How to? - Complex numbers

dVdotdt= @(-(Whalf-1i.*w).*V + alfa.*Uin);

(Whalf-1i.*w) part is complex.

When I put the above in my equation I get 'Warning: Imaginary parts of complex X and/or Y arguments ignored'

I am not sure if my syntax is correct to be honest as my answer should be exponential but it is a straight line; also does ode handle complex well- maybe thats the reason my curve is flawed.

[Adam]

You defined a function of two inputs, t and y yet you don'y use either of them in the function so it will just be a constant, defined by the variables you do give which, I assume, are in the workspace at the time you create this function handle. I have no idea where t and y are supposed to fit into your equation though.

Answer 1

[James Tursa]

E.g., for an anonymous function you need to give the input argument list first. E.g., for a generic derivative function that takes t and y as inputs, the syntax would be this:

dydt = @(t,y) some_expression_involving_t_and_y_goes_here;

[STP]

this is the complex equation ; should I be putting it as it is or in e^power form ?

[Torsten]

omega_half = ...;
delta_omega = ...;
beta = ...;
U_in = ...;
dVdt = @(t,V) -(omega_half-1I*delta_omega)*V + 2*beta/(1+beta)*U_in;

[STP]

Yup; I tried that. But I get my output like :

instead of something like this --

[madhan ravi]

Just reflect it.

[STP]

It doesn\t work like that. It should come originally like that with a peak value of around 1.67 at 4.2 ; and here its around 5 (if I reflect it )

[Torsten]

Analytical solution is

V(t)=V0 - b/a ( 1- exp(a*(t-t0)) )

with

a = -(omega_half-1I*delta_omega)

b = 2*beta/(1+beta)*U_in

and V(t0)=V0.

Compare to your solution and check the constants.

[STP]

Perfect got it! Thanks. I wish I could accept ur answer

[STP]

If I am doing it with the analytical solution the curve is a little sharper (the bold curve) whereas with the equation it is flatter ; is it a matlab thing?

Attaching the code for the equation and solution both:

beta=5;
alfa = 2.*beta/(beta+1);
tau1=2.0;
tau2=2.4;
Tc=2.0;
gamma=alfa.*(2-exp(-tau1));
% Time frames
t1=0:0.01:4.2;
t2 = 4.2:0.01:5;
t3 = 5:0.01:8;
dEeA = @(t,y) alfa/Tc * exp(-t/Tc);
tspan = t1;
y0 = 0;
[T1 EeA1] = ode15s(dEeA,tspan,y0);
plot(t1,EeA1)
hold on 
figure(1)
dEeB = @(t,y)-gamma *exp(-(t - tau1)/Tc)/Tc ;
tspan = t2;
y0 = EeA1(end);
[T1 EeB1] = ode15s(dEeB,tspan,y0);
plot(t2, EeB1)
Ee2 = EeB1(end);
dEeC = @(t,y) -Ee2*exp(-(t-tau2)/Tc)/Tc;
tspan = t3;
y0 = Ee2;
[T1 EeC1] = ode15s(dEeC,tspan,y0);
plot(t3, EeC1)
hold off

omega_half = 0.090;
delta_omega = 0.16;
beta = 5;
U_in = 0.30;
y1 = 0;
alfa = 2*beta/(1+beta);
t1= [0 4.2]
dV1dt = @(t,V) -(omega_half-1i*delta_omega)*V +alfa*U_in;
[t1 V1] = ode15s(dV1dt, t1, y1)
plot(t1, V1)
hold on 
t2 = [4.2 5]
U_in1 = -0.30;
y2= V1(end);
dV2dt = @(t,V) -(omega_half-1i*delta_omega)*V +alfa*U_in1;
[t2, V2] = ode15s(dV2dt, t2, y2)
plot(t2, V2)
hold on
t3 = [5 8]
U_in3 = 0;
y3 = V2(end);
dV3dt = @(t,V) -(omega_half-1i*delta_omega)*V +alfa*U_in3;
[t3, V3] = ode15s(dV2dt, t3, y3) 
plot(t3, V3)

[Torsten]

The differential equations and their solutions are completely different. So you can't expect that the solution curves coincide.

[STP]

Its the same equation ; only diff being in one i am taking the equation and applying ode solver ; and in the other I have back drafted from solution to find df and then applied ode solver.

[Torsten]

No.

The first equation has solution

y(t) = 5/3*(1-exp(-t/2)),

the second equation has solution

y(t)=(0.045/0.0337+0.08/0.0337*i)*(1-exp(-0.09*t)*(cos(0.16*t)+i*sin(0.16*t)))

(both only for the t1-range).

Doesn't look the same to me.