Question

moving at 300 m/s and transmitting at 10 GHz, what subcarrier spacing ensures the induced Doppler spread is 10 percent of the subcarrier spacing? Answer in Hz.

Answer 1

Let us begin with the calculation of the subcarier spacing

is Subcarier Spacing

fs is sample transmission frequency

N is sample spacing

The fs is given as 10 GHz

The N is generally for FFT is taken as 64

So

= 10 GHz / 64

= 15.62 MHz

Now,

Taking Doppler Band

It is 10 percent of

So ,

Doppler difference is 1.562 MHz

f' = 1.562 Mhz

Using Doppler effect

f = (v/(v+vs) )f'

v is speed of sound waves in air

v = 343 m/s

f = (343/(343+300))1.562 MHz

f = 0.833 MHz