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Does cubic interpolation (or any other) have any advantages over linear for the specific case of audio?
You'd use neither for audio. The reason is simple: The signal models you typically assume for audio signals are very "Fourier-y", to say, they assume that sound is composed of weighted harmonic oscillations, and bandlimited in its nature.
Neither linear ...
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There is no such proof because it's not always true. It's a rule of thumb, because I guarantee that you could come up with a situation- an infinite number of situations actually- where higher order splines would do better than cubic splines. The optimal spline order for any given situation is the exact same order as the system you are trying to model. If ...
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The difference between cubic interpolation as described in your question and cubic spline interpolation is that in cubic interpolation you use 4 data points to compute the polynomial. There are no constraints on the derivatives. Cubic spline interpolation computes a third order polynomial only from two data points with the additional constraint that the ...
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This is not the case. First of all, a second-order hold would use three sample points to compute an interpolation polynomial, but your suggested impulse response $\text{tri}(t)\star\text{tri}(t)$ is non-zero in an interval of size $4$ (assuming a sample interval of $T=1$, as you do in your question). However, the impulse response corresponding to a second-...
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This plot depicts how to convert your digital signal back to the analog one, using $\mathrm{sinc}$ functions. The nice property of these functions used in this process, is that maximum of each function always occurs at minimums of the other, shifted function:
Now the process of the D/A conversion that is depicted on your plot is basically taking a $\mathrm{...
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I don't get your downsample step when you downsampled by factor $M$.
Let me go from scratch with the spectrum visualization below, with time domain, continuous frequency domain and discrete frequency domain from left to right.
When we reduce the sampling frequency by a factor $k$, the signal spectrum is copied to new replicas at $f_s/k$. The discrete ...
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But is this good practice?
Yes, that's a perfectly reasonable approach.
Will this will work for all types of signals I might have ?
There are a LOT of signals out there and there is always an outlier. In your case, I think the most "vulnerable" would be low frequency sine waves. A 40 Hz sine wave has a period of 25 ms and looping one at an ...
7
You want to have an interpolation with a equal number of weights on both sides of the points you want to interpolate in between. So you choose either one or two weights on each side, resulting in an interpolation of two (linear) or four (cubic) points.
An quadratic interpolation would need three points, which would only make sense at the border of a grid, ...
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I don't have a real answer but I have the feeling that this result will help you out: Bernstein's inequality says that, if the signal $x(t)$ is bandlimited to $|f|\leq B$, then $$\left| \frac{\textrm{d}x(t)}{\textrm{d}t}\right|\leq 4\pi B \,\textrm{sup}_{\tau\in\mathbb{R}}|x(\tau)| ,\,\,t\in\mathbb{R}$$ where $\textrm{sup}$ stands for "least upper bound".
I ...
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I do not understand why you have 30Hz resolution so I will focus only to the principle of the question "does interpolation increase resolution?".
Short answer is no, no new data, no new information.
A longer answer needs the spectrum visualization below, with time domain, continuous frequency domain and discrete frequency domain from left to right.
The ...
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Duane Wise and i wrote a paper back in the 90s that we presented to an AES convention that spelled out how to model time-domain polynomial interpolation (of which linear interpolation is an example) in the frequency domain.
i think you can get a copy here: Performance of Low-Order Polynomial Interpolators in the Presence of Oversampled Input
answered Jul 28 '17 at 17:24
robert bristow-johnson
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6
Interpolation (sampling frequency 44.1 kHz ➔ 88.2 kHz)
Your original 44.1 kHz sampled signal has frequencies up to 22.05 kHz, so you should lowpass filter at 22.05 kHz after dilution with zeros. Your filter should have a gain of 2. Otherwise the signal amplitude drops to half because you set half of the samples to zero. Like Jim Clay says, you can combine ...
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Your interpolation operation is an upsampling operation with subsequent low-pass filtering. This translates to zero-padding in the frequency domain. Given your original signal $x[n]$, you upsample it to become
$$y[n] = \begin{cases}x[n/M]& n/M \in \mathcal{Z} \\ 0 & \text{else}\end{cases}$$
to become an M-times upsampled signal. Then, your ...
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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 Summation Formula.
Let's introduce different approach which is more ...
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We need to assume the reader knows some basic stuff to answer that.
Let's give it a try.
Lets understand the sentence - Zero / First Order Hold.
We have the Zero / First Order and the Hold.
Zero / First Order hold means the order of the Taylor Series of the function we use to interpolate. In other words, the degree of the Polynomial we can write the ...
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Three reasons to increasing the sampling rate further are
1) To relax the requirements of the post D/A conversion filtering for image rejection.
2) Increase signal SNR by spreading quantization noise for a fixed number of DAC bits across a wider frequency range.
