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# 大學 計算機概論 網路資訊問題 煩請高手火速解答

1. If the baud rate of a line were 9600 and 16 different is signaling levels were used, what is the equivalent bit rate of the line? What will happen if 4906 different signaling levels were used for the same situation?

2. Approximately, how long does it take to download a compressed file of size 20

MB using a medium with a bandwidth of 512 Kbps?

3. If P_final is one nanoWatt(1* 10的負九次方Watts)and P_ref is one milliWatt(1* 10的負三次方 Watts), what is the gain or loss in decibels? Is this value positive or negative? Does the value represent a gain or a loss in power?

我第二題算出來是320秒，不知道對不對

其他兩題不太會算，請高手火速幫我解答，時間相當急迫!!

第一題是4096

還有可以請P大將第三題的算式列出來一下好嗎?

非常感謝您回答的這麼快

如果可以用中文的話就更好了

### 1 Answer

- prisoner26535Lv 71 decade agoFavorite Answer
(1a) bits/sym = log2(signal levels);

so the bit rate = 9600 * log2(16) = 9600*4

(1b) if the signal level is 4906 (4096?)

bit rate = 9600 * log2(4906) = 9600 * 12.26

(2) if you ignore the followings:

+ signaling errors

+ signaling overhead, such as TCP/IP headers

time t = file_size/medium_rate

=20MB/512Kb/s

=160Mb/512Kb/s

=320s

(3a) -60dB

(3b) negative

(3c) loss of 60dB.

2009-11-10 05:09:05 補充：

(3a) dB = 10 * log10(p_final/p_ref)

= 10 * log10(1x 10^-9/ 10^-3)

= 10 * (-6)

= -60dB