I want to know what does it mean by the term including "laplace" word in Laplace Transform

When I tried to solve an eqn analytically using Laplace Transform, the soln included the term laplace(qpl(t), t, s).

I do not know what does this mean?

The commands I entered as following:

>> clear E

>> syms fplc Rc Nc0 Kgr fplh Rh Nh0 Kel real

>> syms qpl(t) s

>> dqpl(t) = diff(qpl(t),t);

>> dqpl(t)= fplc*Rc*Nc0*exp(Kgr*t)+fplh*Rh*Nh0-Kel*qpl(t);

>> L(t)=laplace(dqpl(t),t,s)

L(t) =

(Nh0*Rh*fplh)/s - Kel*laplace(qpl(t), t, s) - (Nc0*Rc*fplc)/(Kgr - s)

[Image Analyst]

The "laplace" word in Laplace Transform is for the name of the person who invented/discovered it.

Answer 1

[Star Strider]

The Symbolic Math Toolbox denotes ‘laplace(qpl(t), t, s)’ as the Laplace transform of the ‘qpl(t)’ function. If you want to denote it in a more traditional form, use the subs function:

L(s) = laplace(dqpl(t),t,s)
L(s) =
(Nh0*Rh*fplh)/s - Kel*laplace(qpl(t), t, s) - (Nc0*Rc*fplc)/(Kgr - s)

L(s) = subs(L(s), {laplace(qpl(t), t, s)}, {QPL(s)})
L(s) =
(Nh0*Rh*fplh)/s - Kel*QPL(s) - (Nc0*Rc*fplc)/(Kgr - s)

This replaces ‘laplace(qpl(t), t, s)’ with ‘QPL(s)’. It is usually good to do this, expecially if you want to simplify your equation or solve for ‘QPL(s)’ later in your code, for example to calculate a transfer function.

[Esraa Abdelkhaleq]

Thanks for your help. But what the command used to solve this eqn for qpl(s).

[Star Strider]

My pleasure.

I understand what you would like to do, but for some reason, solve is getting confused by my call to it.

That said, this is a relatively simple hand calculation:

L(s) = (Nh0*Rh*fplh)/s - Kel*QPL(s) - (Nc0*Rc*fplc)/(Kgr - s)

then solve for ‘QPL(s)’:

QPL(s) = ((Nh0*Rh*fplh)/s - L(s) - (Nc0*Rc*fplc)/(Kgr - s))/Kel

[Esraa Abdelkhaleq]

I have a problem ..

When I retype the commands, the last step (subs) gives the result:

Empty sum: 0_by_1

Not:

L(s) = (Nh0*Rh*fplh)/s - Kel*QPL(s) - (Nc0*Rc*fplc)/(Kgr - s)

What should I do?

[Star Strider]

I did the last step by hand. I couldn’t get solve to work with it for some reason. I went through the documentation for solve to see if it had any name-value pair options that could produce a result, and none did.

It mystifies me that solve misses such an obvious result.

Maybe Walter Roberson will see this and suggest something. I’m out of ideas, except for my hand solution.

[Esraa Abdelkhaleq]

Really thanks for your efforts.

What I understood that I should either write qpl(s) or Qpl(s) in the L(s) command:

"L(s) = subs(L(s), {laplace(Qpl(t), t, s)}, {Qpl(s)})"

OR

"L(s) = subs(L(s), {laplace(qpl(t), t, s)}, {qpl(s)})"

NOT

"L(s) = subs(L(s), {laplace(qpl(t), t, s)}, {Qpl(s)})"

I reached to the same result of you. But there is still a gap or a missing part. The result that I want to reach is:

s*Qpl(s) - q0 = fplc*Rc*Nc0/(s-Kgr) + fplh*Rh*Nh0/s -Kel*Qpl(s)

The author wrote that the original eqn can be solved analytically for qpl(t) by taking Laplace Transform and wrote the previous solution then solved the eqn for Qpl(s) giving:

Qpl(s) = -fplh*Rh*Nh0*Kgr - (Kgr*q0 - fplh*Rh*Nh0 - fplc*Rc*Nc0)*s + q0*s^2/s*(s-Kgr)*(s+Kel)

I do not know what should I do?

[Star Strider]

My pleasure.

I don’t know what you’re doing. I don’t know the equations you began with.

My only thought at this point is that you may not have set your original equations up correctly. Check those, and then re-code your equations. If those are correct, the derivation should be correct.

[Esraa Abdelkhaleq]

I want to know the solution steps of the equation as shown in the attached fig. I hope you can see the equations.

[Star Strider]

For whatever reason, the solve function refuses to solve ‘Laplace_Eqn’ for ‘QPL’ in the commented-out lines.

So, I decided to just directly solve it using the dsolve function:

syms fplc Rc Nc0 Kgr fplh Rh Nh0 Kel qpl(t) s QPL(s) qpl0 real

dqpl(t) = diff(qpl(t),t);

Eqn = dqpl(t) == fplc*Rc*Nc0*exp(Kgr*t)+fplh*Rh*Nh0-Kel*qpl(t);

qpl = dsolve(Eqn, qpl(0) == qpl0);

qpl = simplify(qpl, 'steps', 10)

% Laplace_Eqn = laplace(Eqn);
% 
% Laplace_Eqn = subs(Laplace_Eqn, {laplace(qpl(t), t, s)}, {QPL(s)});
% 
% Soln = solve(Laplace_Eqn, QPL, 'IgnoreAnalyticConstraints',true, 'IgnoreProperties',true);

The integrated (and simplified) differential equation:

qpl =

(Kel*Nh0*Rh*fplh + Kgr*Nh0*Rh*fplh + Kel*Nc0*Rc*fplc*exp(Kgr*t))/(Kel*(Kel + Kgr)) - exp(-Kel*t)*((Nh0*Rh*fplh)/Kel - qpl0 + (Nc0*Rc*fplc)/(Kel + Kgr))

Answer 2

[syed ali hasan kibria]

y¨(t) + 2 ˙y(t) + 10y(t) = 20 cos(6t) y(0) = 1 y˙(0) = 5