How can I resolve this equation using matlab

Foufou · Dec 23, 2007

matlab solve equation for variable

How can I resolve this equation using matlab where
e is the unknown variable with
t0 ta and ta as known ones

eTa=1-exp(-e(t0-te)

exp denote the exponentional function

saeidj · Dec 23, 2007

matlab solve equation variable

Dear friend

a = Ta;
b = T0 - Te;
x = e;

a*x = 1 - exp(-b*x);

In this equation the number of roots depends on the value of parameters a and b. If a,b>0 or a,b<0 it may have 2 roots else has one root. I think you should substitute the known value a and b in the equation, then solve it.

For example:

% a = Ta = 1
% b = T0-Te = 0.4
R = solve('1*x = 1 - exp(-0.4*x)')

Regards

Foufou · Dec 24, 2007

roots of an equation using matlab

thank friend but when I try this exemple
t0=1;
te=0.55;
ta=0.04;

the solve function give me 1 root equal to 0 althougth there are 2 roots seen when we plot the functionf=x-(1/a)*(1-exp(-b*x)); (see figure atached

thanks

saeidj · Dec 24, 2007

matlab roots of equation

Dear friend

% a = Ta = 0.04
% b = T0 - Te = 0.45
syms a b;
f1 = 'a*x = 1 - exp(-b*x)';
f2 = subs(f1,{a,b},{'0.04','0.45'});
R = solve(f2)

Result:
R =
[ 24.999674769963571780041109258934]
[ 0 ]

Plotting the functions a*x and 1-exp(-b*x) :

x = -1:0.001:30;
a = 0.04; b = 0.45;
plot(x,1-exp(-b*x)); grid
hold on;
plot(x,a*x,'r')

Best regards

Foufou · Dec 25, 2007

matlab resolve

thank you very much friend

Added after 49 minutes:

Dear friend I will be verry grateful if you help me to resolve my problem explained in details in the attached doc file

Thank you for helping me

saeidj · Dec 25, 2007

how to solve integrals using matlab

Dear friend

This is an M-file for CASE 1, you can change the time constants for other CASEs.
________________________________________________
clc; clear; close all;
%Parameters of Case 1:
t0 = 1;
tp = 0.41;
te = 0.55;
ta = 0.04;
tc = 0.58;
% ******************************************
% Finding e: { e*ta = 1 - exp(-e*(tc-te)) }
syms e;
f = e*ta - 1 + exp(-e*(tc-te));
R_e = numeric(solve(f));
E = R_e(1) + R_e(2) % E = nonzero root of f (one root is zero)
% ******************************************
% Finding alpha = x:
k1 = (pi/tp)^2;
k2 = (pi/tp)/sin(pi*te/tp);
k3 = (pi/tp)/tan(pi*te/tp);
k4 = t0 - te;
syms x;
y = (1/(x^2+k1))*(k2*exp(-te*x) + x - k3) - k4/(exp(E*k4)-1) + 1/E;
alpha = numeric(solve)
_______________________________________________

Results:

CASE 1:
Parameters: t0 = 1; tp = 0.41; te = 0.55; ta = 0.04; tc = 0.58;
Results: E = -18.3400 , alpha = -1.3604
CASE 2:
Parameters: t0 = 1; tp = 0.48; te = 0.59; ta = 0.027; tc = 0.72;
Results: E = 36.7243 , alpha = 5.7369
CASE 3:
Parameters: t0 = 1; tp = 0.46; te = 0.66; ta = 0.027; tc = 0.77;
Results: E = 36.3583 , alpha = 1.9533
CASE 4:
Parameters: t0 = 1; tp = 0.50; te = 0.80; ta = 0.08; tc = 1;
Results: E = 11.1581 , alpha = 0.3510

Best regards

Foufou · Dec 25, 2007

matlab equation

thank friend, but i don't have the function numeric under matlab 7, can you send me the m file of this fuction
thanks

Added after 15 minutes:

thank you friend I have replaced the function numeric by eval and it work, tahnk you very mutch

Added after 10 minutes:

I have one last question friend,
the equation (*) in the document (the result of the integration ) can I do symbolic integration with matlab (using int function) to assure my integrated result and about the 0 , I have replaced it integration by 0, is it true ??? (normaly it is an unkwon arbitery constant) if it is not the case what should I do.

Thank you

saeidj · Dec 25, 2007

solve for variable matlab

Dear friend

You've made a mistake in equation (**), on the right hand t0 must be tc. The correct equation is:

See this M-file for Case1:
___________________________________________
clc; clear; close all;
%Parameters of Case 1:
t0 = 1;
tp = 0.41;
te = 0.55;
ta = 0.04;
tc = 0.58;
% ******************************************
% Finding e: { e*ta = 1 - exp(-e*(tc-te)) }
syms e;
f = e*ta - 1 + exp(-e*(tc-te));
R_e = eval(solve(f));
E = R_e(1) + R_e(2) % E = nonzero root of f (one root is zero)
% ******************************************
% Finding alpha = x:
k1 = (pi/tp)^2;
k2 = (pi/tp)/sin(pi*te/tp);
k3 = (pi/tp)/tan(pi*te/tp);
k4 = tc - te;
syms x;
y = (1/(x^2+k1))*(k2*exp(-te*x) + x - k3) - k4/(exp(E*k4)-1) + 1/E;
alpha = eval(solve)
% ******************************************
% is {int(g(t))} equal zero ?
Ee = 1; % an arbitrary value for Ee
syms t;
E0 = -Ee/(exp(alpha*te)*sin(pi*te/tp));
g1 = E0*exp(alpha*t)*sin(pi*t/tp); % for interval: [0 te]
Ig1 = eval(int(g1,0,te))
g2 = -(Ee/(E*ta))*(exp(-E*(t-te))-exp(-E*(tc-te))); % for interval: [te tc]
Ig2 = eval(int(g2,te,tc))
IG = Ig1+Ig2
___________________________________________
Results:
E = -18.3400
alpha = 3.9462
Ig1 = 0.0164
Ig2 = -0.0164
IG = 7.6328e-017

The value of IG is less than eps (In MATLAB eps=2.2204e-016 (Floating-point relative accuracy)), so IG={int(g(t)} is zero.

