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# Lecture 6 Second-Order Circuits

What if we have both L and C in the circuit

## Series RLC Circuits

when $i_L(0_-) = 0, v_c(0_-) = 0$ find $V_c(t)$
$$\left { \begin{array}{lr} i_L(t) = c \frac{dv_c(t)}{dt}\ V_s - R\cdot i_L(t) - L\frac{di_l(t)}{dt}-v_c(t) = 0 \end{array} \right.\Rightarrow RC\frac{dv_c(t)}{dt}+LC\frac{d^2v_c(t)}{dt^2}+v_c(t) = v_s$$

$$\frac{d^2v_c(t)}{dt^2}+\frac{R}{L}\frac{dv_c(t)}{dt}+\frac{1}{LC}v_c(t) = \frac{v_s}{LC}$$

### Initial and Final condition

Find $v_c(0_-),i_c(0_-),i_L(0_-),v_L(0_-),\v_c(0_+),i_c(0_+),i_L(0_+),v_L(0_+),\v_c(\infty),i_c(\infty),i_L(\infty),v_L(\infty)$

$t = 0_-$ steady state capacitor likes open circuit inductor like short circuit

$v_c(0_-)=v_s\i_c(0_-)=0\i_L(0_-)=0\v_L(0_-)=0$

$t = 0_+$电容电压不突变，电感电流不突变

$v_c(0_+)=v_s\i_c(0_+)=0\i_L(0_+)=0\v_L(0_+)=v_s$

$t = \infty$

$v_c(\infty)=0\i_c(\infty)=0\i_L(\infty)=0\v_L(\infty)=0$

#### Example

$t = 0_-$

$i(0_-) = 12/6 = 2A \ v(0_-) = 2\cdot2 = 4V$

$t = 0_+$

$i(0_+) = i(0_-) = 2A \ v(0_+) = v(0_-) = 4V$

$\frac{di(0_+)}{dt} = \frac{1}{L}V_L(0_+) = 12-2*4-4 =0A/s$

$\frac{dv(0_+)}{dt} = \frac{1}{c}i_C(0_+) = 2/0.1 = 20V/s$

$t = \infty$

$i(\infty) = 0A\v(\infty) = 12V$

## Natural Response自然响应，零输入响应

### series RLC circuit

$V_c(0_-) = V_0,i_L(0_-) = I_0$ Find $v_c(t),(t\geq 0)$
$$\left { \begin{array}{lr} i_L(t) = c \frac{dv_c(t)}{dt}\ • R\cdot i_L(t) - L\frac{di_l(t)}{dt}-v_c(t) = 0 \end{array} \right.\Rightarrow RC\frac{dv_c(t)}{dt}+LC\frac{d^2v_c(t)}{dt^2}+v_c(t) =0$$

$$\frac{d^2v_c(t)}{dt^2}+p\frac{dv_c(t)}{dt}+qv_c(t) =0,\text{ where }p =\frac{R}{L},q = \frac{1}{LC}$$

Let $V_C(t) = Ae^{st}$, then $Ae^{st}(s^2+ps+q)=0$.   Solve equation $s^2+ps+q =0$ about $s$
$$s = \frac{-p\pm \sqrt{p^2-4q}}{2}=\frac{R}{2L}\pm\sqrt{ (\frac{R}{2L})^2 - \frac{1}{LC} }$$
So we have 3 cases here:

overdamped 过阻尼 $s$ have two different real solution

where $(\frac{R}{2L})^2 - \frac{1}{LC}>0$, let $\alpha =\frac{R}{2L},\omega _0 = \frac{1}{\sqrt{LC}}$ so the equation about the circuit can be interpret into $s^2+2\alpha s+\omega 0^2 = 0$, so that $s{1,2} = -\alpha \pm \sqrt{\alpha^2-\omega _0^2}$.

