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by: Cindy Nguyen

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# ELEG 310 Week 1 Notes ELEG310

Cindy Nguyen
UD

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First week of notes
COURSE
Random Signals and Noise
PROF.
Dr. Daniel Weile
TYPE
Class Notes
PAGES
11
WORDS
CONCEPTS
Probability, random, eleg, electrical engineering, Engineering, eleg310, eleg 310, random signals and noise, Math
KARMA
Free

## 15

1 review
"Clutch. So clutch. Thank you sooo much Cindy!!! Thanks so much for your help! Needed it bad lol"
Darrel Hahn

## Popular in Electrical Engineering

This 11 page Class Notes was uploaded by Cindy Nguyen on Sunday February 7, 2016. The Class Notes belongs to ELEG310 at University of Delaware taught by Dr. Daniel Weile in Spring 2016. Since its upload, it has received 231 views. For similar materials see Random Signals and Noise in Electrical Engineering at University of Delaware.

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## Reviews for ELEG 310 Week 1 Notes

Clutch. So clutch. Thank you sooo much Cindy!!! Thanks so much for your help! Needed it bad lol

-Darrel Hahn

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Date Created: 02/07/16
ELEG 310 – 2/9/2016 Lecture 1: Introduction to Probability Two Types of Models Deterministic Stochastic -For example, circuit theory. -You are given values, and -Used when part of system is somehow unpredictable calculate the outputs -for example: NOISE -Some inputs produce the -Quantifies uncertainty same outputs -But inevitably, this model is not exact; it neglects certain factors -Example: Ohm’s laws, V=IR So I =V/R What is Probability? Take, for example, the idea of balls in an urn. The balls are labeled 1, … , k -Likelihood of choosing ball k from the urn? DefineN (k) , where N kis the # of times we picked ball knais the # of experiments By definition, we kn0 ≤ N kn) ≤ n Divide by n0 ≤ f kn) ≤ 1 Symbols: ∈ element of ∉ Not an element of ∩ Intersects (“and”) ∪ Union (“or”) ⊂ Subset ⊄ Not a subset c A “A-complement”; “not” in A Ø Null; empty set Example: Assume k = 5 (There are 5 balls) What is the probability of pulling a perfect square out? (Perfect squares are 1 and 4) N ps) = N (1)+ N (4) fpsn) = f1(n) + f4(n) f (n) = f (n)  Relative frequency of getting perfect cube ps 1 You can never carry out ∞ (infinity) number of experiments – it’s impossible. Therefore, we adopt an axiomatic approach e.g. Getting heads in a coin flip: probability = ½ or 50% Side Note: Not all errors are statistical. -no amount of math tells you someone is cheating -e.g. Truman vs. Duey election, where sampling was wrong -no statistics saves a wrong sampling; math only goes so far. Axiomatic Approach The Axioms of Probability 1. 0 ≤ P[A] ≤ 1 2. P[S] = 1 This just means “Something happens.” All possible outcomes are in set S 3. If A ∩ B = Ø, then P[A ∪ B] = P[A] + P[B] This means: A intersects B If there is no overlap, this is “All elements in A and B” true; they are mutually Equal the null set. exclusive, disjoint subsets. If these are true, then we can do probability. Definitions to know: Set S Outcomes to an experiment A ⊂ S Subset of S; e.g. an event is a subset For every event, associate a number. Probability P[A] Z Zeta, which represents an outcome Sample space the space of possible outcomes; can be discrete or continuous Subset of a sample space An event Sample points set members Examples: Consider Random Experiments: -Result varies even when experiment is fixed -Must specify procedures and outcomes Examples from the book we’ll use: E1: Select a ball from an urn containing balls #1-50 Side Note: E3: Toss a coin 3 times. Note sequence of heads and tails. E3 and E4 are the same experiment, but w/ E4: Toss a coin 3 times. Note the # of heads different sample sets E13: Pick a number x ∈ [0,1] (meaning x is between 0 and 1) And y ∈ [0, x] (meaning y is between 0 and x) Sample Space: space of possible outcomes  Outcome of E1: ball #23 = Z  Outcome of E3: HTH = Z  Outcome of E4: 2 = Z  Outcome of E13: x=1/π , y= 1/963 …. Ordered pair (1//π , 1/963)= Z Set of all outcomes:  S1 = {1, 2, 3, … , 50}  S3 = {TTT, TTH, THT, THH, HTT, HTH, HHT, HHH}  S4 = {0, 1, 2, 3}  S13 = {}(x,y): 0 ≤ y ≤ x ≤ 1} A sample space can be discrete or continuous. (Side note: discrete doesn’t mean finite, necessarily). Subset of a sample space is an event: A ⊂ S ↔ Z∈A  Z∈S (subset A of S) (zeta is a member of A) (zeta is a member of set S) So in E1: Different events (also sets), Examples: A1 = {x, x is even} A2 = {x, x is prime} 2 Events in Probability Certain Event S Impossible event Ø (null) Review of Set Theory A ⊂ B = {x, x∈A or x∈B} A ∩ B = {x, x∈A and x∈B} A = {x, x∈A or x∈B} A ∩ B = Ø (null) A ⊂ B (A is a subset of B) A – B = {x; x∈A and x∉B} x∈A Implies  x∈B Some Rules: Subset Rule: A = B ↔ A ⊂ B ↔ B ⊂ A (Means A is a subset of B, and B is a subset of A) Commutative Rules: A ∪ B = B ∪ A (union) A ∩ B = B ∩ A (intersect) A ∪ (B ∪ C) = (A ∪ B) ∪ C (A ∩ B) ∩ C = A ∩ (B ∩ C) Distributive Rules: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) PROOF BY VENN DIAGRAM Left Hand Side: A ∪ (B ∩ C) Right Hand Side: (A ∪ B) ∩ (A ∪ C) DeMorgan’s Rules (A ∪ B) = A ∩ B c c c c (A ∩ B) = A ∪ B Can be proven with the following definition: c c (A ) =A c c c • Proof: x ∈ (A )  x ∉ A  x ∈ A st Proof of 1 DeMorgan’s Rule: Therefore, c x ∉ (A ∪ B) If x ∈ (A ∪ B) , then x ∉ A by defintion x ∉ B Therefore, Therefore, Therefore, x ∈ A c x ∈ A , x ∉ A x ∈ A c∩ Bc x ∈ Bc x ∈ B , x ∉ B Therefore, Therefore, x ∉ A ∪ B (A ∪ B) = A ∩ B c so x ∈ (A ∪ B) c 1st DeMorgan Rule nd Proof of 2 DeMorgan’s Rule: c c c c c c c If (A ∪ B ) , then (A ) ∩ (B ) A ∩ B c c c c c c c (A ∩ B) = A ∪ B ((A ∪ B ) ) 2nd DeMorgan Rule Lecture2 - 2/11/2016 Probability is a number assigned to an event Probability law: ∀ (for all) A⊂S  P[A] P[A] is a number. Mapping P[A] must have some properties. P[A] must satisfy: I. 0 ≤ P[A] II. P[S] = 1 III. If A∩B ≠ Ø (disjoint), then P[A∪B] = P[A] + P[B] Side Note: If there is an infinity set of events, then III is not sufficient. So III. If Ai Aj… is a sequence of events, and A ∩ A ≠ Øi and ij≠ j, COROLLARIES: C Corollary #1: P[A ] = 1 – P[A] Corollary #2: P[A] ≤ 1 Corollary #3: P[Ø] = 0 Corollary #4 (using I-III): If Ai∩ A ≠jØ ∀ i ≠ j, then Union of finite number of sets (not infinite) Corollary #5: *very important P[A∪B] = P[A] + P[B] – P[A∩B] (not disjoint) Corollary #5’: P[A∪B∪C] = P[A] + P[B] + P[C] – P[A∩B] – P[A∩C] – P[B∩C] – P[A∩B∩C] PROOF BY VENN DIAGRAM: P[A∪B∪C] = P[A] + P[B] + P[C] – P[A∩B] – P[A∩C] – P[B∩C] + P[A∩B∩C] Regions A, B, and C are in union and are all Since each of these This middle accounted for by pairwise intersections section has not addition. This includes their overlaps (football-shaped) been accounted regions are added for; thus it is (pairwise twice (2X), they each added in. intersections) must be subtracted once so that they are only accounted for once. Corollary #6: Corollary #7: If A⊂B then P[A] ≤ P[B] (if A happened, we know B happened.) Example of “Space is Discrete” Let S = {a1, a2, 3 , … an} Suppose we know the probabilities of each atomic outcome (an atomic outcome is the most basic, unbreakable event). B is a Subset of S: B = {b 1 b2, b3, … bn} ⊂ S Then P[B]= P[{b }]1+ P[{b }]2+ P[{b }]3… P[{b }]n Example: An Urn has 10 Balls Balls 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 each equally likely chosen, 1/10 Let A: {selected ball is odd} This is an event. Let B: {selected ball is a multiple of 3} This is another event. A = {1, 3, 5, 7, 9} P[A] = 5/10 = ½ B = {0, 3, 6, 9} P[B] = 4/10 = 2/5 A∪B = {0, 1, 3, 5, 6, 7, 9} P[A∪B] = 7/10 (odd OR multiple of 3) A∩B = {3, 9} P[A∩B] = 2/10 = 1/5 (odd AND multiple of 3) Check Corollary #5: P[A∪B] = P[A] + P[B] – P[A∩B] 7/10 = 5/10 + 4/10 – 2/10 7/10 = 7/10

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