作者查詢 / sqrt1089
作者 sqrt1089 在 PTT [ Grad-ProbAsk ] 看板的留言(推文), 共18則
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12F推:樓上..你是說AV那邊最後面嗎?如果是..他是以Vo為點,流03/16 21:06
13F→:出=0,而Vi/(re+0.175),是E端電流↓。再*阿法(re*gm),變03/16 21:08
14F→:Ic↓,如果要用Rpi的話,變成vi/(Rpi+(1+背他)0.175)03/16 21:09
15F→:再*背他也是Ic電流...03/16 21:10
1F推:4:我是令D^4+2D^3+3D^2+2D+1=(D^2+aD+b)(D^2+cD+d)03/15 17:04
2F→:乘開後比較係數,abcd=1==>(D^2+D+1)^2..結果發現有人回03/15 17:06
3F→:文了....03/15 17:06
3F推:其實我解的M不是用公式..我是用迴路解出來的..03/15 16:55
4F→:M=(gm+1/ro)/(1/ro+1/RL)跟公式只差分子多(1/ro)一瞇瞇..03/15 16:57
5F→:恩..英文單字....03/15 17:07
1F推:沒RL吧...RL那端不是相依...03/15 00:07
13F推:我算出來的M是2667.33,所以拆橋後RO下=50/(1-2667.33)03/15 01:20
14F→:= -0.018752,然後在跟re(約0.012376)並聯(硬並)=03/15 01:22
15F→:算出來是0.0364跟公式解0.0374很接近了...03/15 01:23
11F推:s^3+2s^2+4s+k...我用螺絲表倒出0<8<k,K=8臨界穩定02/28 22:12
21F推:X(s)/R(S)=1/s^3+2s^2+4s+k= kG(s)/(1+KG(s))02/28 22:38
22F→:可得G(s)=1/(s^3+2s^2+4s)====>可用根軌跡,不過上面的02/28 22:41
23F→:等式的左式分母就是特性方程式了,所以沒必要算出G(s)..02/28 22:42
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