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The given figure illustrates the simplest ripple filter. A voltage $V=V_0(1+\cos(\omega t))$ is fed to the left input.
Find the output voltage V'(t).
enter image description here

As it can be seen that the voltage fed to the circuit has two components, one d.c. and another a.c.
Now suppose only the a.c. component was fed to the circuit. Then the current will lead the voltage and hence the current equation becomes $$I=\frac{V_0}{\sqrt{R^2+(\frac{1}{\omega C})^2}}\cos(\omega t-\phi)$$ where $\tan(\phi)=\frac{1}{\omega RC}$.
Now suppose only the d.c. component was fed to the circuit then the equation of current would become $$I=\frac{V_0}{R}e^{\frac{-t}{RC}}$$.
And so we can find the voltage across the capacitor in the respective cases discussed above.
But if both the the capacitors are fed simultaneously then how are the currents going to superimpose.Will the each component behave independently or there is something else which I am missing.
The answer is given as $V'=V_0+V_m\cos(\omega t-\alpha)$ where $V_m=\frac{V_0}{\sqrt{1+(\omega RC)^2}}$, $\alpha=arctan(\omega RC)$.
Here I also did not understand how the d.c. component(Voltage $V_0$) fed across the circuit would be equal to the voltage across the capacitor.

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Because the system is linear the response to the two driving voltages [$V_0$ and $V_0cos(\omega t)]$ will equal the sum of each acting alone (superposition).

Yes, each component behaves independently and the output will be their sum.

Think of it like this (intuition) - the dc source will charge the capacitor up to $V_0$ after some time (after 5 time constants it is practically fully charged). Now, to get charge to flow into the capacitor you will have to raise the applied voltage above $V_0$. This will be done by the ac component of the source. This will charge the capacitor above $V_0$, and then when the ac source goes negative, the capacitor will discharge back into the source. After the initial transient period, this charge/discharge oscillation will be centered around $V_0$ as I show below,

enter image description here

I used R=2$\Omega$, C=100µF, f=1000Hz, and $V_0$=25V for the above plot with the circuit being closed at t=0.

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    $\begingroup$ @ relayman357. Thank you very much for you crystal clear explanation, and also thanks for providing a link on what are linear systems. $\endgroup$ – sameed hussain Jun 2 at 3:55
  • $\begingroup$ My pleasure @sameedhussain $\endgroup$ – relayman357 Jun 2 at 4:02
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Good Voltage V0V0) fed across the circuit would be equal to the voltage across the capacitor. Because the system is linear the response to the two driving voltages [V0V0 and V0cos(ωt)]V0cos(ωt)] will equal the sum of each acting alone (superposition).

Yes, each component behaves independently and the output will be their sum.

I used R=2ΩΩ, C=100µF, f=1000Hz, and V0V0=25V for the above plot with the circuit being closed at t=0.

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