Ch7 First Order Circuits

Terminologies


The Source-Free RC Circuit

Capacitor initially connected in a circuit with source, which is suddenly disconnected.

Energy stored in capacitor released to resistor.

What is circuit’s reaction to excitation, i.e., circuit response.


Voltage can’t suddenly change $\Rightarrow$ choose voltage to analyze.

$C\frac{\mathrm d v}{\mathrm d t}+\frac{v}{R}=0\\ \frac{\mathrm d v}{v}=-\frac{1}{RC}\mathrm d t$

Voltage response of RC circuit.

The natural response of a circuit is due to initial energy stored and physical characteristics of circuit itself, with no external sources of excitation.

$v(t)=Ae^{-t/RC} \Rightarrow v(t)=V_0e^{-t/RC}$

$\tau := RC \quad v(t)=V_0e^{-t/\tau}$

Time required for response to decay to a factor of 36.8% of $V_0$, expressing rapidity with which voltage decreases, measured in seconds.

Capacitor is assumed fully discharged after $5\tau$ of the circuit takes $5\tau$ to reach its final state or steady state.

$v(t+\tau)=0.368v(t)=\frac{1}{e}v(t)$, regardless $t$.

The smaller the $\tau$, the more rapidly the voltage decreases.


Natural Response.

How to determine the time constant?

1. Combining several capacitors to a single $C_{\mathrm{eq}}$.
2. Finding Thevenin equivalent $R_{\mathrm{Th}}$ at terminals of $C_{\mathrm{eq}}$.
3. Determining $\tau=R_{\mathrm{Th}}C_{\mathrm{eq}}$.

Current, Power, and Energy

$v(t)=V_0e^{-t/\tau}\Rightarrow i_R(t)=\frac{v(t)}{R}=\frac{V_0}{R}e^{-t/\tau}$

$p(t)=vi_R=\frac{V_0^2}{R} e^{-2t/\tau}$

$w_R(t)=\int_0^t p\mathrm d t=\int_0^t \frac{V_0^2}{R}e^{-2t/\tau}\mathrm d t=\frac{1}{2}CV_0^2(1-e^{-2t/\tau})$

$w_R(\infin)\rightarrow\frac{1}{2}CV_0^2 =w_c(0)$, which means that energy initially stored in $C$ is eventually dissipated in $R$.

Analyzing source-free $RC$ circuit includes:

• Find $v(0)=V_0$ and $\tau$.
• Obtain $v(t)=V_0e^{-t/\tau}$.
• Determine $i_C,v_R,i_R$.
• Determine $p$ and $w$.

Chapter 7, Problem 1.



Chapter 7, Problem 2.


Thevenin Equivalent.

Chapter 7, Problem 4.


$v_0=40$.

Chapter 7, Problem 7.


Using

$v_o(t)=v_o(\infin)+[v_o(0)-v_o(\infin)]e^{-t/\tau}=7.2+0.8e^{-t/24}\mathrm{~V}$

Chapter 7, Problem 10.


Decay to $1/n$.

$Ae^{-t_2/\tau}/Ae^{-t_1/\tau}=e^{(t_1-t_2)/\tau}=1/n \Rightarrow \boxed{t_2-t_1=\tau\ln(n)}$

The Source-Free RL Circuit

• Inductor initially connected in a circuit with source, which is suddenly disconnected.
• Energy stored in inductor released to resistor.


$i(t)=Ae^{-Rt/L} \overset{i(0)=A=I_0}{=\!=\!=\!=\!\Rightarrow}i(t)=I_0e^{-Rt/L}$

$\tau:=\frac{L}{R} \quad i(t)=I_0e^{-t/\tau}$

Summary


Notes

• Source-free $RC$ or $RL$ circuit is due to sudden change of a circuit.

