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Absolutely! Conjugates are mentioned in textbooks because conjugation has no effect on real signals, but it does on complex ones. This way, formulations are more general and apply to both real and complex valued signals. Complex numbers don't exist themselves, they are a mathematical construct. Having said that, their mathematical properties can be ...

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How can a signal be imaginary? It's not imaginary in the everyday sense of meaning something that doesn't exist. "Real" and "imaginary" have technical meanings in signal processing, and more widely in mathematics, that are different from their everyday meanings. In signal processing "real" denotes a signal component that is in ...

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Software-defined radio (SDR) models real band-pass signals as complex baseband signals. All signals and filters operate on complex numbers.

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As mentioned in the comments you can just take the integral of the squared magnitude. Notice that the magnitude of $|e^{a+bj}| = |e^a||e^{bj}| = e^a$ for $a,b \in \mathbb{R}$, comparing with your function we see that $|x_i(t)|=1$. We conclude that $x_i(t)$ is a power signal, not an energy signal, i.e. you must define a time interval to integrate. Then the ...

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Depending on the context, the use of the complex form could be for mathematical convenience or for a no-kidding need for both real and imaginary parts. When you factor the expression, you get $$u(t) = e^{{\eta}t^2}e^{j{\beta}t^2}$$ Where the first exponential is a generic magnitude envelope, in this case Gaussian. The second exponential is the chirp itself ...

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To demodulate BPSK and QPSK as described by the OP is to simply take the sign of the waveform (decision), which has trivial computational complexity. The real complexity lies in everything that occurs prior to decision in a practical receiver: There is typically an equalizer to remove multi-path distortion, a matched filter to optimize the signal to noise ...

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$i$ is the symbol for $\sqrt{-1}$ There is a very important formula called Euler's Equation. $$e^{i\theta}=\cos(\theta) + i \sin(\theta) = (e^i)^\theta$$ "$e^i$" is a point on the unit circle one radian along the circumference. Any point on the unit circle raised to a power will stay on the unit circle and its distance along the circumference ...

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and then discard the phase data What's gone is gone. You can't reconstruct the phase of the original data unless you have some additional information. in which the Hilbert transform is applied to this magnitude-only data to estimate the analytic signal, from which instantaneous phase is extracted. That's a just mathematical process. You shove data in, and ...

