# How can i calculate forced response of $V_c$ across the capacitor used transfer funtion? $$v= |V| u(t) \quad [V]$$ 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 ?

• Vc(t)= i (t) * Zc, where Zc= 1/sC – 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.

• So, the complete response of $$v_c$$ is only $v_c(t)= |V|$ or $v_c = V \cos(\frac{t}{\sqrt{LC}}) - V}$ – PCat27 Aug 21 '16 at 18:32
• 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). – rioraxe Aug 21 '16 at 21:01