# Tag Info

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 ...

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 ...

8

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 ...

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 ...

7

Using a sinusoid is a bad idea. In order to precisely detect the time of arrival, you want a signal with wide bandwidth. Here's a brief description of a good approach: Transmit a known direct-sequence spread spectrum (DSSS) waveform (typically a phase-shift-keyed digital signal, either BPSK or QPSK). You should pick a spreading code that has good ...

5

This is a behaviour that is commonly outlined in some textbooks and tutorials on FEC, but usually in the form of an observation. For instance, Turbo Codes are sometimes characterised by their Threshold $E_b/N_o$, which can be considered as the minimum SNR per bit from where the BER starts to decrease. I have not seen a formal explanation for this property, ...

5

Two successive symbols in the demodulator are $Z_1 = (X_1,Y_1)$ and $Z_2 =(X_2,Y_2)$ where $X$ is the output of the I branch and $Y$ the output of the Q branch of the receiver. The hard-decision DBPSK decision device considers the question: Is the new symbol $Z_2$ closer to the old symbol $Z_1$ or to the negative $-Z_1$ of the old symbol? and thus ...

5

A PLL on its own will not work on the direct PSK modulation, assuming the symbols are equi-probable as that results in a nulled carrier, so there is nothing the PLL can track! Costas-Loops are effective for BPSK and QPSK implementations, and as Dilip had suggested, for BPSK you can square your signal and then use a PLL to lock onto the 2F frequency that ...

5

An "integrate and dump" filter is a common approach for this application, which is essentially an averaging filter averaged optimally over your symbol period. To implement this, you simply reset your accumulator at the symbol boundaries, accumulate your output over the symbol period and then "dump" your result by taking the accumulated output as the result ...

5

The final statement is correct, for PSK with proper pulse shaping the baud rate and the bandwidth are the same (the bandwidth will typically be 20-30% higher than the symbol rate, but read on). The baud rate is the symbol rate or the rate of change from one symbol to the next in the modulated constellation. Your earlier figures suggest unfiltered waveforms ...

5

These stuffs $E_s/N_0, E_b/N_0 \textrm{ and SNR}$ are convertible. \begin{align} E_s/N_0 &= E_b/N_0 + 10\log_{10}(k) \\ E_s/N_0 &= 10\log_{10}(T_{sym}/T_{samp}) + \mathrm{SNR} \end{align} where $k$ is the number of information bits per symbol, $T_{sym}$ is the signal's symbol period and $T_{samp}$ is the signal's sampling period. More details can ...

4

The mixer/LPF filter that you describe as the receiver is typically one that is designed (possibly for optimal performance e.g. matched filter) using a mathematical model that assumes its input is the desired signal plus noise (often AWGN) and nothing else. If other signals are present, they are assumed to be orthogonal to the signal of interest (as in CDMA ...

4

Well, I took a look at your code, and spent a lot of time on it, and I have discovered some mistakes, some practicals and some theoretics. Here are my answers: (1) You can use the function filter. Just remember that it returns a vector of the same length of the input, while conv returns the $N_x + N_h - 1$, where $N_x$ is the length of your input and $... 4 The energy per bit$E_b$means one thing and one thing only: how many joules is the transmitter spending, on average, per information bit transmitted. Note that the channel has absolutely nothing to do with$E_b$. However, your question involves the SNR too. In a channel with non-unit gain, you need to decide where you'll measure the SNR. You can define ... 3 Zero-mean noise by itself can't modify the decision boundary. However, the number of things that can go wrong in your system is large. Is your channel flat or frequency-selective? You can think of the channel as the sound card and its driver, the loudspeaker, the air, the microphone, and the receiver's sound card and its driver. In my experience, there's ... 3 OBPSK in the way that you describe it is also called$\pi/2$-BPSK meaning that the signal constellation is different from bit to bit, rotating by$\pi/2$after it is used. There are two variations of this format. Use$\pm \cos(\omega_0 t)$to transmit one bit and$\mp \sin(\omega_0 t)$to transmit the next bit. Repeat this over and over again. Note the$\...

3

This problem is typically solved by preceding your actual payload data with a synchronization pattern, or sync word. This is a sequence of symbols that are known to the receiver ahead of time and are used to aid synchronization with the payload data frame. It could be detected in a couple ways: Sync pattern matched filtering: When looking for the ...

3

This question apparently is not at all about negative frequencies which is the point I addressed earlier in a separate answer but about the signal at the image frequency that occurs as a result of the It appears from the comments by the OP on an earlier answer of mine that his question is about an entirely different problem. The OP apparently creates a ...

