# Tag Info

18

Basics The amplitude of an IQ signal is just the vector magnitude, $\sqrt{I^2 + Q^2}$. The power of an IQ signal is the squared magnitude, $I^2 + Q^2$. When you see a logarithmic (dB) meter, it is usually measuring the log of the power, i.e. $10 \log_{10}(I^2 + Q^2)$. (This can also be calculated as $20 \log_{10}$ of the amplitude, but unless you already ...

9

if the two audio signals are totally uncorrelated, the the squares of the two crossfade gains should add to 1. so, to prevent that dip, your crossfade function will look more like $\sqrt{\frac{1}{2}(1+t)}$ and $\sqrt{\frac{1}{2}(1-t)}$ rather than $\frac{1}{2}(1+t)$ and $\frac{1}{2}(1-t)$ for $-1 \le t \le +1$. note, for constant power crossfades, at the ...

8

Your code is bit unclear, especially generation of your signal. Python allows for vectorized operations so it is good to use it. What's more, it is good to clearly specify the sampling frequency of your signal and use it then. Also please remember to normalize your FFT by length of your signal (in this particular case) and multiply by 2 (half of spectrum is ...

8

Creating the Frequency Vector The arrangement of the output of fft() depends on whether you use an odd or even number of points for your fft. I think this post nicely summarises how the frequencies are arranged. Have a look at it. Since you are using an even number of points, the Nyquist frequency, $F_N = F_s/2$, is present in the output of your fft, and is ...

8

The OP's opening statement is incorrect: $f_s > f_{max}/2$ prevents frequency aliasing for a bandlimited signal, but not amplitude aliasing $f_s > 2 f_{max}$ prevents aliasing. It's as simple as that. There is no such distinction as "amplitude aliasing". Since the OP has stated the signal is band-limited; as long as we can assume that means ...

7

An exponentially decaying envelope $a\exp(-b x)$ is a good choice, and is used for example in vintage Yamaha FM synthesizers. It has the favorable property that over any constant length time interval, by the end of the interval the envelope has decayed to a constant fraction of what it was at the beginning of the interval. Damped oscillation (with some ...

6

I am thinking of two numbers. They add to the number 15. Tell me what the two numbers are.

