digital encoding

We allocating different systems in the same environment different frequency bands; by doing this, we are able to communicate pack information more effectively to prevent interference.

“how do we take a sequence of bits 10100…. and map it to a continuous-time signal \(X(t)\) such that the spectrum of this system is limited to \([0, B]\)”?

sinc digital encoding

IDEA: recall sinc sampling theorem, which (even if under sampled), will recover the source points exactly. As such, we can write:

High Frequency Signal

Frequency content of a signal would be symmetric around a target frequency \(f_{c}\), and most of the energy will go from \(f_{c} \pm \frac{1}{T}\). Strictly speaking, you maybe leaking some energy outside the band.

energy of signal

\begin{equation} \varepsilon_{1} = \frac{1}{T} \int_{0}^{T} y^{2}(t) \dd{t} \end{equation}

if \(\varepsilon_{s} > \varepsilon_{T}\), then decode \(1\). Otherwise, \(\epsilon_{1} < \epsilon_{T}\), decode \(0\).

Hamming Distance

the Hamming Distance between two sequences is the number of positions in which these two sequences differ from each other

an is a collection of binary strings such that the minimum between any two codes is some distance \(d\).

This code allows you to correct up to:

• \(t = \left\lfloor \frac{d_{c}-1}{2}\right\rfloor\) one bit errors
• and can detect up to \(d-1\) errors (otherwise we can’t get up to a minimum size codeword

minimum length

what is the largest \(M\) we can have for a code with each codeword of length \(L\) and minimum inter-codeword \(d_{c} = 3\).

Convolutional Code

A Convolutional Code, at each time \(j\), takes bit \(b_{j}\) and output two bits \(p_{2j-1}\), \(p_{2j}\), by using \(b_{j}\) and the previous two its \(b_{j-1}\) and \(b_{j-2}\).

Working Example

Consider that if you have \(k\) bits to communicate to the receiver:

\begin{equation} b_1, b_2, \dots, b_{k} \end{equation}

Codewords/Output sequence:

\begin{equation} p_1, p_{2}, \dots \end{equation}

Let we have some sequence of \(k\) input bits

\begin{equation} j = 1, \dots, k \end{equation}

Convolutional Code

The output sequences of Convolutional Code behaves like a five bit difference in Hamming Distance.

Decoding

• brute force decoding: precompute, for a sequence length of \(k\), compute \(2^{k}\) sequneces and what they should correspond to in our target code — of course, this is not computationally feasible
• Virtirbi Algorithm: time complexity — \(4(k+2)\), \(8(k+2)\) 4 possible blocks of 2, and 8 comparisons for Hamming Distance: in general \(k_0\)

in general

• \(k_0\) of source symbols entering the decoder
• \(n_0\) of symbols produced by decoder at each step
• constrain length \(m_0\), how many bits are considered

for the Convolutional Code setup we discussed, we have \(k_0=1\), \(n_0=2\), and \(m_0 = 3\) (one bit produces 2 bits, and we consider 3 bits per step.)