From ba611f56e7d4b05e225187089dbe597a05745e15 Mon Sep 17 00:00:00 2001
From: Dane Sabo <dane.sabo@pitt.edu>
Date: Thu, 9 Jan 2025 18:10:27 -0500
Subject: [PATCH] vault backup: 2025-01-09 18:10:27

---
 .obsidian/plugins/colored-tags/data.json |  3 +-
 2025-01-09 Sampling Theory.md            | 46 +++++++++++++++++++++++-
 2 files changed, 47 insertions(+), 2 deletions(-)

diff --git a/.obsidian/plugins/colored-tags/data.json b/.obsidian/plugins/colored-tags/data.json
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@@ -158,7 +158,8 @@
     "Unstructured-multiplicative-uncertainty": 146,
     "NPIC2025": 147,
     "Thesis": 148,
-    "Organization": 149
+    "Organization": 149,
+    "Frequency": 150
   },
   "_version": 3
 }
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diff --git a/2025-01-09 Sampling Theory.md b/2025-01-09 Sampling Theory.md
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-a
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+# Impulse Sampling
+How do we represent a sequence of numbers?
+Impulse sampling does it by 
+1. having a continuous signal
+2. having an impulse train (impulses at sampling frequency)
+3. multiply em together
+
+>[!info] Functionals
+>Laurent Schwartz (1950):
+>$$\int_{-\infty}^\infty \phi(x) f(x) dx = z$$
+>Shift Property:
+>$$\int_{-\infty}^\infty \phi(t-\tau) f(t) dx = FINISH$$
+>Laplace Transform
+
+**Pulse Train $\delta_t(t)$**
+$$\delta_T(t) = \sum_{k=-\infty}^\infty \delta(t-kT)$$
+$$x^*(t) = x(t)\delta_T(t) = \sum_{k=-\infty}^\infty x(t)\delta(t-kT)$$
+Where the sampled signal is $x^*$
+What about in Laplace domain?
+$$X^*(t) = \int \left[ \sum_{k=-\infty}^\infty x(t)\delta(t-kT)\right] e^{-st} dt$$
+$$X^*(t) =\sum_{k=-\infty}^\infty \left[\int   x(t)\delta(t-kT) e^{-st} \right] dt$$
+$$X^*(t) =\sum_{k=-\infty}^\infty \left[\int   x(t)e^{-st} \delta(t-kT) \right] dt$$
+Now using the shift property...
+$$X^*(t) =\sum_{k=-\infty}^\infty x(kT)e^{-kTs} $$
+## Some Observations
+If we change the variable $z = e^{Ts}$
+>[!important] **The Z-Transform** 
+> $$X^*(t) = X(z) = \sum_{k=-\infty}^\infty x(kT) z^{-k} $$
+> Z transform can be viewed as short hand of the Laplace transform
+> 
+> Sampling is a time varying process. If x(t) is time shifted by a small amount, the sampled signal x(kT) will be different. 
+
+# Frequency Domain Interpretation
+$\delta_T(t)$ is periodic, so we can turn it into a Fourier series...
+$$\delta_T(t) = \sum_{N=-\infty}^N C_N e^{j(\frac{2\pi}{T})Nt}$$
+$$C_n = \frac{1}{T}\int_{-T/2}^{T/2} \delta_T(t) e^{-j(\frac{2\pi}{T})Nt} dt$$
+Apply a shift and do some stuff...
+
+$$C_n = \frac{1}{T} \int_{-T/2}^{T/2} e^{-j \frac{2\pi}{T} Nt} dt $$
+$$C_n = \frac{1}{T}$$
+So then...
+
+$$\delta_T(t) = \frac{1}{T} \sum_{N=-\infty}^N e^{j(\frac{2\pi}{T})Nt}$$
+$$\delta_T(t) = \frac{1}{T} \sum_{N=-\infty}^N e^{j \omega_s T Nt} $$
+Then, this can be used to get all the way to $X(z)$.
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