5

In $2^{2n}$-QAM with a square constellation, there are $4$ "corner" points and $4(2^n-2)$ "edge" points, and $(2^n-2)^2$ "interior" points. The conditional symbol error probabilities given that each type of point is transmitted, are \begin{align} P_e(\text{corner}) &= 2Q(x) - Q^2(x)\\ P_e(\text{edge}) &= 3Q(x) - 2Q^2(... 5 The correct way to group multiple bins together is to multiply each complex fft bin output by its complex conjugate (which gives the bin power) then add all the bin powers together and divide by the number of bins in the group. If you want to display in db (which is the conventional approach) then take 10*log10() of the result. Negative db values are normal ... 5 This is a continuation of the (music-dsp) thread started by Element Green titled: "Algorithms for finding seamless loops in audio" As far as I know, it is not published anywhere, although I have recently been informed that a similarly-themed paper by Marco Fink, Martin Holters, and Udo Zölzer has been presented at the 2016 DAFx conference, which is a while ... 5 The usual way to estimate the amplitude of a particular frequency is to use the Goertzel algorithm. There is a good write-up by Rick Lyons here. Even though Rick's writeup is about single tone detection, it can be applied when multiple tones are present, too. 5 Both amplitude and frequency modulated radio signals nowadays use quadrature modulation and demodulation as a mean to transfer and receive radio signals. This question is ill-defined – a signal doesn't use quadrature modulation/demodulation, transmitters/receivers do. So, your question is, if we try to "rescue" it, is Do modern FM receivers/... 4 Generally, the frequency estimation problem decouples from the amplitude + phase estimation problem. As I said in the comments, you could just do: \hat{A} = \left|\sum_{n=0}^{N-1} x_n e^{-j 2\pi\hat{f} n} \right| $$where \hat{f} is your frequency estimate. Regarding your issues: because the number of points isn't high enough to hit the right ... 4 Regarding your Edit #2: It is not true that phase estimation requires that frequency be at a bin center. However if a frequency estimate requires interpolation between FFT result bins, so does any phase estimation at that frequency estimate. Using a Sinc interpolation kernel on the complex FFT results, plus successive approximation for the magnitude peak,... 4 Not really, as the transform is real. However, one could interpret the sign as a poor man's phase, being "quantized" or restricted to values 0 or \pi. In other words, 1 = 1.e^{0.\imath} and -1 = 1.e^{\pi.\imath}. [EDIT] There are some instances where people use the sign of DCT or (real) wavelet coefficients, for subpixel image registration or ... 4 A DCT is equivalent to a DFT of real data that is doubled and mirrored, thus rendering it symmetric. The DFT of any symmetric real signal has a phase of zero (its all cosines, no antisymmetric sine components). 4 Know that each bin in the FFT is the value for the exponential e^{j\omega t}, not sine or cosine. So for the OP's case 5sin(2\pi 3 x-2) in terms of exponentials (see Euler's identity) this is:$$\frac{5}{2j}e^{j(6\pi x -2)} - \frac{5}{2j}e^{-j(6\pi x -2)}$$Each is a DFT bin with that magnitude and phase given by the exponentials above, as long as ... 4 The claim is wrong. Sampling of a pure sinusodial whose frequency is below but arbitrarily close to the Nyquist frequency (half the sampling frequency) is a perfectly valid operation, as long as you can create ideal (zero width transition band) brickwall lowpass filters to be used at the reconstruction interpolation of the continuous waveform from its ... 4 In order to resample without aliasing, the resampling process needs to apply an anti aliasing filter at the new Nyquist Frequency (or thereabouts). So chances are, your resampling process applied a low pass filter at 256 Hz or so and that removed A LOT of the energy. 3 As people in the other answer have been asking for a code snippet (and as I don't have enough reputation to comment on answers), here's the other answer implemented in C# (which should be easy enough to convert to other languages) /* returns a float array with two indexes representing the volumes of the left (index 0) and right (index 1) channels when t is -... 3 Maybe you have a look at Goertzels Algorithm. 3 I don't see the question here. But you're right. On the DFT map of the image, the magnitude / amplitude, etc. are powers of the length of the vector of the complex number. On more useful visualization to help working with large Dynamic Range is using the log(1 + abs()) on the DFT. 3 Try adding a gain of 16 before your low pass filter, or equivalently using a low pass filter with a passband gain of 16 instead of 1. 3 That's correct, 0.4053 is the approximate magnitude of the frequency response of linear interpolation at the Nyquist frequency. With the definition:$$\operatorname{sinc}(x) = \frac{\sin(\pi x)}{\pi x}$$the exact value is:$$\operatorname{sinc}^2\left(\frac{1}{2}\right)= \left(\frac{\sin\left(\frac{\pi}{2}\right)}{\frac{\pi}{2}}\right)^2 = \left(\frac{...

3

There is no issue of frequency resolution in a typical modulation and demodulation scheme. (unless you want to implement some frequency/signal detectors) Putting two message signals in adjacent channels requires that their spectrums do not overlap. In your case you should be able to specify the associated bandwidths of those two slowly changing analogue ...

3

For baseband real signals the frequency spectrum is conjugate symmetric; i.e., $$X(f) = X(-f)^*$$ This translates to an even magnitude spectrum; i.e., $$|X(f)| = |X(-f)|$$ and an odd phase spectrum. Because of the symmetry it's also called as the double side band spectrum in commnnications terminology. For the real basband signal, a single side band is also ...

3

I'm wondering how can i get the frequency and the amplitude from a 16 bits sample That makes no sense. A single sample is just a number. A number doesn't have a frequency. The number itself is the amplitude.

3

$2^k$-ary PAM maps $k$ bits (which can be expressed as integers in the range $[0, 2^k-1]$ to $2^k$ levels that are equally-spaced and symmetric about $0$. ($0$ is NOT one of the levels).. The levels are odd integers, and so for $k=2$, the levels are $-3, -1, +1, +3$ which are equally spaced and symmetric about $0$. As to which integer in $[0,2^k-1]$, ...

3

To extract only the carrier with a band-pass filter, the filter pass-band has to be narrower than the modulating frequencies, or else some of the envelope modulation will "leak" through the filter. Also note that a very narrow filter has a long settling time, so you will need a long enough test signal.

3

To a first approximation, a plucked string has an exponential decay, so your envelope will look like $$x(t) = a \exp -bt.$$ The physical explanation for this is simple: when you pluck the string, you put a certain amount of energy into the system. Over time, energy is lost. At any given time, the amount of energy lost is proportional to the amount of energy ...

Only top voted, non community-wiki answers of a minimum length are eligible