Timeline for Participation of DC component in total average power of signal
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 4, 2014 at 22:37 | comment | added | etf | So contribution of DC component is 75% | |
| Oct 4, 2014 at 19:41 | comment | added | etf | 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? | |
| Oct 4, 2014 at 15:48 | vote | accept | etf | ||
| Oct 4, 2014 at 12:34 | comment | added | etf | Thanks for reply! It's repeating, I just didn't draw it :) I will try to solve it now using your suggestions | |
| Oct 4, 2014 at 5:05 | history | edited | The Photon | CC BY-SA 3.0 |
added 399 characters in body
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| Oct 4, 2014 at 4:59 | history | answered | The Photon | CC BY-SA 3.0 |