circuit in frecuency domain

$$ v= |V| u(t) \quad [V]$$ u(t) is : t<0 and 1 to t>0

\$ i_L(t) \$ = inductor current \$ v_c(t) \$= voltage capacitor

This is the transfer funtion : $$ \frac{sC}{1+s^2LC} \\ w= \frac{1}{\sqrt{LC}} $$ How do i find \$ a_2 \$ constant of \$i_L\$?

Transient response : \$ i_{nL}= a_2 \sin(\frac{t}{\sqrt{LC}})\$

Then to calculate forced response of \$v_c\$ across the capacitor used the same transfer funtion, What do i do ?

  • \$\begingroup\$ Vc(t)= i (t) * Zc, where Zc= 1/sC \$\endgroup\$ – Sunnyskyguy EE75 Aug 20 '16 at 14:03

You have given a transfer function without defining what it is.

Here is the same transfer function with definitions: $$ \frac{I_L(s)}{V(s)} = \frac{1}{Z_C+Z_L} = \frac{sC}{1+s^2LC} = H_{I_L}$$

Using phasor analysis, \$V(s)\$ is only non-zero when s=0 such that \$V(s=0)=|V|\$. But \$H_{I_L}(s=0) = 0\$

So the forced response of \$I_L\$ is just 0. Interpreting the title question literally, this is the answer.

Here is a relevant transfer function for \$V_C\$: $$ \frac{V_C(s)}{V(s)} = \frac{Z_C}{Z_C+Z_L} = \frac{1}{1+s^2LC} = H_{V_C}$$

And \$H_{V_C}(s=0) = 1\$

So the forced response is just: \$V_C(s=0)=|V|\$ which translates to \$v_C(t) = |V|\$ for t>0.

The forced responses given here do not include any natural response.

As you have pointed out, there is an oscillatory transient/natural response that never dies out. One way to obtain the constant for the natural response is to recognize that current through an inductor and voltage across a capacitor are continuous over time. Therefore the initial conditions are \$v_C(t=0)=0\$ and \$i_L(t=0) = 0\$. The first condition can be translated to \$v_L(t=0_+)=|V|\$. Combine with the basic relationship \$v_L = L \frac{di_L}{dt}\$, you can figure out the constant.

  • \$\begingroup\$ So, the complete response of $$v_c$$ is only \$v_c(t)= |V|\$ or \$ v_c = V \cos(\frac{t}{\sqrt{LC}}) - V} \$ \$\endgroup\$ – PCat27 Aug 21 '16 at 18:32
  • \$\begingroup\$ The complete response would be like the second part, \$v_C=V(1-\cos(\omega t))\$. We did not explicitly define the direction of \$v_C\$, but adhering to other calculations, the answer should be the negative of what you have written (btw, the formatting of your answer is off and it is hard to read). \$\endgroup\$ – rioraxe Aug 21 '16 at 21:01

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