Ch8 Second Order Circuit

Terminologies


Introduction

Containing resistors and two energy storage elements and characterized by second-order differential Eq.


Cannot be reduced.


$\frac{\mathrm d ^2 y}{\mathrm d t^2}+p\frac{\mathrm d y}{\mathrm d t}+qy=0$

Characteristic Eq: $s^2+ps+q=0$.

1. $s_1\not=s_2$ and real, $y=A_1e^{s_1t}+A_2e^{s_2t}$. $y_1=e^{s_1t},y_2=e^{s_2t}$.
2. $s_1=s_2=s$ and real, $y=(A_1t+A_2)e^{st}$. $y_1=te^{st},y_2=e^{st}$.
3. $s_1,s_2=\beta\pm j\beta$, $y=e^{\beta t}(A_1\cos \gamma t+A_2\sin\gamma t)$. $y_1=e^{\beta t}\cos \gamma t,y_2=e^{\beta t}\sin \gamma t$.

Where $A_1$ and $A_2$ determined from $y(0)$ and $\mathrm d y(0)/\mathrm d t$.

We assume $A_1$ and $A_2$ and solve them by the initial value of $y$ and the initial derivative of $y$.

Finding Initial and Final Values

Finding

1. $i_L(t=0^-)\quad v_C(t=0^-)$
2. $i_L(t=0^+) \quad v_C(t=0^+)$
3. $\displaystyle \frac{\mathrm d i_L}{\mathrm d t}=\frac{v_L}{L}(t=0^+)(\mathrm{Use ~ KVL})\quad \frac{\mathrm d v_C}{\mathrm d t}=\frac{i_C}{C}(t=0^+)(\mathrm{Use ~ KCL})$
4. $i_L(t\to\infin)\quad v_c(t\to \infin)$


For $\boldsymbol{t=0^-}$


$i(0^-)=1\mathrm{~A},v(0^-)=2\mathrm{~V}$.

For $\boldsymbol{t=0^+}$

$v(0^+)=v(0^-)$ $i(0^-)=i(0^+)$

$i(0^+)=\frac{1}{20}\underbrace{\frac{\mathrm d v(0^+)}{\mathrm d t}}_{0\mathrm{~V/s}}+\frac{v(0^+)}{2}$

For $\boldsymbol{t=\infin}$

$v(\infin)=12 \mathrm{~V}\quad i(\infin)=6\mathrm{~A}$

$i_L(0^+),v_C(0^+)$ cannot change suddenly, thus equal to $i_L(0^-),v_C(0^+)$. However $v_R(0^+)\not=v_R(0^-)$

Circuit for $\boldsymbol {t=0^-}$


$i_L(0^+)=i_L(0^-)=-6\mathrm{~A}$, $v_C(0^+)=v_C(0^+)=0\mathrm{~V}$

Circuit for $\boldsymbol {t=0^+}$


Write KCL Eqn.

$i_R(0^+)=i_L(0^+)+6 \Rightarrow i_R(0^+)=0\Rightarrow v_R(0^+)=5\times i_R(0^+)=0$

For $\frac{\mathrm d i_L(0^+)}{\mathrm d t}$, Write KVL Eqn.

$2\frac{\mathrm d i_L(0^+)}{\mathrm d t}+v_R(0^+)=v_c(0^+)$

For $\frac{\mathrm d v_C (0^+)}{\mathrm d t}$, Write KCL Eqn.

$4=\frac{1}{5}\frac{\mathrm d v_C (0^+)}{\mathrm d t}+i_R(0^+)$

For $\frac{\mathrm d v_R(0^+)}{\mathrm d t}$, write KVL/KCL?

$\frac{\mathrm d v_C(0^+)}{\mathrm d t}=\frac{\mathrm d v_R(0^+)}{\mathrm d t}+{\color{blue}\frac{\mathrm d v_L(0^+)}{\mathrm d t}}$

$v_R(0^+)/5=i_R(0^+)=0$

$t \to \infin$, 电流源串联并不能叠加电流，而是需要满足 KCL.


