Hi!
Here is my circuit and graph of waveforms:
My question is, how can I find expression for output voltage vout(t)? If I use superposition method, for active AC source (DC source is short) I will get vout_ac(t)
\(=Vrms\sqrt{2}\sin{(314t+\varphi )}\), where \(\varphi\)
is phase angle. For DC analysis (AC source is short, capacitor is removed), I will get Vout_DC=constant, so
vout(t) will be \(=Vrms\sqrt{2}\sin{(314t+\varphi )}+constant\). But it is not expression for our output waveform.
Here is my circuit and graph of waveforms:
My question is, how can I find expression for output voltage vout(t)? If I use superposition method, for active AC source (DC source is short) I will get vout_ac(t)
\(=Vrms\sqrt{2}\sin{(314t+\varphi )}\), where \(\varphi\)
is phase angle. For DC analysis (AC source is short, capacitor is removed), I will get Vout_DC=constant, so
vout(t) will be \(=Vrms\sqrt{2}\sin{(314t+\varphi )}+constant\). But it is not expression for our output waveform.
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