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 Mik256 April 25th, 2017 04:06 PM

integral of absolute value of a Fourier transform

Hi guys,

I am going to calculate the following integral:
$$\int_0^{f_c+f_m} |Y(f)|^2\, df$$ where:

$$Y(f)=\frac{\pi}{2} \alpha_m \sum_{l=1}^{L} \sqrt{g_l}\left [ e^{-j(\omega \tau_l - \theta_m)} \delta(\omega - \omega_0) + e^{-j(\omega \tau_l + \theta_m)} \delta(\omega + \omega_0) \right ]$$

with $$\omega_0= 2\pi (f_c + f_m), \ \ \alpha_m=constant, \ \ f_c,f_m: frequencies, \ \ \theta_m: initial \ phase$$.

Then, the integral weare looking for will get the following form:

$$\int_0^{f_c+f_m} |Y(f)|^2 df= \int_o^{f_c + f_m} (\pi \alpha_m)^2 \Big|\sum_{l=1}^L \sqrt{g_l}e^{-j \omega \tau_l} \Big|^2 cos^2[2 \pi (f_c + f_m) + \theta_m]df =\\ (\pi \alpha_m)^2\int_0^{f_c+f_m} \sum_{l=1}^L g_l e^{-2j \omega \tau_l} \Big[cos^2[2 \pi (f_c + f_m) + \theta_m]\Big]df =\\ (\pi \alpha_m)^2 \Big(\sum_{l=1}^L g_l e^{-j2(2\pi) \tau_l}\Big) \Big[cos^2[2 \pi (f_c + f_m) +\theta_m] \Big] \int_0^{f_c+f_m}e^f df$$

Using a delta's Dirac property: ##\delta(\omega - \omega_0)f(\omega)= f(\omega - \omega_0)## (please correct me if it is wrong, because I have doubts about it), I got:

$$Y(f)=\frac{\pi}{2} \alpha_m \sum_{l=1}^{L} \sqrt{g_l}\left [ e^{-j[(\omega - \omega_0 )\tau_l - \theta_m)]} + e^{-j[(\omega - \omega_0) \tau_l + \theta_m)]} \right ] =\\ =\frac{\pi}{2} \alpha_m \sum_{l=1}^{L} \sqrt{g_l} e^{-j \omega \tau_l} \left [ e^{j(\omega_0\tau_l + \theta_m)} + e^{-j( \omega_0 \tau_l + \theta_m)]} \right ] =\\ =(\pi \alpha_m) \Big(\sum_{l=1}^{L} \sqrt{g_l} e^{-j \omega \tau_l} \Big) cos [2 \pi (f_c + f_m)\tau_l + \theta_m]$$

So, finally:

$$|Y(f)|^2=(\pi \alpha_m)^2 \Big|\sum_{l=1}^L \sqrt{g_l}e^{-j \omega \tau_l} \Big|^2 cos^2[2 \pi (f_c + f_m) + \theta_m]$$.

Being ## \int_0^{f_c+f_m}e^f df = e^{f_c+f_m} - 1\approx e^{f_c+f_m} ##, then:

$$\int_0^{f_c+f_m} |Y(f)|^2 df= (\pi \alpha_m)^2 \Big(\sum_{l=1}^L g_l e^{-j4 \pi (f_c + f_m) \tau_l}\Big) \Big[cos^2[2 \pi (f_c + f_m) +\theta_m] \Big]$$

My supervisor told me I am supposed to find a solution proportional to: ##\Big|\sum_{l=1}^L \sqrt{g_l}e^{j 2 \pi (f_c + f_m)\tau_l} \Big|^2##.