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It is actually not distorted, it is sampled at high enough rate. What fools you is the straight lines drawn between sample points, it gives you a false impression of the waveform. It shows you a linear interpolation of the signal. It does not represent how the signal would actually look like. A sampled signal exists only at the sample points, and to convert ...


9

The actual requirement is to sample at GREATER then twice the bandwidth, not at a rate equal to it... So only your 80Hz same set actually meets the requirement, because the 60Hz case is ambiguous in general, consider if you were sampling sin (2PiFt) instead then you would get a flat line at zero amplitude.... And changing the angle between sin and cos would ...


6

There is no aliasing as 𝑓 = 30 Hz is less than or equal to the folding frequency, 30 Hz and 40 Hz, respectively. Yes and no. There isn't significant aliasing when you're sampling at 80Hz, because the resulting signal has frequency components at 30Hz and 50Hz. The result is thus unambiguous as long as you take that 50Hz signal into account. There is ...


4

This could happen as discriminator gain is increased with a filter discriminator approach since in many of those approaches the gain would be maximum and linear for small signals only and then the slope of the discriminator slowly goes down coinciding with the results in your plot (such that you no longer get a perfect sine wave out for a sine wave in—- so ...


3

MP3 is a lossy compression format, so encoding an uncompressed lossless WAV file to MP3 will lose some quality in the encoding process. Decoding a MP3 file back to WAV file has at most the same quality as the MP3 file has, it is just as good or bad as the contents in the MP3 file was, it is not in any way different to decode it for saving into a file or ...


3

My first inclination is to say this is a meaningless question. The concept of "instantaneous" frequency really only pertains to a single pure tone with a slightly varying frequency. In this light, one may construct a definition saying "The instantaneous frequency at time $t$ is the same as that of a pure tone which matches the function (sum) in the ...


3

Eventhough Laurent has given a broader sense of the answer, let me put here the communications theory sense ot it. The concept of instantaneous freqency emerges when you consider Frequency Modulation or Phase Modulation systems, where the message is embedded into the change of the frequency or phase of a carrier signal. This carrier is typically a single ...


3

How can we know the used sample frequency fs by looking at y(t)? You can't. The sampling theorem states that any sampling higher than twice the highest signal frequency allows for perfect reconstruction. In you case any sampling frequency higher than 4 would result in the same $y(t)$.


3

Remembering from my 1970 Signal Processing lectures we have ... The crucial thing is the filter used to reconstruct the signal. Let's do the theory first for ideal sampling a perfect sine wave at 2x its frequency and filtering with an ideal low pass filter. The samples are infinitely thin - they are delta functions separated by time t. The filter is an ...


3

Frequency is mathematically defined as the number of cycles per second. So it is a more strict word mathematically. It is represented numerically by the unit called Hertz. $f=1/T$, where $T$ represents the one-period length of a waveform. This makes frequency quantifiable. Pitch on the other hand, is a perceptual characteristic of a sound frequency, so it'...


3

@Man. Hi. Measuring frequency in terms of $\omega$ (radians/second) or in terms of $f$ (cycles/second, Hz) is the same as measuring speed in miles/hour or kilometers/hour. People measure frequency in terms that are convenient for them. In algebra equations it's convenient to represent frequency in terms of $\omega$ because it's easier to write the single $\...


3

Consider a wheel rotating counter-clockwise at one revolution per second. Its frequency of rotation is 1 Hz. If it rotates clockwise, its frequency of rotation is -1 Hz. It's as simple as that.


2

I think the main point you are missing is that the DFT occurs over a span of time and is not instantaneous. There is an inherent contradiction there that you can never resolve. The pure tone case will "bridge" the two concepts, but it is still not the same. I have written several blog articles on finding the instantaneous-as-possible frequency in the time ...


2

If $T$ is a positive real number such that $$x(t) = x(t+T) ~\text{for all}~t \in \mathbb R\tag{1}$$ for continuous-time signals, or if $T$ is a positive integer and $$x[n] = x[n+T]~\text{for all}~n \in \mathbb Z \tag{2}$$ for discrete-time signals, then $x$ is said to be periodic with period $T$ or to have period $T$. A signal $x$ that has period $T$ also ...


2

This is to explain what actually sampling is and why and how it's done. Here I've generated a simple continuous-like sinusoidal signal with frequency fm = 10kHz. In order to make it appear as a continuous signal when plotting, a sampling rate of fs=500kHz is used Pretending the above-generated signal as a “continuous” signal, we would like to convert the ...


