深綠色箭頭為什麼那邊要用Re1不能用Rπ 淺綠色螢光筆框框處為什麼有負號

如圖所示的三級串接負回授结 構,已知Emiz=100mA/V, ign = 20 mA/V, h% = 100 台 「如,試求其迴路增益。 [4] =r=h g = 100 100=1kQr₁ = h 電子學第四冊 IV V=0 A₁B₁ = 例題 14-8 =-Bz2 關於迴路增益 AA 測試,其相關電路 關閉信號源 其中,因Q 共射放大器為單向,該處的輸入電阻與其他無關,所以 最佳切開點選在「Q的輸入端」 R = R | = R || 1+h -V =100× R9kQ X T R1002 x x x 關聯V跳 vi I I x X 分析步驟: (a)切開閉迴路,切口看人等效電阻R=r=k2 10002 R R₂ + R₁₁ 5 5+13.75 R. 392 {R x101x -Rex(1+h)x R (b) 在響應端並聯一電阻R,並在測試端加上電壓源V ey (C)為了便於分析,於是先定義出各處等效電阻,加以整理,如圖所示 者,分別是: 1₁ I V V bi x 0.1 0.649 + 0.1 R T V co Re5kn 第十四章 負回授放大器 1- R₁ = ₁; +(1+h) (R₁ + R₂)||R₁|=5+ 101×00.649||0.1) rex 21 =13.75k2 (d)迴路增益分析:算法如下所列 Q方面:採用轉導放大,將信號轉成電流信號,再計算進化的 分流。不過,對於Q 而言,此分流為倒流者 Q方面:為射極追隨器,採用電流放大論點,再以分流觀念,計算 流入Q輸入端的電流 V R, www 6402 =h/g=100 20=5k$2. 1000 AI -=100|| 392 Q方面:為共基放大器,因輸入端有分支,遂用轉導放大觀念牙計 算出響應端電壓V RE Rask R₂ www 6402 (R+R) + R₂ Rea0.6k 1002 Re × 0.009×100X R 0.6k25R2 9×1 9+1 V 1000 Re ×R ×g (RRY = 291.27 說明: (1)此迴路增益A3與R 無關,這是因為不考慮r 之故。 (2)若考慮時,則將會干擾R 與此刻的人,而干擾的程度與尺, 門 1. 21 有關。因此,這時的迴路增益A 值將有關於R,所幸,此關聯 性甚弱,遂可忽略之。

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

請問要怎麼把圖一的表格匯入到phpmyadmin裡呢？我結構是設定圖二這樣，也有試過load data infile了是圖三那樣但都錯誤，拜託大家幫我解惑，想了兩天還是沒有頭緒。

