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 error-correction code an error-correction code is a collection of binary strings called “codewords” size of the code: numer of codewords M (M=2 for repetition codes), we want this to be large because — the number of bits that the code can encode is \log_{2} M, because that’s the number of combinations of a binary bit \log_{2} M which you can encode with a thing of size M length of the code: the length of each codeword L (M =3 for repetition codes), we want this to be small to be efficient minimum distance of the code: the minimum distance between any two codewords in the code d_{c} (measure by Hamming Distance, repetition codes its 3), we want this high minimum codeword size Key question: *assume we want to correct t bit flips; what is the minimum distance d_{c} we will need? Note that if we had t bit flips, we would have a Hamming Distance of t bitflips. So, we will need at least d_{c} = 2t + 1 away to disambiguate exactly (if you have a distance of just t+1, you will not be able to tell a code that’s exactly in line with your error, t+1 away, if your code error had a Hamming Distance of t). formally: an error correction code can correct t errors its minimum distance is d_{c} \geq 2t+1; meaning if you have a code error d_{c}, we can correct up to t = \left\lfloor \frac{d_{c}-1}{2}\right\rfloor a code can detect t errors d_{c} \geq t+1, meaning we can’t accidentally end up with a random codeword linear code In a linear code, if you XOR two codes, you will get another codeword. rate (error correction code) \begin{equation} R = \frac{\log_{2} M}{L} \end{equation} We want the rate to be as high as possible. repetition code Literally send it multiple times Message Code 0 0 0 0 1 1 1 1 Then, you look at the three received bits and try to see what the most nearest legitimate codeword is. By doing nearest neighbor (closest value) decoding, we