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REVISION MODE
MODUL 6 : FIBER OPTIC
EXAMPLES 1:

 Penyambungan kabel gentian optik pada jarak 20km
 mempunyai kehilangan sebanyak 0.5dB/km.

  Setelah diperiksa dengan menggunakan meter kuasa,
 didapati jumlah keluaran,Po adalah 20W manakala,
 kuasa masukan,Pi adalah separuh daripada kuasa
 keluaran. Kirakan jumlah kuasa yang hilang dalam
 gentian.                                (5 Markah)
ANSWER 1:
 Po = 20W
 Pi = ½(20W)
    = 10W
Kehilangan kabel = 20km x 0.5dB/km
                   = 10 dB
                dB = 10log10 Po/Pi
                dB = 10log10 20W/10W
                dB = 3.01 dB
Total link loss    = 10 dB + 3.01 dB
                   = 13.01 dB
EXAMPLES 2:
 Suatu talian gentian optik dipasang dengan kaedah
 suapbalik sepanjang 10km dan diketahui bahawa
 kehilangan pada patch cord adalah sebanyak 0.5dB
 manakala, kehilangan suapbalik bernilai separuh
 daripada kehilangan patch cord. Kirakan nilai
 kehilangan yang disebabkan oleh peranti perantaraan
 dalam sistem tersebut jika gentian tersebut
 mempunyai kehilangan 1.5 dB/km.
ANSWER 2:
Kehilangan kabel / gentian =     10km x 1.5dB/km
                           =     15dB

Kehilangan patch cord     =      0.5 dB
Kehilangan suap balik     =      0.5/2 dB

Kehilangan disebabkan oleh peranti perantaraan
= 15dB – 0.5 dB – 0.5 dB /2
= 14.25 dB

**Kehilangan kabel / gentian = Kehilangan patch cord
                               (peranti penyambung) +
                               Kehilangan suap balik +
                               Kehilangan peranti perantaraan
EXAMPLES 3:
 Satu gentian optik telah di pasang dengan
 menggunakan kaedah hujung ke hujung (end to end).
 Gentian tersebut mempunyai kehilangan sebanyak 8.1
 dB dan dan patch cord berkehilangan sebanyak 0.5dB
 maka dapatkan jumlah kehilangan talian tersebut?
ANSWER 3:

Dengan kaedah pengiraan “hujung ke hujung”,

Jumlah Kehilangan = Kehilangan dalam gentian +
                    Kehilangan patch
                  = 8.1 dB + 0.5 dB
                  = 8.6 dB
EXAMPLES 4:
 Penyambungan kabel gentian optik pada jarak 15km
 mempunyai kehilangan sebanyak 0.5dB/km. Setelah
 diperiksa dengan menggunakan meter kuasa, didapati
 jumlah keluaran adalah 20W manakala, kuasa
 masukan adalah separuh daripada kuasa keluaran.
 Kirakan jumlah kuasa yang hilang dalam gentian?
ANSWER 4:
Kehilangan kabel         :    15km x 0.5 db     =   7.5 dB

Kuasa keluaran:          Po   =       20W
Kuasa masukan:           Pi   =       ½ x Po    =   10W

Maka :

  dB     =    10 log Po/Pi

  dB     =    10 log 20W/10W

  dB     =    10 log 2

  dB     =    3.01 dB

Jumlah kehilangan        =    (3.01 + 7.5) dB
                         =    10.6 dB
EXAMPLES 5:
 Diberi kabel gentian optik jenis mod tunggal, di mana
 kehilangan pada gentian optik sahaja adalah 1.5dB/km.
 Manakala, kehilangan pada penyambungan pula bernilai
 0.5dB dan pada peranti penyambung (patch cord) pula
 mempunyai nilai kehilangan yang normal sebanyak 0.5 dB.
 Dapatkan jumlah kehilangan sekirannya panjang gentian
 optik sepanjang 10km dan menggunakan dua patch cord?
ANSWER 5:
 Kehilangan pada gentian : 10 km x 1.5 dB
                          = 15 dB

 Kehilangan pada patch cord : 2unit x 0.5 dB
                         = 1 dB

 Kehilangan pada splicing: 2 unit splicing x 0.5dB
                          = 1 dB