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請教我這一題這題有點難

5. O 根據表1~表2, (1)請求出管制圖中的R、又 (2)請求出管制上、下界與中心線 (3) 繪出X管制圖 表1 電腦製造過程之觀測值 組別 數據 1 6 10 2 10 3 7 10 4 8 5 9 10 6 12 10 7 16 10 8 7 10 9 9 10 15 16 10 N00 TTTT 3489111571 6 15 11 5 13 13 10 库116116718181 OO oww ON OL 9 NO Olo 4 12 13 WNI 表2 建立計量值管制圖因子 ** 反三均售重 OS R全拒害 警告三 =美的三表 89 三 B1 B2 B3 B4 d2 d3 D1 D2 D3 D4 #min 1.843 0 3.267 1.128 0.853 0 3.686 3.257 A A1 A2 C2 2.121 3.760 1.880 0.5642 3 1.732 2.394 1.023 0.7236 1.500 1.890 0.729 0.7979 1.342 1.596 0.577 0.8407 1.858 0 2.568 1.693 0.888 0 4.358 2.575 loooo 0 1.808 0 2.266 2.059 0.880 4.698 2.282 O 1.756 2.089 2.326 0.864 4.918 2.115 1.225 1.4100483 0.8586 0.026 1.711 0.030 1.970 2.534 0.848 5.078 0 2.004 7 1.134 1.277 0.419 0.8882 0.105 1.672 0.118 1.882 2.704 0.833 0.205 5.203 0.076 1.924 8 1.061 1.175 0.373 0.9027 0.167 1.638 0.185 1.815 2.847 0.820 0.387 5.307 0.136 1.864 9 1.000 1.094 0.337 0.9139 0.210 1.609 0.239 1.761 2.970 0.808 0.546 5.394 0.184 1.816 10 0.949 1.028 0.308 0.9227 0.262 1.584 0.284 1.716 3.078 0.797 0.687 5.469 0.233 1.777 110.905 0.973 0.285 0.9300 0.299 1.561 0.321 1.679 3.173 0.787 0.8125.534 0.256 1.744 120.866 0.925 0.266 0.9359 0.331 1.5410.3541.645 3.258 0.778 0.924 5.592 0.284 1.716 13 0.832 0.884 0.249 0.9410 0.359 1.523 0.382 1.618 3.336 0.770 1.026 5.646 0.308 1.692 14 0.802 0.848 0.285 0.9453 0.3841.507 0.405 1.594 3.407 0.762 1.121 5.693 0.329 1.671 150.775 0.816 0.223 0.9490 0.406 1492 0.428 1.572 3.472 0.755 1.207 5.737 0.348 1.652 16 0.750 0.788 0.2120.9523 0.4271.478 0.448 1.552 3.532 0.749 1.285 5.779 0.364 1.636 170.728 0.762 0.203 0.9551 0.4451.465 0.465 1.5343.588 0.743 1.359 5.817 0.3791.621 180.707 0.738 0.1940.9576 0.461 1.454 0.4821.5183.640 10.738 1.426 5.854 0.392 1.606 190.688 0.717 0.1870.9599 0.4771.443 0.4971503 2.689 0.733 1.49 5.888 0.404 1.596 20 0.671 0.697 0.1800.9619 0.491 1.433 0.510 1.490 3735 0.729 1548 5.922 0.414 1.586 21 0.655 0.679 0.173 0.9638 0.504 1.424 0.523 1.477 3.778 0.724 1.605 5.950 0.425 1.575 22 0.6400 0.662 0.167 0.9655 0.516 1.415 0.534 1.456 3.819 0.720 1.659 5979 0.434 1.586 23 0.525 0.647 0.162 0.9070 0.527 1.427 0.545 1.455 3.885 0.716 1710 6.006 0.443 1.557 24 0.612 0.632 0.157 0.9584 0.538 1.399 0.5551445 3.895 0.712 17596.031 0.452 1.548 25 0.500 0.9190.153 0.9696 0.548 1.392 0.5651.435 3.931 0.709 1 904 6.058 0.459 1.541

大學水文學 有人可以幫忙解答嗎？ 謝謝

| | b3 bi b2 di d3 d6 d5 (1)有一河川断面,呈測流速位置資料如下表,流速儀公式為V(m/s) =0.3 +0.2N(轉/s),試以平均 斷面法推算該斷面之流量Q(cms)? (20%) *十年到小数第2位 星期 E N J Vi KÉ 布 bi(m) di(m) (禁 (m/s) (m/s) Ai (m2) Qi ( cms) 0 0.00 1 1.2. 0.5 b4 b5 b6 b7 2 更 2. 0.5 1 1.5 0.8 4 3 1.8 0.8 5 2 0.5 6 0.2 0.00 å åt Q (cms) (2)有一集水區 400 公頃,其連續兩場各4hr (兩組圖如右圖)之親測出口流 美如下表蓋扣除其流会到中直接庭DR日,並推三日平均入文 4 (cm/hr)? (20%) (hr) 12 意三 (cms) 4 3 NN NJ | 2 24 Rim 2 2:03 下午yrp-wysr-ctu