By Steven Karris
This article is an creation to the fundamental rules of electric engineering and covers DC and AC circuit research and Transients. it's meant for all engineering majors and presumes wisdom of first 12 months differential and vital calculus and physics. The final chapters contain step by step tactics for the options of straightforward differential equations utilized in the derivation of the traditional and forces responses. Appendices A, B, and C are introductions to MATLAB, Simulink, and SimPowerSystems respectively. Appendix D is a evaluate of advanced Numbers, and Appendix E is an creation to matrices and determinants. for more information. please stopover at the Orchard courses website
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Extra info for Circuit Analysis I with MATLAB Computing and Simulink/SimPowerSystems Modeling
75) and setting it equal to zero. 2 Response of Parallel RLC Circuits with AC Excitation The total response of a parallel RLC circuit that is excited by a sinusoidal source also consists of the natural and forced response components. The natural response will be overdamped, critically damped or underdamped. The forced component will be a sinusoid of the same frequency as that of the excitation, and since it represents the AC steadystate condition, we can use phasor analysis to find the forced response.
26 a known as a Multiple Feed Back (MFB) active lowpass filter. For this circuit, the initial conditions are v C1 = v C2 = 0 . Compute and sketch v out t for t 0 . 26. 83) 130 Circuit Analysis II with MATLAB Computing and Simulink / SimPowerSystems Modeling Copyright © Orchard Publications Other Second Order Circuits We observe that v 2 = 0 (virtual ground). 92). syms s; y0=solve('s^2+2*10^3*s+2*10^6') % Must have Symbolic Math Toolbox installed y0 = [-1000+1000*i] [-1000-1000*i] that is, s 1 ,s 2 = – j = – 1000 j1000 = 1000 – 1 j1 We cannot classify the given circuit as series or parallel and therefore, we should not use the damping ratio S or P .
You should follow this practice with the problems in all chapters of this book. 140 Circuit Analysis II with MATLAB Computing and Simulink / SimPowerSystems Modeling Copyright © Orchard Publications Solutions to EndofChapter Exercises 1. 2 8 10 dt 2 dv C d vC --------- + 625 v C = 62500 ---------+ 50 2 dt dt From the characteristic equation 2 s + 50s + 625 = 0 we obtain s 1 = s 2 = – 25 (critical damping) and S = R 2L = 25 The total solution is v C t = v Cf + v Cn = 100 + e –S t k 1 + k 2 t = 100 + e – 25 t k 1 + k 2 t (1) With the first initial condition v C 0 = 0 the above expression becomes 0 0 = 100 + e k 1 + 0 k 1 = – 100 and by substitution into (1) we obtain v C t = 100 + e – 25 t k 2 t – 100 (2) Circuit Analysis II with MATLAB Computing and Simulink / SimPowerSystems Modeling 141 Copyright © Orchard Publications Chapter 1 Second Order Circuits To evaluate k 2 we make use of the second initial condition i L 0 = 0 and since i L = i C , and dv i = i C = C --------C- , we differentiate (2) using the following MATLAB script: dt syms t k2 % Must have Symbolic Math Toolbox installed v0=100+exp(25*t)*(k2*t100); v1=diff(v0) v1 = -25*exp(-25*t)*(k2*t-100)+exp(-25*t)*k2 Thus, dv C – 25t – 25t --------- = k 2 e – 25e k 2 t – 100 dt and dv --------Cdt dv dt i C = k 2 + 2500 (3) t=0 i C Also, --------C- = ---C- = ---L- and at t = 0 dv --------Cdt t=0 iL 0 = --------------= 0 (4) C From (3) and (4) k 2 + 2500 = 0 or k 2 = – 2500 and by substitution into (2) v C t = 100 – e – 25 t 2500t + 100 (5) We find i L t = i C t by differentiating (5) and multiplication by C .