3

An actual BPSK signal can be expressed as \begin{align} s(t) &= \operatorname{Re}\left[\sum_{n=-\infty}^\infty (-1)^{b_n}p(t-nT)e^{j(2\pi f_c t + \theta)}\right]\\ &= \left[\sum_{n=-\infty}^\infty (-1)^{b_n}p(t-nT)\right]\cos(2\pi f_c t + \theta) \end{align} where $b_n \in {0,1}$ is the $n$-th data bit (note that $(-1)^{b_n} = \pm 1$), $p(t)$ ...

3

While I do not have proof, it seems pretty clear, given the "Characteristics of Compass signals reported as of May 2008 compared to GPS-L1CA" table at the Wikipedia page that you offered as an example, that the "(n)" notation refers to the chip rate. You will notice that there is one BPSK(1) signal that has a chip rate of 1.023 Mchips/s. There are four ...

3

There are three reasons to avoid ISI through pulse shaping rather than correcting it via channel correction methods. As Bruce pointed out, it is a simpler solution and requires fewer computations. Channel correction usually amplifies the received noise. Received noise will corrupt the channel model, making the ISI removal imperfect to some degree.

3

The asterisk refers to a complex conjugate. One typical method for soft decoding of differential modulations is the delay, conjugate, multiply technique: $$S_i = D_i D_{i-1}^*$$ where $D_i$ and $D_{i-1}$ are two consecutive differentially-encoded symbols and $S_i$ is the differentially-decoded result. This general formula will work for DBPSK or DQPSK (...

3

A random sequence of +1 and -1 is not really a BPSK signal; it's just a sequence of bits. As such, its FFT has not much meaning, at least not in the sense you're thinking of. To create an actual BPSK signal, first denote your sequence of $N-1$ bits by $a_k$, where $k=0,\ldots,N-1$ is an integer. Second, choose a bit rate $R_p=1/T_p$. Then, choose a pulse ...

3

The energy per bit, $E_b$, is independent of the coding rate. Note that $E_b$ measures the energy per transmitted information bit, not per transmitted symbol. Let's say you're willing to spend one joule per information bit, so $E_b=1$. You use uncoded BPSK, so that each transmitted symbol carries one bit of information and so it also has energy one. Let us ...

3

I have a PRN generator that I have validated with live captured signals that is available on the Mathworks Exchange site at this address and equally runs in Octave (Update: I also pasted the core of this in a code block below): https://www.mathworks.com/matlabcentral/fileexchange/14670-gps-c-a-code-generator The two tap coder is as given in the diagram in ...

2

1) A typical approach would be to have a frame alignment word (and by "word" I mean bit sequence) at the beginning of every frame. You could time synchronize by looking for that FAW through cross-correlations. Barker codes are a good choice because they have good autocorrelation properties. The problem with the approach is that the data, being random, can ...

2

This is the analytic signal. See wikipedia for better details. Basically you are taking the carrier correlation, $f(x)$, and its Hilbert transform, $(h \ast f)(x)$ and combining them into a single complex valued signal, the analytic signal $$f_A(x) = f(x) - i (h \ast f)(x)$$ which can be written as the complex exponential $$f_A(x) = A(x) e^{i \phi(x)}$$ ...

2

The reason cyclic prefix is used in OFDM systems is to avoid complex equalizers in receivers. Cyclic prefix converts linear convolution of fading channel(h) & Tx data(x) into circular convolution. Without cyclic prefix: symbol at receiver y = h*x; *- linear convolution Here is a useful link for your question.

2

In BPSK, BER is equal to $Q(\sqrt{2E_b/N_0})$, so you need $E_b/N_0>11.3$ to have a BER no larger than $10^{-6}$. Since the bandwidth is 200 kHz, the maximum rate is 400 kb/s. So, since $SNR\approx136$, $$\frac{Eb}{N_0}=\frac{S}{N}\frac{W}{R}=136\frac{200\cdot10^3}{400\cdot10^3}=\frac{136}{2}=68$$ Since $68>11.3$, you can be sure that you can ...

2

I figured it out. It turned out that I don't really need that 'Phase Accumulator', and I should reverse the control signal in the NCO. So, I use -PhErr(i-1) as a control signal to NCO. Here is the modified code: % Siraj Muhammad % 25/3/2015 % BPSK Demodulator %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% load RRC.mat fc = 0.05000001; phase_offset = pi/7; N = length(...

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