The circuit is given below.
I know that until the Vin reaches the first peak, the voltage across R (output voltage) will be 0 since capacitors draw current to charge.
After that, I couldn't figure it out.
Please help me understand.
\$\begingroup\$
\$\endgroup\$
2
-
\$\begingroup\$ For Steady state only or both (transient +steady assuming source applied at t=0 from graph)? \$\endgroup\$– user215805Commented Mar 16, 2021 at 9:01
-
\$\begingroup\$ steady state only. \$\endgroup\$– sakxam devCommented Mar 16, 2021 at 9:07
Add a comment
|
2 Answers
\$\begingroup\$
\$\endgroup\$
5
Well, for steady state analysis, this is essentially a HPF (high pass filter), so it will only allow siganls with frequencies above a certain threshold (cut-off frequency) to be passed through the capacitor (aka the capacitor will act as short circuit).
There is a simple way to see this, which is the equation of the capacitive reactance
So as the frequency increases, the reactance decreases, aka the "resistance" of the cap decreases, hence it acts as a short circuit.
And there is also a different approach using the frequency domain analysis as well
Quick calrification, the cut-off frequency is the frequency at which the output power is half of that input power, and sometimes is called teh 3-dB cut-off frequency.
answered Mar 16, 2021 at 9:34
-
\$\begingroup\$ If someone can tell me how to post equations that would be great, thank you. \$\endgroup\$ Commented Mar 16, 2021 at 9:34
-
1
-
1\$\begingroup\$ Hi abd- you can learn from here , how to write equation -math.meta.stackexchange.com/questions/5020/… \$\endgroup\$ Commented Mar 16, 2021 at 9:39
-
\$\begingroup\$ Thank you very much indeed JRE & user215805. \$\endgroup\$ Commented Mar 16, 2021 at 9:46
-
\$\begingroup\$ Your opening sentence is very naïve; there is no cut-off point at which signals get through or are rejected. \$\endgroup\$– Andy akaCommented Mar 16, 2021 at 11:26
\$\begingroup\$
\$\endgroup\$
until the Vin reaches the first peak, the voltage across R (output voltage) becomes 0 since capacitors draws current to charge
Wrong! Use equation written below and then think how charging and discharging takes place.
Hint- $$I={VSin(\omega t +\phi)/Z}$$ where $$Z^2=R^2+X_c^2$$ and $$Cos \phi=R/Z$$ $$V_R=IR$$
answered Mar 16, 2021 at 9:32