電力系統複數疑問 - 考試
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By James
at 2014-10-27T21:56
at 2014-10-27T21:56
Table of Contents
[問題] 應考資格、各種國考疑難雜症等,以有正確作法、答案者為主
(不包括書裡的疑問)。若問題如人生規劃、讀書計畫等,無一
定作法、答案者,請用閒聊選項。
各位前輩好,
小魯日前在寫李進雲的電力系統導論時,
發現詳解部分的複數變換如下:
0.0298+j0.3172=e^(0.0298)∠18.17°
吾資駑鈍只知複數與極座標轉換的公式為X+jY=Z∠θ°
Z=(X^2+Y^2)^0.5 θ=tan^(-1)(Y/X)
照此做法應為0.3186∠84.63°
不知是否有高手可為小魯解惑?
另外若是遇到像此題的e^(0.0298)∠18.17°
或是e^(j0.46)之類的數字,
請問該如何使用計算機算出實際數值呢?
還是只能用e^(j0.46)=cos(0.46)+jsin(0.46)=Z∠θ°
再套用公式慢慢算出Z與θ值呢?
--
(不包括書裡的疑問)。若問題如人生規劃、讀書計畫等,無一
定作法、答案者,請用閒聊選項。
各位前輩好,
小魯日前在寫李進雲的電力系統導論時,
發現詳解部分的複數變換如下:
0.0298+j0.3172=e^(0.0298)∠18.17°
吾資駑鈍只知複數與極座標轉換的公式為X+jY=Z∠θ°
Z=(X^2+Y^2)^0.5 θ=tan^(-1)(Y/X)
照此做法應為0.3186∠84.63°
不知是否有高手可為小魯解惑?
另外若是遇到像此題的e^(0.0298)∠18.17°
或是e^(j0.46)之類的數字,
請問該如何使用計算機算出實際數值呢?
還是只能用e^(j0.46)=cos(0.46)+jsin(0.46)=Z∠θ°
再套用公式慢慢算出Z與θ值呢?
--
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考試
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By Olga
at 2014-10-29T21:48
at 2014-10-29T21:48
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