Determine the maximum and minimum spectral frequencies recei

Determine the maximum and minimum spectral frequencies received from a stationary GSM transmitter that has a center frequency of exactly 1950.000000MHz, assuming that the receiver is traveling at speeds of: (a) 1km/hr; (b)5km/hr;(c) 100km/hr, and (d) 1000km/hr.

Solution

Answer:

Given: Center Frequency ,fc=1950 MHz

fmin=? fmax=?

a) V =1 Km/hr    

= c/fc =3*10⁸/ 1950* 10⁶

   =2/13 m

V = 1 *( 1000/(60*60))                         1km =1000m

   = 5/18 m

fd =v/ *cos

fd max = v/ = (5/18)/(2/13)

             =1.81 Hz

fmin =fc – fd max

          = 1950*10⁶ – 1.81

          = 1949.999998M Hz

fmax = fc + fd max

          =1950 * 10⁶ + 1.81

          = 1949.000002 M Hz

b) V =5 Km/hr           

= c/fc =3*10⁸/ 1950* 10⁶

   =2/13 m

V = 5 *( 1000/(60*60))                        

   = 25/18 m

fd =v/ *cos

fd max = v/ = (25/18)/(2/13)

             =9.03 Hz

fmin =fc – fd max

          = 1950*10⁶ – 9.03

          = 1949.999991M Hz

fmax = fc + fd max

          =1950 * 10⁶ + 9.03

          = 1950.000009M Hz

c) V =100 Km/hr        

= c/fc =3*10⁸/ 1950* 10⁶

   =2/13 m

V = 100 *( 1000/(60*60))                         1km =1000m

   = 250/9 m

fd =v/ *cos

fd max = v/ = (250/9)/(2/13)

             =181 Hz

fmin =fc – fd max

          = 1950*10⁶ – 181

          = 1949.999819M Hz

fmax = fc + fd max

          =1950 * 10⁶ + 1.81

          = 1950.000181 M Hz

d) V =1000Km/hr      

= c/fc =3*10⁸/ 1950* 10⁶

   =2/13 m

V = 1000 *( 1000/(60*60))                         1km =1000m

   = 2500/9 m

fd =v/ *cos

fd max = v/ = (2500/9)/(2/13)

             =1805.6 Hz

fmin =fc – fd max

          = 1950*10⁶ – 1805.6

          = 1949.998194M Hz

fmax = fc + fd max

          =1950 * 10⁶ + 1805.6

          = 1950.001806 M Hz