• Method I Short-Cut Approach: find $\tau$ and initial value

  $v(t)=V_0e^{-t/\tau} \quad i(t)=I_0e^{-t/\tau}$

• Method II Straightforward Approach. Write KCL or KVL $1^{\mathrm{st}}$-order differential Eqs., and solve

  $\frac{\mathrm d x}{\mathrm d t}+ax=0 \Rightarrow x(t)=X_0e^{-at}$

Examples

$\frac{\mathrm d v}{\mathrm d t}+2v=0 \Rightarrow \int\frac{\mathrm d v}{v}=\int-2\mathrm d t \Rightarrow v=Ae^{-2t}\overset{v(0)=-1\mathrm V}{=\!=\!=\!=\!=\!}-e^{2t}$

$2\frac{\mathrm d i}{\mathrm d t}-3i=0 \Rightarrow\frac{\mathrm d i}{i}=1.5t\Rightarrow i=Ae^{1.5t}\overset{i(0)=2\mathrm A}{=\!=\!=\!=\!=\!}2e^{1.5t}$


$R_\mathrm{eq}=20//5=4$ $C=0.1F$ $\tau=0.4$ $v_C(t)=15e^{-2.5t}$ $v_x(t)=\frac{12}{20}v_C(t)=9e^{-2.5t}$ $i_x=v_x(t)/12=0.75e^{-2.5t}$


$\tau=L/R=1/50 \quad R=v/i=3\quad L=3/50=60\mathrm{~mH} \quad \tau=20\mathrm{~ms}\\W(0)=\frac{1}{2}Li(0)^2=\frac{1}{2} 60m\times 30^2=27\mathrm{~J}\quad W(0.01)=\frac{1}{2}60m\times(30e^{-0.5})^2=W(0)/e$


$v=L\frac{\mathrm d i}{\mathrm d t}\\ i+\frac{v}{2}+\frac{v-3i}{4}=0\Rightarrow i=-3v\\ -2/3 i=\frac{\mathrm d i}{\mathrm d t}\\ \frac{\mathrm d i}{i}=-2/3 \mathrm d t\\ \Rightarrow i=10e^{-2/3 t}\mathrm{~A}\\ i_x=\frac{v}{2}=-\frac{1}{6}i=-\frac{5}{3}e^{-2/3t}\mathrm{~A}$



Draw the circuit before $t=0$

$v_0=24\mathrm{~V}\times\frac{3}{6+3}=8\mathrm{~V}$

$w_c(0)=\frac{1}{2}Cv_0^2=\frac{1}{12}\times 64=\frac{16}{3}$

Draw the circuit after $t=0$

$R_{eq}=3\mathrm{~\Omega},C=\frac{1}{6} \mathrm{~F} \Rightarrow v(t)=8e^{-t/\tau}=8e^{-2t}\mathrm{~V}$




Before $t=0$

$v_0=12\mathrm{~V}$

Between $t=0$ and $t=1$

$RC=500k \times 2m=1000 s\Rightarrow v_t=12e^{-t/1000}$

After $t=1$

$v_1=12e^{-1/1000}\approx 11.988\\ RC=1k\times 2m=2\\ v_t=v_1 e^{-(t-1)/2}=11.988e^{-(t-1)/2}$


Before $t=0$

$v=20\frac{9}{9+3}=15\mathrm{~V}$

After $t=0$

$v_t=15e^{-t/(10\times20m)}=15e^{-5t}$

$E=\frac{Cv^2}{2}=\frac{20m 15^2}{2}=2.25$

Chapter 7, Problem 11.


对于电容，找电压关系，对于电感，找电流关系，为了找电流关系，变换电压源为电流源。


Chapter 7, Problem 13.


(b) The energy dissipated in the resistor is

$\begin{aligned} W & =\int_0^t p d t=\int_0^t 80 \times 10^{-3} e^{-2 \times 10^3} d t=-\left.\frac{80 \times 10^{-3}}{2 \times 10^3} e^{-2 \times 10^3 t} \right| \begin{array}{c} 0.5 \times 10^{-3} \\ 0 \end{array} \\ & =40\left(1-e^{-1}\right) \mu \mathrm{J}=25.28 \mu \mathrm{J} \end{aligned}$

注意负号

Chapter 7, Problem 16.



Chapter 7, Problem 18.


Chapter 7, Problem 19.


最好用的方法是直接解微分方程，利用 Supermesh.

