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Lecture 6 Second-Order Circuits

Lecture 6 Second-Order Circuits

. 9 min read
封面是同济的🐱
What if we have both L and C in the circuit

Series RLC Circuits

image-20200413031358589

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

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

这是一个二阶方程,所以也是二阶电路。second order circuit

Initial and Final condition

image-20200413032545657

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

image-20200415094453102

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

$t = 0_+$电容电压不突变,电感电流不突变

image-20200415094620569

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

$t = \infty$

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

Example

image-20200415095143574

$t = 0_-$

image-20200415095227497

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

$t = 0_+$

image-20200415095400580

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

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

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

$t = \infty$

image-20200415095816171

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

Natural Response自然响应,零输入响应

series RLC circuit

image-20200415100151558

$V_c(0_-) = V_0,i_L(0_-) = I_0$ 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
    $$

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

解上述二阶常系数微分方程($2^{nd}$ order ordinary differential equation/ODE)就可以解除这个电路。

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
$$
v_c(t) = e^{-\alpha t}[A_1cos\omega_dt+A_2sin\omega_d t]
$$

Summary series RLC network

Case$\alpha = \frac{R}{2L},\omega_0 = \sqrt{\frac{1}{LC}}$1overdamped$\alpha >\omega \Leftrightarrow R>2\sqrt{\frac{L}{C}}$$V_C(t) = A_1 e^{s_1t} +A_2 e^{s_2t} \ s_{1,2} = -\alpha \pm \sqrt{\alpha^2-\omega _0^2}$2critically damped$\alpha =\omega \Leftrightarrow R=2\sqrt{\frac{L}{C}}$$V_C(t) = (A_1t+A_2)e^{st}\s = -\alpha$3underdamped$\alpha <\omega \Leftrightarrow R<2\sqrt{\frac{L}{C}}$$V_C(t) = e^{-\alpha t}[A_1cos\omega_dt+A_2sin\omega_d t] \ \omega_0 = \sqrt{\omega _0^2-\alpha^2}$

Image

Example

image-20200420230050791
image-20200422214418503
image-20200422214758942

More Notes

Case$\alpha = \frac{R}{2L},\omega_0 = \sqrt{\frac{1}{LC}}$1overdamped$\alpha >\omega \Leftrightarrow R>2\sqrt{\frac{L}{C}}$$V_C(t) = A_1 e^{s_1t} +A_2 e^{s_2t} \ s_{1,2} = -\alpha \pm \sqrt{\alpha^2-\omega 0^2} \ \left {\begin{array}{lr}  s_1A_1 +s_2A_2 = \frac{i(0+)}{c} \A_1+A_2 = v(0_+) \end{array}\right.$2critically damped$\alpha =\omega \Leftrightarrow R=2\sqrt{\frac{L}{C}}$$V_C(t) = (A_1t+A_2)e^{st}\s = -\alpha \ \left {\begin{array}{lr}  A_1 +sA_2 = \frac{i(0_+)}{c} \A_2 = v(0_+) \end{array}\right.$3underdamped$\alpha <\omega \Leftrightarrow R<2\sqrt{\frac{L}{C}}$$V_C(t) = e^{-\alpha t}[A_1cos\omega_dt+A_2sin\omega_d t] \ \omega_0 = \sqrt{\omega 0^2-\alpha^2} \ \left {\begin{array}{lr}  \omega_0A_2 -\alpha A_1 = \frac{i(0+)}{c} \A_1 = v(0_+) \end{array}\right.$

parallel RLC circuit

image-20200420231540904

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

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

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
$$
i_L(t) = e^{-\alpha t}[A_1cos\omega_dt+A_2sin\omega_d t]
$$

Case$\alpha = \frac{1}{2RC},\omega_0 = \sqrt{\frac{1}{LC}}$1overdamped$\alpha >\omega \Leftrightarrow \frac{1}{R}>2\sqrt{\frac{L}{C}}$$V_C(t) = A_1 e^{s_1t} +A_2 e^{s_2t} \ s_{1,2} = -\alpha \pm \sqrt{\alpha^2-\omega 0^2} \ \left {\begin{array}{lr}  s_1A_1 +s_2A_2 = \frac{v(0+)}{L} \A_1+A_2 = i(0_+) \end{array}\right.$2critically damped$\alpha =\omega \Leftrightarrow \frac{1}{R}=2\sqrt{\frac{L}{C}}$$V_C(t) = (A_1t+A_2)e^{st}\s = -\alpha \ \left {\begin{array}{lr}  A_1 +sA_2 = \frac{V(0_+)}{L} \A_2 = i(0_+) \end{array}\right.$3underdamped$\alpha <\omega \Leftrightarrow \frac{1}{R}<2\sqrt{\frac{L}{C}}$$V_C(t) = e^{-\alpha t}[A_1cos\omega_dt+A_2sin\omega_d t] \ \omega_0 = \sqrt{\omega 0^2-\alpha^2} \ \left {\begin{array}{lr}  \omega_0A_2 -\alpha A_1 = \frac{v(0+)}{L} \A_1 = i(0_+) \end{array}\right.$

