[ Tugas 3 SKSO ] Tugas nomor genap

 

1.2 A WDM optical transmission system is designed so that each channel has aspectral width of 0.8 nm. How many wavelength channels can be used in the C-band?

Jawab :

Sumber : wikipedia

dari tabel, terlihat bahwa rentang C band adalah 1530 – 1565 atau 35 nm. jadi banyaknya kanal yg lebar spektralnya 0.8 adalah 35/0.8 = 43.75 atau 43 kanal.

1.4  A sine wave is offset 1/6 of a cycle with respect to time zero. What is its phase in degrees and in radians?

Jawab :

sebuah siklus penuh tentu saja sudutnya adalah 360. jika diketahui bahwa sebuah gelombang sinusoidal 1/6 dari siklus, maka

1/6 * 360 = 60 derajat

lalu kita ubah ke bentuk radian

60 * 2 π /360 = 1.04 radian

1.6 What is the duration of a bit for each of the following three signals which have bit rates of 64 kb/s, 5 Mb/s, and 10 Gb/s?
jawab :

a. bitrate 64 kbps -> bit duration = 1/bitrate = 1/64000 = 0.000015625 s = 15.62 μs

b. bitrate 5 Mbps -> bit duration = 1/5000000 = 0.0000002 s = 0.2 μs

c. bitrate 10 Gbps -> bit duration = 1/(10*10^9)  = 0.000000001 s = 1 ns

1.8 (a) Convert the following absolute power levels to dBm values: 1pW, 1nW, 1mW, 10 mW, 50 mW.
(b) Convert the following dBm values to power levels in units of mW: -13 dBm, -6 dBm, 6 dBm, 17 dBm.
jawab : 

1.10 A signal passes through three cascaded amplifiers, each of which has a 5 dB gain. What is the total gain in dB? By what numerical factor is the signal amplified?

jawab :

total gain = 5 dB + 5 dB + 5 dB

                    = 15 dB

maka faktor numeriknya  = 10^(15/10)

                                                    = 31.622

1.12 A transmission line has a bandwidth of 2 MHz. If the signal-to-noise ratio at the receiving end is 20 dB, what is the maximum data rate that this line can support?

jawab : 

rumus dari maksimum data rate adalah

dengan signal to noise ratio nya

maka, maksimum data ratenya adalah

1.14 To insert low-speed signals such as 64 kb/s voice channels into a SONET frame, 84 columns in each SPE are divided into seven groups of 12 columns. Each such group is called a virtual tributary.
(a) What is the bit rate of such a virtual tributary?
(b) How many 64 kb/s voice channels can a virtual tributary accommodate?
(c) What is the payload efficiency?

jawab : 

a. diketahui, virtual tributary: 84 = 7 grup dari 12 kolom pada sebuah SPE

     –      setiap grupnya adalah virtual tributary

     –      12 x 9 x 8000 x 8 = 6.912 Mbps

b. dapat mengakomodasi 4 T1 sinyal ->   4 x 1.544 = 6.176 dan masih lebih kecil dari 6.912

c. efisiensi payload adalah rasio antara P (payload) dengan data yang terkirim (D)