Hot answers tagged

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 ...


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

No, because this is a sufficient condition (for regularly sampled signals), and not a necessary one. This condition restricts the space of all possible continuous signals to a subspace of discrete sequences that contain the same information. Suppose that you can constrain the signal space, eg limited band-width, positivity, parametric models, sparsity, etc....


4

Is the rate of 2B exclusive? Yes. The sampling theorem states that the signal must be band limited to half the sample rate. That implies that the energy at the Nyquist frequency must be zero. In practice you need a healthy margin between the highest usable frequency and the Nyquist frequency. There is always some "transition band" that you need to get the ...


4

The sampling theorem requires a perfectly bandlimited signal, bandlimited to below twice the sampling frequency. The problem with this is that only an infinite length signal (e.g. exists before the big bang) can be perfectly bandlimited. This is from the Fourier theorem regarding any domain with finite support. Thus all real-world signals are ...


4

Looks like a potential application for blue noise also known as Poisson disk sampling, which is random placement of samples but with a guaranteed minimum distance between sample locations. I think that would give more accurate transform results than independently located random samples. Various spherical harmonic transform algorithms with specific sampling ...


4

I'm a little confused about what you are asking and how it relates to the set of spherical basis functions. Those are the same one that electron orbitals are based on, right? I have never understood them to be constrained to the surface of a sphere. Then again, that is not my usual stomping grounds. Nevertheless, I think I can generalize your question ...


4

Answer: You will see residual images of $X(f)$ at multiples $f_s$, $2f_s$ and $3f_s$, and distorted image of $X(f)$ at non-zero multiples of $4f_s$, when sampling in the manner you explained. Depending on value $e$, the size of residual will change. I have explained how in detail below. Ideally, sampling at $4f_s$ would have completely cancelled those ...


4

Yes the OP is correct in that you can implement pulse shaping in less than 2 samples per symbol for exactly the reasons that was outlined. However importantly we must also keep in mind having excess bandwidth to simplify subsequent filtering required (such as after the DAC on the transmitter side). The Nyquist criteria is the sampling rate must be twice the ...


3

It really boils down to aliasing. In continuous-time, if you have any two signals $x_1(t) = \sin(2 \pi F_1 t)$ and $x_2(t) = \sin(2 \pi F_2 t)$, then as long as $F_1$ and $F_2$ are distinct, the signals are, too. But consider sampling at some time interval $T_s$, so that the sampled signals are $x_1(k) = \sin(2 \pi F_1 T_s k)$ and $x_2(k) = \sin(2 \pi F_2 ...


3

I think you're confusing two different (but related) terms. Nyquist says that in a channel of bandwidth $B$ you can transmit up to $2B$ orthogonal pulses per second. So, $R_p \leq 2B$, where $R_p$ is the pulse rate. To achieve $R_p = 2B$, the pulses need to be sinc-shaped. Other, more practical pulses achieve slightly less than that. For example, raised ...


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

From an ADC perspective, it is just taking a sample of the voltage in time. I fail to see how a "misinterpretation" could be made since there is no "turning car wheel" to take pictures of at the wrong time. Do the harmonics alias in such a way that the wave shape is preserved? You can reason this out yourself, in the time domain. Consider a square wave ...


2

Considering basic audio applications, the digital to analog conversion reconstruction filter (aka interpolation filter) is a low pass analog filter that removes all the image spectrum at the output before it goes to loudspeakers and retains only the baseband spectrum that resides in the filter's passband: inside its cutoff frequency of the lowpass filter. ...


2

A quick answer, but as human hearing does not go past 20kHz bandwidth, 44.1 kHz is enough for storing and transmitting audio. The problem is that the analog antialiasing filter before ADC must be extremely sharp to pass 20 kHz enough and block 22.05 kHz enough and this is just needs many components with good performance and tolerance. When sampling at higher ...


2

When you add 2 or more sinusoids at the same frequency $f_o$ but with different phase shifts you get a sinusoid at same frequency but an additional attenuation term. Mathematically, you will have following: $$cos(2\pi f_ot + \phi)+cos(2\pi f_ot + \theta) = 2cos(\frac{\phi - \theta}{2}).cos(\frac{2\pi f_ot + \phi + 2\pi f_ot + \theta}{2})$$$$= 2cos(\frac{\phi ...


