(a) Directly towards the transmitter

(b) Directly towards away from the transmitter

(c) In the direction which is perpendicular to the direction of arrival of the transmitted signal

Mumbai University > Electronics and Telecommunication > Sem 7 > Mobile Communication

Marks: 10 M

Year: Dec 2013

Given:
1) velocity of vehicle v=60miles/hr

2) Carrier frequency f=1850 MHz

Solution:

1 mile = 1609.344m

60 mile = 96560.64m/hr = 26.8224m/sec

$f = \frac{2π}{λ}$

$∴λ = \frac{2π}{f} = \frac{2π}{1850 ×10^6} = 0.1621 m$

We know Doppler shift $f_d$ is given by,

$f_d = \frac{v cosθ}{λ} ……. θ=0°$

$∴ f_d = \frac{26.8224×cos0}{0.1621}$

$ ∴ f_d = 165.40 Hz$

Case 1: when mobile moving directly toward the transmitter

Carrier frequency

$f1 = f + f_d = 1850 Mhz + 165.4 Khz = 1850.000165Mhz$

Case 2: when mobile moving directly away from the transmitter

Carrier frequency

$f2 = f - f_d = 1850 Mhz - 165.4 Khz = 1849.999835 Mhz$

Case 3: when mobile moving in the direction which is perpendicular to the direction of arrival of the transmitted signal

θ = 90°; cosθ=cos⁡ 90° = 0

$ ∴ f_d = \frac{26.8224 × cos⁡ 90}{0.1621} = {26.8224 × 0}{0.1621} = 0$

Thus carrier frequency $f3 = f + f_d = 1850 Mhz + 0 = 1850 Mhz$