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We have this signal:

$$x(t)= 2 + 2\cos(2\pi f_0 t) + \frac{2}{T_0} \operatorname{sinc}\left(\frac{2 t}{T_0}\right)e^{j2 \pi 4/ T_0} + \operatorname{sinc}\left(\frac{2 t}{T_0}\right)e^{-j2 \pi 4/ T_0}.$$

I have calculate Fourier trasformed and have found: $$ X(f)=\delta(f)+ \delta(f-f_0) + \delta(f+f_0) + \operatorname{rect}\left(\frac{f-\frac 4{T_0}}{\frac 2{T_0}}\right) + \operatorname{rect}\left(\frac{f+\frac 4{T_0}}{\frac 2{T_0}})\right) $$

I must find the average power. If use Parseval's have: \begin{align} P_x&=\frac{1}{T_0} \int_{-\infty}^{\infty}\lvert X(f)\rvert^2df \\ &= \frac 1{T_0}\left(\int\lvert\delta(f)+\delta(f+f_0)+\delta(f−f_0)+\operatorname{rect}+\operatorname{rect}\rvert^2df\right). \end{align} How can I solve this?

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  • $\begingroup$ The Fourier transform doesn't help here. You have to compute the power from the time domain signal, as shown in Maximilian Matthé's answer. $\endgroup$ – Matt L. Mar 28 '17 at 11:39
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A signal either has finite energy, finite power or even infinite power. If it has finite energy, it will have zero average power, according to your definition $$P_x=\lim_{T_0\rightarrow\infty}\frac{1}{T_0}\int_{-\frac{T_0}{2}}^{\frac{T_0}{2}}|x(t)|^2dt.$$

Knowing that the sincs are orthogonal to each other, as well as the cos function, the average power is given by

$$P_x=P(2)+P(2\cos(2\pi f_0t))+P(\operatorname{sinc})$$

where $P(...)$ is the power of the components in brackets. The sinc function has finite energy, hence it has zero power on average. Hence, we have

$$P_x=P(2)+P(2\cos(2\pi f_0 t)$$ with $$P(2)=4$$ and $$P(2\cos(2\pi f_0 t)=2^2\cdot\frac{1}{2}=2.$$

Therefore, the overall power is given by

$$P_x=6.$$

For the calculation of the $\cos$, consider $$P_{\cos}=\frac{1}{T}\int_{0}^{T}\cos^2(2\pi t/T)dt=\frac{1}{2}$$

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    $\begingroup$ Your first equation $P_x=\ldots$ should have the integration limits $-T_0/2$ and $T_0/2$. Furthermore, the first sentence is not true. There are also signals with infinite power. $\endgroup$ – Matt L. Mar 28 '17 at 11:37
  • $\begingroup$ @MattL. You are right, I've corrected the presentation. Thank you! $\endgroup$ – Maximilian Matthé Mar 28 '17 at 11:52
  • $\begingroup$ please note that the symbol $T_0$ in electrical engineering (and signal processing) notation refers to period of signals in general. This is important in the sense that engineering tradition place a strong emphasis on established symbols, unlike mathematicians (I think you're a mathematician?) who can quite be free in selecting theirs own. So in your first equation a careless reader can infer a wrong meaning. Especially when the context is about periodic signals. So It's better to replace it with $T$. $\endgroup$ – Fat32 Mar 29 '17 at 21:19

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