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Since this question has multiple sub-questions in edits, comments on answers, etc., and these have not been addressed, here goes. Matched filters Consider a finite-energy signal $s(t)$ that is the input to a (linear time-invariant BIBO-stable) filter with impulse response $h(t)$, transfer function $H(f)$, and produces the output signal $$y(\tau) = \int_{-... 25 To demodulate a phase-shift keyed signal, of which BPSK is the simplest, you have to recover the carrier frequency, phase, and symbol timing. Bursty Signals Some signals are bursty and provide a known data sequence called a preamble or mid-amble (depending on whether it shows up at the beginning or middle of the burst). Demodulators can use a matched ... 14 This is a very common communications problem. Look in a textbook for "frequency synchronization"; whole books have been written on these and related topics. The technique that you would choose is a function of the specifics of your system. There are two common sources of frequency offset: Differences in frequency between the reference oscillator at the ... 11 ISI, or intersymbol interference, means different things in the context of PSK and OFDM signals. In PSK signals the symbols almost always have tails that extend, in the time-domain, into the times of other symbols. This is what they mean by "intersymbol interference". Unfortunately they have to do this to reduce the bandwidth of the signal. They ... 11 Assuming a channel whose input at each time is a continuous random variable X and its output is Y=X+Z, where Z\sim\mathcal{N}(0,N) and Z is independent of X, then$$C_{\text{CI-AWGN}}=\frac{1}{2}\log_2\left(1+\frac{P}{N}\right)$$is the capacity of the continuous-input channel under the power constraint$$\mathsf{E}X^2\le P$$The mutual information ... 10 This is called biphase mark code, and you have to focus on the zero-crossings instead of the pulse amplitudes. You have multiple zero crossings per pulse, though, because of the low-cut filters inherent to the pickup and the phone's mic input. Yours drop farther than this between transitions, and cross zero: You could restore a more pulsy shape by using a ... 10 I will try to give a relatively simple answer to a complex question. I will not give any exact expressions for the error rates for two reasons. The first I stated in my initial sentence, the second is that from your formulas I can see that you only looked for (or at least found) approximate expressions. Judging whether an expression for the error rate is ... 9 If you have a system that causes 80% of its input power to be lost, then 20% of the input power will remain at the output. The power gain of the network is therefore:$$ G = \frac{P_{out}}{P_{in}} =0.2  G|_{dB} = 10 \log_{10}(G) = 10 \log_{10}(0.2) \approx -7 \text{ dB} $$9 The signal always changes phase at bit transition times. This makes it easier to achieve bit synchronism at the receiver. In plain BPSK, long runs of 0 or 1, with no phase transitions occurring at the boundary between two identical bits, the phase-lock loop that is generating the signal telling the receiver where the bit transitions are, can lose lock; or ... 9 The distortion you are seeing in the first figure is due to phase noise; the further you get from the origin, the further span the constellation will have for a given phase. QAM constellations are not shaped like the second graph as the points in the constellation are not all equi-distant. Having all points in the constellation the same distance from each ... 9 You can also think of delta encoding as linear predictive coding (LPC) where only the prediction residual (x[n]-\hat{x}[n] in @robertbristow-johnson's notation) is stored and the predictor of the current sample is the previous sample. This is a fixed linear predictor (not with arbitrary coefficients optimized to data) that can exactly predict constant ... 8 You've got a pretty good set of circumstances here; you should be able to meet your goal without too much trouble. I don't see anything in your description that would eliminate a whole class of modulation (e.g. phase-shift keying, frequency-shift keying, etc.). Some of the factors that would go into the choice of a suitable format would include: The ... 8 All implementation aspects aside, the constellation you propose performs worse than QPSK in an additve white gaussian noise (AWGN) channel. I claim this based on simulations that I have run with Matlab calculating the symbol error rate (SER) as a function of signal-to-noise ratio (SNR). Here is the result: As you can see, for a given SNR, the proposed ... 8 The time span of the zero crossings increases after the final RRC filtering (and the symbol sampling locations converge which is the goal for the benefit of zero ISI but the zero crossing increase in the process is to the detriment of timing recovery!). So if you are using a Gardner TED which is sensitive to this, it is better to have TED prior to RRC ... 8 Phase (or carrier) Recovery for BPSK can be done over the entire sequence using the information from every sample. Here are common approaches to doing Carrier Recovery: Frequency Doubling (squaring): If you square a BPSK signal (multiply it by itself) a strong tone will be created at twice your carrier frequency. The Squaring operation strips the data ... 8 Like other comments, I don't really understand your question. What I am trying to do is to list basic things so that other ones can suggest edit because I find it is much easier to write in answer part :). Please let me know if it makes you less confused. The general communication system is At transmitter, before modulator is digital, modulator is the ... 8 Phase shift keying is linear modulation. Digital phase modulation is not necessarily linear. 8 There are very few m-sequences of any given length with good cross-correlation properties. Their autocorrelation properties are excellent, but the cross-correlation properties are variable. For example, there are 18 m-sequences of period 127 but to have good cross-correlation properties, one must choose a set of no more than \require{cancel}\cancel{\text{... 8 What you need is carrier phase synchronization. This is a complicated topic with many different approaches. The approach that you'll choose could depend on things like: Data-aided versus blind: Does the underlying sequence contain any known data (e.g. a training or sync sequence of some kind) that you can use to divine the phase offset? Or, do you have to ... 7 Regarding you're general question about how symbol sychronization is done in OFDM systems: One of the most popular and frequently used techniques is the transmission of one or several pilot symbols that are known in the receiver. A pilot symbol is a complete OFDM symbol where the value of each subcarrier is predefined and known in transmitter and receiver. ... 7 The filter that you're referring to is called a preselection filter. Its purpose is to filter out everything but the desired signal of interest before mixing to baseband. Unwanted components could include other signals that are nearby in frequency, or just noise that lies outside the desired signal's bandwidth. Preselection can serve multiple purposes: It ... 7 The cyclic prefix is to make sure that any multi-path interference (or other process similar to a linear time-invariant filter in the transmission channel) acts as a circular convolution on the FFT data frame, thus not affecting the orthogonality of the data channels within the FFT bins. It also allow some slop in the receiver's symbol clock. Thus the ... 7 It's important to note that from a practical point of view QAM has two significant advantages: The in-phase and quadrature components are independent PAM signals with \sqrt{M} levels each (where M=2^{2k} is the number of points in the constellation, 2k being the (even) number of bits per symbol). This makes the design of the coder very simple. The ... 7 The capacity formula$$C = 0.5 \log (1+\frac{S}{N}) \tag{1}$$is for discrete time channel. Assuming you have a sequence of data \left\lbrace a_n \right\rbrace to send out, you need an orthonormal waveform set \left\lbrace \phi_n(t) \right\rbrace for modulation. In linear modulation, whom M-ary modution belongs to, \phi_n(t) = \phi(t - nT) where T ... 7 At long distances, your transmit signal loses a lot of power. As a result of that, the SNR at the receiver is possibly relatively low. A low SNR means that you cannot transport many bits per second over that channel (Shannon limit). Not having many bit/s means you can't push much data across, which means you must reduce the account of data in the digitized ... 7 To answer this you need to understand what is a pole and what is a zero of a transfer function. Let's look at a simple 2 poles 2 zeros filter (also called biquad filter) transfer function :$$ H(z) = \frac{b_0+b_1 z^{-1}+b_2 z^{-2}}{1+a_1 z^-1 +a_2 z^{-2}} $$This can be factored as :$$\begin{align} H(z) &= \frac{ b_0 \, (1-q_1 z^{-1})(1-q_2 z^{-1})}{(...

