New answers tagged discrete-signals
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Other then the endpoints, this is really just an FIR filter. You can easily assess the impact of smoothing on the spectrum by looking at the Fourier Transform of the impulse response of the filter . In this case it's a notch filter which seems like an odd choice.
In general smoothing is in many cases similar or equivalent to a lowpass filter operation and ...
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$\displaystyle \hbox{ DTFT}(\alpha x_1 + \beta x_2) = \alpha\cdot \hbox{ DTFT}(x_1) + \beta \cdot\hbox{ DTFT}(x_2)$
Linearity property cannot applied here.
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this depends a lot how "good" this estimate needs to be. Here is a "rough & dirty" method.
Put your scope in spectrum analyzer mode. Make sure you see all harmonics.
Look where the highest harmonic is and that's still visibly sticking out over the noise floor.
Quadruple that frequency and use this as your sample rate.
Sample your signal for a sizable ...
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The correlation is used to find the time shift between signsls while convolution represents system response to predefined input.
Since the system response depends on previous input, rether then future input, the sign of the time distance is negative.
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The plot as shown makes sense as nothing has limited the time axis to be going from $-\infty$ to +$\infty$. So we are showing the finite duration sequence as it would appear on such a time axis.
We can limit the time axis to a finite number of samples as used in the Discrete Fourier Transform (DFT) or the an unlimited time axis as used in the Discrete Time ...
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Bottom Line
$$(\theta_2-\theta_1) = 2\pi f(T_2-T_1)n -(\phi_2[n]-\phi_1[n]) \tag{1}$$
$f$: frequency in Hz of two tones of the same frequency and fixed phase offset
$(\theta_2-\theta_1)$: phase difference in radians of tones being sampled
$T_1$: period of sampling clock 1 with sampling rate $f_{s1}$ in seconds
$T_2$: period of sampling clock 2 with ...
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First of all, note that the discrete-time Fourier transform (DTFT) of a sequence is always periodic with period $2\pi$:
$$X(\omega)=\sum_{n=-\infty}^{\infty}x[n]e^{-jn\omega}\tag{1}$$
This is obvious because $e^{-jn\omega}$ is $2\pi$-periodic in $\omega$.
Next, consider the inverse DTFT:
$$x[n]=\frac{1}{2\pi}\int_{-\pi}^{\pi}X(\omega)e^{jn\omega}d\omega\...
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For the first part of your question, perhaps this will shed some light:
Phase difference measurement of a signal sampled with two different sampling frequencies
The answer to your second part of your question is yes for a single pure tone. It will appear as a lower frequency alias in the DFT, but if you know the actual frequency range, you can calculate ...
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The link in the comment is for exact answers for real tones.
Intuition (derived from understanding the theory) is pretty simple.
A real valued pure tone is actually the sum of two complex pure tones. I like to use the cosine function for pure real tones. I use $\alpha$ in my articles for the radians per sample frequency value. $\omega$ is commonly ...
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There is a circular waveform discontinuity at sample 0 of an FFT input (to sample N-1), if it isn’t exactly integer periodic in aperture. However, if the waveform is continuous at sample N/2, then the phase can be measured at that point.
You can measure phase at the halfway point by doing an FFTshift, or by flipping the phase of every odd numbered FFT ...
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SOLUTION
Bottom Line
$$(\theta_2-\theta_1) = 2\pi f(T_2-T_1)n -(\phi_2[n]-\phi_1[n]) \tag{1}$$
$f$: frequency in Hz of two tones of the same frequency and fixed phase offset
$(\theta_2-\theta_1)$: phase difference in radians of tones being sampled
$T_1$: period of sampling clock 1 with sampling rate $f_{s1}$ in seconds
$T_2$: period of sampling clock 2 ...
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You are right.
In the solution, the second line which performs the long division
$$ H(z) = -4 + \frac{ 5 + \frac{7}{2} z^{-1} }{1 - \frac{3}{4}z^{-1} + \frac{1}{8} z^{-1} } $$
is wrong and should be corrected as:
$$ H(z) = -4 + \frac{ 5 - 3 z^{-1} }{1 - \frac{3}{4}z^{-1} + \frac{1}{8} z^{-1} } .$$
However, the partial fraction expansion at the following ...
