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Topic: Problem with inverse laplace
Replies: 5   Last Post: Mar 19, 2014 11:05 AM

 Messages: [ Previous | Next ]
 spv Posts: 1 Registered: 12/6/12
Re: Problem with inverse laplace
Posted: Dec 6, 2012 3:07 PM

that the roots of the polynomial are complex is quite obvious because this kind of equation represent the transient of an electric system (active and reactive parts), the problem is that matlab is perfectly able to compute ilaplace over a (polynomial of degree <=2) divided by a (polynomial of degree <=2) but when the function on the denominator grows one degree it return this kind "result"

for example:

syms s;
t = 10:0.01:50;

% Data
M1 = -3.5;
D1 = 1;
M2 = 4;
D2 = 0.75;
T = 6.283;
T_G1 = 0.01;
T_G2 = 0.05;
T_CARGA1 = 10;
T_CARGA2 = 20;
inc_P_L1 = 0.2;
inc_P_L2 = 0.0;

%equation
inc_w2= inc_P_L1*((-T)/((M1*s+D1)*((M2*s+D2)*s+T)+(M2*s+D2)*T));
inc_w2_i = ilaplace(inc_w2);

the result of this is:

-(25132*sum(exp(r4*t)/(11000*r4 - 168000*r4^2 + 15566), r4 in RootOf(s4^3 - (11*s4^2)/112 - (7783*s4)/28000 - 6283/8000, s4)))/5

Completely unusable, if I apply vpa() to this result I get a sum of exponentials, it might be ok but it is not, for this result to be usable, we have to apply euler's equation:
exp(x+iy)=exp(x)*(cos(y) + i*sin(y))
which is not possible because the result is a symbolic type. (at least I couldn't)

but if we downgrade the equation:

inc_w2= inc_P_L1*((-T)/((M1*s+D1)*((M2*s+D2)+T)+(M2*s+D2)*T));
inc_w2_i = ilaplace(inc_w2)

the result is now something quite usable :

(25132*2712531089^(1/2)*sinh((2712531089^(1/2)*t)/56000)*exp((9033*t)/56000))/13562655445

a function that effectively represents the transient of a sinusoidal wave.

So after googling a lot I have not seen a proper answer to this issue.

Date Subject Author
2/13/11 Chaitanya Vichare
2/13/11 Walter Roberson
12/6/12 spv
12/7/12 Steven Lord
3/19/14 vansac
3/19/14 vansac