# Lecture 5 : Inflation

• 30
• 322
• 1

## Tài liệu Modulation and coding course- lecture 5 doc

... z = ( z1 , z , , z N ) are independent Gaussian random variables Its pdf is ⎛ z − si ⎞ ⎟ pz ( z | s i ) = exp⎜ − N /2 ⎜ N0 ⎟ (πN ) ⎝ ⎠ Lecture 5 Detection Optimum decision rule (maximum a posteriori ... Z i z Lecture i 14 Example for binary PAM pz (z | m2 ) pz (z | m1 ) s2 − Eb s1 ψ (t ) Eb ⎛ s1 − s / ⎞ ⎟ Pe (m1 ) = Pe (m2 ) = Q⎜ ⎜ N /2 ⎟ ⎝ ⎠ ⎛ Eb PB = PE (2) = Q⎜ ⎜ N ⎝ Lecture ⎞ ⎟ ⎟ ⎠ 15 Union ... deterministic Elements of noise vector n = (n1 , n2 , , nN ) are i.i.d Gaussian random variables with zero-mean and variance N / The noise vector pdf is ⎛ n2⎞ ⎟ pn (n) = exp⎜ − (πN )N / ⎜ N ⎟...
• 21
• 223
• 0

## Tài liệu LECTURE 5: SOFTWARE PROJECT MANAGEMENT - Software Engineering Mike Wooldridge pdf

... for many projects, this is the raison d’etre Mike Wooldridge Lecture Software Engineering • Activities in software project management: project planning; – project scheduling; – risk management; ... management cannot guarantee success, but poor management on signiﬁcant projects always leads to failure Mike Wooldridge Lecture Software Engineering Software projects have several properties that make ... computers meant that larger software projects were tackled with techniques developed on much smaller projects • Techniques were needed for software project management Good project management cannot guarantee...
• 18
• 336
• 2

## lecture 5 hplc

... – Normal phase HPLC – Reverse phase HPLC http://www.waters.com/WatersDivision/Contentd.asp?watersit=JDRS-6UXGZ4 Design & Operation of an HPLC Instrument Design & Operation of an HPLC Instrument ... faster runs Ex: mobile phase composition can be programmed to vary from 75% water: 25% acetonitrile at time zero to 25% water: 75% acetonitrile at the end of the run • More polar components will tend ... components will elute later in the gradient Design & Operation of an HPLC Instrument Design & Operation of an HPLC Instrument 4) HPLC pump: 5) Injector: • Fill stroke: mobile phase is pulled from the...
• 6
• 208
• 1

## CMOS VLSI Design - Lecture 5: DC & Transient Response doc

... and 5: DC and Transient Response CMOS VLSI Design 4th Ed Pass Transistor Ckts VDD VDD VDD VDD VDD VDD Vs = VDD-Vtn VDD-Vtn VDD-Vtn VDD VDD-Vtn VDD-Vtn Vs = |Vtp| VDD VDD-2Vtn VSS 5: DC and Transient ... 3C 3C 3C 5: DC and Transient Response CMOS VLSI Design 4th Ed 7C 3C 3C 35 Layout Comparison  Which layout is better? VDD A VDD B Y GND 5: DC and Transient Response A B Y GND CMOS VLSI Design 4th ... 6RC 5: DC and Transient Response CMOS VLSI Design 4th Ed 27 Delay Model Comparison 5: DC and Transient Response CMOS VLSI Design 4th Ed 28 Example: 3-input NAND  Sketch a 3-input NAND with transistor...
• 36
• 362
• 0

## Digital Communication I: Modulation and Coding Course-Lecture 5 potx

... we talked about:  Receiver structure  Impact of AWGN and ISI on the transmitted signal  Optimum filter to maximize SNR  Matched filter and correlator receiver  Signal space used for detection ... deterministic  Elements of noise vector n = (n1 , n2 , , nN ) are i.i.d Gaussian random variables with zero-mean and variance N / The noise vector pdf is  n2  pn (n) = exp − ( πN ) N /  N ... z = ( z1 , z , , z N ) are independent Gaussian random variables Its pdf is  z − si   pz ( z | s i ) = exp − N /2  N0  ( πN )   Lecture 5 Detection  Optimum decision rule (maximum a posteriori...
• 21
• 234
• 0

