7

The unit circle on the z-plane represents the frequency axis, similar to the imaginary axis $j\Omega$ on the s-plane for the Laplace Transform in the continuous time case. So the frequency response of the system is given by $H(z)$ when $z= e^{j\omega}$ with $\omega$ going from $0$ to $2\pi$ representing the normalized fractional radian frequency (which is ...


6

Answer : No, any causal LTI system with frequency response $H(f)$ cannot produce the output $y(t)$ in advance. And, the answer lies in the causality of input signal $x(t)$ being applied to $h(t)$. Any causal input $x(t)$ which has an identifiable beginning cannot truly be Narrow-Band or Band-Limited. It will have non-zero frequency content at all frequencies....


6

The reason is Euler's formula, from which you get $$\cos(\omega)=\frac12\big(e^{j\omega}+e^{-j\omega}\big)\tag{1}$$ and $$j\sin(x)=\frac12\big(e^{j\omega}-e^{-j\omega}\big)\tag{2}$$ If you have symmetric or anti-symmetric coefficients, the corresponding frequency response can always be decomposed in purely real-valued cosine terms $(1)$ or purely ...


5

It is the objective of the receiver to make the best estimate for each symbol as to what was transmitted. This is often done by ultimately determining a decision time in each sample (through timing recovery) on the waveform after it has been processed by the receiver (equalization and matched filtering) in which to sample the waveform and make a decision as ...


5

Well this goes to show that Fourier series is just approximation that gets more and more correct when you add more harmonics. Take a look at this: $\dfrac{4\sin\theta}{\pi}$ is just first harmonic. Harmonics are integer multiple of base frequency as you can see: $\sin3\theta$, $\sin5\theta$ etc. And the more of them you add to the first harmonic the more ...


5

So, from the discussion in the comments it's clear you know most you need to know. The window method for FIR filter design is based on this idea: We know the "ideal" frequency response $H(f)$ we want. Often, that's something like a rectangle in frequency design. Well, the easiest thing to achieve that shape would simply be transforming $H$ to time domain,...


5

Yes, the square of the $L_2$ norm of a signal is also by definition its energy $\mathcal{E}_x$. There's nothing surprising, unbelievable, or mysterious in that though? The concept of signal energy : $$ \mathcal{E}_x = \int_{-\infty}^{ \infty } x(t)^2 dt\tag{1} $$ is fundamentally based on the concept of energy (or work) in physics as the Kinetic Energy of a ...


5

From physics, energy is a term often used as a quantitative property. In other words, energy is a quantity that is preserved under some actions, transformations, etc. In signal processing (where physics vanish), this often takes the shape of a sum or an integral of a squared quantity for reals, or its modulus for complex data. We can write it symbolically ...


4

For two non-zero sequences $x_n$ and $h_n$ given that their convolution is zero, At least one of them must be infinite length.


4

The only signal, that really has just one frequency component, is an infinite sine signal. Limiting the signal duration in any way is bound to produce other frequency components, as time limiting can be thought of as multiplying with a rectangle window, which translates to convolution with an si-function in the frequency domain, thus introducing new ...


4

This is a valid question within the context of GSP although, it might not sound like much of a digital signal processing question. To understand that point, you need to understand what the eigenvalues/eigenvectors tell us (about a matrix), then how is the Laplacian constructed (and what it means) and then relate it to DSP. The Eigendecomposition of a ...


4

For traditional signal processing, time and frequency are dual variables, linked through Fourier transformations. This strong linkage yields a balance, sometimes called Heisenberg–Pauli–Weyl uncertainty inequalities. In everyday words, one could not be arbitrarily precise in time location and frequency determination. One cannot determine the frequency of a ...


4

Here's Octave code for a (digital) RRC filter: pkg load signal; % Octave needs this; MatLab doesn't Fs = 16000; % sample rate Rs = 400; % symbol rate sps = Fs/Rs; % samples per symbol % % Root raised cosine pulse filter % https://www.michael-joost.de/rrcfilter.pdf % r = 0.22; % bandwidth factor ntaps = 8 * sps + 1; % Pulse duration is 8 symbols st = [-...


4

The answer can be given without being at all specific to filters. In fact, one has to answer that in terms of engineering in general: Whenever you have alternative solutions, the only measure of quality that really matters, is fulfillment of the requirements. So, you'll have to know what you'll use that filter for, and evaluate how well the alternatives ...


4

You have already read those Oppenheim's Signals & Systems, and Discrete-Time Signal Processing books. I'm not sure what you mean by foundations but in some sense these two are also the foundations on signal processing. In other words, there are no (popular & successful) graduate level DSP books that discuss at an advanced level the same topics that ...


4

Firstly, I am confused if I am supposed to filter my signals to get rid of any frequencies above the Nyquist frequency. My sampling frequency is 32Hz and my time series is somewhat noisy and has some artifacts. I am also unsure of which filter to select for this. That ship has sailed. Let $S=\left\{\left.\alpha e ^{i(\omega t+\varphi)}\right|\alpha > 0, ...


