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Chapter Data & Signal


Tugas  Chapter Data & Signal, saya kebagian membahas 4 soal ini.

Questions :

  • Draw an example signal (similar to those shown in figure 2-12) using NRZI in which the signal never changes for 7 bits. What does the equivalent differential manchester encoding look like ?
  • Twenty four voice signals are to be transmitted over a single high speed telephone line. what is the bandwidth required (in bps) if the standard analog to digital sampling rate is used and each sample is converted into an 8-bit value?
  • Draw or give an example of a signal for each of the following conditions: the baud rate is equal to the bit rate, the baud rate is greater than the bit rate, and the baud rate is less than the bit rate.
  • Using shannon’s theorem, calculate the data transfer rate given the following information : signal   frequency = 10000 hz, signal power = 5000 watts, noise power = 230 watts.

  1. untuk NRZ-I, pemetaan data biner menjadi physical signal dengan persyaratan sebagai berikut :
    1. Bernilai 1 apabila ada transisi physical signal (low ke high atau high ke low)
    2. Bernilai 0 apabila tidak ada transisi physical signal

Sehingga untuk signal yang tidak berubah selama 7 bit adalah :

Untuk encoding Differential Manchester akan selalu ada transisi pada pertengahan interval dengan prasyarat adalah tidak mempersoalkan antara bit 1 dan bit 0 namun lebih menekankan pada transisi di awal interval, apabila ada transisi di awal interval maka nilai 0 akan ditransmisikan, apabila tidak ada transisi diawal interval, maka nilai 1 akan ditransmisikan.

5.  24 buah sinyal suara dikirimkan melalui sebuah line telepon high speed, Satu sinyal disampling menjadi 8 bit,

jumlah 24 sinyal menjadi :24  x 8 = 192 bit
1   bit     ~ 1 second
192 bit     ~ 192 second
Maka besar bandwidth-nya : 192 bps

6.   a. Baud rate = Bit rate

Baud rate = 8 ;  Bit rate = 8

        b. Baud rate >   bit rate

Baud rate = 16; Bit rate = 8

   c. Baud rate <  bit rate

Baud rate = 1; Bit rate = 8

22.

Signal frekuensi = 10000 hz
Signal power  = 5000 Watt
Noise power  =  230 Watt
Data rate =  f . Log2 ( 1 + S/N)
= (10000) * (log2 (1 + 5000/230)
= (10000) * ( log2 22,74)
= 10000 * ( log 22,74)/(0,301)
= 10000 * 1,36/0,301
= 43767,4 bps
= 43,77 kbps

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