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

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Esraa Abdelkhaleq
回答： syed ali hasan kibria 2020-12-15
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)

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采纳的回答

Star Strider
Star Strider 2016-5-7
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.
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Star Strider
Star Strider 2016-5-9
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))

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

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