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CPM Modulation and Demodulation, How to determine the values for normalized bandwidth?

Here’s a step-by-step procedure to find Bnorm: Determine the highest frequency fmax​ in the signal. Set the initial sampling frequency fs​≥2fmax​. You can start with fs=2.5fmax​ or fs=3fmax to ...
Naveed Ahmed's user avatar
2 votes

s to z domain transformation

Before I detail the mapping approaches further let me comment since the OP mentions using a Bessel filter, that a Bessel filter is an analog filter that has a maximally-flat group delay and very nice ...
Dan Boschen's user avatar
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Compressed sensing and Logan's Theorem

We should at least read the original sources (Shannon 1949, Communication in the Presence of Noise): Theorem 1: If a function contains no frequencies higher than $W$ cps, it is completely determined ...
Marcus Müller's user avatar
1 vote

Meaning of the phase of a sample

I think there is a common confusion with "phase" as used in signal processing, and it is often incorrectly conflated with "time delay". I believe this originates in our ...
Dan Boschen's user avatar
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Why do the lengths of the sampled signals $x_1, \: x_2$ have to be $\text{length}(x_1)+\text{length}(x_2)-1$?

Convolution can be realized as a polynomial multiplication. A sequence of samples can be represented as a polynomial of order one less than the length of sequence, whereas, the coefficients of ...
learner's user avatar
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Why do the lengths of the sampled signals $x_1, \: x_2$ have to be $\text{length}(x_1)+\text{length}(x_2)-1$?

Let $x_1[k]$ and $x_2[k]$ be two discrete time signals with respective lengths $N_1$ and $N_2$. Let $X_1[r]$ be the DFT of $x_1[k]$ and $X_2[r]$ be the DFT of $x_2[k]$. Then it holds that the circular ...
Carl's user avatar
  • 416
1 vote

does extension to reals of an arbitrary signal with a zero tail eventually cross a threshold

does there exist a time $t$ such that for all $u \in \mathbb{R}, u >= t$, we have $|S(u)| < \varepsilon$?$ Not as I interpret the question. For simplicity we can just flip the signal in time ...
Hilmar's user avatar
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