Graduationwoot

Dragon Notes

i

\( \newcommand{bvec}[1]{\overrightarrow{\boldsymbol{#1}}} \newcommand{bnvec}[1]{\overrightarrow{\boldsymbol{\mathrm{#1}}}} \newcommand{uvec}[1]{\widehat{\boldsymbol{#1}}} \newcommand{vec}[1]{\overrightarrow{#1}} \newcommand{\parallelsum}{\mathbin{\|}} \) \( \newcommand{s}[1]{\small{#1}} \newcommand{t}[1]{\text{#1}} \newcommand{tb}[1]{\textbf{#1}} \newcommand{ns}[1]{\normalsize{#1}} \newcommand{ss}[1]{\scriptsize{#1}} \newcommand{vpl}[]{\vphantom{\large{\int^{\int}}}} \newcommand{vplup}[]{\vphantom{A^{A^{A^A}}}} \newcommand{vplLup}[]{\vphantom{A^{A^{A^{A{^A{^A}}}}}}} \newcommand{vpLup}[]{\vphantom{A^{A^{A^{A^{A^{A^{A^A}}}}}}}} \newcommand{up}[]{\vplup} \newcommand{Up}[]{\vplLup} \newcommand{Uup}[]{\vpLup} \newcommand{vpL}[]{\vphantom{\Large{\int^{\int}}}} \newcommand{lrg}[1]{\class{lrg}{#1}} \newcommand{sml}[1]{\class{sml}{#1}} \newcommand{qq}[2]{{#1}_{\t{#2}}} \newcommand{ts}[2]{\t{#1}_{\t{#2}}} \) \( \newcommand{ds}[]{\displaystyle} \newcommand{dsup}[]{\displaystyle\vplup} \newcommand{u}[1]{\underline{#1}} \newcommand{tu}[1]{\underline{\text{#1}}} \newcommand{tbu}[1]{\underline{\bf{\text{#1}}}} \newcommand{bxred}[1]{\class{bxred}{#1}} \newcommand{Bxred}[1]{\class{bxred2}{#1}} \newcommand{lrpar}[1]{\left({#1}\right)} \newcommand{lrbra}[1]{\left[{#1}\right]} \newcommand{lrabs}[1]{\left|{#1}\right|} \newcommand{bnlr}[2]{\bn{#1}\left(\bn{#2}\right)} \newcommand{nblr}[2]{\bn{#1}(\bn{#2})} \newcommand{real}[1]{\Ree\{{#1}\}} \newcommand{Real}[1]{\Ree\left\{{#1}\right\}} \newcommand{abss}[1]{\|{#1}\|} \newcommand{umin}[1]{\underset{{#1}}{\t{min}}} \newcommand{umax}[1]{\underset{{#1}}{\t{max}}} \newcommand{und}[2]{\underset{{#1}}{{#2}}} \) \( \newcommand{bn}[1]{\boldsymbol{\mathrm{#1}}} \newcommand{bns}[2]{\bn{#1}_{\t{#2}}} \newcommand{b}[1]{\boldsymbol{#1}} \newcommand{bb}[1]{[\bn{#1}]} \) \( \newcommand{abs}[1]{\left|{#1}\right|} \newcommand{ra}[]{\rightarrow} \newcommand{Ra}[]{\Rightarrow} \newcommand{Lra}[]{\Leftrightarrow} \newcommand{rai}[]{\rightarrow\infty} \newcommand{ub}[2]{\underbrace{{#1}}_{#2}} \newcommand{ob}[2]{\overbrace{{#1}}^{#2}} \newcommand{lfrac}[2]{\large{\frac{#1}{#2}}\normalsize{}} \newcommand{sfrac}[2]{\small{\frac{#1}{#2}}\normalsize{}} \newcommand{Cos}[1]{\cos{\left({#1}\right)}} \newcommand{Sin}[1]{\sin{\left({#1}\right)}} \newcommand{Frac}[2]{\left({\frac{#1}{#2}}\right)} \newcommand{LFrac}[2]{\large{{\left({\frac{#1}{#2}}\right)}}\normalsize{}} \newcommand{Sinf}[2]{\sin{\left(\frac{#1}{#2}\right)}} \newcommand{Cosf}[2]{\cos{\left(\frac{#1}{#2}\right)}} \newcommand{atan}[1]{\tan^{-1}({#1})} \newcommand{Atan}[1]{\tan^{-1}\left({#1}\right)} \newcommand{intlim}[2]{\int\limits_{#1}^{#2}} \newcommand{lmt}[2]{\lim_{{#1}\rightarrow{#2}}} \newcommand{ilim}[1]{\lim_{{#1}\rightarrow\infty}} \newcommand{zlim}[1]{\lim_{{#1}\rightarrow 