 Maka jumlah keseluruhan : 15 dB + 1dB + 1 dB
                           = 17 dB
EXAMPLES 6:
      Satu rangkaian gentian optik sepanjang 8km disambung di antara
      sumber cahaya dan pengesan cahaya. Kuasa pada sumber cahaya
      ialah 120 μW dan kuasa pada pengesan cahaya ialah 3 μW. Kirakan:
i.      Jumlah kehilangan dalam dB jika tiada splice dan penyambung di
        sepanjang kabel tersebut. (2 markah)
ii.     Kadar kehilangan kabel dalam unit dB/km. (2 markah)
iii.    Jumlah kehilangan jika kabel gentian optik sepanjang 10km yang
        digunakan dipisahkan oleh penyambung bagi setiap jarak 1km.
        Setiap penyambung mempunyai kehilangan 1dB. (3 markah)
iv.      Kuasa masukan jika diberi kuasa keluaran = 1mW. (3 markah)
ANSWER 6:
 i. dB = 10 log10 (Po / Pi)
            = 10 log10 (3uW / 120uW)
            = -16.0206 dB
 ii. Total link loss = length fiber in km x fiber loss in dB/km
      -16.0205 dB = 8km x fiber loss in dB/km
      fiber loss in dB/km = -16.0206 dB / 8km
                            = -2.0026 dB/km
 iii. Connector loss = 10 x 1dB = 10dB
       Fiber loss = -2.0026 dB/km x 10km = -20.026 dB
       Total link loss = Connector loss + Fiber loss
                        = 10 dB + -20.026 dB = -10.026 dB
ANSWER 6:
 iv. dB = 10 log10 (Po / Pi)
   -16.0206 dB = 10 log10 (1mW / Pi)
   Pi = 0.04 W
EXAMPLES 7:
 A fiber optic communication system using
 four connectors with loss of 0.5dB each. The
 bend loss is 0.2dB and fiber loss is 2dB/km.
 Minimum power received is -20dBm.
 Calculate total link loss & power light source
 in dBm and watt, if the transmission distance
 is 15km.
ANSWER 7:
 Pt = Pr + Total Losses
 Pt = Power transmitter = Power Light source (dBm)
 Pr = Power received (dBm)

 Fiber Losses      = 15km x 2dB/km
                   = 30dB
 Bend Loss         = 0.2dB
 Connector loss = 4 x 0.5dB
                   = 2dB
 Total link losses = Fiber Losses + Bend Losses +
                     Connector loss
                   = 30dB + 0.2dB + 2dB
                   = 32.2dB
ANSWER 7:
 Pt = Pr + Total Losses
 Pt = -20dBm + 32.2dB
 Power Source Light in dBm, Pt = 12.2dBm


 Pt = 10 log10 xW /1mW
 12.2dBm = 10 log10 xW /1mW
 log10 xW /1mW = 12.2 /10

 xW /1mW = Antilog 1.22
 xW = 16.6 x 1m
  Power Source Light in watt, xW = 16.6 m W
EXAMPLES 8:
 Suatu sistem komunikasi gentian optik menggunakan
 2 pengganding yang menyebabkan kehilangan 0.5dB
 tiap satu.
 Kehilangan gentian ialah 1dB/km. Kuasa minima yang
 perlu diterima ialah -20dBW.
 Kirakan kuasa pancaran sekiranya jarak penghantaran
 ialah 20km.
ANSWER 8:
 Pt = Pr + Total Losses
 Pt = Power transmitter = Power Light source (dBm)
 Pr = Power received (dBm)

 Fiber Losses      = 20km x 1dB/km
                   = 20dB
 Coupling Loss = 0.5dB x 2
                    = 1dB
 Total link losses = Fiber Losses + Coupling loss
                   = 20dB + 1dB
                   = 21dB
ANSWER 8 :
 Pt = Pr + Total Losses
 Pr = Power received in dBm = (-20dBW + 30) dBm = 10dBm
 Power Source Light in dBm, Pt = 10dBm + 21dB = 31dBm

 Pt = 10 log10 ( X /1mW)

  31dBm = 10 log10 (X /1mW)
 log10 (X/1mW) = 31/10

 (X/1mW) = Antilog 3.1
 X = 1258.9 x 1m
  Power Source Light in watt, x = 1.2589W
EXAMPLES 9:
  Determine the optical power received in dBm and watts for
  a 20km optical fiber link with the following parameters:

 LED output power of 30mW

 Loss of optical cable is 0.5dB/km

 Three cable to cable connectors with a loss of 2 dB each

 No cable splices

 Light source to fiber interface loss of 1.9dB

 Fiber to light detector loss of 2.1dB

 No losses due to cable bends
ANSWER 9:
 LED output power = 30mW
 Pt= 10 log10 (30mW / 1mW)
    = 14.7712dBm
 The cable loss is simply the product of total cable length in km and
  the loss in dB/km.
 Total cable loss = 20km x 0.5 dB/km = 10dB
 Total connector loss = 3 x 2dB = 6dB
 Total loss = Total cable loss + Total connector loss +
              Light source to cable loss + fiber to light detector loss
            = (10 + 6 + 1.9 + 2.1) dB = 20 dB
ANSWER 9:
 Pt = Power transmitter = Power Light source
    = LED output power (dBm)
 Pr = Power received (dBm)
 Total Losses = sum of losses (dB)
 Pr = Pt - Total Losses
    = 14.8dBm – 20dB = -5.2dBm