## Lecture 5: Interface Design pptx

... ); Extending interfaces • Interfaces support multiple inheritance – an interface can extend more than one interface • Superinterfaces and subinterfaces Example public interface SerializableRunnable ... – In the subinterface, explicitly use the superinterface name to refer to the constant of that superinterface E.g B.val); interface A int val } interface B int val } { = 1; { = 2; interface C ... (2) • If a superinterface and a subinterface contain two constants with the same name, then the one belonging to the superinterface is hidden in the subinterface – access the subinterface-version...
• 19
• 208
• 0

## Lecture 5: Phân tích làm rõ yêu cầu (Eliciting Requirements) ppt

... chủng học Phân tích ngôn từ Phân tích đàm thoại Phân tích lời nói Phương pháp xã hội học Phân tích hệ thống mềm Kỹ thuật nhận thức Phân tích công việc Phân tích giao thức Các kỹ thuật làm tri ... (để thuyết phục bạn công việc thực người!) Phân tích yêu cầu phần mềm Các kỹ thuật làm yêu cầu Kỹ thuật truyền thống Tự xem xét Đọc tài liệu Phân tích “dữ liệu cứng” Phỏng vấn Không cấu trúc ... kích thước mẫu - e.g 20 giao dịch Phân tích yêu cầu phần mềm Ví dụ liệu “cứng” (Hard data) Câu hỏi : Dữ liệu cung cấp cho bạn ? Bạn làm với liệu ? Phân tích yêu cầu phần mềm (3) Kỹ thuật vấn (Interviews)...
• 16
• 218
• 1

## CS193P - Lecture 5 doc

... system ■ 0, 55 0 View A 200, 100 200 0, 400 250 View B View A frame: origin: 0, size: 55 0 x 400 View A bounds: origin: 0, size: 55 0 x 400 View B frame: origin: 200, 100 size: 200 x 250 View B bounds: ... UIImages in -drawRect: - [UIImage drawAtPoint:(CGPoint)point] - [UIImage drawInRect:(CGRect)rect] - [UIImage drawAsPatternInRect:(CGRect)rect] • You can draw NSString in -drawRect: - [NSString ... session is a special, super-mega office hour ■ featuring Troy and Paul • To sign up for cs193p- auditors@lists.stanford.edu: ■ https:/mailman.stanford.edu/mailman/listinfo /cs193p- auditors • AT&T Big...
• 55
• 84
• 0

## LECTURE 5: MORE APPLICATIONS WITH PROBABILISTIC ANALYSIS, BINS AND BALLS ppsx

... Nguyen Probability for Computing 19 Balls into Bins We have m balls that are thrown into n bins, with the location of each ball chosen independently and uniformly at random from n possibilities What ... Agenda Review: Coupon Collector’s problem and Packet Sampling Analysis of Quick-Sort Birthday Paradox and applications The Bins and Balls Model © 2010, Quoc Le & Van Nguyen Probability ... does the distribution of the balls into the bins look like    “Birthday paradox” question: is there a bin with at least balls How many of the bins are empty? How many balls are in the fullest...
• 24
• 109
• 0

## Kinh tế môi trường - Lecture 5 ppsx

... khụng s dng Giỏn tip VND USD 108,200 86,400 15, 000,000 132,000 132 000 7.07 5. 65 980.39 8.63 63 2,860,000 18 ,50 0 La chn - - 15, 000 0.98 Tn ti - - - USD - Lu truyn 186.93 1.21 Dc tho Du lch S DNG ... 60.134.000 63. 957 .000 11.336. 650 34.620.100 12.932.720 14.4 45. 000 64. 050 .000 47.420.000 40.093.000 70.286.800 6 .58 1.240 58 1 240 10.249. 750 31 .56 5.720 12.022.700 31.1 25. 200 13.688. 450 43.192.100 ... 704 600 11.336. 650 34.620.100 12.932.720 35. 208 .50 0 16.882 .50 0 47.420.200 70.286.800 - Xó ng Rui (Tnh Tin Yờn) - Ca sụng Vn c -Ca sụng Ba Lt (Tnh Nam nh) -Bói ly Kim Sn -Tnh Ngh An -Tnh Cn Gi, Thnh...
• 18
• 147
• 0

