11

Hilbert envelope, also called Energy-Time Curve (ETC), only works well for narrow-band fluctuations. Producing an analytic signal, of which you later take the absolute value, is a linear operation, so it treats all frequencies of your signal equally. If you give it a pure sine wave, it will indeed return to you a straight line. When you give it white noise ...


6

The baseband QAM signal is complex, and the only way to draw it is by doing two drawings, one for the in-phase (real) component, and one for the quadrature (imaginary) component. The passband QAM signal, though, is real, and it is a pulse-shaped carrier whose amplitude and phase depend on each symbol. Myself, I would "draw" it using Matlab or some other ...


6

Well of course you can apply it and many people try. If anyone has been really successful in doing this they probably are sitting now on a nice warm beach sipping Martinis, so this information may not be easy to come by.


5

The financial market data normally available does not comply with the Nyquist sampling criteria. Any information potentially in the data samples available to high frequency traders has to be taken advantage of before it is reflected in the prices, which leads to some profits in making exotic FPGA-based custom signal processing hardware. According to ...


4

At least f2 and f3 seem to work appropriate on a pure chirp signal, but all methods, including f2 and f3 seem to fail horrible, when it comes to more than one frequency in the signal. In reality having more than one frequency in a signal is rather always the case. So how can one get the (more or less) correct instantaneous frequency? as Hilmar suggests, the ...


4

Not really an answer but maybe helpful: Personally I found that the concept of instantaneous frequency is only useful for sufficiently narrow band signals. Consider the simple example of two steady sine waves, say 100Hz and 934Hz. In this case you can certainly define and calculate the instantaneous frequency (in whatever way you want) but what should the ...


4

Here's an attempt to answer your first question: "An example of a 94% overlap transform processing could be demonstrated by a 279 sample frame with 262 samples overlapped from each frame into the next. A spectrogram is really just a sequence of spectra. In this case, each individual spectrum is taken of 279 time samples (which I will call a frame). ...


3

one thing about a non-minimum phase system (with a rational transfer function), is that it can be thought of as the series concatenation (or cascade) of a minimum-phase system, having identical magnitude response as the given non-min-phase filter, with an all-pass filter. the APF will have a poles that cancels specific zeros of the min-phase system that are ...


3

You can choose any mapping that you desire from your unique data sequences to the symbol locations on the QAM constellation. It is common and usually best to Gray-code the mapping such that only one bit changes between adjacent constellation locations (those with minimum Euclidean distance). This way if there is a symbol error due to noise crossing an ...


3

I will not address the question whether or not it is meaningful to compute the instantaneous frequency of a speech signal. Instead I will show you a better method for computing the instantaneous frequency from a given analytic signal. This method avoids the phase unwrapping problem by directly computing the instantaneous frequency from the real and imaginary ...


3

Your problem seems to be finding the analytic signal to a given real signal. What you apparently need is the Hilbert transform. This will allow you to "recreate" the imaginary part of $y(t)$ from the real part alone, which is essentially your measurement $x(t)$. Note that the transform can only be approximated as it is a acausal filter, and only works best ...


3

First answer deleted since I completely misunderstood the question. The problem here is that apparently you are missing one frequency point. From the question it's not clear what is missing. You say you have the points from 0 to N/2. That's actually N/2+1 (if 0 and N/2 are included) and all you need. If that's not the case, there are gew possibilities Go ...


3

This question apparently is not at all about negative frequencies which is the point I addressed earlier in a separate answer but about the signal at the image frequency that occurs as a result of the It appears from the comments by the OP on an earlier answer of mine that his question is about an entirely different problem. The OP apparently creates a ...


3

An actual BPSK signal can be expressed as $$\begin{align} s(t) &= \operatorname{Re}\left[\sum_{n=-\infty}^\infty (-1)^{b_n}p(t-nT)e^{j(2\pi f_c t + \theta)}\right]\\ &= \left[\sum_{n=-\infty}^\infty (-1)^{b_n}p(t-nT)\right]\cos(2\pi f_c t + \theta) \end{align}$$ where $b_n \in {0,1}$ is the $n$-th data bit (note that $(-1)^{b_n} = \pm 1$), $p(t)$ ...


3

[Update: I mentioned a possible +3dB processing gain by including the Hilbert Transform prior to DDC for the case of a real IF signal in the first version of this post, which @MattL questioned so I dug into this further and confirmed that there is no such processing gain, so the only advantage to doing that would be to simplify filtering since it would ...


