Real part and imaginary part

[匿名]

2011-11-26 3:40 pm

So H(f) = 1/(1-ae^-jw) , which a is a constant, e is exponential e, j is imaginary number and w is frequency "omega"

I don't understand why the real part of H(f) is: 1 -a*cos w

and

the imaginary part of H(f) is: -a* sin w

How do I know how to calculate the real part and imaginary part of function?

Thanks. I really need help with these .

更新1:

To 自由自在 : But it will be different than what my book written. So I take an image of what my book did: http://i40.tinypic.com/2cgizc.jpg I don't know if the book is incorrect or ? Thanks

回答 (1)

用戶9112

2011-11-26 7:16 pm

✔ 最佳答案

http://i1090.photobucket.com/albums/i376/Nelson_Yu/int-196.jpg

圖片參考：http://i1090.photobucket.com/albums/i376/Nelson_Yu/int-196.jpg

2011-11-26 14:47:42 補充：
My answer is correct.
But what the book means is phase angle = arctan(Im part / Real part)
Since the denominator is the same for Im and Re, they just cancel out leaving
arctan [-a sin w / (1 - a cos w)]

👍 0 👎 0