Lecture03-Tolerant-retrieval

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Lecture3

A a2 b2
b arg tan( ) a
Signal space analysis
• M signal ebtries, si (t ), i 1,, M • with finite energy within [0, T) • there exists a set of real-valued and orthogonal functions, , such that
2 1 (t ) cos(2f c t ) T
T
2 2 (t ) sin(2f c t ) T
T

0
2 j
(t )dt 1 , j 1, 2

0
1
(t ) 2 (t )dt 0
y(t ) a1 (t ) b2 (t )
Complex number space versus signal space of modulated signals
• PSD of BPSK, Rectangular filter
1 Py ( f ) [ Pm ( f f c ) Pm ( f f c )] 4
Pm, BPSK ( f ) 2Eb [sin c(fTb )]2
Typical linear modulation techniques
• Quadrature PSK (QPSK)
y (t ) Es cos[( k 1) ]1 (t ) Es sin[( k 1) ]2 (t ) 2 2
yQPSK 2 (t ) E s cos[( k 1) ]1 (t ) E s sin[( k 1) ] 2 (t ) 2 4 2 4
Modulation waveform design

Circuit Analysis Lecture 3

• And that
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Power and Energy
• And we know that Current is the rate of flow of charge (C/s or Amps)
• Gives,
17
Topic 2.10
POWER
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Power
By direct substitution of Ohm’s law the equation for power can be obtained in two other forms:
IS1 = 6 A
IS2 = 4 A
IS = 10 A
RT
Same Polarity!
IS1 = 5 A
IS2 = 4 A
IS = 1 A
RT
Opposite Polarity!
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Topic 2.9
POWER AND ENERGY
14
Power and Energy
• Power is an indication of how much work can be done in a specified time • Power is the rate of doing work • The rate at which energy is delivered
Vs 5 6 30V
RL experiences the same current flow (IL) and therefore the same voltage drop (VL) when connected between ‘a’ and ‘b’ in both circuits
RS is constant Note: the direction of the arrow and positive terminal indicate the direction of current flow.

lecture_3

5 i3 = ia − ib = A 7 3 i = −i = − A 4 b 7
- 36 -
Further, if necessary, other unknown voltages and currents can also be determined, and the circuit is completely solved; If necessary, answers can be checked by power balance.
-2-
Part 1: Resistive Circuit Analysis
1. 2.
Circuit Variables and Circuit Elements Simple Resistive Circuit Analysis
3. Techniques of Circuit Analysis
4.
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Example:
Determine the currents of every branch.
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Solution:
Step 1: Select mesh current ia and ib
ia
ib
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Step 2: Build KVL equations for all meshes
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Node3-1 Node-Voltage Analysis Method What is node voltage? How is node voltage analysis method implemented?
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What is Node Voltage?
Node voltage is defined as the voltage rise from the reference node; Voltages across all branches can be represented by node voltages.

Lecture notes for Chapter 3

3.8 Linear small-signal equivalent circuits
1. DC resistance RD RD=VDQ/IDQ Q2 Q1 RD changes with the change of the current flowing through the diode. Think over: What is the relationship between the DC resistances for forward bias and reverse bias? 2. Small signal resistance rd
Chapter 2 Diodes and Diode Circuits
Outline Diode characteristics Load-line analysis The ideal-diode model Rectifier circuits Wave-shaping circuits Linear small-signal equivalent circuit Basic semiconductor concepts Physics of the junction diode
Example
Think about:
1. 2. 3. For the left part, using KVL write an equation for iD and vD. For the right part, the volt-ampere characteristics of the diode is given. If we want to solve for the current iD flowing through and the voltage vD across the diode, what does that mean?

