自動控制 求解

Design a compensator such that the damping ratio < = 0.5 and the undamped natural frequency Wn 2 rad/sec. R(s) + G.(s) Compensator (a) Dominant closed-loop poles? (b) Angle deficiency? (show the diagram) (c) Lead or lag compensator? 1 0.1s+1 1 52 G(s) → C(s)

左邊紅框處不是應該放summationX*×Y嗎為何是summationXY阿？

例題 2017 另一種形式的衡量誤差 (10%) The parameter ß is defined in the model Y₁ = BX + u₁, where u; is independent of |X₁, E(u₁) = 0, E(u²) = 0², the observables are (Y₁, X₁) where X₁ = 2 random measurement error. Assume that v; is independent of X and Vi Uj. Also assume that X; and X are non-negative and real-valued. Consider the least-squares estimator for B. "Jon at Y ni 10713 2. (5%) Find a condition under which ß is consistent for ß? 1. (5%) Find the probability limit of ß, expressed in terms of ß and moments of (X (Xi, Vis Uj). 《108 台大經研》 ΣX; Y; ΣX² 《解》 | 已知 X = X*v, 將 X* = X/v 代入得] Y = BX* +u = BX/v+u。此爲無截距模型, 故 2X* ( ) Σ₁ X₁Y;/n_p_E(XY) B 將 X = X*v,且 v - X*, v lu 代入可得到 BP B ΣX²/n i=1 ¸E [(X*v)² /v]_E[X*vu] E(v) = BE(v²) E(X²) U = (S), X)-(SA EXB- + u V E (X²) atlo) co Beta, u) + Cov (Beta, e lo) 1mm wod2 (2) € Xv₁ and v₁ > o is 4)] E(X²/v)_ E(Xu) = B- E(X²) E(X²) 区一这三 E[(X*)²] E(v) __E(X*)E(v)E(u) + = E[(X*v)²] E [(X*v)²]E[(X+)²] E(v²) E[(X*)²] E(v²) = ß ↔ E(v) = E(v²) + +

請問空英周考卷有辦法拿到嗎？2023年1～6月，有償

四中英语歌室 Studio Classroom 12 ntary DECEMBER 2022 AS TO 請中藥行 www 初級/中級 週考卷 Teacher's & ediate Weekly Test Complimentary Copy 贈品不得到信 StateClatroom 生中英语歌室 Studio Classroom 12 DECEMBER 2022 建議售價 90 元 Teachio 單元卷 Unit Test | 下載更環保 YDR 美国中笑语驶室 Studio Classroom 學習更便料 12 DECEMBER 2022 雜誌補充教材 Grammar Review 25 26 2

求題目的過程與解答，問題四延續問題三，謝謝

Problem 3 Creating a 0.1 Normal Solution Suppose you must determine the amount of hydrogen halides (HCI, HBr and HF) in the flue gas leaving a chemical reactor. The emission sampling train for hydrogen halide determination calls for a total of 200 mL of 0.1 N H₂SO4 as an absorbing solution. The absorbing solution will be located on the impingers of the sampling train. How many grams of H₂SO4 should be added to water to create 200 mL of a 0.1 N H₂SO4 solution? > The chemical reaction of H₂O and H₂SO4 is as follows: H₂SO4+H₂O 2H++SO₁²+H₂O 4 > Approximate atomic weights of elements: H=1, S=32, O=16

陰離子的電子親和力不是跟游離能差一個負號嗎？為何不能選C

16.The electron affinity of fluorine is essentially equal to the negative of the ionization energy F. the ionization energy F. the negative of the ionization energy F™. the ionization energy Ne. the negative of the ionization energy Ne. B A) B) C) D) E)

