Matlab - ODE, find roots of the characteristic equation for the natural response

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SUMMARY

The discussion focuses on solving ordinary differential equations (ODEs) using MATLAB to find the roots of the characteristic equation for the natural response of an electrical circuit. The user defines the polynomial constants based on resistor and capacitor values, calculates the roots using the roots function, and derives the transfer function. The challenge lies in plotting the natural response function over time with an initial voltage of 5V, which requires understanding the inverse Laplace transform and determining the constants from the initial conditions.

PREREQUISITES
  • Familiarity with MATLAB programming and syntax
  • Understanding of ODEs and their applications in electrical circuits
  • Knowledge of Laplace transforms and inverse Laplace transforms
  • Basic concepts of transfer functions and natural response in control systems
NEXT STEPS
  • Learn how to use MATLAB's residue function for partial fraction decomposition
  • Study the process of finding inverse Laplace transforms in MATLAB
  • Explore MATLAB plotting functions to visualize time-domain responses
  • Understand how to apply initial conditions to solve for constants in ODE solutions
USEFUL FOR

Electrical engineering students, control system engineers, and MATLAB users looking to solve ODEs and analyze circuit responses.

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Homework Statement



I need to use MATLAB to solve these problems.

http://users.bigpond.net.au/exidez/IVDP.jpg

Homework Equations



MATLAB

The Attempt at a Solution



a)

R1=3.6;
R2=R1;
C1=33*10^-6;
C2=22*10^-6;

% defining the polynomial constants
Vs=[R1*R2*C1*C2 (2*R2*C2)+(R1*C1) 1];

'the roots of the equation are';
roots(Vs)

ans =

-25253
-4209


b) Largest time constant:

numf=1;
denf=Vs;

%Converting to Partial fraction to find the inverse Laplace transform

[r,p,k]=residue(numf,denf)

r =

1.0e+003 *

-5.0505
5.0505


p =

1.0e+004 *

-2.5253
-0.4209


Transfer function:

\frac{-5051}{s + 25253} + \frac{5051}{s + 4209}


Based on the inverse Laplace of the partial fractions, the eqn then looks like below:

Vo(t) = r(2)*exp(p(2)*t) +r(1)*exp(p(1)*t)

Vo(t) = 5051e^(-4209t)-5051e^(-25253t)

I believe that is my natural response?

c) obtain and plot the natural response function versus time (use the first 1ms) for an initial constant Vs(t=0) = 5V

I have no idea, this is all new to me. The initial constant has me fooled and i don't know how to do it with the roots from part a. Can someone help me out with part c please
 
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example:-
 

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that still doesn't help me much...

so i know the natural response in S domain. I can find the inverse Laplace to find what it is in time domain so i can plot it against t. but the initial constant Vs(t=0) = 5V still confuses me. Also, this doesn't use the root directly obtained in part a!

Do i just modify the graph so it starts at 5V ?
 
Last edited:
The question seems to ask you to set Vs(t) = 5v.
 
well this is my understanding of it...

basically i assume a natural response in the form y= Ce^(at)

given the roots:

y= C1e^(-25253t) + C2e^(-4209t)

i am given Vs(0)= 5V

however to solve for the constants C1 and C2 don't i need the value of the derivative of Vs??
 

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