4

I couldn't quite follow your code (you calculate two different versions of the ACF?), but I believe the problem with the plot being shifted toward zero is that xcov calculates the cross covariance not correlation. Remember that cross covariance is subtracting the mean, so there should be a shift! You should be calling xcorr instead. The following code ...


3

@Hustler. Hi. I suggest you use all 48000 vibration samples. As a general rule: In mathematical analysis of measured data it's preferred that you use all available data to estimate some physical quantity. If you perform 48000-point DFTs (discrete Fourier transforms) your first DFT bin frequency will be zero Hz and your DFT bin spacing will be (as jithin said)...


3

It is common in DSP practice to define some convenient center for a filter as being at time 0, even though we cannot build non-causal systems in practice. You see this most when you're designing a symmetrical filter, and you define t = 0 as the filter center, but it happens elsewhere. You do this because it makes the analysis easier, and you justify it by ...


3

Actual channels are always causal (like everything else in the physical universe). Actual (discrete-time) channels also sometimes have one tap that is considerably larger than the rest; an example impulse response would be h = [0.1, 1.5, 0.2]. Some authors prefer to define h[0] as the largest tap; in my example, we'd have h[-1] = 0.1, h[0] = 1.5, and h[1] = ...


3

Let me put a practical answer with the following Matlab / Octave Code : L = 2*1000; % signal sample count n = 0:L-1; % discrete-time index Fs = 44100; % sampling frequency am = [1, 0.5, 2, 0.5, 0.3, 0.6, 0.1, 0.2]; % magnitudes of 8 components fm = [882, 2646, 4410, 6615, 8820, 10000, 13230, 15876]; % frequencies % Time domain ...


3

Consider the case $\ f(t) = 2 \cos(\omega_0 t) = e^{+i \omega_0 t} + e^{-i \omega_0 t}.\ $ Then $$ F(\omega) = \int\limits_{-\infty}^{+\infty} e^{i (-\omega + \omega_0) t} \ dt + \int\limits_{-\infty}^{+\infty} e^{i (-\omega - \omega_0) t} \ dt\\ $$ When $|\omega| \ne |\omega_0|$, both integrands oscillate around zero, and the integrals are effectively zero....


3

This seems like more of a semantics problem. A signal is periodic with time $T$ if $$x(t+n\cdot T) = x(t), n \in \mathbb{Z}$$ So the signal is periodic in $0.5$ since the for $T = 0.5 \cdot n$ the argument of the cosine is an integer multiple of $2 \pi$. Since it's periodic in $0.5$ it's also periodic in all integer multiples of $0.5$, i.e $1$, $1.5$, $2$...


3

Here is a actual example with negative group delay that will provide further insight: Below is a plot of the output and input of a pulse through a realizable filter that has negative group delay: It seems like a complete violation of causality, but it is just a clever DSP magic trick. Let's explore further: The filter above that did this had the following ...


3

If you know the frequency (and it seems that you do), simply construct a basis set of vectors from 1) DC (all ones) 2) Half a Sine wave 3) Half a Cosine wave These form an orthogonal basis just like a DFT. In fact you can think of it as a "one half bin and DC bin" specialized DFT. Discard the DC reading (unless you care) and use the Sine and Cosine ...


3

If you use the Euler's formula, you can simplify like this: $$ [d]_{k,n} = \frac{\sqrt{2}}{N}\left( \cos{\left[ \frac{(k-1)(2n-1)\pi}{2N} \right]} e^{j\frac{2 \pi nk}{N}} \right) $$ I think we can't simplify more. PS: If you use your expressions of $A$ and $B$ and again the Euler's formula, you will get the same result.


3

My first swing at the answer had some very incorrect claims. I do not have access to the article, so I am inferring some things from the portion posted in the question. NOTA BENE: My arguments assume that the eigenvectors of $\mathbf{R}$ are arranged so that the first $n$ belong to the signal subspace and that the last $m-n$ belong to the noise subspace. ...


3

You are referring here to sparse signal sampling and Reconstruction. You do not require the sampling matrix to be sparse. Infact we require the sampling matrix to follow the restrictive isometry property. And sparse matrices do not follow this property. Random matrices do. So no point having a sparse sampling matrix


3

If you are "observing" the source, this implies there is some sort of information you are looking to get out of it, whether it be the total background noise, interference levels etc. Do you find "value and use" in the RF signal's magnitude versus time? What about the RF signal's phase versus time? The IQ representation gives us both of ...


3

Fat32's answer is correct and shows a common pitfall. The reason that you must do this is because recall that the output of the autocorrelation of a signal of length $N$ is $2N - 1$. You were performing the FFT with the original sample size of $N = 1000$, effectively destroying necessary information to retrieve the autocorrelation of size $2N - 1 = 1999$.


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