Best

Foufou · Dec 26, 2007

how to resolve sin(0)/0 in matlab?

thanks friend,

Added after 32 minutes:

Dear friend can i receive your Email adress,
thanks

AhSa2003 · Dec 27, 2007

root of an equation matlab

you don't need to set value for t0 tp ...
use:
syms params
witch params is t0 tp and so on.
then write your function(not equation I mean all in one hand) like this:
f=e*Ta-1+exp(-e*(t0-te) );
and finally use solve:
solve(f)
here is the code:

syms e Ta t0 te
f=e*Ta-1+exp(-e*(t0-te) );
solve(f)

adventure · Dec 27, 2007

root equation in matlab

Hi AhSa2003

I think Saeidj is true. In Matlab, "solve(eq,var)" solves the equation eq=0 for the variable var. In Fofou equation: eTa=1-exp(-e(t0-te)) the unknown variable is e, so:
eq = e*ta -1 + exp(-e*(t0-te))
var = e

If you do not set the variabels t0, ta, and te with known values , Matlab only return one root (e=0), while it may have two roots.

syms e ta t0 te;
f=e*ta-1+exp(-e*(t0-te) );
x = solve(f,e)

Matlab returns: x = 0

But if you set the known values:

t0 = 1;
te = 0.55;
ta = 0.04;
syms e;
f=e*ta-1+exp(-e*(t0-te) );
x = eval(solve(f,e))

Matlab returns: x = 24.9997 , 0

layes2 · Dec 27, 2007

known variables and solve equation matlab

use perl

AhSa2003 · Dec 29, 2007

how to solve equation in matlab for a variable

yes i don't test it in matlab
Yes u are right

Foufou · Dec 29, 2007

eval equation matlab

Dear saeidj,
When I compute the g(t) function using the e and alpha value calculated above the trapz function for calculating numeric integrationd doesn't give 0, can you help me to resolve this problem.
Thanks

saeidj · Dec 29, 2007

finding roots of an equation using matlab

Dear Foufou

Trapz(t,g) computes an approximation of the integral of g(t) with respect to t using trapezoidal integration. For example in CASE1 you should:

1- Integrate on [0 0.55] (Ig1 = trapz(t,g1) and t=0:Ts:0.55).
2- Integrate on [0.55 0.58] (Ig2 = trapz(t,g2) and t=0.55:Ts:0.58).
So, IG = Ig1 + Ig2.

The integration intervals are small, so to accurate the integration you should use very small Ts.

If Ts=1e-7 then:

E = -18.34002472196918
alpha = 3.94620051353478
Ig1 = 0.01636861161367
Ig2 = -0.01636861161366
IG = 1.655273140777069e-014 (notice that this is only an approximation).

If you use smaller Ts, IG --> 0. (If Maximum variable size allows it).

The M-file (for CASE1) is:
clc; clear; close all;
%-------------------------------------------------------------------------
%Parameters of Case 1:
t0 = 1;
tp = 0.41;
te = 0.55;
ta = 0.04;
tc = 0.58;
format long;
%-------------------------------------------------------------------------
% Finding e: { e*ta = 1 - exp(-e*(tc-te)) }
syms e;
f = e*ta - 1 + exp(-e*(tc-te));
R_e = eval(solve(f));
E = R_e(1) + R_e(2) % E = nonzero root of f (one root is zero)
%-------------------------------------------------------------------------
% Finding alpha = x:
k1 = (pi/tp)^2;
k2 = (pi/tp)/sin(pi*te/tp);
k3 = (pi/tp)/tan(pi*te/tp);
k4 = tc - te;
syms x;
y = (1/(x^2+k1))*(k2*exp(-te*x) + x - k3) - k4/(exp(E*k4)-1) + 1/E;
alpha = eval(solve)
%-------------------------------------------------------------------------
% is {int(g(t))} equal zero ? (Using TRAPZ function)
Ee = 1; % an arbitrary value
E0 = -Ee/(exp(alpha*te)*sin(pi*te/tp));
%................
t = 0:1e-7:te;
g1 = E0*exp(alpha*t).*sin(pi*t/tp); % for interval: [0 te]
Ig1 = trapz(t,g1)
%................
t = te:1e-7:tc;
g2 = -(Ee/(E*ta))*(exp(-E*(t-te))-exp(-E*(tc-te))); % for interval: [te tc]
Ig2 = trapz(t,g2)
%................
IG = Ig1+Ig2

Best, (My Email: wri2t A.T Y.A.H.O.O)

Foufou · Dec 29, 2007

matlab resoudre equation solve

thanks friend,
I will sent you a message to your Email adress
Best regard
foufou

Added after 15 minutes:

thanks friend,
I will send you a message to your Email adress
Best regard
foufou

How can I resolve this equation using matlab

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