0}} \newcommand{Pr}[]{\t{Pr}} \newcommand{prop}[]{\propto} \newcommand{ln}[1]{\t{ln}({#1})} \newcommand{Ln}[1]{\t{ln}\left({#1}\right)} \newcommand{min}[2]{\t{min}({#1},{#2})} \newcommand{Min}[2]{\t{min}\left({#1},{#2}\right)} \newcommand{max}[2]{\t{max}({#1},{#2})} \newcommand{Max}[2]{\t{max}\left({#1},{#2}\right)} \newcommand{pfrac}[2]{\frac{\partial{#1}}{\partial{#2}}} \newcommand{pd}[]{\partial} \newcommand{zisum}[1]{\sum_{{#1}=0}^{\infty}} \newcommand{iisum}[1]{\sum_{{#1}=-\infty}^{\infty}} \newcommand{var}[1]{\t{var}({#1})} \newcommand{exp}[1]{\t{exp}\left({#1}\right)} \newcommand{mtx}[2]{\left[\begin{matrix}{#1}\\{#2}\end{matrix}\right]} \newcommand{nmtx}[2]{\begin{matrix}{#1}\\{#2}\end{matrix}} \newcommand{nmttx}[3]{\begin{matrix}\begin{align} {#1}& \\ {#2}& \\ {#3}& \\ \end{align}\end{matrix}} \newcommand{amttx}[3]{\begin{matrix} {#1} \\ {#2} \\ {#3} \\ \end{matrix}} \newcommand{nmtttx}[4]{\begin{matrix}{#1}\\{#2}\\{#3}\\{#4}\end{matrix}} \newcommand{mtxx}[4]{\left[\begin{matrix}\begin{align}&{#1}&\hspace{-20px}{#2}\\&{#3}&\hspace{-20px}{#4}\end{align}\end{matrix}\right]} \newcommand{mtxxx}[9]{\begin{matrix}\begin{align} &{#1}&\hspace{-20px}{#2}&&\hspace{-20px}{#3}\\ &{#4}&\hspace{-20px}{#5}&&\hspace{-20px}{#6}\\ &{#7}&\hspace{-20px}{#8}&&\hspace{-20px}{#9} \end{align}\end{matrix}} \newcommand{amtxxx}[9]{ \amttx{#1}{#4}{#7}\hspace{10px} \amttx{#2}{#5}{#8}\hspace{10px} \amttx{#3}{#6}{#9}} \) \( \newcommand{ph}[1]{\phantom{#1}} \newcommand{vph}[1]{\vphantom{#1}} \newcommand{mtxxxx}[8]{\begin{matrix}\begin{align} & {#1}&\hspace{-17px}{#2} &&\hspace{-20px}{#3} &&\hspace{-20px}{#4} \\ & {#5}&\hspace{-17px}{#6} &&\hspace{-20px}{#7} &&\hspace{-20px}{#8} \\ \mtxxxxCont} \newcommand{\mtxxxxCont}[8]{ & {#1}&\hspace{-17px}{#2} &&\hspace{-20px}{#3} &&\hspace{-20px}{#4}\\ & {#5}&\hspace{-17px}{#6} &&\hspace{-20px}{#7} &&\hspace{-20px}{#8} \end{align}\end{matrix}} \newcommand{mtXxxx}[4]{\begin{matrix}{#1}\\{#2}\\{#3}\\{#4}\end{matrix}} \newcommand{cov}[1]{\t{cov}({#1})} \newcommand{Cov}[1]{\t{cov}\left({#1}\right)} \newcommand{var}[1]{\t{var}({#1})} \newcommand{Var}[1]{\t{var}\left({#1}\right)} \newcommand{pnint}[]{\int_{-\infty}^{\infty}} \newcommand{floor}[1]{\left\lfloor {#1} \right\rfloor} \) \( \newcommand{adeg}[1]{\angle{({#1}^{\t{o}})}} \newcommand{Ree}[]{\mathcal{Re}} \newcommand{Im}[]{\mathcal{Im}} \newcommand{deg}[1]{{#1}^{\t{o}}} \newcommand{adegg}[1]{\angle{{#1}^{\t{o}}}} \newcommand{ang}[1]{\angle{\left({#1}\right)}} \newcommand{bkt}[1]{\langle{#1}\rangle} \) \( \newcommand{\hs}[1]{\hspace{#1}} \)