3) Minimize passband droop in the D/A reconstruction.
Reason 1 is the most dominant one in my ...
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Let's call the window length $M$, and $N$ is the order of the polynomial. One important thing to realize is that you use a polynomial of order $N$ to approximate $M$ data points. This means that you use $M$ points to compute $N+1$ polynomial coefficients (an $N^{th}$ order polynomial has $N+1$ coefficients). So if $N+1=M$ then you do not smooth at all but ...
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There are various methods on 2d interpolation (this one, and this one). But most of them considered at least 4 points rather than 2. The simplest 2d interpolation is 3 1d interpolation, in which you interpolate the points between (x1-d1, y1-d2) and (x1+d1, y1-d2) as (x2,y2), then you interpolate the points between (x1-d3, y1+d4) and (x1+d3, y1+d4) as (x3,y3)....
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You are looking at the wrong metric of "correctness". Nearest neighbor is introducing significant discontinuities that are showing up as massive quantities of noise in the result.
The problem is that you should be comparing to the result you would have gotten if you had sampled at 26.25MHz in the first place.
Let's try it (sample a 12Hz sine wave at 26 Hz ...
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Here pretty good explanation.
I'll start by consider the 1D case of cubic interpolation (because its easier to explain) and then go onto the 2D case.
The basic idea of cubic interpolation is to estimate the values between any two points by as a cubic function, $f(x)$, with first derivative, $f'(x)$
$$f(x) = ax^3 + bx^2 + cx + d$$
$$f'(x) = 3ax^2 + 2bx + c$...
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Since Sinc based Interpolation requires you to know the data at any point it can not be done.
You might do a Truncated Sinc Interpolation.
The artifacts you're seeing can be caused by a kernel which is too short or the parameters aren't good.
In order to create a good Sinc kernel you need to know things about the Band Width of the signal and the Sampling ...
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Yes, you can interpolate and decimate at the same time. This is called "resampling". If you google resampling you will find lots of information about it. And yes, your reasoning about resampling is mostly correct.
When thinking about resampling theoretically you usually put the interpolation first to avoid Nyquist issues. An interpolator is upsampling ...
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The reason is aliasing. Because the modulation is digital, on the discrete signal, the sampling frequency should be high enough that the channel's frequency band does not extend beyond the Nyquist frequency = sampling frequency / 2. If the sampling frequency is too low, the frequencies beyond the Nyquist frequency will alias to other frequencies, meaning ...
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so this is why i think an $n$-th order hold is a $\operatorname{rect}\left( \frac{t - T/2}{T} \right)$ convolved against itself $n$ times.
Wikipedia isn't the final reference of all things, but there is something that i sniffed from there. consider sampling and reconstruction (the Shannon Whittaker whatever formula). if the original bandlimited input is $...
answered Mar 26 '16 at 7:52
robert bristow-johnson
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5
Observations
I have used +1 and -1 in the sequence instead of your 1 and 0. With $\alpha=1$, the band-limited continuous function $f_m(T)$ in your first two figures (with the above mentioned modification) is:
$$f_m(T) = \sum_{k=1-m}^m \operatorname{sign}\left(\operatorname{sinc}(\pi k - \pi/2)\right)\operatorname{sinc}(\pi T-\pi k),\tag{1}$$
where:
$$\...
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First a demonstration that the squares of both
$$\begin{align}&[\dots, 0, 0, 1,\hphantom{-}1, 0, 0, \dots] \text{ and}\\
&[\dots, 0, 0, 1, -1, 0, 0, \dots]\end{align}$$
equal
$$[\dots, 0, 0, 1, \hphantom{-}1, 0, 0, \dots]\hphantom{\text{ and}}$$
but the squares of their sinc interpolations differ (Fig. 1):
Figure 1. Squares of sinc interpolations ...
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Take a look at the cubic Hermite spline. The interpolated function is continuous at the data points and the first derivative is also continuous. Away from the data points all of the derivatives are continuous.
Let's say that the function $f(x)$ is defined by equally-spaced data points for all $x$ that is an integer. This means you know the values of $f(0)...
answered Dec 13 '16 at 6:21
robert bristow-johnson
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5
There are 2 main factors here:
Aligned timeline
Namely each sample has time stamp taken with the same reference time.
The lower sampling rate is high enough according to the sampling theorem
Namely even the lowest sampling rate, in your case 32 [Hz] is high enough to reconstruct the original signal.
If both (1) and (2) hold you'll be able easily to compare ...
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Audio signals
An audio special-purpose analog-to-digital converter (ADC) normally has an internal or external analog low-pass filter and samples the analog filtered signal at a multiple of the target sampling frequency. This high-rate digital signal is then low-pass filtered by a digital decimation filter and decimated to the final sampling frequency. If we ...
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