通訊系統歷屆考題數題 - 考試
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By Carol
at 2013-08-05T21:43
at 2013-08-05T21:43
Table of Contents
[考題] 國考歷屆考題與考題觀念討論(書裡看到的選這個)請附上想法、出處
以下解答請幫我檢驗一下是否正確,謝謝。
101調特:
以下連結為考選部考畢試題
http://wwwc.moex.gov.tw/ExamQuesFiles/Question/101/101140_40950.pdf
第三題:
(一)
X(f)=rect(f/2)
Y(f)=rect(f/2)*[-jsgn(f)]
=jrect(f+0.5)-jrect(f-0.5)
y(t)=jsinc(t)exp(-jπt)-jsinc(t)exp(jπt)
=jsinc(t)[exp(-jπt)-exp(jπt)]
=jsinc(t)[-2jsin(πt)]
=2sin(πt)sinc(t)
=[1-cos(2πt)]/(πt)
=============================================================================
97調特:
以下連結為考選部考畢試題
http://wwwc.moex.gov.tw/ExamQuesFiles/Question/097/026306300.pdf
第一題:
(二)
由於打出係數太雜亂,因此只敘述我解題的想法。
題目給的是一個週期性的方波,高值為1,低值為-1,週期1/3000。
我把x(t)表示為振幅為2(即高值為2,低值為0)的週期性方波,然後整體減一(即下降1,
成為高值為1,低值為-1)。化為頻域則為「無限個sinc函數相加-δ(f)」,接下來當然
是乘上|H(f)|∠H(f),但是「無限個sinc函數相加」我還真不知道怎麼處理...@@"
請問是有其他的作法來解嗎?謝謝。
--
以下解答請幫我檢驗一下是否正確,謝謝。
101調特:
以下連結為考選部考畢試題
http://wwwc.moex.gov.tw/ExamQuesFiles/Question/101/101140_40950.pdf
第三題:
(一)
X(f)=rect(f/2)
Y(f)=rect(f/2)*[-jsgn(f)]
=jrect(f+0.5)-jrect(f-0.5)
y(t)=jsinc(t)exp(-jπt)-jsinc(t)exp(jπt)
=jsinc(t)[exp(-jπt)-exp(jπt)]
=jsinc(t)[-2jsin(πt)]
=2sin(πt)sinc(t)
=[1-cos(2πt)]/(πt)
=============================================================================
97調特:
以下連結為考選部考畢試題
http://wwwc.moex.gov.tw/ExamQuesFiles/Question/097/026306300.pdf
第一題:
(二)
由於打出係數太雜亂,因此只敘述我解題的想法。
題目給的是一個週期性的方波,高值為1,低值為-1,週期1/3000。
我把x(t)表示為振幅為2(即高值為2,低值為0)的週期性方波,然後整體減一(即下降1,
成為高值為1,低值為-1)。化為頻域則為「無限個sinc函數相加-δ(f)」,接下來當然
是乘上|H(f)|∠H(f),但是「無限個sinc函數相加」我還真不知道怎麼處理...@@"
請問是有其他的作法來解嗎?謝謝。
--
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考試
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