2

This paper may be of interest: David Vakman, "On the Analytic Signal, the Teager-Kaiser Energy Algorithm, and Other Methods for Defining Amplitude and Frequency." IEEE Trans. Signal Processing. (1996) Summarising from the paper: $$\Psi(u) = a^2w^2 = [u'(t)]^2 - u(t)u''(t)$$ $$\Psi'(u') = a^2w^4 = [u''(t)]^2 - u'(t)u'''(t)$$ where $u(t)$ is the signal, ...


2

As with most things in Engineering the answer is "depends on your requirements". In general you can't do a perfect Hilbert transform unless the signal is strictly periodic (see http://andrewduncan.net/air/ ). Since you can't do perfect you need to define what's good enough for your particular application. For most typically audio applications, that it's ...


2

Your complex baseband signal before modulation is given by $$s(t)=\sum_{m=-\infty}^{\infty}a_mh(t-mT)\tag{1}$$ where $a_m$ are the complex symbols, $h(t)$ is the impulse response of the transmit filter, and $T$ is the symbol period. As you correctly assumed, $h(t)$ is usually real-valued, so you need to filter the real and imaginary parts of your symbols ...


2

This is a wierd thing you are trying to do. What i the purpose behind this signal generation? What particular application needs this? I think of two ways, but maybe one is not correct. 1 - Generate Uniform White noise, and then filter it to your desired bandwith of interest. But that will surely affect the probability distribution of the amplitudes and ...


2

Let me put in an unconnected form things about audio codecs. Audio can be music or speech (discluding ultrasound, sonar etc). Music is wideband and requires high-fidelity. Speech has a lower bandwidth and does not require fidelity but intelligibility. A Codec can be lossy or lossless, the choice depends on the source type, the purpose of application and ...


2

A digital system with complex input and complex output has a block diagram in which the various arrows carry complex signals, each register holds a complex number, each multiplier computes the product of two complex numbers, each adder computes the sum of two complex signals, etc. Consider, for illustration, a simple FIR filter with complex input $x$ and ...


2

The problem in your derivation is the way you discard the high frequency components. Note that $\hat{s}(t)$, the Hilbert transform of the bandpass signal $s(t)$, has frequency components around $\omega_c$ as well as around $-\omega_c$. So you can't just discard one of the two terms $\hat{s}(t)e^{j\omega_ct}$ and $\hat{s}(t)e^{-j\omega_ct}$. The low pass ...


1

instead of calculating angle (from the real and imaginary parts), unwrapping the wrapped angle, and computing a difference, the better way is to compute the difference from the real and imaginary parts directly. this (computing instantaneous frequency from the hilbert transform) is very similar to the same phase unwrapping problem in computing group delay (...


1

Since I don't really know what you're using software-wise, I just whipped up a really short demo in GNU Radio Companion. This uses a PLL in a live processing demo: If you're inclined to play around with it, here is the flow graph ready to be run if you've got a running installation of GNU Radio. The flow graph looks like this: The left part up to the ...


1

Hilbert transform method from your earlier question and Jason R's answer should still work. What you are interested in is extracting the instantaneous phase of your function. For a signal $x(t)$ this is given by $\phi(t) = \angle x_a(t)$ where $x_a(t)$ is the analytic signal of $x(t)$. Digitally you can do this using a Hilbert transform too. Here's a code ...


1

With constant amplitudes and phases, this becomes a simple exercise: First, go to the spectral domain $$ X(f) = \mathcal{F}\{x(t)\}=\sum A_i \frac{1}{2j} (\delta(f-(f_c+f_i)-\delta(f+(f_c+f_i))\exp(j2\pi\tau_if) $$ with $$ \tau_i=\frac{\theta_i}{2\pi (f_c+f_i)} $$ being phase offset translated to the equivalent time shift. Now, you can shift this such ...


1

@user85858302: If designing a complex-valued bandpass filter is the solution to your problem, you can design such a filter without using the Hilbert transform. Start by using your favorite method for designing a real-valued N-tap FIR lowpass filter whose one-sided bandwidth is B Hz. That will give you N real-valued coefficients. Next, multiply those ...


1

Sorry to provide an answer a year after the fact, but I stumbled across this post while searching for articles on this very topic. Your questions reflect the widespread disagreements and interpretations of "instantaneous frequency" that have plagued the field since its inception. Numerous people will tell you, as some of the answers here, that IF is only ...


1

Wow, what a huge question. I'm going to answer the not-so-important question first: And one last, not so important question, why is it that most papers I find on instantaneous frequencies are from the area of geography, especially in calculating seismographic events like earthquakes. Barne's paper also takes that as an example. Isn't the instantaneous ...


1

One counterexample is two wave packets of identical spread and amplitude, well separated in time so that their envelopes do not overlap, and of high enough (differing) well-separated frequencies such that their analytic signals can be considered Gaussian-enveloped complex exponentials. The Hilbert envelope will thus consist of two identical well-separated ...


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