# 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 Commented Aug 20, 2016 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 $v_c = V \cos(\frac{t}{\sqrt{LC}}) - V}$ Commented Aug 21, 2016 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). Commented Aug 21, 2016 at 21:01