2

I have some very short signals in the range of 8 to 16 samples. These represent a bandlimited signal, sampled at or slightly above the Nyquist rate. Nope. A signal can't be limited in time and in frequency at the same time. If it's very short, than chances are the bandwidth is a lot higher than you think it is and that you've already picked up some ...


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

You may be interested in the simple experiment using matlab. https://poweidsplearningpath.blogspot.com/2020/04/ch4-adcdac-how-to-simulate-adcdac.html Reconstruction is essentially a kind of interpolation or so called digital to analog conversion (DAC). Detail descriptions are introduced in chapter 4.8.3 of the DSP Bible 1. However, we all understand the ...


2

Answer : What you are considering as $\Omega_{N_x}$ is equal to $\frac{\Omega_N}{2}$ according to question. So, what you are saying is same as what answer mentions given we are considering Baseband Samping of $y_a(t)$. I think you are confused because of the terms Nyquist rate and Nyquist Frequency. Nyquist rate and Nyquist frequency are two different ...


2

I'd say that this is not only "similar to a cross-domain equivalent to Nyquist's Sampling Theorem", but it simply is the sampling theorem. The sampling theorem does not specify the domains of the signals involved; it is rather a mathematical condition that a function of a continuous variable needs to satisfy such that it is perfectly represented by ...


2

I think that considering the DFT from a linear algebraic point of view has some value, so I will attempt to introduce the foundations. We will assume that our signal is a vector of $N$ complex entries. $\mathbb{C}^N$ is the vector space of vectors with $N$ complex entries. Let $\mathbf{u}_0,\mathbf{u}_1,\ldots,\mathbf{u}_{N-1}$ be vectors in $\mathbb{C}^{N}...


2

Well! all signals in this world are made up of sum of different rotations(sinusoidals) - different in three senses: a. how big is the amplitude (A) b. how fast is the rotation ($\omega$) c. where is the starting point of the rotation (phase $\phi$) Fourier made this very clear. How do we measure rapidness of the rotating signals(sinusoidals) : by their ...


1

One possibility is to use a "discrete sinc interpolation", which uses a compact support version of a sinc (which is not a truncated sinc). Otherwise there are methods based on the discrete cosine transform (DCT) and discrete sine transform (DST). Another interesting approach is based on "sinc-lets". These are reviewed in this paper. In particular, look at ...


1

A bandlimited signal is infinite in duration. Even a low pass filtered signal for anti-aliasing implies a long duration. So if you don’t have signals off the ends, try generating them. Add a Monte Carlo shotgun of points to each end generated using anything known about the legal distribution of the signal. Reject the random end extension points that ...


1

I don't think I've seen capacity defined like that before. In the "go-to" information theory book by Thomas Cover, capacity is defined as $C=\frac{1}{2}log_2(1+SNR)$ bits per channel use or $C=Wlog_2(1+SNR)$ bits per second. The bandwidth is the symbol rate so you could have a symbol represent multiple bits which is what happens in all digital communication ...


1

If your frequency increase linearly, then you have a chirp signal with linearly increasing frequency. https://en.wikipedia.org/wiki/Chirp_spectrum There is a closed-form solution for the Fourier transform. I'm not sure how that will help you. Why not pick a sampling frequency $\omega_s \gt 2 \omega_{max}$ Since your highest frequency is $\omega_{max}$, ...


1

Approaching The Sampling Theorem as Inner Product Space Preface There are many ways to derive the Nyquist Shannon Sampling Theorem with the constraint on the sampling frequency being 2 times the Nyquist Frequency. The classic derivation uses the summation of sampled series with Poisson SummationFormula. Let's introduce different approach which is more ...


1

Square waves in a practical (and analog) video signal should always be bandlimited. May be they seem infinetely sharp at first, but if you zoom in you would see that their edges are actually rounded, indicating bandlimitedness. So if you use high enough sampling rate then you will avoid aliasing without an anti-aliasing filter. However, for a bandlimited ...


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