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For a sine wave of amplitude $A$ and frequency $f=1/T$, its energy from $-T/2$ to $T/2$ is $A^2T/2$. Thus the power depends only on amplitude. The energy over one period does depend on frequency, but in the same observation duration, two sines having same $A$ should have the same energy (assuming observation duration is multiple of their periods). In ...

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You are talking of capacity of two different types of channel. In one case, the channel inputs and outputs are discrete in time. At the $i$-th time instant, the received signal is $X_i+N_i$ where $X_i$ is the received symbol of average energy $E$ and $N_i$ is the noise (typically modeled as a sequence of i.i.d. $\mathcal N(0,\sigma^2)$ random variables)....

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The sinusoid test signal is given by $$s(t) = V_0\cos(2\pi f_0)$$ and its Fourier transform by $$S(f) = \frac{V_0}{2}\big[\delta(f-f_0) + \delta(f+f_0)\big].$$ Inserted in $S_\mathrm{AM-DSB-C}$ it yields  S_\mathrm{AM-DSB-C}(f) = \frac{AV_0}{4}\big[\delta(f-f_c-f_0) + \delta(f-f_c+f_0) + \delta(f+f_c-f_0) + \delta(f+f_c+f_0) \big] + \frac {Ac}{2} \big[\...

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The OFDM signal as a whole is affected by frequency selective filtering. It is usually designed such that the subcarrier bandwidth is smaller than the channel coherence bandwidth. This yields you frequency flat fading for each subcarrier which can be described by a single complex multiplication and equalization can equally done with just one tap. ...

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