-3\hfill \end{array}$

Thus, the sequence is translated into an analog waveform by initiating a scaled pulse at the symbol time $kT$ , where the amplitude scaling is proportional to the associated symbolvalue. Ideally, the pulse would be chosen so that

• the value of the message at time $k$ does not interfere with the value of the message at other sample times(the pulse shape causes no intersymbol interference ),
• the transmission makes efficient use of bandwidth, and
• the system is resilient to noise.

Unfortunately, these three requirements cannot all be optimized simultaneously, and so the design of thepulse shape must consider carefully the tradeoffs that are needed. The focus in Chapter [link] is on how to design the pulse shape $p\left(t\right)$ , and the consequences of that choice in terms of possible interference betweenadjacent symbols and in terms of the signal-to-noise properties of the transmission.

For now, to see concretely how pulse shaping works, let's pick a simple nonrectangular shape and proceedwithout worrying about optimality. Let $p\left(t\right)$ be the symmetrical blip shape shown in the top part of [link] , and defined in pulseshape.m by the hamming command. The text string in str is changed into a 4-level signal as in Example [link] , and then the complete transmitted waveform is assembled by assigning an appropriatelyscaled pulse shape to each data value. The output appears in the bottom of [link] . Looking at this closely, observe that the first letter T is represented by the four values $-1\phantom{\rule{4pt}{0ex}}-1\phantom{\rule{4pt}{0ex}}-1\phantom{\rule{4pt}{0ex}}-3$ , which corresponds exactly to the first four negative blips, three small and one large.

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I think
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