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Power signal calculation using Parseval's Theorem

  1. Jun 5, 2017 #1
    1. The problem statement, all variables and given/known data
    Hi guys,
    I have the following transmitted power signal:

    $$x(t)=\alpha_m \ cos[2\pi(f_c+f_m)t+\phi_m],$$
    where: ##\alpha_m=constant, \ \ f_c,f_m: frequencies, \ \ \theta_m: initial \ phase.##

    The multipath channel is:
    $$h(t)=\sum_{l=1}^L \sqrt{g_l} \ \delta(t-\tau_l).$$

    The received signal is the convolution between ##x(t)## and ##h(t)##, i.e.:
    $$y(t)=x(t)*h(t)$$
    where ##*## represent the convolution symbol.

    I need to calculate the received power, which is:
    $$R=(f_c+f_m) \int_0^{\frac{1}{f_c+f_m}} |y(t)|^2 dt.$$

    2. Relevant equations
    In order to simplify the calculation of ##y(t)##, I decided to calculate the Fourier-transform of ##x(t)## and ##h(t)##, so that the time-domain convolution will become a multiplication in frequency-domain; then I got:

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

    3. The attempt at a solution
    According the Parseval's theorem, the integral ##\int_0^{\frac{1}{f_c+f_m}} |y(t)|^2 dt## will be equivalent to ##\int_0^{f_c+f_m} |Y(f)|^2 df##, so that:

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

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

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

    We can rewrite ##R## in this form:
    $$R=(f_c+f_m) \Big(\frac{\pi}{2} \alpha_m\Big)^2 \sum_{l=1}^{L} {g_l} \Big[ \Big|e^{-j(\omega_0 \tau_l - \theta_m)} \Big|^2 + \Big| e^{j(\omega_0 \tau_l - \theta_m)} \Big|^2 \Big]$$

    My supervisor suggested told me the solution is about correct and it should be proportional to ##|H(f_c+f_m)|^2##.

    Actually I am stucked here and I cannot find the errors; so, your help will be really appreciated.

    Thanks so much.
     
  2. jcsd
  3. Jun 5, 2017 #2

    Charles Link

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    Homework Helper

    I question whether it is mathematically sound to have ## h(t) ## as a delta function so that you wind up taking the square of delta function(s) for power received. It might be necessary to make it rectangles of finite width. Additional item is it may be necessary to have the signal ## x(t) ## last for a finite time so that you don't have a delta function ## \delta(\omega-\omega_o) ## that winds up getting squared, which again, I'm not sure that such a thing is workable. Just a couple of inputs=perhaps others have additional inputs that might help to get the solution.
     
    Last edited: Jun 5, 2017
  4. Jun 5, 2017 #3

    marcusl

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    Gold Member

    I see two additional problems:
    1. Your upper limit of integration is too small. Under the usual condition that fc >> fm, your time integral runs over just a single cycle of the carrier, which excludes contributions from the delayed paths. The limit should be at least as big as the biggest [itex]\tau[/itex] term.
    2. You have incorrectly taken the absolute value squared in section 3. The squared term must include the summation and [itex]\sqrt{g_l}[/itex] inside of it so that you get a lot of cross terms.
     
  5. Jun 6, 2017 #4
    Firstly, thanks Charles Link and Marcusl for your reply.
    Unfortunately it was the text of the exercise, then I guess my supervisor made some assumptions.

    The condition that the biggest ##\tau## is bigger than ##f_c+f_m## is always verified.

    I then wrote again the expression of ##R##, and considering that ##|Y(f)|^2=Y(f) \ Y^*(f)##, I got this:

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

    Now the terms which present the multiplication between ##\delta(\omega-\omega_0)## and ##\delta(\omega+\omega_0)##, according the distribution theory, are equal to ##0##; so we got:

    $$R=(f_c+f_m) (\frac{\pi}{2}\alpha_m)^2 \int_0^{f_c+f_m} 2 \sum_{l=1}^L g_l \ df= \frac{\pi^2}{2}{\alpha_m}^2 (f_c+f_m)^2 \sum_{l=1}^L g_l $$

    Can you see any mistakes in this?
     
  6. Jun 6, 2017 #5

    marcusl

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    Gold Member

    1. This makes no sense since time and frequency have different units. You need to integrate over a time longer than [itex]\tau_{max}[/itex] . In fact, you have mis-applied Parseval's (Plancheral's) theorem altogether, since it is true only for infinite limits.

    2. Your square is still wrong. You must write it out like this $$\int_{-\infty}^\infty \Big| \sum_{l=1}^{L} c_l(\omega) \Big|^2 d\omega = \int_{-\infty}^\infty \sum_{l=1}^{L} c_l(\omega) \sum_{m=1}^{L} c^*_m(\omega) d\omega$$
    where c is all the stuff in the summand.
     
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