Therefore $V_C(t) = A_1 e^{s_1t} +A_2 e^{s_2t}$,where $A_1,A_2$ can be found using two initial conditions.

critically damped 临界阻尼 $s$ have only one real solution

where $(\frac{R}{2L})^2 - \frac{1}{LC}=0$, let $\alpha =\frac{R}{2L},\omega _0 = \frac{1}{\sqrt{LC}},\alpha = \omega_0$ $s=s_1 = s_2 = -\alpha$

Therefore $V_C(t) = (A_1t+A_2)e^{st}$,where $A_1,A_2$ can be found using two initial conditions.

underdamped 欠阻尼 $s_1,s_2$ are complex

where $(\frac{R}{2L})^2 - \frac{1}{LC}<0$, let $\alpha =\frac{R}{2L},\omega _0 = \frac{1}{\sqrt{LC}}$ so the equation about the circuit can be interpret into $s^2+2\alpha s+\omega 0^2 = 0$, so that $s{1,2} = -\alpha \pm \sqrt{\omega _0^2-\alpha^2}i$, seeing that $V_C(t) \in \R$
$$c_1e^{s_1t}+c_2e^{s_2t} = c_1e^{-(\alpha+j\omega_d i)t}+c_2e^{-(\alpha-j\omega_d i)t}$$

$$(c_1e^{s_1t}+c_2e^{s_2t})^* = c_1^*e^{-(\alpha-j\omega_d i)t}+c_2^e^{-(\alpha+j\omega_d i)t}$$
so that $c_1 = c_2^$, then
$$v_c(t) = e^{-\alpha t}[A_1cos\omega_dt+A_2sin\omega_d t]$$

### parallel RLC circuit

$$\left \{ \begin{array}{**lr**} V_c(t) = L \frac{di_L(t)}{dt}\\ \frac{V_c}{R}+i_L +c\frac{dV_c}{dt} = 0 \end{array} \right.\Rightarrow \frac{L}{R}\frac{di_L(t)}{dt}+LC\frac{d^2i_L(t)}{dt^2}+i_L(t) =0$$

$$\Rightarrow \frac{d^2i_L}{dt^2}+\frac{1}{RC}\frac{di_L}{dt}+\frac{i_L}{LC} = 0$$

overdamped 过阻尼 $s$ have two different real solution

where $(\frac{1}{2RC})^2 - \frac{1}{LC}>0$, let $\alpha =\frac{1}{2RC},\omega _0 = \frac{1}{\sqrt{LC}}$ so the equation about the circuit can be interpret into $s^2+2\alpha s+\omega 0^2 = 0$, so that $s{1,2} = -\alpha \pm \sqrt{\alpha^2-\omega _0^2}$.

Therefore $i_L(t) = A_1 e^{s_1t} +A_2 e^{s_2t}$,where $A_1,A_2$ can be found using two initial conditions.

critically damped 临界阻尼 $s$ have only one real solution

where $(\frac{1}{2RC})^2 - \frac{1}{LC}=0$, let $\alpha =\frac{1}{2RC},\omega _0 = \frac{1}{\sqrt{LC}},\alpha = \omega_0$ $s=s_1 = s_2 = -\alpha$

Therefore $i_L(t) = (A_1t+A_2)e^{st}$,where $A_1,A_2$ can be found using two initial conditions.

underdamped 欠阻尼 $s_1,s_2$ are complex

where $(\frac{1}{2RC})^2 - \frac{1}{LC}<0$, let $\alpha =\frac{1}{2RC},\omega _0 = \frac{1}{\sqrt{LC}}$ so the equation about the circuit can be interpret into $s^2+2\alpha s+\omega 0^2 = 0$, so that $s{1,2} = -\alpha \pm \sqrt{\omega _0^2-\alpha^2}i$, seeing that $V_C(t) \in \R$
$$c_1e^{s_1t}+c_2e^{s_2t} = c_1e^{-(\alpha+j\omega_d i)t}+c_2e^{-(\alpha-j\omega_d i)t}$$

$$(c_1e^{s_1t}+c_2e^{s_2t})^* = c_1^*e^{-(\alpha-j\omega_d i)t}+c_2^*e^{-(\alpha+j\omega_d i)t}$$

so that $c_1 = c_2^*$, then
$$i_L(t) = e^{-\alpha t}[A_1cos\omega_dt+A_2sin\omega_d t]$$

### parallel RLC Circuit

$$\frac{d^2I}{dt^2}+\frac{1}{RC}\frac{dI}{dt}+\frac{I}{LC} = \frac{I_s}{LC}\tag{*}$$
Solution :$I(t) = \underbrace{I_t(t)}{\text{transient response}}+\underbrace{I{ss}(t)}_{\text{steady state response}}$

Statement: specifically, for step response,$I_t(t)$ is the same format as in natural response, $I_{ss} = I_s$, $v(t)$ is the solution to equation$(*)$