Calculation Of Effective Isotropic Radiated Power And Bandwidth Efficiency In RF Communication

1. Determine the wavelength for each of the following frequencies 

1. 702kHz (AM radio)

       Wavelength= Speed/ frequency

                     = 

                    =4.2735m

1. 1800MHz (Cellular phone) 

Wavelength=   

             =0.167m 

• 2.4GHz (Wi-Fi) 

1GHz=1000000000 cycles (Smalling, 2010, p. 456)

2.4=2.4 x 1000000000

=2400000000cycle

=

wavelength=1.25m

1. 72GHz (planned 5G band)

           =

             Wavelength=0.04166m

1. What is the EIRP (dBm) for transmitter antenna.

         Using EIRP Calculator

         Rada Output=150

        Cable loss =0

         Antenna gain= 9.2

         Result= Effective Isotropic Radiated Power 159.2dB 

1. Maximum possible bit rate without errors

                      Bandwidth=20MHz

                       SNR=18Db

         Assume the noise is addictive white Gaussian noise (Kramer, 2015, p. 126)

                        From NY Quist theorem

                     C= 2 B log₂ 2ⁿ

                                    C is the capacity in bps and B bandwidth in Hz

                                18=2 x 20 log 20x

                            = 32 

1. A 16 QAM modulated signal of bandwidth 10MHz 

1. What is baud rate of signal?

=10000 divide by 16

=625 bps 

1. What is the data rate of signal? 

                                                N is the number of significant modulation and T is the unit interval

                                                  = log₂ 16 divide by 10

                                                      = 32 divide by 10

                                                           =3.2

1. What is the bandwidth efficiency?

                 Bandwidth efficiency=

                                                    =

                                                     =625bps

1. Minimum number oscillation points           

Data rate= 30Mbps (Smalling, 2010, p. 123)

Bandwidth =10MHz

              =   

                   Oscillations = 300 oscillations 

1. Determine the received power distance between transit and antenna in free space communication.

                  Received distance= 900 divide by (8dBi +2dBi)

                                                 =900 divide by 10

                                            =90m (Adwera, 2013, p. 45) 

1. What is the loss in dB at a distance of 3km, reference distance 100km at 45dB and path loss index 3.8? 

             = 1ookm +3 km

                  = 103km divide by 2

             = 51.5 km

     Log 51.5 equals to 1.711807

                  1.711807 x 3.8

              =6.5018 

2. Determine Eb/No 

SNR (Db) =Eb x bit rate+10 log base 10

Eb no = 0.4 divide with 0.5

             = 0.8

1. Determine bit error rate from the diagram

Error rate=0.3

1. What is the probability that the received power is less than 2uW? 

                     == log₂( 1 +  )

                        = log₂ 11 = 3.46

                              = 3.46

Bibliography

Adwera, P., 2013. Networking. s.l.:OLMA Media Group.

Kramer, J., 2015. Communication. s.l.:Adventure Works Press.

Smalling, C., 2010. Radio Signals. s.l.:China Publishing Company.

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