Lecture 7 网易课教程字幕有机化学可汗学院

Lecture 7I’ve pre-drawn some molecules here. Let’s see if we can use what we know to name them. So what do we have? This first molecule right here, I have a bunch of rings. This is a one, two, three, four, five, six carbon ring. These are each four carbon rings: one, two, three, four, so the largest ring is essentially going to be our backbone. It’s going to be this six-carbon ring right here: one, two, three, four, five, six carbons. So that is a cyclohexane. All double bonds. That’s where we get the –ane. Six carbons hex-, it’s in a cycle, cyclo-, and then we have two of these four-carbon rings. So four carbons, we’re dealing with the prefix but-, but- for four. We’ve got two of them, so both of these are butyl groups. But they’re in a cycle, so they’re actually cyclobutyl groups. We have two things attached to this ring right there. If we only had one thing attached to it, you wouldn’t have to number it. But when you have two things, you start numbering at one of them. Let’s say we start numbering here, and you go in the direction so that the next group has the lowest number. So in this case, you want to go in the counterclockwise direction. If we went this way, it would be one, two, three, four, five. This guy would be a five. If we go in the counterclockwise, it would be one, two, three. This guy will only be a three. So this right here, we have two cyclobutyls, so it’s dicyclobutyl. We have two of them and they are at the one and three positions. So at the one and three position, I have two cyclobutyls on my cyclohexane main ring.Let’s try this one right here. I have a five-carbon ring, one, two, three, four, five right there, and then I have a one, two, three, four, five, six-carbon ring right there, so this is going to be the main ring. That is a cyclohexane. It has six carbons on them in a cycle, all single bonds. Attached to that, I have a cyclopentyl group, -yl for the group. This is a cyclopentyl group on it. We don’t have to number it because it’s only one group attached to the main ring. If there was another group, we would have to number it like we did up here. This is cyclopentyl. That’s this part right here. Cyclopentyl attached to cyclohexane: cyclopentylcyclohexane.Let’s try this one over here. The first thing we want to do, there’s no cycles here, but we have to identify the longest chain. To do that, let’s just count it out. It could be one, two, three, four, five, six, seven. That’s not the longest chain. Maybe it’s one, two, three, four, five, six, seven, eight. That looks like the longest chain, so let’s make that the longest chain. Let’s make that the longest chain right over there. We want to start numbering in the direction so we encounter the first attached groups first. We do want to start numbering down here, because we have groups attached right on the two carbon. If we started over here, we’d have to to pretty far until something’s attached. So we go one, two, three, four, five, six, seven, eight. We know we’re dealing with an octane, all single bonds. It’s not a cyclooctane. It’s not in a cycle, so we know we’re dealing with an octane. Now we just have to add the groups to it. So what do we have here? This right here is just one carbon attached to the main chain. This is another carbon attached to the main chain. Both of these right here are methyl groups. Meth- is for one carbon. Those are methyl groups. If you look at all of them, these are the only methyl groups. These two up here aren’t methyl groups, so we have two methyl groups on our entire chain, so it’s going to be dimethyl. Both of the methyl groups our at our two position, so this is going to be 2,2-dimethyl. This part right here is 2,2. That right there is 2,2-dimethyl. We’re going to decide what order to write it after we figure out what these are called because it has to be in alphabetical order. This is 2,2-dimethyl. The whole chain is an octane. What are these over here? How many carbons do we have here? One, two, three. They actually look the same. We have one, two, three here. We have one, two, three there, so these areboth propyl groups. If we deal with common names, this is kind of that Y shape. You could call it sec-propyl because this carbon right here that’s attached to the main chain is attached to two other carbons. But the more common one, because it forms this Y shape, is isopropyl. We have two isopropyl groups. These are both isopropyl. We could have disopropyl. They ‘re occurring at the four and five positions, two isopropyls at four and five. This is 4,5-disopropyl. That’s that group and that group right there are accounted for with this. Now we have to just figure out the order that we write it in. You ignore the di- or the tri- out front and you just look at them in alphabetical order. We have an I for isopropyl. We have an M for methyl. Let’s write the isopropyl first. I’ve actually seen some people want to go for the p, but the main thing I ignore is just the di- or the tri- in front of the isopropyl. You shouldn’t involve that. But everything after that, you do involve. So I’ll write the isopropyl first. I comes before M, so this is going to be—if we were going to write the whole thing, this is going to be 4,5-disopropyl, 2,2-dimethyl. Actually, this should be a comma here, 4,5-disopropyl-2,2-dimethyloctane. And we’re done. But this was just the common name. you might remember that when we deal with iso- or sec- or tert- butyl or propyl, that’s the common name. if we want the systematic name, we can start at where we are attached to the main chain and view that as one and then make the longest chain with that so one. And so you could say that we have a chain there and this would be both of these cases. So this is one, two carbons. two carbons, we’re dealing with an ethyl. And on the first carbon, you have a methyl attached to it. So you could also call each of these groups a 1-methylethyl instead of an isopropyl. So you can either say isopropyl for each of these groups or you could call each of them a methylethyl if you do systematic naming. We have two of these 1-methylethyl groups, just like we had two isopropyl groups. If you’re using common naming, you can say disopropyl to say you have two of these groups. When you’re using systematic naming, you don’t say di-1-methylethyl, although that would probably get the point across. You use bis-. Since we have two of them, instead of writing di-, you write bis-. That means you have two of these things right there, and it’s still in the four and the five position. If you look at it in alphabetical order now, methylethyl comes after methyl, right? So the order will now change. So now if you want to write it with systematic naming, it would be written as 2,2-dimethyl. That’s these two guys, 2,2-dimethylethyl. And then you would write this guy, so the order changed for the two groups just based on how they’re named. Bis-, and then over here you have two 1-methyl ethyl groups. I know it’s confusing, but when you just break it down, it actually makes a reasonable amount of sense. You have two of these methyl ethyl groups. Oh, sorry, I forgot where they’re located. So we have them at the four and five position, so in the four and five position, we have two, so bis-(1-methylethyl) groups. I know it’s a little daunting now, but it all makes sense when you break it down. Methylethyl groups. And then we can just add the octane at the end. Let me scroll over to the right a little bit. Octane. Now this might seem more confusing, but when you break it down, it makes sense. We have octane as be backbone. We have two methyls. They’re both sitting on the two position. So you have two methyls sitting on the two position, and then you have two 1-methylethyl groups sitting at the four and five positions. So in the four and five position, 1-methylethyl. You have an ethyl, and in the one position you have a methyl, so that’s all it’s saying. Or another way to think about it, just in case this doesn’t confuse you enough, you could call it that, or you could say 4,5-disopropyl. These two things are the same thing: common naming, systematic. Hopefully, you found that useful.。