 -5.2dBm = 10 log10 (X / 1mW)
 X = 0.302mW

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Revision Mode Fiber Optic - E5122 - Sistem Komunikasi

  • 1. REVISION MODE MODUL 6 : FIBER OPTIC
  • 2.
  • 3. EXAMPLES 1:  Penyambungan kabel gentian optik pada jarak 20km mempunyai kehilangan sebanyak 0.5dB/km. Setelah diperiksa dengan menggunakan meter kuasa, didapati jumlah keluaran,Po adalah 20W manakala, kuasa masukan,Pi adalah separuh daripada kuasa keluaran. Kirakan jumlah kuasa yang hilang dalam gentian. (5 Markah)
  • 4. ANSWER 1:  Po = 20W  Pi = ½(20W) = 10W Kehilangan kabel = 20km x 0.5dB/km = 10 dB dB = 10log10 Po/Pi dB = 10log10 20W/10W dB = 3.01 dB Total link loss = 10 dB + 3.01 dB = 13.01 dB
  • 5. EXAMPLES 2:  Suatu talian gentian optik dipasang dengan kaedah suapbalik sepanjang 10km dan diketahui bahawa kehilangan pada patch cord adalah sebanyak 0.5dB manakala, kehilangan suapbalik bernilai separuh daripada kehilangan patch cord. Kirakan nilai kehilangan yang disebabkan oleh peranti perantaraan dalam sistem tersebut jika gentian tersebut mempunyai kehilangan 1.5 dB/km.
  • 6. ANSWER 2: Kehilangan kabel / gentian = 10km x 1.5dB/km = 15dB Kehilangan patch cord = 0.5 dB Kehilangan suap balik = 0.5/2 dB Kehilangan disebabkan oleh peranti perantaraan = 15dB – 0.5 dB – 0.5 dB /2 = 14.25 dB **Kehilangan kabel / gentian = Kehilangan patch cord (peranti penyambung) + Kehilangan suap balik + Kehilangan peranti perantaraan
  • 7. EXAMPLES 3: Satu gentian optik telah di pasang dengan menggunakan kaedah hujung ke hujung (end to end). Gentian tersebut mempunyai kehilangan sebanyak 8.1 dB dan dan patch cord berkehilangan sebanyak 0.5dB maka dapatkan jumlah kehilangan talian tersebut?
  • 8. ANSWER 3: Dengan kaedah pengiraan “hujung ke hujung”, Jumlah Kehilangan = Kehilangan dalam gentian + Kehilangan patch = 8.1 dB + 0.5 dB = 8.6 dB
  • 9. EXAMPLES 4: Penyambungan kabel gentian optik pada jarak 15km mempunyai kehilangan sebanyak 0.5dB/km. Setelah diperiksa dengan menggunakan meter kuasa, didapati jumlah keluaran adalah 20W manakala, kuasa masukan adalah separuh daripada kuasa keluaran. Kirakan jumlah kuasa yang hilang dalam gentian?
  • 10. ANSWER 4: Kehilangan kabel : 15km x 0.5 db = 7.5 dB Kuasa keluaran: Po = 20W Kuasa masukan: Pi = ½ x Po = 10W Maka : dB = 10 log Po/Pi dB = 10 log 20W/10W dB = 10 log 2 dB = 3.01 dB Jumlah kehilangan = (3.01 + 7.5) dB = 10.6 dB
  • 11. EXAMPLES 5: Diberi kabel gentian optik jenis mod tunggal, di mana kehilangan pada gentian optik sahaja adalah 1.5dB/km. Manakala, kehilangan pada penyambungan pula bernilai 0.5dB dan pada peranti penyambung (patch cord) pula mempunyai nilai kehilangan yang normal sebanyak 0.5 dB. Dapatkan jumlah kehilangan sekirannya panjang gentian optik sepanjang 10km dan menggunakan dua patch cord?
  • 12. ANSWER 5: Kehilangan pada gentian : 10 km x 1.5 dB = 15 dB Kehilangan pada patch cord : 2unit x 0.5 dB = 1 dB Kehilangan pada splicing: 2 unit splicing x 0.5dB = 1 dB Maka jumlah keseluruhan : 15 dB + 1dB + 1 dB = 17 dB
  • 13. EXAMPLES 6: Satu rangkaian gentian optik sepanjang 8km disambung di antara sumber cahaya dan pengesan cahaya. Kuasa pada sumber cahaya ialah 120 μW dan kuasa pada pengesan cahaya ialah 3 μW. Kirakan: i. Jumlah kehilangan dalam dB jika tiada splice dan penyambung di sepanjang kabel tersebut. (2 markah) ii. Kadar kehilangan kabel dalam unit dB/km. (2 markah) iii. Jumlah kehilangan jika kabel gentian optik sepanjang 10km yang digunakan dipisahkan oleh penyambung bagi setiap jarak 1km. Setiap penyambung mempunyai kehilangan 1dB. (3 markah) iv. Kuasa masukan jika diberi kuasa keluaran = 1mW. (3 markah)
