So a ac filter doesnt allow the dc component to pass through it, so the dc component would take a altetnative path where as the ac component would easily pass through it. My question is, where is dc component in the input wave?

  • \$\begingroup\$ It depends on what kind of signal. Every kind of signal can be represented by a sum of sine, cosine and a DC component. \$\endgroup\$ – Long Pham Sep 11 '18 at 14:26
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    \$\begingroup\$ I'm not sure I understand the question, but if the filter blocks DC, then it doesn't "go" anywhere. It simply appears across whatever component inside the filter that's doing the blocking, such as a capacitor. \$\endgroup\$ – Dave Tweed Sep 11 '18 at 14:49
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    \$\begingroup\$ @LongPham, to be more precise, every periodic signal can be decomposed into a sum of sinusoids and a DC component. Finite, non-periodic signals can be approximately understood by pretending that they were snipped from an infinitely long periodic signal. \$\endgroup\$ – user197845 Sep 11 '18 at 14:57
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    \$\begingroup\$ @besmirched: To be even more precise, nonperiodic signals can be represented with sines and cosines if you allow an infinite number of them. The Fourier Transform does not require periodicity. \$\endgroup\$ – Dave Tweed Sep 11 '18 at 15:04
  • \$\begingroup\$ @DaveTweed I dont know anything about fourier series etc... It will probably start next semester... So you are saying the input sine wave can be expressed as a sum of trig terms, and as we approach infinity the graph of the terms become flatter and flatter? ie a dc component? \$\endgroup\$ – Manav Shetty Sep 11 '18 at 17:31

DC component is the average amplitude over a cycle time period of the signal. So, basically in simple terms you can call the DC component as the bias given to an oscillating signal. For a sinusoidal signal with no bias it would be 0.

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