0

I am trying to figure out the channel impulse of a doppler shift channel. This will lead to a frequency shift. Supposing the input signal is $x(t)$ and its Fourier transform is $X(f)$. Then the corresponding output can be $Y(f)=X(f-f_1)$. How could I get the channel response $H(f)=\frac {Y(f)} {X(f)} = \frac {X(f-f_1)} {X(f)}$?

I need the inverse Fourier transform of $H(f)$, i.e. $h(t)$.

Alex M.
  • 35,207
Iris
  • 1

1 Answers1

0

The doppler effect doesn't work that way. The shift is not a constant frequency, the shift is by a constant factor $f = ({c+v_r\over c+v_s})f_0$. So your output would rather be $Y(f) = X(f/k)$ and the inverse transform would be $h(t) = x(kt)$.

skyking
  • 16,654