Problem 1



Problem 2


Problem 3


Problem 6

The Source-Free Series RLC Circuit

By KVL,

$Ri+L\frac{\mathrm d i}{\mathrm d t}+\frac{1}{C}\int_{-\infin}^0 i\mathrm d t=0$


Take derivative on both sides,

$\frac{\mathrm d ^2 i}{\mathrm d t^2}+\frac{R}{L}\frac{\mathrm d i}{\mathrm d t}+\frac{1}{LC}i=0$

Using $v$ to express:

$Ri+L\frac{\mathrm d i}{\mathrm d t}+v=0$

$i=C\frac{\mathrm d v}{\mathrm d t}\quad \frac{\mathrm d i}{\mathrm d t}=C\frac{\mathrm d v^2}{\mathrm d t^2}$

$RC\frac{\mathrm d v}{\mathrm d t}+LC\frac{\mathrm d v^2}{\mathrm d t^2}+v=0$

$\frac{\mathrm d ^2 v}{\mathrm d t^2}+\frac{R}{L}\frac{\mathrm d v}{\mathrm d t}+\frac{1}{LC}v=0$

Equals to

$\frac{\mathrm d ^2 i}{\mathrm d t^2}+\frac{1}{\tau}\frac{\mathrm d i}{\mathrm d t}+\frac{1}{LC}i=0$

Initial Conditions

$i(0)=I_0$

The current through the inductor cannot change suddenly.

How about the derivative of $i(0)$?

Write KVL Eq. for $t>0$.

$Ri(0^+)+L\frac{\mathrm d i(0^+)}{\mathrm d t}+{\color{blue}v(0^+)}=0$

Characteristic Eq.

$Ae^{st}\left(s^2+\frac{R}{L}s+\frac{1}{LC}\right)=0$

Roots:

$s_1=-\frac{R}{2L}+\sqrt{\left(\frac{R}{2L}\right)^2-\frac{1}{LC}},s_2=-\frac{R}{2L}-\sqrt{\left(\frac{R}{2L}\right)^2-\frac{1}{LC}}$

$\boxed{i(t)=A_1e^{s_1 t}+A_2e^{s_2 t}}$

and $i(0)=A_1+A_2,\mathrm d i(t)/\mathrm d t=s_1 A_1+s_2A_2$.

Introduce $\boldsymbol \alpha$ and $\boldsymbol \omega_0$

$\alpha :=\frac{R}{2L}=\frac{1}{2\tau}$

Neper frequency or damping factor. Np/s

Determines rate at which response is damped. 随时间变化的相应逐渐削弱的过程，叫做阻尼。

$\omega_0:=\frac{1}{\sqrt{LC}}$

Resonant frequency or undamped natural frequency. rad/s

The characteristic eqn. can be changed to:

$s^2+2\alpha s+\omega_0=0$

$s_1=-\alpha+\sqrt{\alpha^2-\omega_0^2},s_2=-\alpha-\sqrt{\alpha^2-\omega_0^2}$

associated with natural response of circuit and measured in nepers per second.

Case 1: $\boldsymbol{\alpha^2>\omega_0^2}$ Overdamping

$i(t)=A_1 e^{s_1 t}+A_2 e^{s_2 t}$

$\alpha>\omega_0 \Rightarrow C>\frac{4L}{R^2}$

$s_{1,2}$ are negative and real.

Case 2: $\boldsymbol{\alpha^2=\omega_0^2}$ Critical damping

$i(t)=(A_1 t+A_2) e^{-\alpha t}$

$\alpha=\omega_0 \Rightarrow C=\frac{4L}{R^2}$

Case 3: $\boldsymbol{\alpha^2<\omega_0^2}$ Underdamping

$s_{1,2}=-\alpha\pm j \omega_d \quad \omega_d=\sqrt{\omega_0^2-\alpha^2}$

Both $\omega_0$ and $\omega_d$ are natural frequencies.

$\omega_0$: undamped natural frequency. $\omega_d$: damped natural frequency.

Solution:

$i(t)=e^{-\alpha t}(A_1 \cos \omega_d t+A_2\sin \omega_d t)$

with time constant $\tau=1/\alpha$ and period $T=2\pi/\omega_d$



Problem 7


Problem 14


Problem 16



注意

$\frac{\mathrm d i(0)}{\mathrm d t}=-\frac{1}{L}(Ri(0)+v_C(0))$

利用KVL计算。

The Source-Free Parallel RLC Circuit


Initial Conditions. $v(0)=v_C,\mathrm d v(0)/\mathrm d t=-\frac{1}{RC}(R I_0+v_C)$.