$10i+40\cdot0.5i+v=0 \quad v=L\frac{\mathrm d i}{\mathrm d t} \Rightarrow i=2e^{-5t}$

也可以利用戴维南等值。


Singularity Functions

奇异函数及其导数都是不连续的。最广泛使用的三种奇异函数是单位阶跃 (unit step)、单位冲激 (unit impulse) 和单位斜坡 (unit ramp) 函数。


Unit Step Function


$v(t)=\left\{ \begin{aligned} &0,&t<t_0\\ &V_0,&t>t_0 \end{aligned} \right.$

can be expressed as

$v(t)=V_0u(t-t_0)$


Unit Impulse / Delta Function

$\delta(t)=\frac{\mathrm d u(t)}{\mathrm dt}=\left\{ \begin{aligned} &0, &t<0\\ &\mathrm{Undefined~or~ 1}, &t=0\\ &0,&t>0 \end{aligned} \right.$

$\delta(t)$ can be regarded as a very short duration pulse of unit area (strength)

$\int_{0^-}^{0^+} \delta(t) \mathrm d t=1$

Sampling or sifting property.

$\int_a^b f(t)\delta (t-t_0)\mathrm d t=f(t_0)$

Proof:

$\int_a^b f(t)\delta(t-t_0)\mathrm d t=\int_a^b f(t_0)\delta(t-t_0)\mathrm d t=f(t_0)\int_a^b \delta(t-t_0)\mathrm d t=f(t_0)$

Special case: $t_0=0$.

Useful in DSP.


Unit Ramp Function

$r(t)=\int_{-\infin}^t u(t)\mathrm d t=tu(t)=\left\{ \begin{aligned} &0,&t\le0\\ &t,&t>0 \end{aligned} \right.$


a. $5(u(t-2)-u(t-4))$. 滤波使用

b. $10u(t)-18u(t-2)+8u(t-5)$

c. $2r(t)-2r(t-5)-10u(t-5)$

d. $5r(t)-10r(t-1)+5r(t-2)$.

e. $2u(t)-2r(t)+2r(t-2)+4u(t-2)-2u(t-3)$


• $v_s=2u(-t)+4u(t)=2+2u(t)$.

• 

• $i=\frac{1}{10m}\int 15\delta(t-2)m \mathrm d t=1.5u(t-2)mA$


The sifting property of delta function.

$(-3^3+5(-3)^2+10)==28$

$\cos 3\pi=-1$


对于电路有变化的情况非常有用。

Step Response of an RC Circuit

What is step response?

Response of circuit due to a sudden application of a dc voltage or current source modeled as step function.

书上这段有问题，应该电流源才能这么写



$\frac{V_s u(t)-v}{R}=C\frac{\mathrm d v}{\mathrm d t}$

Equals to

$\frac{v_s -v}{R}=C\frac{\mathrm d v}{\mathrm d t} \quad t>0$

$\Rightarrow v(t)=V_s+(V_0-V_s) e^{-t/\tau}$


Current through capacitor

$i(t)=C\frac{\mathrm d v}{\mathrm d t}=\left\{ \begin{aligned} &\frac{V_s}{R}e^{-t/\tau}u(t) , &V_0=0\\ &\frac{V_s-V_0}{R}e^{-t/\tau}u(t),&V_0\not=0 \end{aligned} \right.$

Decomposing complete response in terms of sources.

自然相应和强制相应。

$\underbrace{v(t)}_{\mathrm{Complete~Response}}=\underbrace{V_0e^{-t/\tau}}_{v_n:\mathrm{Natural~Response}}+\underbrace{V_s(1-e^{-t/\tau})}_{v_f:\mathrm{Forced~Response}}$


• Natural response is due to internal stored energy.
• Forced response is due to external independent source.
• $v_n$ eventually dies out along with transient component of $v_f$, leaving only steady state component of $v_f$.