Step response 阶跃响应

series RLC circuit

image-20200422231332926


$$
\frac{d^2V}{dt^2}+\frac{R}{L}\frac{dV}{dt}+\frac{V}{LC} = \frac{V_s}{LC}\tag{*}
$$
Solution :$V(t) = \underbrace{V_t(t)}{\text{transient response}}+\underbrace{V{ss}(t)}_{\text{steady state response}}$

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

Case$\alpha = \frac{R}{2L},\omega_0 = \sqrt{\frac{1}{LC}}$1overdamped$\alpha >\omega \Leftrightarrow R>2\sqrt{\frac{L}{C}}$$V_C(t) = A_1 e^{s_1t} +A_2 e^{s_2t}+V_s \ s_{1,2} = -\alpha \pm \sqrt{\alpha^2-\omega _0^2}$2critically damped$\alpha =\omega \Leftrightarrow R=2\sqrt{\frac{L}{C}}$$V_C(t) = (A_1t+A_2)e^{st}+V_s\s = -\alpha$3underdamped$\alpha <\omega \Leftrightarrow R<2\sqrt{\frac{L}{C}}$$V_C(t) = e^{-\alpha t}[A_1cos\omega_dt+A_2sin\omega_d t]+V_s \ \omega_0 = \sqrt{\omega _0^2-\alpha^2}$

example

image-20200423135613369
image-20200423141127718
image-20200423141153252
image-20200423141241866
image-20200423141300661
image-20200423141436684

parallel RLC Circuit

image-20200423155408520


$$
\frac{d^2I}{dt^2}+\frac{1}{RC}\frac{dI}{dt}+\frac{I}{LC} = \frac{I_s}{LC}\tag{*}
$$
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$(*)$

Case$\alpha = \frac{R}{2L},\omega_0 = \sqrt{\frac{1}{LC}}$1overdamped$\alpha >\omega \Leftrightarrow R>2\sqrt{\frac{L}{C}}$$I_L(t) = A_1 e^{s_1t} +A_2 e^{s_2t}+I_s \ s_{1,2} = -\alpha \pm \sqrt{\alpha^2-\omega _0^2}$2critically damped$\alpha =\omega \Leftrightarrow R=2\sqrt{\frac{L}{C}}$$I_L(t) = (A_1t+A_2)e^{st}+I_s\s = -\alpha$3underdamped$\alpha <\omega \Leftrightarrow R<2\sqrt{\frac{L}{C}}$$I_L(t) = e^{-\alpha t}[A_1cos\omega_dt+A_2sin\omega_d t]+I_s \ \omega_0 = \sqrt{\omega _0^2-\alpha^2}$

example

image-20200423161814741

General RLC Circuit

example

image-20200423162208210

Summary

without excitation(Natural Response)自然响应
| Case |                   | $\frac{d^2y(t)}{dt}+2\alpha \frac{dy(t)}{dt}+\omega^2_0y(t)=0$ |                                                              |
| ---- | ----------------- | ------------------------------------------------------------ | ------------------------------------------------------------ |
| 1    | overdamped        | $\alpha >\omega $                                            | $y(t) = A_1 e^{s_1t} +A_2 e^{s_2t} \ s_{1,2} = -\alpha \pm \sqrt{\alpha^2-\omega _0^2}$ |
| 2    | critically damped | $\alpha =\omega $                                            | $y(t) = (A_1t+A_2)e^{st}\s = -\alpha $                      |
| 3    | underdamped       | $\alpha <\omega $                                            | $y(t) = e^{-\alpha t}[A_1cos\omega_dt+A_2sin\omega_d t] \ \omega_0 = \sqrt{\omega _0^2-\alpha^2}$ |

with any excitation
$$
\frac{d^2y(t)}{dt}+2\alpha \frac{dy(t)}{dt}+\omega^2_0y(t)=f(t)
\
y(t) = y_h+y_p\ y_h \text{ is the general solution to the corresponding homogeneous equation}
\
y_p \text{ is a particular solution 特解}
$$

how to find $y_p$

$$
f(t) = e^{rt}P_m(t),P_m(t) \text{是关于t的m次多项式},r\in \R \
\left {\begin{array}{lr}
r\neq s_{1,2},&y_p = Q_m(t)e^{rt}\
r = s_1 \text{ or }s_2(s_1\neq s_2), &y_p = tQ_m(t)e^{rt}\
r =s_1 = s_2, &y_p = t^2Q_m(t)e^{rt}
\end{array}\right.
$$

$$
f(t) = e^{rt} (Mcos\omega t+Nsin\omega t),  r,M,N,\omega \in \R,\omega\neq 0 \
\left {\begin{array}{lr}
r\pm j\omega\neq s_{1,2}, &y_p = e^{rt}(B_1cos\omega t+B_2sin\omega t)\
r\pm j\omega = s_{1,2}, &y_p = te^{rt}(B_1cos\omega t+B_2sin\omega t)
\end{array}\right.
$$

$$
f(t) = f_1(t)+f_2(t)
y_p = y_{p_1} + y_{p_2}
$$



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