21

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 ...


7

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 ...


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 ...


5

You can try tracking Phase-frequency of your $50 Hz$ signal using Costas Loop. Costas loop does not require the signal to be pre-processed in order to expose the desired frequency. I am not giving details of a Costas Loop because it can be found anywhere.It is pretty popular Carrier Recovery technique and a good starting point would be Wikipedia: CostasLoop ...


5

Your interrupt seems to occur at 8 kHz. There are 2 obvious jitter causes that I can see. First of all if $$\frac{f_s}{f} \neq N$$ where N is an integer, $f_s$ is the sampling frequency and $f$ the signal frequency, you will get jitter since the sampling instants will not be "periodic". Try to set f to 80 Hz, you will have 100 samples per period, ...


5

Note that when you compute the DFT, you compute the dot product with a sine term and with a cosine term. In your third example, even though the dot product with the sine term is zero, the dot product with the cosine term isn't, so you do get spectral leakage, as you expected. The dot product with the sine term is zero because you have a whole number of ...


5

Amplitude modulation is a linear operation (excluding the carrier in the classical AM) and does not introduce harmonics due to the modulated signal. Sidebands are not harmonics. They are created as the message signal spectrum is shifted up to carrier frequency. Frequency mixing of a message signal with a sinusoidal carrier, another definition of AM, does not ...


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 ...


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 ...


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

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

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

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

@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.


3

There's a lot of things that influence your choice of sampling rate. First of all, of course, the Nyquist theorem, which says you need to sample at more than twice of the signal's single-sided bandwidth $B$ (for real signals)¹. Then, you often choose a sampling rate that is useful for your system. For example, a lot of 3G/4G/5G systems use sampling rates ...


3

Given that the FFT over that many samples does not show any results, your challenge may be in the overall spectral purity of the 50 Hz tone you seek. The FFT bin at 50 Hz is a correlation to that bin frequency which is the optimum detection in terms of SNR of a 50 Hz signal in the presence of white noise. The issue is the equivalent noise bandwidth of that ...


3

If there are phase discontinuities after all $2\pi$-jumps have been removed, then these discontinuities are usually caused by zeros of the frequency response. The phase jumps by $\pi$ at these frequencies, and the group delay doesn't exist, or, if one prefers, is non-finite. Note that the group delay is meaningless anyway at frequencies where the frequency ...


3

Interrupt driven sampling can be surprisingly jittery all by itself. It depends on a great many things how much time elapses between the desired moment of time and the sample time. The easiest solution would be use an ADC that can be synchronized to a simple clock. One edge tells the ADC to start the sample, then the other edge can drive your interrupt to ...


3

Dot returns zero, but I don't understand why. Why do we get zero here instead of a non-zero number Luck mostly. It's only zero because you chose the right phase difference. If you add any non-trivial phase to either one of the sine waves, you would get a non zero result. That's NOT the case for 5Hz and 6 Hz, where the dot product is zero regardless of the ...


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

I cannot post comments, as I created this account only to add something to this answer and I must have 50 reputation, as I found this post very helpful as a memory refresher (currently writing my PhD thesis). I hope this is not considered necro-posting. I would like to add to Dan Boschen's answer that one must pay A LOT OF ATTENTION to units: It is true ...


2

The ref. unit mentioned is perfectly fine. Some microphone manufacturers use 1 dyne/cm^2. This would translate into -74 dB instead of -94 dB because the reference pressure applied is 0.1 Pa. So you can do the same math, just subtract -20 dB. To resume: to calculate the sensitivity you must: 20.*log10(sens.)-20. In you case, this would imply a sensitivity (...


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

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 ...


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

What does it mean exactly to "exist" vs "just theoretical"? Do we for some reason think that $cos(\omega t)$ exists while $e^{j\omega t}$ does not? Both are equally mathematical constructions that describe our physical world. We somehow conclude that the latter as a complex quantity does not exist but the former as a real quantity does, ...


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