高数Lecture_2

Geometric Series
The sum of the first n terms of this geometric progression is expressed as
qn −1 a( )(q ≠ 1) q −1
Where a is the first term of a geometric progression, and q is the geometric ratio
Mathematics for Economics and Business
Lecture 2: Mathematics of Finance Ⅱ
Terms
Simple interest Deposit Compound interest Principal Future value Geometric progression Geometric ratio Sinking fund Mortgage payments Outstanding debt Instalment 单利存款复利现值(本金） ( 终值等比数列公比偿债基金（零存整取）分期付款有待偿还的债务分期付款
r n S = P(1+ ) 100
where the future value, or the final sum, is denoted by S, the principal, or the original sum of money, is denoted by P, and the interest rate is denoted by r
Lecture Goals
After completing this lecture, you should be able to: Calculate compound interest Apply geometric series to calculate the total investment and instalments

麻省理工大学算法导论lecture02

2
(
5 + 5 2 1 + 16 16
+L geometric series
L2.17
( ) +( )
)
Introduction to Algorithms
The master method
The master method applies to recurrences of the form T(n) = a T(n/b) + f (n) , where a ≥ 1, b > 1, and f is asymptotically positive.
Day 3
Introduction to Algorithms
L2.11
Example of recursion tree
Solve T(n) = T(n/4) + T(n/2) + n2: n2 (n/4)2 T(n/16) T(n/8) (n/2)2 T(n/8) T(n/4)
Day 3
Introduction to Algorithms
(n/8)2
…
Example of recursion tree
Solve T(n) = T(n/4) + T(n/2) + n2: n2 (n/4)2 (n/16)2 Θ(1)
Day 3 Introduction to Algorithms L2.16
n2
5 n2 16 25 n 2 256
…
(n/2)2 (n/8)2 (n/4)2
Pick c1 big enough to handle the initial conditions.
Day 3 Introduction to Algorithms L2.7

学术英语理工详解答案Unit3

第11页/共53页
Unit 3 Listening to Lectures
1 Preparing for listening to a lecture
4 What are the potential effects of global warming?
The effects of global warming may be physical, ecological, social or economic. Evidence of observed climate change includes the instrumental temperature record, an increase of extreme weather events (such as hurricanes, earthquakes, volcanoes, and landslides), rising sea levels, decreased snow cover in the Northern Hemisphere, mass animal extinctions and human migrations.
Unit 3 Listening to Lectures
Unit Contents
1 Preparing for listening to a lecture 2 Paying attention to the introduction 3 Understanding the ideas through examples 4 Following a lecture through signal words 5 Memorizing the points by taking notes
1 What are the definitions of the following terms?

