I can't get the ilaplace of a function

Hello every body, My problem is that I can't get the inverse laplace transform of a function. Is it that the function is too complicated? do I have to use an approximation of ilaplace rather than the ilaplace function that exists is matlab? or it's just something I did wrong? Here is the code: This the function I want to have the ilaplace for:

function y = Egs(s)
%constantes
a  =14;
b=15;
c=25;
L=c-b;
r=c;
phi=15;
V=7.*7.*28;
ro=1000;
RHi=100;
RHb=50;
dRH = RHb-RHi;
nus=0.3;
d=1;
M=8.3.*10.^(-6);
k=1.7;
Es=200000;
nuc=0.2;
lambda=0.3;
tp=1;
nu = k.*(1-exp(-lambda.*tp));

%------------------------

 D1 = s.*Epsa(s)-(s+lambda).*Epsa(s+lambda)/exp(lambda.*tp);
 y = (1./k).*(a-b).*(a+b).*(Es./s).*D1.*s.*(b.^2.*(nuc-1)-c.^2.*(nuc+1))./(4.*b.^2.*(s.*Epsf(s)-(s+lambda).*Epsf(s+lambda))+(b-c).*(b+c).*D1.*((a.^2.*(nus+1)-b.^2.*(nus-1))));

Where Epsa is:

function y = Epsa(s)
syms t;
AUX = -16.7*10.^(-6)*t.^(0.3);
y = laplace(AUX,t,s);

and Epsf is

function y = Epsf(s)
%constantes
a = 14;
b=15;
c=25;
L=c-b;
r=c;
phi=15;
V=7.*7.*28;
ro=1000;
RHi=100;
RHb=50;
dRH = RHb-RHi;
nus=0.3;
D=1;
M=8.3.*10.^(-6);
k=1.7;
Es=200000;
nuc=0.2;
lambda=0.3;
tp=1;
nu = k.*(1-exp(-lambda.*tp));
%-------------
%RH(c,s)
  %Bi

        B1 = b.*sqrt(s)./sqrt(D);
        B2 = c.*sqrt(s)./sqrt(D);
        B3 = r.*sqrt(s)./sqrt(D);
        B4 = -b.*exp(B2).*(b-r).*(b+r).*RHi.*sqrt(s)-3.*c.*exp(B2).*(b-r).*(b+r).*sqrt(s);
        B5 = D.^(7./4).*sqrt(pi).*dRH.*(3.*sqrt(D)-2.*b.*sqrt(s)).*sqrt(-s./D).*(test(sqrt(b).*(s./D).^(1./4))-test(sqrt(r).*(s./D).^(1./4)))./(b.^(3./2).*s.^(9./4));
RH= sqrt(D).*exp(-c.*sqrt(s)./sqrt(D)).*(-b.*c.*s./D).^(3./2).*((D).^(3./2).*sqrt(-b.*c.*s./D).*(-6.*c.^(3./2).*(sqrt(6.*D).*exp(B1)-exp(B3).*sqrt(D.*r)).*dRH-B4)./(b.^2.*c.^2.*s.^2)+B5)./(2.*(3.*c+b).*s.^(3./2));
y = -1.1.*M.*((c.^2-b.^2).*RHi./(2.*s)-RH);

and here is the function I want to evaluate and plot :

function y = Eg(t)
syms s;
y = ilaplace(Egs(s),s,t);

However when I want to have Eg(1) it gives me this : ilaplace(....,s,1) I tried eval, simplify..but it doesn't work... therefore I can't plot Eg(t) :s well, I hope I'll get the answer. Thank you :)

Answer 1

[Paulo Silva]

The documentation for the ilaplace says that ilaplace(L) computes the inverse Laplace transform of the symbolic expression L, on your code you give a function handle to ilaplace and that won't work.

[Mohamed Yassin OUKILA]

I really didn't get it..the function that I want its ilaplace is symbolic..

Egs(s)?

please explain more because I am still new with matlab ;)

[Paulo Silva]

sorry my mistake it's really symbolic

f=@(s) 1/s

syms s

ilaplace(f(s))

ans=1 has expected