Integral of absolute value of a Fourier transform

[Mik256]

Homework Statement

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

2. Homework Equations
Then, the integral weare looking for will get the following form:

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

The Attempt at a Solution

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: $\Big|\sum_{l=1}^L \sqrt{g_l}e^{j 2 \pi (f_c + f_m)\tau_l} \Big|^2$.

Could you please help me to find the right solution and where the error is?

Thank you so much for your help, I would really appreciate that![/B]

 

[eys_physics]

$\delta(\omega-\omega_0)f(\omega)=f(\omega)$

This is not correct, it should be $\int_0^{\infty} d\omega \delta(\omega-\omega_0)f(\omega)=f(\omega_0)$
Are you sure that the upper limit should be $f_c+f_m$?
Because in that case both the delta function will give zero.

 

[Mik256]

This is not correct, it should be $\int_0^{\infty} d\omega \delta(\omega-\omega_0)f(\omega)=f(\omega_0)$
Are you sure that the upper limit should be $f_c+f_m$?
Because in that case both the delta function will give zero.

Yes I am sure about it. Could you briefly explain me why I will get zero?

And, if you had an idea to solve it, would you be so kind to sketch me a solution?

Thanks for your help!

 

[eys_physics]

The statement I gave concerning the delta function in my previous post. The way to treat the delta function is to use (see https://math.stackexchange.com/questions/342743/delta-dirac-function-integral) :
$\delta(x)=1/(2\pi)\sum_{n=-\infty}^{\infty} e^{inx}$
and thus

$\delta(\omega-\omega_0)=1/(2\pi)\sum_{n=-\infty}^{\infty} e^{in(\omega-\omega_0)},$
$\delta(\omega+\omega_0)=1/(2\pi)\sum_{n=-\infty}^{\infty} e^{in(\omega+\omega_0)}.$

I cannot give you the complete solution according to the rules of this forum. But, if you need more help please tell where at the derivation you are stuck.

 

[Mik256]

Alright, this is my attempt of solution:

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

$$ \int_0^{f_c+f_m} |Y(f)|^2df= \Big(\frac{\pi}{2} \alpha_m\Big)^2 \int_0^{f_c+f_m} \Big| \sum_{l=1}^{L} \sqrt{g_l}\left [ e^{-j(\omega \tau_l - \theta_m)} \delta(\omega - \omega_0) + e^{-j(\omega \tau_l + \theta_m)} \delta(\omega + \omega_0) \right ] \Big|^2 df = $$

$$= \Big(\frac{\pi}{2} \alpha_m\Big)^2 \sum_{l=1}^{L} {g_l} \int_0^{f_c+f_m} \left [ \Big( e^{-j(\omega \tau_l - \theta_m)} \delta(\omega - \omega_0) \Big)^2 + \Big( e^{-j(\omega \tau_l + \theta_m)} \delta(\omega + \omega_0) \Big)^2 +\underbrace{2 \Big| e^{-j(\omega \tau_l - \theta_m)} e^{-j(\omega \tau_l + \theta_m)}\delta(\omega - \omega_0)\delta(\omega + \omega_0)}_{=0} \Big| \right ] df $$

The last term is equal to zero because I have the multiplication of 2 delta; then:

$$ \int_0^{f_c+f_m} |Y(f)|^2df = \Big(\frac{\pi}{2} \alpha_m\Big)^2 \sum_{l=1}^{L} {g_l} \Big[ \int_0^{f_c+f_m} \Big( e^{-j(\omega \tau_l - \theta_m)} \delta(\omega - \omega_0)\Big)^2 df + \int_0^{f_c+f_m} \Big(e^{-j(\omega \tau_l + \theta_m)} \delta(\omega + \omega_0) \Big) ^2 df \Big] = $$

$$\Big(\frac{\pi}{2} \alpha_m\Big)^2 \sum_{l=1}^{L} {g_l} \Big[ \Big|e^{-j(\omega_0 \tau_l - \theta_m)} \Big|^2 + \Big| e^{j(\omega_0 \tau_l - \theta_m)} \Big|^2 \Big] $$

Finally: $$ \int_0^{f_c+f_m} |Y(f)|^2df= \Big(\frac{\pi}{2} \alpha_m\Big)^2 \sum_{l=1}^{L} {g_l} \Big( \Big|e^{-j(2 \pi (f_c+f_m) \tau_l - \theta_m)} \Big|^2 + \Big| e^{j(2 \pi (f_c+f_m) \tau_l - \theta_m)} \Big|^2 \Big) $$

Could it be the right solution or is there anything wrong?