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Pros:
Lookup is faster than for a tree: O(1)
Cons:
No easy way to find minor variants: • judgment/judgement No prefix search [tolerant retrieval] If vocabulary keeps growing, need to occasionally do the expensive operation of rehashing everything
a-hu
hy-m
n-z
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Trees
Simplest: binary tree More usual: B-trees Trees require a standard ordering of characters and hence strings … but we standardly have one Pros: Solves the prefix problem (terms starting with hyp) Cons: Slower: O(log M) [and this requires balanced tree] Rebalancing binary trees is expensive • But B-trees mitigate the rebalancing problem
Surviving enumerated terms are then looked up in the term-
document inverted index.
Fast, space efficient (compared to permuterm).
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Processing wild-card queries
CI6226 Information Retrieval & Analysis
Lecture 03
Dictionaries and tolerant retrieval
Slides adapted from /class/cs276/
Recap of the previous lecture
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Tree: binary tree
a-m
Root
n-z
a-hu
hy-m
n-sh
si-z
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Tree: B-tree
Definition:
Every internal node has a number of children in the interval [a,b] where a, b are appropriate natural numbers, e.g., [2,4].
term.
E.g., consider the query:
se*ate AND fil*er
This may result in the execution of many Boolean AND
queries.
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B-trees handle *’s at the end of a query term
Numbers, case folding, stemming, lemmatization
Skip pointers
Encoding a tree-like structure in a postings list
Phrase queries
Biword indexes for phrases Positional indexes for phrases/proximity queries
How do we quickly look up elements at query time?
An array of struct:
char[20] int
20 bytes 4/8 bytes
Postings *
4/8 bytes
5
Dictionary data structures
Two main choices:
The type/token distinction
Terms are normalized types put in the dictionary
Tokenization problems:
Hyphens, apostrophes, compounds, Chinese
Term equivalence classing:
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Query processing
At this point, we have an enumeration of all terms in the
dictionary that match the wild-card query.
We still have to look up the postings for each enumerated
As before, we must execute a Boolean query for each
enumerated, filtered term.
Wild-cards can result in expensive query execution
very large disjunctions…e.g., pyth* AND prog*
If you encourage, “laziness” people will respond!
Search
Type your search terms, use ‘*’ if you need to. E.g., Alex* will match Alexander.
Which web search engines allow wildcard queries?
Query = hel*o X=hel, Y=o Lookup o$hel*
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Permuterm query processing
Rotate query wild-card to the right
Now use B-tree lookup as before.
Permuterm problem: ≈ quadruples lexicon size
Easy with binary tree (or B-tree) lexicon: retrieve all words in
range: mon ≤ w < moo
*mon: find words ending in “mon”: harder
Maintain an additional B-tree for terms backwards. Can retrieve all words in range: nom ≤ w < non. Exercise: from this, how can we enumerate all terms meeting the wild-card query pro*cent ?
2
This lecture
Dictionary data structures
“Tolerant” retrieval
Wild-card queries Spelling correction Soundex
3
Dictionary data structures for inverted indexes
How can we handle *’s in the middle of que could look up co* AND *tion in a B-tree and intersect the two term sets Expensive The solution:
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Bigram index example
The k-gram index finds terms based on a query consisting of
k-grams (here k=2).
$m mo on mace among among madden amortize arond
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This lecture
Dictionary data structures
“Tolerant” retrieval
Wild-card queries Spelling correction Soundex
18
Processing wild-cards
Query mon* can now be run as
$m AND mo AND on
Gets terms that match AND version of our wildcard query.
But we’d enumerate moon. Must post-filter these terms against query.
Hash table Tree
Some IR systems use hashes, some trees
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Hashes
Each vocabulary term is hashed to an integer
(We assume you’ve seen hashtables before)
Empirical observation for English.
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Bigram (k-gram) indexes
Enumerate all k-grams (sequence of k chars) occurring in any
term, e.g., from text “April is the cruelest month” we get the