# Homework Help: Integral of absolute value of a Fourier transform

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1. Apr 25, 2017

### Mik256

1. The problem statement, all variables and given/known data
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. Relevant equations
Then, the integral weare looking for will get the following form:

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

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

Thank you so much for your help, I would really appreciate that!

2. Apr 26, 2017

### eys_physics

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.

3. Apr 26, 2017

### Mik256

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?

4. Apr 26, 2017

### 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.

5. Apr 27, 2017

### 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?