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12

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 ...


9

You mention in a comment that your target platform is a custom IC. That makes the optimization very different from trying to optimize for an already existing CPU. On a custom IC (and to a lesser extent, on an FPGA), we can take full advantage of the simplicity of bitwise operations. In addition, to reduce the area it is not only important to reduce the ...


9

PROLOGUE My answer to this question is in two parts since it is so long and there is a natural cleavage. This answer can be seen as the main body and the other answer as appendices. Consider it a rough draft for a future blog article. Answer 1 * Prologue (you are here) * Latest Results * Source code listing * Mathematical justification for ...


8

Given two complex numbers $z_1=a_1+jb_1$ and $z_2=a_2+jb_2$ you want to check the validity of $$a_1^2+b_1^2>a_2^2+b_2^2\tag{1}$$ This is equivalent to $$(a_1+a_2)(a_1-a_2)+(b_1+b_2)(b_1-b_2)>0\tag{2}$$ I've also seen this inequality in Cedron Dawg's answer but I'm not sure how it is used in his code. Note that the validity of $(2)$ can be checked ...


8

1. Logarithms and exponents to avoid multiplication To completely avoid multiplication, you could use $\log$ and $\exp$ tables and calculate: $$I^2 + Q^2 = \exp\!\big(2\log(I)\big) + \exp\!\big(2\log(Q)\big).\tag{1}$$ Because $\log(0) = -\infty,$ you'd need to check for and calculate such special cases separately. If for some reason accessing the $\exp$ ...


8

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 ...


6

I'm putting this as a separate answer because my other answer is already too long, and this is an independent topic but still very pertinent to the OP question. Please start with the other answer. A lot of fuss has been made about the effectiveness of the initial "early-out" test, which I have been calling the Primary Determination. The base approach ...


5

Schwarz Inequality for continuous-time Complex valued functions is given as follows: $$\left|\int^{\infty}_{-\infty}f(t)^* \cdot g(t) dt \right|^2 \le \int^{\infty}_{-\infty}\left|f(t)\right|^2dt \cdot \int^{\infty}_{-\infty}\left|g(t)\right|^2dt$$ As you can see that, on left hand side magnitude is taken after integration making it real number and the ...


5

The Sigma Delta Argument Test I came up with my own solution with the premise of resolving maximum vector magnitude (including equality) by testing the angle for quadrature between the sum and difference of the two vectors: For the sum $\Sigma$ and difference $\Delta$ of $z_1$ and $z_2$ given as (which is a 2 point DFT) $\Sigma = z_1 + z_2$ $\Delta = ...


5

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


3

It is simply because each sample of $n(t)$ has a random magnitude and phase by definition given as $n(t) = |n(t)|e^{j\phi(t)}$. With real and imaginary components as follows: $$|n(t)|e^{j\phi(t)} =|n(t)|\cos(phi(t))+j|n(t)|\sin(phi(t)) $$ Real: $I(t) = |n(t)|\cos(\phi(t))$ Imag: $Q(t) = |n(t)|\sin(\phi(t))$ Since the phase and magnitude are independent ...


3

This is an unprecedented (for me) third answer to a question. It is a completely new answer so it does not belong in the other two. Dan (in question): max(I,Q) + min(I,Q)/2 Laurent Duval (in question comments): 0.96a + 0.4b a concerned citizen (in question comments): |a1| + |b1| > |a2| + |b2| By convention, I am going to use $(x,y)$ as the point ...


3

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 ...


3

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 ...


2

This can be implemented, without needing to access the real and imaginary parts of $z$ individually, by writing $$\operatorname{Re}(z) + \operatorname{Im}(z) = \frac{(1 + i)(z^* - iz)}{2},$$ where $i$ is the imaginary unit and $z^*$ is complex conjugation. Factoring out a complex constant, $\xi = 1/2 + i/2$, turns this into a slightly tidier expression, $...


2

For $\mu > 0$, $z^L \in \left ( \sqrt[L]{\mu} \right) e^{j 2\pi n/L} \,\forall\, n \in 1 \cdots L $. I.e., the roots are evenly spaced on a circle $\sqrt[L]{\mu}$ in radius, and there's an $L$ of a lot of them.


2

Given a complex time-domain signal is it always implied that this is a QAM-demodulated form of a real time domain signal? No, we generally use complex baseband to describe all kinds of signals, or even passband systems (e.g. frequency-selective or time-variant channels). So, any bandlimited signal has an equivalent complex baseband, and it's not implied ...


2

$c$ need not be real number. For the LHS you would get $$ |\langle s_1,s_2\rangle| = |\langle s_1,cs_1\rangle |=|c|\,|\langle s_1,s_1\rangle |= |c|\|s_1\|^2 $$ For RHS, $$ \|s_1\| \times \|c s_1\| =\|s_1\| \times |c|\|s_1\| = |c|\|s_1\|^2 $$ So $|c|=|1+i|=\sqrt{2}$


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

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

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 ...


2

$i$ 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 ...


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 ...


1

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 ...


1

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 ...


1

While reading the above few answers, it’s been stated like negative frequency has no physical significance, but it is completely wrong. Fourier transform is used to pull out - that component of a signal (under test)which is having a particular repetition frequency( or rotation frequency in that sense). If you agree to the fact that complex exponential ...


1

That is a very poorly written article IMO. In simple rule of thumb terms: For real valued tones, there isn't a qualitative distinction between negative and positive frequencies, so there is no point in using negative frequencies. Choosing not to use them by convention is quite different than they don't exist. For complex valued tones, the negative and ...


1

I think you may be confused about the use of negative frequencies and the sign of the integration coefficient. Let's take a look at the inverse Fourier Transform $$ x(t) = \int_{-\infty}^{+\infty}F(\omega)\cdot e^{ 2 \pi i \omega} d \omega $$ That basically means that you can construct any time domain signal as the from a set of complex exponential. So ...


1

The answer about $c$ being complex has be given already. I would add information on "How is inequality established for complex signals?" Indeed, the inegality is very general, and works in spaces equipped with an inner product (aka "inner-product spaces"). It is a very important inequality, in so many forms, that a whole book is dedicated to it: The Cauchy-...


1

$X(t) = Re\{XX(t)e^{j2\pi ft}\}$. So $XX(t)$ can be any generic complex baseband signal with real I and Q components - $XX(t) = XX_i(t)+j XX_q(t)$ and $e^{j2\pi ft} = \cos(2\pi ft) + j\sin(2\pi ft)$ After the complex multiplication, and taking real part, you get $X(t) = XX_i(t)\cos(2\pi ft) - XX_q(t)\sin(2\pi ft)$. So $XX_i(t)$ and $XX_q(t)$ are indeed ...


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