Decomposing complete response in terms of permanency

$\underbrace{v(t)}_{\mathrm{Complete~Response}}=\underbrace{V_0e^{-t/\tau}}_{v_{ss}:\mathrm{Steady-state~response}}+\underbrace{V_s(1-e^{-t/\tau})}_{v_t:\mathrm{Transient~Response}}$

• Steady-state response $v_{ss}=V_s$: remains after transient response has died out-permanent part.
• Transient response: $v_t=(V_0-V_s)e^{-t/\tau}$ will die out with time - temporary part.
• When $V_s=0$ (source-free), $v_n=v_t,v_f=v_{ss}$.

Find step response of $RC$ circuit, Short-cut Method

$v(t)=v(\infin)+[v(0)-v(\infin)]e^{-t/\tau}$

where

$v(0):$ initial capacitor voltage obtained from circuit for $t<0$.

$v(\infin):$ final capacitor voltage obtained from circuit for $t>0$. (after $5\tau$)

$\tau:$ time constant obtained from circuit from circuit for $t>0$.

Notes

If switching at $t=t_0 \not=0$,

$v(t)=v(\infin)+[v(t_0)-v(\infin)] e^{-{\color{red}(t-t_0)}/\tau}$

where $v(t_0)$ is initial value at $t=t_0^+$.

• $\tau$ only depends on $R,L,C$
• Source-free $RC$ circuit is a special case where $v(\infin)=0$.

Eq. of complete response applies only to step responses (input excitation is constant)

Step Response of an RL Circuit

$\boxed{i(t)=\frac{V_s}{R}+\left(I_0-\frac{V_s}{R}\right)e^{-t/\tau}}$

where $\tau=L/R$.

$i(t)=i(\infin)+[i(0)-i(\infin)]e^{-t/\tau}$

where

$i(0)$: initial inductor current obtained from circuit for $t<0$.

$i(\infin)$: final inductor current obtained from circuit for $t>0$ (after $5\tau$)

$\tau$: time constant obtained from circuit for $t>0$.

For calculating $v(t)$, use $v(t)=L\frac{\mathrm d i(t)}{\mathrm d t}$, we can derive that:

$v(t)=\frac{L}{\tau} (i(0)-i(\infin))e^{-t/\tau}$


Solve for $v$ in the following differential Eqs., subject to the stated initial condition.

$2\frac{\mathrm d v}{\mathrm d t}-v=3u(t) \quad v(0)=-6$

Initial state: $v=-6$, final state: $v=-3$. Homogenous equation:

$2\frac{\mathrm d v}{\mathrm d t}=v \Rightarrow 0.5\mathrm d t=\frac{\mathrm dv}{v}\Rightarrow v=-3e^{0.5t}-3$

Chapter 7, Problem 55

后面 $i_0=0$，相当于导线。

Chapter 7, Problem 62

注意第二个从 $1$ 开始计时。

Chapter 7, Problem 64



关键是 $R_{eq}=2+6//3$

First-Order Op Amp Circuits

RC Type Op Amp Circuit (small)

Chapter 7, Problem 75



注意电流参考方向。

Quiz 4


首先注意到 Op Amp 对应的电压和电流关系。设 $C_1$ 电压为 $v_1$，$C_2$ 电压为 $v_2$，可以得到：

$v_1=v_2+i_2R=v_2 +RC \frac{\mathrm d v_2}{\mathrm d t}=v_2 + \frac{\mathrm d v_2}{\mathrm d t}$

还有对应的电压关系：

$v_2+CR\left(\frac{\mathrm d v_1}{\mathrm d t}+\frac{\mathrm d v_2}{\mathrm d t}\right)=1$

也就是

$v_1+\frac{\mathrm d v_1}{\mathrm d t}=1$

结合 $v_1(0)=0$，得到 $v_1=1-e^{-t}$。

为了求出 $v_2$，还是向上回代，得到

$1-e^{-t}=v_2 + \frac{\mathrm d v_2}{\mathrm d t}$

解得 $v_2(t)=c_1 e^{-t}-e^{-t}t+1$，得到 $c_1=-1$，得出 $v_2(t)=-e^{-t}(t+1)+1$。

$v_o(t)=-\frac{\mathrm d v_2}{\mathrm d t}=-te^{-t}$