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Assuming that you understand the left-hand side of Eq. $(8.47)$, for understanding the right-hand side you need to know that $-1=e^{j\pi}$, and that $e^{j2k\pi}=1$. So in order to obtain all $2N$ roots of $(-1)^{\frac{1}{2N}}$ you rewrite $-1$ as $$-1=e^{j\pi}e^{j2\pi k}\tag{1}$$ from which you get $$(-1)^{\frac{1}{2N}}=e^{j\frac{\pi}{2N}}e^{j\frac{2\pi k}{... 2 You can, if you increase phase between samples slowly enough, using unwrap(angle(signal)). "Slowly enough" means the phase doesn't jump by more than 2 \pi; unwrap works by tracking "total phase". Python equivalent implementation here (note you can configure discont there for jumps greater than \pi (or TOL) but it's pointless with ... 2 HINT: It is based on the fact that$$\sin(x) = \frac{e^{jx} - e^{-jx}}{2j}\quad\text{and}\quad j^2 = -1$$2 All the other responses are excellent, especially Envidia's, so not to take away from those but I want to add this very intuitive view that bottom lines it quickly: Consider the spectrums below that start with a real signal (positive and negative frequencies are complex conjugate symmetric). This is what we could measure with a single scope probe (one stream ... 2 The efficiency is in spectral utilization, at the expense of hardware complexity. Spectrum is such an expensive resource and the demand for data so great that this is a compelling trade to make. Although sampling rates can be reduced in the hardware, it takes two real data-paths (and as the OP mentioned 4 real multipliers and two adders to implement a full ... 2 The bandwidth part is somewhat trivial. If you do eg 1000 complex samples/symbols/... per second, that represents twice as much information as 1000 real samples/symbols/... per second. The convenience part depends of what you find convenient, what operations you want to do, and what capabilities the building blocks available to you have. To me, the «complex ... 1 The simple answer is that$$e^{j\omega_0t}\neq e^{j(\omega_0+2\pi)t}\tag{1},\qquad t\notin\mathbb{Z}$$Since t is a real variable, the inequality is true for uncountably many values of t. Equality is only achieved for countably many integer values of t. From (1) it follows that for real t, the function e^{j\omega_0t} is not 2\pi-periodic in \... 1 HINT: You need to go back to the original definition of the Fourier coefficients via the integral, and treat |n|=1 as a special case. Your mistake happened there already (you divided by zero). 1 The line fscanf(sensorInput, "%lf + j*%lf\t", &x[i]); should be fscanf(sensorInput, "%lf,%lf\t", &x[i].real, &x[i].imag); Notice that C++ provides complex type, they support the common complex operations, and provide an io, and they can read complex numbers writen as (1,3) directly file_in &... 1 Here is one aspect that often creates confusion: In many cases, complex signals are used solely as a mathematical convenience, not because the complex representation is required by anything physical. A good example is circuit analysis in electronics: all physical quantities (voltages, currents, impedances) are real and you can do the analysis with these ... 1 Phasors are useful for the analysis of (real-valued) linear time-invariant (LTI) systems. A phasor is a complex number$$C=|C|e^{j\phi}=A+jB\tag{1}$$which represents a sinusoidal signal:$$x(t)=\textrm{Re}\big\{Ce^{j\omega t}\big\}=A\cos(\omega t)-B\sin(\omega t)=|C|\cos(\omega t+\phi)\tag{2}$$That's already the end of the story. Eq. (2) is the generally ... 1 The method of subtracting the clean signal from the received signal is a good method and will work as long as the scaling is matched up. As an alternative, you can also try the cross-correlation method which takes care the of the amplitude estimate. This previous answer gives the basic idea, How to calculate time domain SNR using known sequence. If x(t) is ... 1 The closed contour C must lie inside the region of convergence, so for the ROC |z|<2 you have no poles inside C for n\ge 0, hence x[n]=0 for n\ge 0. 1 It's because$$ j \cdot ( a + b) = -\frac{a + b}{j} $$which stems from the fact that the imaginary unit j has the property :$$ j = \frac{-1}{j} 

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The bandwidths might be the same. I agree with Jason’s assessment. Is there a difference? Yes. Let’s say we construct both signals such that their frequency domain’s are equivalent from $0-F_s$. For the real signal the frequency domain would be limited to $X(F_s-f) = X^{*}(F_s+f)$. For the complex signal it would be $X(f) = X(F_s+f)$. Can the IQ ...

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The question should be what is the complexity of taking the RF signal and producing a bit estimate for BPSK and QPSK. Read Dan's answer for more information. For each received symbol, there is only one operation that you need to do and it is a comparison. You need to figure out which region the received symbol lies in, and that will give you a bit estimate. ...

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You can't. $y = e^{ix}, x \in \mathbb{R}$ is a many to one relationship , that means it's not uniquely invertible. Angles are periodic in $2\pi$ so there is no reason to. $10000$ and $10000+2*pi$ are the SAME anle.

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Here is an example I already had in Python demonstrating the Wiener-Hopf equations detailed here Compensating Loudspeaker frequency response in an audio signal to solve for the channel equalizer. This application by the OP is ideal for this given the static channel condition as long as deep spectral nulls don't exist, over a iterative LMS or RLS equalizer ...

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Are these Tx chain and Rx chains the correct way to implement this transmission? No. When you want to transmit a real cosine in passband, then you need a single complex tone in baseband. Not a real one. Otherwise, you'll have two real tones on the air, one at the mixing frequency + cosine frequency, and one at the mixing frequency - cosine frequency. But ...

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You should explore how are these complex time-domain symbols transmitted over channel (atmosphere or wire) using modulated waveforms. Also, a good starting point would be to figure out that complex numbers are nothing but 2 orthogonal/perpendicular dimensions. When we say $x = 3 + 3i$, we are basically saying we have a pair of numbers which lie in ...

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Your question makes sense. The difference between a complex network and a regular network with twice the amount of channels, on a mechanical level, is the multiplication operation which ties pairs of channels together. This can be viewed as a restriction of the hypothesis class the network can express compared to a real-valued network with the same parameter ...

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The FFT probably had to be ran on a square image that was padded out, the inverse has returned the padded out image.

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