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

1. Sep 4, 2009

### exidez

1. The problem statement, all variables and given/known data

I need to use MATLAB to solve these problems.

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

2. Relevant equations

MATLAB

3. 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 dont know how to do it with the roots from part a. Can someone help me out with part c please

Last edited by a moderator: Apr 24, 2017
2. Sep 4, 2009

example:-

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3. Sep 7, 2009

### exidez

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 doesnt use the root directly obtained in part a!

Do i just modify the graph so it starts at 5V ????????

Last edited: Sep 7, 2009
4. Sep 7, 2009

### NJunJie

The question seems to ask you to set Vs(t) = 5v.

5. Sep 8, 2009

### exidez

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 dont i need the value of the derivative of Vs??