英文 想請問17題，複合代名詞究竟該如何判斷意思及該選的選項呢？謝謝🙏

:20.透透組 While once known for its people's good looks and slim body figures, Brazil is now confronting surging obesity rates. Almost one in four Brazilian adults is obese, a doubled rate in the past 20 years. It has led to a movement to protect overweight people 16. being discriminated. This issue is also at the center of the heated debates on one of the Brazil's most popular TV programs. This country, in media, has been featuring 17. women in ads and on catwalks. In addition to the fashion industry, other organizations and businesses are taking action. Hospitals are purchasing larger beds and MRI machines. Movie theaters are offering wider seats. At school, a day is 18. to promote the rights of overweight 這一天被預留下来 people. Laws are changing as well. Across the country, ese people are now 19. seats on public transportation and priority at certain places like banks. Besides, a comedian was fined $1,000 for making jokes about a dancer's weight. Though obese people are being treated more and more kindly, some experts urge the government to address the issues triggered by obesity from the perspective of health. 20. so have related health problems, such as diabetes, stroke, and sleep apnea. Luckily, Brazilian governments have launched campaigns encouraging a healthier diet, financed exercise spaces in Brazilian towns, and promoted more nutritious meals in school cafeterias. 16. (A) not (A) full-figuring 18. (A) lining up 19. (A) deprived of BRO CR (B) for (B) full-figured (B) kicked off B) characteristic of (C) with adi. slim-figuring (C) set aside entitled to D) from (D) slim-figured (D) following suit BIF (D) withdrawn from

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各位大神 求解 拜託🙏🆘

CA 梁柱 一截面積是原來的兩倍 22. Two girders are made of the same material. Girder A is twice as long as girder B and Bo DE C B³ has a cross-sectional area that is twice as great. The ratio of the mass density of girder A to the mass density of girder B is: 質量密度 A) 4 B) 2 C) 1 D) 1/2 E) 1/4 29. During a short interval of time the speed v in m/s of an automobile is given by v=at² + bt³, where the time t is in seconds. The units of a and b are respectively: A) m-s²; m-s4 s³/m; s4/m dv=at+bt² B) C) m/s²; m/s³ D) m/s³; m/s4 E) m/s4; m/s5 34. The position y of a particle moving along the y axis depends on the time t according to the equation B) C) D) E) 質固定,体2倍 #%# 40. The coordinate of an object is given as a function of time by x = 7t-31², where x is in meters and t is in seconds. Its average velocity over the interval from t = 0 to t = 2 s is: A) 5 m/s X₂=0, X₁₂=14-12=2 B) -5 m/s 11 m/s C) D) -11 m/s (2-0)/2=1 E) -14.5 m/s 41. The coordinate of a particle in meters is given by x(t) = 16t-3.01³, where the time t is in seconds. The particle is momentarily at rest at t = A) 0.75 s A y = at -bt². The dimensions of the quantities a and b are respectively: A) L²/T, L³/T2 B) L/T², L²/T C) L/T, L/T² D) L³/T, T²/L E) none of these 1.3 s 5.3 s 7.3 s 9.3 s V(t) = 16-9+² =- [9t²-16) =-(3t+4) 13t-4) =+= 0.75 45. The velocity of an object is given as a function of time by v = 4t-312, where v is in m/s and t is in seconds. Its average velocity over the interval from t=0 to t=2 s: A) is 0 B) is -2 m/s C) is 2 m/s D) is -4 m/s E) cannot be calculated unless the initial position is given 49. A particle moves along the x axis according to the equation x = 61² where x is in meters and t is in seconds. Therefore: E B) C) D) E) A) the acceleration of the particle is 6 m/s² t cannot be negative the particle follows a parabolic path each second the velocity of the particle changes by 9.8 m/s none of the above

請問b跟d 🙏

Chan 1.18 Problems Perform subtraction on the given unsigned binary numbers using the 2's complement of the subtrahend. If the result of subtraction is negative, its decimal equivalent is formed by taking the 2s complement of the result and including a minus sign.SE.F (a) 11001 10010 bra (b) 1100 - 111100 (d) 1100011 (c) 10101 - 11011 10001 55 grow rorrorOFFIC 1.19* The following decimal numbers are shown in signed-magnitude form: +9,286 and +801. Convert them to signed-10's-complement form and perform the following operations (note that the sum is +10,627 and requires five digits and a sign). (a) (+9,286) + (+801) id (b) (+9,286) + (−801) (d) (-9,286) + (−801)

請問17 然後 小數的1、2補數跟9、10補數如何計算呢thx!

TUTTI (e) 11000011 1.15 Find the 9's and the 10's complement of the following decimal (a) 65,234,035 (c) 87,000,367 1.16 (a) Find the 8's complement of (2360). (b) Convert (2360), to binary. (b) 56,783,223 (d) 99,999,000 (c) Find the 2's complement of the result in (b). Minib mol (d) Convert the answer in (c) to octal and compare with the answer in (a). 1.17 Perform subtraction on the given unsigned numbers using the 10's complement of the subtrahend. Where the result should be negative, find its 10's complement and affix a minus sign. Verify your answers. (a) 7,523-4.567 (c) 224-712 raison (b) 230 - 1,204 (d) 2,390 - 945