  UNDER CONSTRUCTION

Phasor Circuit Analysis

Method 1:

  • [1] Transform circuit and signal into phasor domain
  • [2] Apply circuit analysis to obtain \(IV\)-relations
  • [3] Inverse-transform circuit into the time domain

\(\begin{align} \ds A\cos{(\omega t)} & \Leftrightarrow A\\ A\cos{(\omega t + \phi)} & \Leftrightarrow Ae^{j\phi} \end{align}\) \(\begin{align} \ds A\sin{(\omega t)} & \Leftrightarrow -jA \\ A\sin{(\omega t + \phi)} & \Leftrightarrow Ae^{-j\pi/2+j\phi} \end{align}\)



Method 2:

  • [1] Obtain integrodifferential \(IV\)-relations via circuit analysis
  • [2] Convert obtained relations into ordinary differential equations
  • [3] Transform the ODEs into phasor domain, solve for \(I/V\) of interest
  • [4] Inverse-transform into the time domain

RESISTOR CAPACITOR INDUCTOR
\(v(t)\) \(\ds Ri(t)\) \(\ds \frac{1}{C}\int_{t_0}^{t}i_C(t)dt+i_C(t_0)\) \(\ds L\frac{di_L(t)}{dt}\)
\(i(t)\) \(\ds v(t)/R\) \(\ds C\frac{dv_C(t)}{dt}\) \(\ds \frac{1}{L}\int_{t_0}^{t}v_L(t)dt + i_L(t_0)\)
\(\vphantom{\int_{t}^{t}} Z\) \(\vphantom{\int_{t}^{t}} R\) \(\vphantom{\int_{t}^{t}} 1/j\omega C\) \(\vphantom{\int_{t}^{t}} j\omega L\)

\(\begin{align} \ds A\cos{(\omega t)} & \Leftrightarrow A\\ A\cos{(\omega t + \phi)} & \Leftrightarrow Ae^{j\phi} \end{align}\) \(\begin{align} \ds A\sin{(\omega t)} & \Leftrightarrow -jA \\ A\sin{(\omega t + \phi)} & \Leftrightarrow Ae^{-j\pi/2+j\phi} \end{align}\)

[Ex]: RLC series/parallel driven by Sinusoid

GIVEN:
\(v_s(t)=5\cos{(\omega t)}\ \t{V},\ R=100\ \Omega,\ C=5\ \mu\t{F},\ L = 10 m\t{H},\ \omega=1,000\ \t{rad/s}\)

DETERMINE: \(i(t), i_C(t), i_R(t)\)
SOLUTION 1:

Step 1: The phasor domain equivalent circuit is
Step 2 & 3: Combine the \(RC\) in parallel, then apply current division to find \(\bf{I}_C\) and \(\bf{I}_R\):
\(Z_{EQ'} = Z_C\|Z_R\)
\(\begin{align} \ds & = \vpl (1/j\omega C)\|R \\ & = \frac{(1/j\omega C)R}{1/j\omega C+R} \\ & = \frac{-jR/\omega C}{R - j/\omega C}\cdot \frac{R+j/\omega C}{R+j/\omega C} \\ & = \frac{R/\omega^2 C^2 - jR^2/\omega C}{R^2+1/\omega^2 C^2} \\ & = \frac{R}{\omega^2 R^2C^2 + 1}(1-j\omega RC) \\ & = \frac{R\sqrt{\omega^2 R^2C^2 +1}}{\omega^2 R^2C^2 +1}e^{j\atan{-\omega RC}} \end{align}\)
\(\ds =\ \frac{R}{\sqrt{\omega^2 \tau^2 +1}}e^{j\atan{-\omega \tau}},\ \ \tau=RC.\)
Plugging in, we find \(\tau = RC=(100\ \Omega )(5\ \mu\t{F})=500\ \mu \t{s}\), and \(\tau \omega = (500\ \mu \t{s})(1,000\ \t{rad/s})=0.5\ \t{rad}\) - and thus:

\(\begin{align} \ds Z_{\t{EQ'}} & = \frac{100}{\sqrt{.25+1}}e^{j\atan{-0.5}} \\ & = 89.4\ e^{-j\deg{26.57}} \\ & = 89.44(.894-j.447)\Rightarrow \\ Z_{\t{EQ'}}& = (80.0-j40.0)\ \Omega \end{align}\)
Combining with the inductor, the (total) equivalent resistance is
\(\begin{align} \ds Z_{\t{EQ}}& =Z_L+Z_{\t{EQ'}} \\ & =j(1,000\ \t{rad/s})(10 m\t{H})+ (79.97-j39.96)\\ & = 80.0 - j40.0 + j10.0 \\ & = 80.0-j30.0 \Rightarrow \end{align} \)
\(\ds \boxed{Z_{\t{EQ}} = 85.4\adegg{-20.54}\ \Omega}\)
Applying Ohm's law and inver-transforming into the time domain, the total current is \(\vpl\)
\(\begin{align} \ds i(t) & = v(t)/Z_{\t{EQ}} \\ & =\frac{5\Cos{\omega t}\ \t{V}}{85.4\adeg{-20.54}\Omega}\\ & = 0.05855\ \Real{\frac{e^{j\omega t}}{e^{-j\deg{20.54}}}}\ \t{A} \\ & = 58.6\ \real{\lrg{}e^{j\omega t +j\deg{20.54}}}\ m\t{A} \Rightarrow \end{align} \)
\(\vpl \ds \bxred{i(t)=58.6\Cos{\omega t+\deg{20.54}}\ m\t{A}}\)
Applying current division,
\(\begin{align} \ds i_C(t) & = \frac{Z_C\|Z_R}{Z_C}i(t) \\ & = \frac{Z_R}{Z_C+Z_R}i(t) = \frac{R}{1/j\omega C+R}i(t) \\ & = \frac{1}{1-j/\omega RC}i(t) \\ & = \frac{1}{1-j/(.5\t{rad})}\cdot \frac{1+j/(.5\t{rad})}{1+j/(.5\t{rad})}i(t) \\ & = \frac{1+j2}{1+4}i(t) \\ & = \vphantom{\frac{A}{A}}{\sqrt{5}}^{-1} \Real{e^{j\deg{63.43}}\cdot 58.6 e^{j\omega t+j\deg{20.54}}} m\t{A} \end{align} \)
\(\vpl \ds \bxred{i_C(t) = 26.2 \Cos{\omega t + \deg{84.0}}\ m\t{A}}\)
\(\begin{align} \ds i_R(t) & = \frac{Z_C\|Z_R}{Z_R}i(t) \\ & = \frac{Z_C}{Z_C+Z_R}i(t) = \frac{1/j\omega C}{1/j\omega C+R}i(t) \\ & = \frac{1/j\omega RC}{1+1/j\omega RC}i(t) \\ & = \frac{-j2}{1-j2}\cdot \frac{1+j2}{1+j2}i(t) \\ & = \frac{4-j2}{1+4}i(t) \\ & = 2{\sqrt{5}}^{-1}\ \Real{e^{-j\deg{26.57}}\cdot 58.6 e^{j\omega t+j\deg{20.54}}}\ m\t{A} \end{align} \)
\(\ds \vphantom{\frac{A^A}{A}}\bxred{i_R(t)=52.4 \Cos{\omega t-\deg{6.01}}\ m\t{A}}\)


SOLUTION 2:

Step 1 & 2: Applying node-voltage KCL at node \(A\) (shown below) and differentiating to obtain an ODE, we obtain