2

It's a spurious artifact due to not referencing your phase generation to the center of your data window, and not doing an fftshift before the fft to place your phase reference point at fft input index 0. e.g. For varying frequencies, either you have to know, generate, or want the phase at location T/2 of your continuous sinusoid. The phase result of an ...


2

In principal, you get $\text{Ws}=\text{J}$ for energy and $\text{W}$ for power, but as you are calculating these measures from a wav-file, you do not get any unit. What you get is just a ratio to the maximum possible value of a wav-file. A square wave with amplitude one has a level (power) of $0\text{ dBFs}$, which means dB Fullscale. Your signal has roughly ...


2

The best way to accomplish your desired result is to estimate the parameters of the interfering tone(s) and remove it(them) from the signal in the time domain. To get you started on how to go about this, I recommend you read these two articles of mine: Two Bin Exact Frequency Formulas for a Pure Real Tone in a DFT Phase and Amplitude Calculation for a Pure ...


2

Neither. To me, filter classes using the notion of frequency bands (low-pass, high-pass, etc.) can be used safely in the linear case. And the bilateral filter is nonlinear. Edges are not really high-frequency: they often have sharp variations across the edge, but slow variation along it. I would consider the bilateral filter as an edge-preserving smoother, ...


2

how do you even measure the frequency of a digital signal? This may be the source of your confusion. In digital communications, data is transmitted using an analog, continuous-time waveform. This is necessary because only this kind of waveform can exist in the physical world. Like all waveforms, it has a bandwidth, and its Fourier Transform determines what ...


2

There is a huge gap between the formalism and construction for the continuous wavelet analysis (Mexican hat or Morlet wavelet, the latter being only approximate or complex Gaussian wavelet) and the discrete wavelet transforms (DWT). Let us make a long story short. Continuous wavelet transforms (CWT) hinge on a continuous wavelet $\psi$, for which for any (...


2

Without more details, when a signal is properly recorded at $F_n$ Hz, one is entitled to expect that all information below the half of it, namely $F_n/2$ Hz, can be recovered, somehow. This is an application of the classical Shannon Sampling Theorem. Twice the maximum frequency is sufficient to sample a signal... in theory. The question says: max frequency ...


2

If I compute the Fourier Transform of Z(f), what will I get? If you apply the inverse Fourier Transform, you get the impulse response of the system which is indeed a time domain signal. You can also apply the forward transform and would get the impulse response time reversed and scaled since the forward and inverse transform are quite similar. Which is ...


2

The notion of instantaneous frequency is (hopefully) consistent with the monochromatic wave model: $$x(t)=a \cos 2\pi \nu t\,,$$ where $a$ is the amplitude and $ \nu$ the frequency. It would be tempting to compute a similar formula for evolving amplitude and frequency cases, something like: $$x(t)=a(t) \cos \left(\phi(t)\right)\,.$$ However, this is not ...


2

The DFT Matrix for Non Uniform Time Samples Series Problem Statement We have a signal $ x \left( t \right) $ defined on the interval $ \left[ {T}_{1}, {T}_{2} \right] $. Assume we have $ N $ samples of it given by $ \left\{ x \left( {t}_{i} \right) \right\}_{i = 0}^{N - 1} $. The samples time $ {t}_{i} $ is arbitrary and not necessarily uniform. We're ...


2

Well to clear your doubt ,i would like you to ask you a question. What would be the value of angular frequency × Time period = ?? Okay,so in eq - $\omega \times t$ . if we put $T$ in place place of $t$, given that the wave is sinusoidal, the value to that eq would be $2\pi$. By definition $T$ is the time taken to complete one oscillation so when we ...


2

Since this is a pure sinusoid, it has a bandwidth of 0 Hz. You can multiply it by a carrier signal of the same frequency, pass it through a low pass filter then take only a few samples. What matters is NOT the frequency of the signal, rather the bandwidth. Consider for example a voice signal modulating a 1 GHz carrier. It will be very costly, to sample this ...


2

Check this figure. Consider the signal $x[n]=A$ which is constant but can be written as $x[n]=A\cos(0n)$, so its frequency is zero. There is no oscillation. Let's increase the frequency. For the signal $x[n]=A\cos(\pi n/4)$, it oscillates at frequency $\omega_0=\pi/4$. The oscillation frequency increases. For the frequency $\omega_0=\pi$, the signal is $x[...


2

The digital frequency represented with $\pi$ is the Normalized Angular Frequency, and has units of frequency given as radians/sample. This signal can be represented as a rotating phasor on the complex IQ plane: $Ae^{j\omega n}$, where $A$ is the amplitude and $\omega$ is the normalized angular frequency. With this representation the normalized frequency is ...


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