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Your analog transfer function looks OK. For the sake of clarity - and to reduce the chance of making errors - I'd just rewrite it as
$$H_a(s)=G\cdot\frac{s^2+as + b}{s^2+cs + d}\tag{1}$$
with
$$\begin{align}G&=\frac{2R_g}{R_d+2R_g}\\a&=\frac{R_d}{L}\\b&=\frac{1}{LC}\\c&=G\left(a+\frac{1}{2R_gC}\right)\\d&=G\cdot b\frac{}{}\end{align}$$
...
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You have to realize that all sequences in DFT relations are inherently periodic. This is usually understood, but it can be made explicit, for instance by that subscript. So both sources are correct, but one explicitly notes the periodicity of the sequence.
If you write down the DFT relation
$$\delta[n]=\frac{1}{N}\sum_{k=0}^{N-1}e^{j2\pi nk/N}\tag{1}$$
...
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This is a homework type problem, so I'll give you a few hints to help you solve it yourself (and learn something while doing so).
From what is given, we can assume that the given difference equation describes a causal discrete-time system. Let $h[n]$ denote the impulse response. It must satisfy
$$h[n]=h[n-1]+\frac{1}{N}\big(\delta[n]-\delta[n-N]\big),\...
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The real-valued amplitude function of a rectangular window of length $N$ is
$$A(\omega)=\frac{\sin\left(\frac{N\omega}{2}\right)}{\sin\left(\frac{\omega}{2}\right)}\tag{1}$$
So what you get is exactly what you would expect according to $(1)$, because in your case you have $N=3$. The constant $M$ in your formula is related to $N$ by $N=2M+1$ or $M=(N-1)/2$.
...
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I have come across my ways.
$$\begin{align} P(e^{j\omega})&=1+2\cos(\omega) \\ &=\sum_{k=-\infty}^\infty \sum_{k=-1}^1 \delta[n-k]e^{-j\omega n} \end{align}$$
Knowing that non-zero only when $n=k$,
$$P(e^{j\omega})=\sum_{k=-1}^1 e^{-j\omega n}$$
Let $M=n+1$,
$$\require{cancel} \begin{align} P(e^{j\omega}) &=\sum_{k=0}^2 e^{-j\omega (M-1)} \\ ...
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Arnold Tustin was my great uncle. I'd be grateful if someone can give me a quick layman's explanation of the bilinear transform (also known as Tustin's method). Unfortunately, I'm not a mathematician so can't really understand the explanation given on Wikipedia. I hope someone can tell me why this calculation is significant, whether it is still used and for ...
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I understand, that this involves a lot of math.
Not so much, in principle. The basic idea behind linear (or nonlinear) filtering is to remplace a inaccurate or noisy sample $s[n]$ by a combination of other samples, assuming that their values or location is somehow close to $s[n]$ (cf. local vs non-local filters).
At a low level, when the filter is both ...
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It may help you to look at each term separately if you are not used to the math as in:
$$x[n] = 4\sum_{m=-\infty}^{\infty}\delta[n-4m] + 8\sum_{m=-\infty}^{\infty}\delta[n-1-4m]$$
Knowing that $\delta[x]$ is only 1 when x=0, and 0 everywhere else, ask yourself what does n need to be to make those terms go to zero given that m can be any integer? Sketch ...
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I assume you'r talking about digital interpolators.
An interpolator is basically an L times expander followed by an anti-imaging lowpass filter with gain L. The cutoff frquency of the lowpass filter is defined in the discrete-time frequency domain as
$$ w_c = \frac{\pi}{L}. $$
The analog equivalent of this cutoff frequency is
$$ f_c = F'_s \frac{ w_c }{ ...
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Consider a moving average over N samples- this is a simple FIR filter where each new output is the average of the past N samples. It is easy to see how high frequency noise can be filtered out (so is a low pass filter), and the longer time duration we include in the averaging window the lower will be the frequency cut off (just compare a stock market 30 day ...
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You effectively want a interpolated waveform that is interpolated by 200,000 such that for each new sample with the 5 ppm offset you can select one additional offset to induce that time offset, for example
x[1], x[200,002], x[400,003], x[600,004]....
(Or equally one less if your sampling clock increased in frequency 5ppm).
One very simple approach to do ...
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To perform carrier recovery for FSK, if the data is random and can assumed to be equiprobable, then you can take the mean of the derivative of the phase versus time to determine the carrier offset.
Optionally the phase change from one sample to the next can be estimated using using a complex conjugate product of successive samples (for small angles, the ...
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