\(\begin{align} \ds & \\ 0 & =i_C(t)+i_R(t)-i(t) \\ & = C\frac{dv_A(t)}{dt} + \frac{v_A(t)}{R}-\frac{1}{L}\int_{t_0}^{t}(v_s(t)-v_A(t))dt \Rightarrow \\ & = C\frac{d^2v_A(t)}{dt^2} + \frac{1}{R}\frac{dv_A(t)}{dt}-\frac{1}{L}(v_s(t)-v_A(t)) \Rightarrow \\ \frac{1}{LC}v_s(t) & = \frac{d^2v_A(t)}{dt^2} + \frac{1}{RC}\frac{dv_A(t)}{dt}+\frac{1}{LC}v_A(t)\ \ \bf{[1]} \end{align}\)
Step 3 & 4: Next we transform the ODE into the phasor domain, and solve for \(\bn{V}_A\):
\(\begin{align} \require{cancel} \ds \frac{1}{LC}\bn{V}_{\t{s}} & = [(j\omega)^2\bn{V}_{\t{A}}-(j\omega)\cancelto{0}{v_{\t{A}}(0^{-})}]+\frac{1}{RC}(j\omega)\bn{V}_{\t{A}}+\frac{1}{LC}\bn{V}_{\t{A}} \\ & =\bn{V}_{\t{A}}\left[-\omega^2 + j\omega/RC + 1/LC \right] \\ & =\bn{V}_{\t{A}}\left[-(1000\ \t{rad/s})^2 + j(1000\ \t{rad/s})/(100\ \Omega \cdot 5\ \mu\t{F}) + 1/(10\ m\t{H}\cdot 5\ \mu\t{F}) \right] \Rightarrow \\ (20\cdot 10^6\ {[\t{rad/s}]}^{-2}) \bn{V}_{\t{s}} & =\bn{V}_{\t{A}}\left[19 +j2\right]10^6\ {[\t{rad/s}]}^{-2} \\ \frac{20}{19+j2}\cdot\Frac{19-j2}{19-j2}\bn{V}_{\t{s}} & = \bn{V}_{\t{A}} \\ \frac{4}{73}(19-j2)\bn{V}_{\t{s}} & = \bn{V}_{\t{A}} \\ \frac{4\sqrt{365}}{73}e^{-j\deg{6.01}}\cdot 5 & = \bn{V}_{\t{A}} \Rightarrow \end{align}\)
\(\boxed{ \bn{V}_{\t{A}}=5.24\lrg e^{ -j\deg{6.01}}\ \t{V}}\)
To find the currents, we apply Ohm's law in the phasor domain and inverse-transform:\(\vpl\)
\(\begin{align} \ds \vpL \bn{I}_C &= \frac{\bns{V}{A}}{1/j\omega C} \\ & = j\omega C\ 5.24\ \t{V}\lrg e^{ -j\deg{6.01}} \\ &= (1000\t{ rad/s}\cdot 5\ \mu\t{F})\cdot 5.24e^{j\deg{90}}e^{ -j\deg{6.01}} \Rightarrow \\ \bn{I}_C &= 26.15\ e^{ + j\deg{83.99}} \Rightarrow \\ i_C(t) &= 26.15\ \real{e^{j\omega t + j\deg{83.99}}}\ m\t{A} \end{align} \)
\(\vpl \ds \bxred{i_C(t) = 26.2 \Cos{\omega t + \deg{84.0}}\ m\t{A}} \hspace{115px}\)


\(\hspace{3px} \begin{align} \ds \bn{I}_R &= \frac{\bns{V}{A}}{R} \\ &= \frac{5.24\ \t{V}}{100\ \Omega}e^{-j\deg{6.01}} \Rightarrow \\ \bn{I}_R &= 52.4\ e^{-j\deg{6.01}} \Rightarrow \\ i_R(t) &= 52.4\ \real{e^{j\omega t-j\deg{6.01}}}\ m\t{A} \Rightarrow \end{align} \)
\(\vpl \ds \bxred{i_R(t) = 52.4 \Cos{\omega t - \deg{6.01}}\ m\t{A}}\)
\(\hspace{3px} \bn{I}_C\) and \(\bn{I}_R\) sum to \(\bn{I}\), thus,

\(\hspace{3px} \begin{align} \ds \bn{I} &= \bn{I}_C + \bn{I}_R \\ &= 26.15e^{ + j\deg{83.99}} + 52.4e^{-j\deg{6.01}} \\ &= 26.15(.1047+j.9945)+52.4(.9945-j.1047) \\ &= 54.75 + j20.52 \Rightarrow \\ \bn{I} &= 58.58\ e^{j\deg{20.54}} \Rightarrow \\ i(t) &= 58.6\ \real{e^{j\omega t +j\deg{20.54}}}\ m\t{A} \Rightarrow \end{align} \)
\(\vpl \ds \bxred{i(t) = 58.6 \Cos{\omega t +\deg{20.54}}\ m\t{A}}\)






Dragon Notes,   Est. 2018     About

By OverLordGoldDragon