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Signal Processing

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    $\begingroup$ Great answer. There's a significant portion of a digital communication theory course shoved into one package. $\endgroup$ Commented Jun 2, 2013 at 12:39
  • $\begingroup$ Thank you for the great answer! I have a question regarding the graph of the matched filter output. What's the output of that graph (I'm trying to hand draw it)? I'm assuming that it's $y(t) = r(t) * r(-t)$ where $y(t)$ is the output signal (third graph) and $r(t)$ is the received signal (first graph). If I correlate the received signal with a box function tall enough to pass encompass both the logic level 1 and 0 states, then I get that output---the only thing is, there is never a 100% match between the correlated signal and the correlation signal, perhaps that is to be expected. $\endgroup$ Commented Jun 7, 2013 at 6:46
  • $\begingroup$ With $p(t)$ denoting a rectangular pulse lasting from $0$ to $T$, the received signal is $r(t)=\sum_{n=0}^\infty (-1)^{b_n}p(t-nT)$, the matched filter has impulse response $h(t)=p(-t)$ and the filter output is $$(r\star h)(t)=\int_{-\infty}^\infty r(u)h(t-u)\,du=\int_{-\infty}^\infty r(u)p(u-t)\,du=\int_{t-T}^t r(u)\,du,$$ that is, the matched filter output at any time $t$ is the integral of the received signal (plus noise, of course) over the past $T$ seconds. Note that except at the sampling instants, the filter output has receiver-induced intersymbol interference (ISI) in it! (continued) $\endgroup$ Commented Jun 7, 2013 at 10:57
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    $\begingroup$ Isn't SNR defined as $\dfrac{\int s^2(t) dt}{N_0/2}=\dfrac{2E}{N_0}$ rather than $\sqrt{\dfrac{2E}{N_0}}$? $\endgroup$ Commented Jan 4, 2017 at 16:30
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    $\begingroup$ @Mostafa SNR can be defined in any way that you want. What is important is: What is the function $g$ such that $BER = g(SNR)$, and Is $g$ a decreasing function of its argument (and what is the rate of decay?). For my definition of SNR, $g(x) = Q(x)$. For you, $g(x) = Q\left(\sqrt x\right)$. For those who call $\frac{E}{N_0}$ the SNR, it is $g(x) = Q\left(\sqrt{2x}\right)$. So, whatever pleases you.... $\endgroup$ Commented Jan 5, 2017 at 16:58