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.obsidian/plugins/colored-tags/data.json
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.obsidian/plugins/colored-tags/data.json
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"Unstructured-multiplicative-uncertainty": 146,
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"Unstructured-multiplicative-uncertainty": 146,
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"NPIC2025": 147,
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"NPIC2025": 147,
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"Thesis": 148,
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"Thesis": 148,
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"Organization": 149
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"Organization": 149,
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"Frequency": 150
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},
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},
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"_version": 3
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"_version": 3
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}
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}
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a
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# Impulse Sampling
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How do we represent a sequence of numbers?
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Impulse sampling does it by
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1. having a continuous signal
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2. having an impulse train (impulses at sampling frequency)
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3. multiply em together
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>[!info] Functionals
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>Laurent Schwartz (1950):
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>$$\int_{-\infty}^\infty \phi(x) f(x) dx = z$$
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>Shift Property:
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>$$\int_{-\infty}^\infty \phi(t-\tau) f(t) dx = FINISH$$
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>Laplace Transform
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**Pulse Train $\delta_t(t)$**
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$$\delta_T(t) = \sum_{k=-\infty}^\infty \delta(t-kT)$$
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$$x^*(t) = x(t)\delta_T(t) = \sum_{k=-\infty}^\infty x(t)\delta(t-kT)$$
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Where the sampled signal is $x^*$
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What about in Laplace domain?
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$$X^*(t) = \int \left[ \sum_{k=-\infty}^\infty x(t)\delta(t-kT)\right] e^{-st} dt$$
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$$X^*(t) =\sum_{k=-\infty}^\infty \left[\int x(t)\delta(t-kT) e^{-st} \right] dt$$
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$$X^*(t) =\sum_{k=-\infty}^\infty \left[\int x(t)e^{-st} \delta(t-kT) \right] dt$$
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Now using the shift property...
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$$X^*(t) =\sum_{k=-\infty}^\infty x(kT)e^{-kTs} $$
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## Some Observations
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If we change the variable $z = e^{Ts}$
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>[!important] **The Z-Transform**
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> $$X^*(t) = X(z) = \sum_{k=-\infty}^\infty x(kT) z^{-k} $$
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> Z transform can be viewed as short hand of the Laplace transform
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>
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> Sampling is a time varying process. If x(t) is time shifted by a small amount, the sampled signal x(kT) will be different.
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# Frequency Domain Interpretation
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$\delta_T(t)$ is periodic, so we can turn it into a Fourier series...
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$$\delta_T(t) = \sum_{N=-\infty}^N C_N e^{j(\frac{2\pi}{T})Nt}$$
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$$C_n = \frac{1}{T}\int_{-T/2}^{T/2} \delta_T(t) e^{-j(\frac{2\pi}{T})Nt} dt$$
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Apply a shift and do some stuff...
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$$C_n = \frac{1}{T} \int_{-T/2}^{T/2} e^{-j \frac{2\pi}{T} Nt} dt $$
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$$C_n = \frac{1}{T}$$
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So then...
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$$\delta_T(t) = \frac{1}{T} \sum_{N=-\infty}^N e^{j(\frac{2\pi}{T})Nt}$$
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$$\delta_T(t) = \frac{1}{T} \sum_{N=-\infty}^N e^{j \omega_s T Nt} $$
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Then, this can be used to get all the way to $X(z)$.
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