  • 14. ANSWER 6: i. dB = 10 log10 (Po / Pi) = 10 log10 (3uW / 120uW) = -16.0206 dB ii. Total link loss = length fiber in km x fiber loss in dB/km -16.0205 dB = 8km x fiber loss in dB/km fiber loss in dB/km = -16.0206 dB / 8km = -2.0026 dB/km iii. Connector loss = 10 x 1dB = 10dB Fiber loss = -2.0026 dB/km x 10km = -20.026 dB Total link loss = Connector loss + Fiber loss = 10 dB + -20.026 dB = -10.026 dB
  • 15. ANSWER 6: iv. dB = 10 log10 (Po / Pi) -16.0206 dB = 10 log10 (1mW / Pi) Pi = 0.04 W
  • 16. EXAMPLES 7: A fiber optic communication system using four connectors with loss of 0.5dB each. The bend loss is 0.2dB and fiber loss is 2dB/km. Minimum power received is -20dBm. Calculate total link loss & power light source in dBm and watt, if the transmission distance is 15km.
  • 17. ANSWER 7: Pt = Pr + Total Losses Pt = Power transmitter = Power Light source (dBm) Pr = Power received (dBm) Fiber Losses = 15km x 2dB/km = 30dB Bend Loss = 0.2dB Connector loss = 4 x 0.5dB = 2dB Total link losses = Fiber Losses + Bend Losses + Connector loss = 30dB + 0.2dB + 2dB = 32.2dB
  • 18. ANSWER 7: Pt = Pr + Total Losses Pt = -20dBm + 32.2dB Power Source Light in dBm, Pt = 12.2dBm Pt = 10 log10 xW /1mW 12.2dBm = 10 log10 xW /1mW log10 xW /1mW = 12.2 /10 xW /1mW = Antilog 1.22 xW = 16.6 x 1m Power Source Light in watt, xW = 16.6 m W
  • 19. EXAMPLES 8: Suatu sistem komunikasi gentian optik menggunakan 2 pengganding yang menyebabkan kehilangan 0.5dB tiap satu. Kehilangan gentian ialah 1dB/km. Kuasa minima yang perlu diterima ialah -20dBW. Kirakan kuasa pancaran sekiranya jarak penghantaran ialah 20km.
  • 20. ANSWER 8: Pt = Pr + Total Losses Pt = Power transmitter = Power Light source (dBm) Pr = Power received (dBm) Fiber Losses = 20km x 1dB/km = 20dB Coupling Loss = 0.5dB x 2 = 1dB Total link losses = Fiber Losses + Coupling loss = 20dB + 1dB = 21dB
  • 21. ANSWER 8 : Pt = Pr + Total Losses Pr = Power received in dBm = (-20dBW + 30) dBm = 10dBm Power Source Light in dBm, Pt = 10dBm + 21dB = 31dBm Pt = 10 log10 ( X /1mW) 31dBm = 10 log10 (X /1mW) log10 (X/1mW) = 31/10 (X/1mW) = Antilog 3.1 X = 1258.9 x 1m Power Source Light in watt, x = 1.2589W
  • 22. EXAMPLES 9: Determine the optical power received in dBm and watts for a 20km optical fiber link with the following parameters:  LED output power of 30mW  Loss of optical cable is 0.5dB/km  Three cable to cable connectors with a loss of 2 dB each  No cable splices  Light source to fiber interface loss of 1.9dB  Fiber to light detector loss of 2.1dB  No losses due to cable bends
  • 23. ANSWER 9: LED output power = 30mW Pt= 10 log10 (30mW / 1mW) = 14.7712dBm The cable loss is simply the product of total cable length in km and the loss in dB/km. Total cable loss = 20km x 0.5 dB/km = 10dB Total connector loss = 3 x 2dB = 6dB Total loss = Total cable loss + Total connector loss + Light source to cable loss + fiber to light detector loss = (10 + 6 + 1.9 + 2.1) dB = 20 dB
  • 24. ANSWER 9: Pt = Power transmitter = Power Light source = LED output power (dBm) Pr = Power received (dBm) Total Losses = sum of losses (dB) Pr = Pt - Total Losses = 14.8dBm – 20dB = -5.2dBm -5.2dBm = 10 log10 (X / 1mW) X = 0.302mW