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My task is to calculate participation (in percent) of DC component in total average power of signal from photo:

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I represented this signal as complex Fourier series: $$u(t)=\frac{1}{2}+\sum_{n=-\infty,n\neq 0}^{n=\infty}\frac{1}{2n\pi}e^{j\frac{\pi}{2}}e^{jn*500t}.$$

Average power of signal is calculated using formula $$P=\lim_{T->\infty}\frac{1}{T}\int_{\tau}^{\tau+T}(f(t))^{2}dt$$ but how should I use it here? And how to calculate participation of DC component? Thanks in advance.

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This question is likely asking you to recall Parseval's identity, which is stated as

$$\sum_{n=-\infty}^\infty |c_n|^2 = \dfrac{1}{2\pi}\int_{-\pi}^\pi |f(x)|^2 \, dx$$

where \$c_n\$ are the coefficients of the Fourier series of a periodic function \$f(x)\$ with period \$2\pi\$.

You can see that the right-hand side of the identity is proportional to the average power of the signal. So, with some manipulation of the scaling factors, you can use this to assign a portion of the power of the single to each of the fourier components.

Note 1: I assume that your signal repeats on every interval of time T (that is, \$f(t+T)=f(t)\$). The way you've drawn it,it looks like the signal is 0 outside the region \$0 < t < T\$, but that would make it a non-periodic signal and it wouldn't make sense to talk about its Fourier series representation or average power.

Note 2: Notice that in your formula for the average power, you are have two different variables named T. One which you take to the limit of infinity, and one built in to the definition of the function f. Since f is periodic with period T, you don't need to take a limit and you can just use

$$ P = \dfrac{1}{T}\int_\tau^{\tau+T} f^2(t)\, dt$$

for any \$\tau\$ of your choice.

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  • \$\begingroup\$ Thanks for reply! It's repeating, I just didn't draw it :) I will try to solve it now using your suggestions \$\endgroup\$
    – etf
    Commented Oct 4, 2014 at 12:34
  • \$\begingroup\$ Here is how I tried it: Average power of periodic signal can be calculated using relation $$Ptotal=\sum_{n=-\infty}^{n=\infty}|Fn|^{2}.$$ Using Parseval's theorem, it is equal to $$\frac{1}{T}\int_{\tau}^{\tau +T}(f(t))^{2}dt.$$ For my signal, $$Ptotal=\frac{1}{T}\int_{\tau}^{\tau +T}(f(t))^{2}dt=\frac{1}{T}\int_{0}^{T}(\frac{Et}{T})^{2}dt=\frac{E}{T^{3}}\frac{T^{3}}{3}=\frac{E}{3} (E=1)=\frac{1}{3}.$$ For DC component I have:$$\frac{1}{T}Pdc=\int_{0}^{T}(\frac{1}{2})^{2}dt=\frac{1}{4}.$$ Is it ok? \$\endgroup\$
    – etf
    Commented Oct 4, 2014 at 19:41
  • \$\begingroup\$ So contribution of DC component is 75% \$\endgroup\$
    – etf
    Commented Oct 4, 2014 at 22:37

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