## slide bài giảng suất chiết khấu tài chính lecture 5 - financial discount rates

... nhập doanh nghiệp!) ngân lưu TIPV, suất chiết khấu thích hợp là: WACC = %D.rd + %E.re USAID, 2009, pp.93 Lựa chọn suất chiết khấu tài Mức độ rủi ro (nguồn tài trợ) NCFEPV > NCFTIPV > NCFAEPV ... Doanh thu 1200 Đầu tư -1 000 NCFTIPV -1 000 1200 1200 − 1000 ρ= = 20% 1000 Nguồn: Joseph Tham (20 05) Chi phí sử dụng vốn Năm -1 000 1200 Ngân lưu tài trợ, rd = 8% 400 -4 32 NCFEPV -6 00 768 NCFTIPV 768 ... thuế) = (1) – (2) = (3) – (4) … n Chi phí sử dụng vốn  Có thể gọi suất chiết khấu (discount rates) lãi suất rào cản (hurdle rates)  Trong kinh tế học, chi phí sử dụng vốn (cost of capital) hiểu...
• 48
• 257
• 0

## Quan tri mang - lecture 5

... Permissions Các nhóm xây dựng sẵn     Built-in Global Groups Built-in Domain Local Groups Built-in Local Groups Built-in System Groups Built-in Global Groups   Chứa thành viên khởi tạo ... Local Groups Windows 2003 Server Windows 2003 Server (Member or Stand-Alone Server) (Member or Stand-Alone Server) Built-in Built-in Local Local Groups Groups Windows 2003 Professional Windows 2003 ... Built-in Built-in Domain Domain Local Local Groups Groups Backup Operators Backup Operators Server Operators Server Operators Administrators Administrators Guests Guests Users Users Built-in Local...
• 43
• 171
• 0

## Econometrics – lecture 5 – assumptions

...  Example:  Exp^ = 1 .5 - 0.2Income +0.1Wealth Wrong sign  Variable C P PA Coefficient Std Error t-Statistic Prob 1207.06 157 5.06 0.77 0. 45 -146.90 479. 15 -0.31 0.76 0. 15 6.34 0.02 0.98 R2 =0.91 ... Coefficient Std Error t-Statistic Prob C 3.28 0. 05 61. 35 0.00 PA 0.01 0.00 38.19 0.00 R-squared 0.99 Mean dependent var 5. 24 Log likelihood 26.27 F-statistic 1 458 .58 Durbin-Watson stat 1.27 Prob(F-statistic) ... BASIC ASSUMPTIONS E(ui) = Var(ui) = σ2 cov(ui, uj) = for i #j ui~ N(0, σ2) No perfect collinearity among independent vars Xj: non random, Y: random Assumptions     Multicollinearity...
• 20
• 57
• 0

## Chapter 5 lý thuyết mạch 1 Lecture 5 Mạch khuếch đại (OPAMP) (chapter 5)

... áp dòng điện OPAMP tưởng  Có khả phân tích mạch OPAMP tưởng đơn giản  Hiểu mạch có chứa OPAMP: khuếch đại đảo, khuếch đại tổng hợp, khuếch đại không đảo ngược, khuếch đại khác biệt  Hiểu ... vô hạn Chúng ta bắt đầu với giả sử trạng thái tưởng… Các mô hình thực tế nghiên cứu cuối chương… 5. 1 cực khuếch đại thuật toán The 7 41 Op Amp  7 41 OPAMP khởi đầu OPAMP nay(tốc độ nhanh, tiếng ... negative, vo giảm, giảm feedback loop http://www.ecircuitcenter.com/Circuits/opfeedback1/opfeedback1.htm Khuếch đại đảo  Các tín hiệu vào đảo ngược nhân với Rf/Rs ngõ ra: v p    Rf vs v0 At inv...
• 14
• 379
• 0

Xem thêm