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

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Homework Help Overview

The discussion revolves around using MATLAB to solve problems related to the natural response of a circuit, specifically focusing on finding roots of the characteristic equation and plotting the natural response function over time.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss defining polynomial constants and finding roots using MATLAB. There is an exploration of the natural response in the time domain and the relationship to the initial condition Vs(t=0) = 5V. Questions arise about how to incorporate this initial condition into the response function and the role of the constants derived from the roots.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and understanding. Some guidance has been offered regarding the initial condition, but there is no explicit consensus on how to proceed with incorporating it into the natural response function.

Contextual Notes

Participants express confusion regarding the initial condition and its application to the natural response, indicating a need for further clarification on how to use the roots obtained in part a effectively.

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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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