$\boxed{\alpha=\frac{1}{2RC}=\frac{1}{2\tau}\quad \omega_0=\frac{1}{\sqrt{LC}}}$






$\alpha=\frac{1}{2RC} \quad \omega_0=\frac{1}{\sqrt{LC}}$


$\alpha=\frac{R_{eq}}{2L}=\frac{1}{\sqrt{LC}}$


$i(0^-)=i(0^+)=5\mathrm{~A}$


$i(0^-)=i(0^+)=12\mathrm{~A} \quad v(0^-)=v(0^+)=0\mathrm{~V}$

Special Case: Undamping.

$v(t)=A\sin(\omega t)$

$C\mathrm d v(t)/\mathrm d t=A\omega \cos (\omega t)$

when $t=0$, $i=12$, $A\omega=12$. and $\omega=1/\sqrt{LC}=1/2$.


串联情况下，电压和电流的方程的情况是一样的。

Assuming $R=2\mathrm{~k\Omega}$, design a series and parallel RLC circuit that has the characteristic equation.

$\tau=L/R=1/100\quad 1/LC=10^6$

$L=20,C=50n$

Problem 18



Problem 19


Problem 20



Problem 35

Problem 40

注意要求电流，然后注意 $\mathrm d v(0)/\mathrm d t$ 的初值条件。

Problem 42

Problem 48

注意 $i(0)$ 正负，得到 $i(0)=-2\mathrm{~A},v(0)=2\mathrm{~V}$。

Problem 55

Step Response of a Series RLC Circuit

Math Background

$\frac{\mathrm d ^2 y}{\mathrm d t}+p\frac{\mathrm d y}{\mathrm d t}+qy=C$

General solution: $y=y_{ss}+y_t$.

• $y_{ss}$ is steady-state part, $y_{ss}=y(\infin)=C/q$.
• $y_t$ is a transient part depending on roots of $s^2+ps+q=0$. 和上面的情况一样。



Solve the following differential equations subject to the specified initial conditions.

• $\frac{\mathrm d ^2 v}{\mathrm d t^2}+4v=12,v(0)=0,\mathrm d v(0)/\mathrm d t=2$

  $v(\infin)=3,s_{1,2}=\pm2i$

  $v(t)=3+A_1\cos 2t+A_2\sin2t$

  $A_1=-3,2A_2=2\Rightarrow A_2=1$

• $\frac{\mathrm d ^2 v}{\mathrm d t^2}+2\frac{\mathrm d v}{\mathrm d t}+v=3,v(0)=5,\mathrm d v(0)/\mathrm d t=1$

  $s^2+2s+1=0,s_{1,2}=-1$. $v_t=(A_1t+A_2)e^{-t}$.

  $v_{ss}=3,v(t)=v_{ss}+v_t=3+(A_1t+A_2)e^{-t}$

  $v(0)=5=3+A_2 \Rightarrow A_2=\cdots$

• $\frac{\mathrm d ^2 i}{\mathrm d t^2}+2\frac{\mathrm d i}{\mathrm d t}+5i=10,i(0)=4,\mathrm d i(0)/\mathrm d t=-2$

  两个复根。

  $i(t)=2+e^{-\alpha t}(A_1\cos \omega_d t+A_2\sin\omega_d t)$


首先是 DC Circuit. Either Ch.7, Ch.8

分析是 Second Order Circuit. 其中 $i$ 和 $v$ 确定一个就好。

$i=C\frac{\mathrm d v}{\mathrm d t},-24+Ri+L\frac{\mathrm d i}{\mathrm d t}+v=0$


确定类型是 Overdamped. 利用

$i=C\frac{\mathrm d v}{\mathrm d t}$

可以得到 $C=150\mathrm{~nF}$.


General Second-Order Circuits



$\frac{\mathrm d ^2 x}{\mathrm d t^2}+p\frac{\mathrm d x}{\mathrm d t}+q=C$


首先确定是 Series.


解得

$\frac{\mathrm d ^2 i}{\mathrm d t^2}+7\frac{\mathrm d i}{\mathrm d t}+10i=30$

解得

$s_1=-2,s_2=-5$

$i=3+A_1 e^{-2t}+A_2e^{-5t}$