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This may sound like a very simple question that would not need help, but I am having a hard time wrapping my head around the idea why we short DC voltages during AC analysis or transfer function analysis.

I do understand that DC sources do not assert any AC voltage therefore do not drive any AC current either. I don't see why they appear as a short in any non-zero frequency. Why don't they just appear as constants in the functions or equations we derive from the said analyses?

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  • \$\begingroup\$ The AC impedance of an ideal DC voltage source is zero. (THe AC impedance of an ideal current source is infinite - so they would be shown as open circuit. \$\endgroup\$
    – Russell McMahon
    Jun 20 at 10:52
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In simple terms they have zero impedance and can therefore sink or source infinite current. So for AC purposes they are the same as a short,

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A DC voltage source always has a constant DC voltage across it, and hence a zero AC voltage.

Consider an AC voltage source and a DC source connected by a resistor. The AC voltage across the AC source will be as specified. The AC voltage across the DC source will be zero. It's just as if the resistor's end was shorted to ground, when looking only at AC signals.

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I would add to the earlier, zero impedance, answer that DC sources are always decoupled (you do, don't you!) with bypass capacitors which, from an analysis point of view, are infinitely large. Hence zero impedance. Sorry, this should be a comment but I have insufficient reputation.

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A "voltage source" (see the headline of the question) - when it can be regarded as "ideal" - provides this voltage independent on the connected load. This is identical to the assumption of a an internal source impedance of zero; otherwise the provided voltage would drop when a load is connected (withna corresponding current).

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This provides the same information as several other answers, but more explicitly.

  • The AC impedance of an ideal DC voltage source is zero. So as the DC voltage is irrelevant the zero impedance is shown as a short circuit.

  • The AC impedance of an ideal current source is infinite - so they would be shown as open circuit.

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