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Engineering 43. MTE2 Review 2 nd Order Transient Ckt. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]. 33 μ F. 300 mH. t = 0. 125 mA. 110 Ω. 8 V. 180 Ω. The Problem. Find for the Circuit Below v C (t) i o (t). + v C (t) −. i o (t). - PowerPoint PPT Presentation
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ENGR-25_Prob_9_3_Solution.ppt1
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Engineering 43
MTE2MTE2ReviewReview22ndnd Order OrderTransient Transient
CktCkt
ENGR-25_Prob_9_3_Solution.ppt2
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
The ProblemThe Problem
Find for the Circuit Below• vC(t)
• io(t)
110 Ω
33 μF 300 mH
125mA
8 V180 Ω
t = 0
+ vC(t) −
io(t)
ENGR-25_Prob_9_3_Solution.ppt3
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ENGR-25_Prob_9_3_Solution.ppt4
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ENGR-25_Prob_9_3_Solution.ppt5
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ENGR-25_Prob_9_3_Solution.ppt6
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ENGR-25_Prob_9_3_Solution.ppt7
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ENGR-25_Prob_9_3_Solution.ppt8
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ENGR-25_Prob_9_3_Solution.ppt9
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ENGR-25_Prob_9_3_Solution.ppt10
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ENGR-25_Prob_9_3_Solution.ppt11
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ENGR-25_Prob_9_3_Solution.ppt12
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ENGR-25_Prob_9_3_Solution.ppt13
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ENGR-25_Prob_9_3_Solution.ppt14
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
MATLAB Code - 1MATLAB Code - 1% Bruce Mayer, PE% ENGR43 * MTE2 Review Problem% File = E43_MTE2_2ndOrder_Vc_100402.m%% Component ValuesR1 = 110R2 = 180L = 300e-3C = 33e-6%%% Calulate the Roots of the Characteristic Equation by Quadratic Formula% * Yields the Time Constants for Exponential Decay of OverDamped CktSplus = (-(R1+R2)*C + sqrt(((R1+R2)*C)^2 - 4*L*C))/(2*L*C)% in Rads/secSminus = (-(R1+R2)*C - sqrt(((R1+R2)*C)^2 - 4*L*C))/(2*L*C)% in Rads/sectauPlus = 1000/Splus % in mStauMinus = 1000/Sminus % in mS%% use Ax = b to solve for K1 & K2A = [1 1; 8.39 1.18] % unitless coefficientsb = [-8; 0]; % units are voltsK = A\b % in K1 = K(1)K2 = K(2)%% Define function for the total solutionvtot = @(y) K1*exp(y/tauMinus) + K2*exp(y/tauPlus) + 22.5t = linspace(0, 60, 500); % in mSvc = vtot(t);%% graph Solutionplot(t,vc), grid, xlabel('time (ms)'), ylabel('vc (V)')%disp('Showing FULL-time plot. Hit AnyKey to show the 2nd-Order IC CHECK-Plot')disp(' ‘)
ENGR-25_Prob_9_3_Solution.ppt15
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
MATLAB Code - 2MATLAB Code - 2pause%% make the [dvc/dt](0+) =0 check plottdv = linspace(0, 0.5); % in mSdvcdt = vtot(tdv);plot(tdv,dvcdt), grid, xlabel('time (ms)'), ylabel('vc (V)')%disp('Showing SHORT-time plot. Hit AnyKey to show the FULL-TIME io(t) Plot')disp(' ')pause%% the solution for io(t)Kio = 36.61; % in mAio = @(t) Kio*(exp(Splus*t) - exp(Sminus*t));tio = linspace(0, 0.04, 1000); % in SECONDSioplot = io(tio);plot(tio*1000, ioplot), xlabel('time (mS)'), ylabel('io (mA)'), title('MTE2 2nd-Order Problem Exam Review'), grid%%% Find ioMax and tioMax using MAX command[ioMax,ItioMax] = max(ioplot);disp('Maximum value for io in mA = ')disp(ioMax)disp(' ')tioMax = tio(ItioMax);disp('TIME for Maximum value for io in mS = ')disp(1000*tioMax)%disp('Showing FULL-time plot. Hit AnyKey to show the Short-TIME io(t) Plot')disp(' ')pausetio = linspace(0, 0.008, 1000); % in SECONDSioplot = io(tio);plot(tio*1000, ioplot, 1000*tioMax, ioMax, 'h'), xlabel('time (mS)'), ylabel('io (mA)'), title('MTE2 2nd-Order Problem Exam Review'), grid
ENGR-25_Prob_9_3_Solution.ppt16
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
vvCC(t) MATLAB Plot(t) MATLAB Plot
0 10 20 30 40 50 6014
15
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18
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21
22
23
time (ms)
vc (
V)
ENGR-25_Prob_9_3_Solution.ppt17
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
CHK: dvCHK: dvCC(t)/dt|(t)/dt|t=0t=0+ = 0+ = 0
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.514.5
14.51
14.52
14.53
14.54
14.55
14.56
14.57
14.58
14.59
time (ms)
vc (
V)
Slope at t=0 looks FLAT; thus 1st Order IC is CONFIRMED
ENGR-25_Prob_9_3_Solution.ppt18
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
iioo(t) MATLAB Plot(t) MATLAB Plot
Note io,max at about 3 mS
0 5 10 15 20 25 30 35 400
5
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15
20
25
time (mS)
io (
mA
)
MTE2 2nd-Order Problem Exam Review
ENGR-25_Prob_9_3_Solution.ppt19
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
iio,maxo,max(2.683 mS) = 22.82 mA (2.683 mS) = 22.82 mA
0 1 2 3 4 5 6 7 80
5
10
15
20
25
time (mS)
io (
mA
)
MTE2 2nd-Order Problem Exam Review
