100電子工程技師-通訊系統 - 考試
By Susan
at 2013-08-02T21:30
at 2013-08-02T21:30
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
以下皆為不確定結果的題目,附上解題過程,敬請不吝給予指教,謝謝。
以下連結為考選部考畢試題:
http://wwwc.moex.gov.tw/ExamQuesFiles/Question/100/100230_01230.pdf
第二題:
(一)Ry(τ)=E{[X(t)-X(t-T)][X(t+τ)-X(t-T+τ)]}
=E[X(t)X(t+τ)]-E[X(t)X(t-T+τ)]-E[X(t-T)X(t+τ)]
+E[X(t-T)X(t-T+τ)]
=E[X(t)X(t+τ)]-E[X(t)X(t-T+τ)]-E[X(t-T)X(t-T+T+τ)]
+E[X(t-T)X(t-T+τ)]
=Rx(τ)-Rx(τ-T)-Rx(τ+T)+Rx(τ)
(二)並不知道該怎麼在微分狀況下求相關性,無從下筆。
第三題:
位元速率Rb=64 kbps,而4/8/16-level PAM可將2/3/4個位元綁在一起送出,所以
Rs=32/21.3/16 kbps,但是要如何計算傳輸頻寬呢?以往寫的題目大多都是以
「不產生ISI」所需的最小頻寬,也就是BW=2Rs,不過本題沒有說明,所以有疑問。
謝謝。
--
以下皆為不確定結果的題目,附上解題過程,敬請不吝給予指教,謝謝。
以下連結為考選部考畢試題:
http://wwwc.moex.gov.tw/ExamQuesFiles/Question/100/100230_01230.pdf
第二題:
(一)Ry(τ)=E{[X(t)-X(t-T)][X(t+τ)-X(t-T+τ)]}
=E[X(t)X(t+τ)]-E[X(t)X(t-T+τ)]-E[X(t-T)X(t+τ)]
+E[X(t-T)X(t-T+τ)]
=E[X(t)X(t+τ)]-E[X(t)X(t-T+τ)]-E[X(t-T)X(t-T+T+τ)]
+E[X(t-T)X(t-T+τ)]
=Rx(τ)-Rx(τ-T)-Rx(τ+T)+Rx(τ)
(二)並不知道該怎麼在微分狀況下求相關性,無從下筆。
第三題:
位元速率Rb=64 kbps,而4/8/16-level PAM可將2/3/4個位元綁在一起送出,所以
Rs=32/21.3/16 kbps,但是要如何計算傳輸頻寬呢?以往寫的題目大多都是以
「不產生ISI」所需的最小頻寬,也就是BW=2Rs,不過本題沒有說明,所以有疑問。
謝謝。
--
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考試
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at 2013-08-06T21:21
at 2013-08-06T21:21
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