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The canonical baseband version of BPSK is that we wish to transmit a sequence $\{a_k\}$ of bits ($0$ or $1$) at a rate of $1$ bit every $T$ seconds using a pulse $g(t)$ of duration $T$ and so the transmitted baseband signal is $$s_{\text{baseband}}(t) = \sum_{k=-\infty}^\infty (-1)^{a_k}g(t-kT).$$ This is a sequence of nonoverlapping pulses of the form $\pm ... 0 According to Wikipedia: The forward transformation is: $$F(x) = \newcommand{\sgn}{\operatorname{sgn}} \sgn(x)\frac{\ln(1+\mu |x|)}{\ln(1+\mu)}$$ Where$F(x)$and$x$both lie between -1 and 1. After this conversion is it necessary to quantize$F(x)$appropriately. The reverse transformation is given by: $$F^{-1}(y) = \sgn(y)\frac{(1+\mu)^{|y|}-1}{\mu}$$ ... 2 In the context of wireless communications, the channel impulse response (CIR) is often estimated indirectly via the time-varying transfer function (TVTF)$H(t, f)$, defined by: $$H(t, f) = \mathcal F_\tau [ h(t, \tau)]$$ where$\mathcal F_\tau [ \cdot ]$denotes the Fourier transform with respect to the$\tau$(lag) variable. For example, the air interface ... 0 I finally derive a solution for my problem. I believe that this explanation is consistent. Let$\mathbf{a} = [a_0,a_1,...,a_N]$is the uncoded bit vector. Without loss of generality, let us assume that the convolutional encoder is of rate$R=1/n$. Therefore, each$a_k$will be encoded to a$\mathbf{c}_k = [c_{0,k},c_{1,k},\dots,c_{n-1,k}]$. Here, the overall ... 3 What you have studied doesn't seem to include a definition of Average Power. For deterministic signal$x(t)$, the instantaneous power delivered at time$t$is$x^2(t){\big (}$yeh, yeh, nitpickers should choose$R=1$in their beloved formula$\dfrac{V^2}R\big )$which varies from instant to instant. Thus, the average power delivered by$x(t)$can be ... 2 is there any difference between DC power and average power Yes. DC power is the square the of the mean of the signal . Average power is the mean of the square of the signal. It looks your teacher is using poor terminology. In this case "average power" and "total power" are the same. Neither one is a great term. 3 IEEE ICASSP is probably still the largest international conference on communications and signal processing. IEEE Transactions on Signal Processing and Communications are well-regarded journals. 1 Your questions are quite broad and hence difficult to answer. If the communication system can afford the bandwidth for a pilot signal, then a pilot can be used. Yes, channel estimation would be needed. Typically, SNRs would be calculated in the digital domain. However, these answers won't help you and you'll need to delve into the theory more deeply. If you ... 2 A DC offset in the frequency domain will be an impulse at$f=0$. The interpretation of this is a single tone interference: "DC" is no different spectrally than any other frequency when viewed as the Fourier Transform of a complex signal with positive and negative frequencies as the baseband spectrum centered about$f=0$is identical to that ... 2 I think you misunderstand what the book is saying. There is nothing like a "unit-energy" that you can compute. There is, however, a "unit-energy [...] pulse", which is a pulse with energy equal to$1$. So if you have some pulse$p(t)$with energy $$E_p=\int_{-\infty}^{\infty}|p(t)|^2dt\tag{1}$$ and you want to normalize it such that its ... 1 Each IFFT complex result bin represents an amplitude and a phase for some frequency. You can use the sum of the a sine and a cosine in the appropriate ratios to produce a sinusoid of any phase (see trig identity). The IQ inphase and quadrature modulators produce the sine and cosine in the appropriate ratios to produce the transmit phase as needed. This ... 1 Because the output of the IFFT operation is a vector of complex numbers. An RF signal is a real thing, so you can't just multiply a sine wave by a complex number and get something sensible. However, you can map a complex number onto an RF carrier by multiplying the real part by$\cos \omega t$, and the imaginary part by$\sin \omega t$. This is I/Q ... 0 I can also not really understand what your question is exactly. In terms of I and Q component, you can simply think of a complex receive symbol, where the real part corresponds to the I and the imaginary part corresponds to the Q value. However, if you want to demodulate a signal (e.g. 4QAM) you have to use a defined mapping from the modulator to demodulate ... 0 The script below is my solution for my case. I hope that it could be useful for anyone facing the same issue. The code can be directly pasted in a "Embedded Python Block". I wrote some comments to explain the mechanism. I hope you find well. Any kind of improvement is welcome! import numpy as np from gnuradio import gr import pmt class ... 2$E_s$would be the energy of the channel symbols at the output of the matched filter as scaled by the path loss between transmitter and receiver as long as all timing and carrier offsets have been corrected for and channel equalization completed - consider how after matched filtering with carrier and timing offsets removed we use just one sample per symbol ... 2 Did Dialup Modem use closed or open loop power/volume control? Yes. How did they determine Tx level? "Dialup Modem" is a very big term that spans > 50 years of technological development. Early modems for landline telephone very pretty plain binary FSK. So, there wasn't much use for any fine-grained volume control. V.90 modems sense the ... 1 Q1: no, not in general. If the sum of multiple sequences is one of the sequences that are in your codebook$x^i\$ (or a multiple), then this will become impossible to resolve. The sequences need to be mutually orthogonal, or the channels will "crosstalk". Look up DSSS, this is exactly what you're doing. Q2: because OOK doesn't use the channel half ...