Removing transient part in equation and solving for a general form

I have the following code that I run:

In:=Clear["Global*"];
Ia = LaplaceTransform[((78)/(10^5))*Sin[2*Pi*5*t], t, s];
Ib = LaplaceTransform[((78)/(10^5))*Sin[2*Pi*5*t], t, s];
Ic = LaplaceTransform[((78)/(10^5))*Sin[2*Pi*5*t], t, s];
R1 = 1/(s*C1);
R2 = 1/(s*C2);
R3 = 1/(s*C3);
R4 = 1;
C1 = ((1124)/10^3)*10^(-9);
C2 = ((1124)/10^3)*10^(-9);
C3 = ((1124)/10^3)*10^(-9);
FullSimplify[
InverseLaplaceTransform[
I4 /. FullSimplify[
Solve[{Ia == I1 + I4, I2 == I1 + I5, I3 == I2 + I6,
Ic == I3 + I4, Ic == Ib + I6, Ib == Ia + I5,
I1 == (V1 - V2)/R1, I2 == (V2 - V3)/R2, I3 == V3/R3,
I4 == V1/R4}, {I1, I2, I3, I4, I5, I6, V1, V2, V3}]][], s,
t]]

Out=(58500000 (281 E^(-750000000000 t/281) \[Pi] -
281 \[Pi] Cos[10 \[Pi] t] +
75000000000 Sin[10 \[Pi] t]))/(5625000000000000000000 +
78961 \[Pi]^2)

I want to do two things to the output of this code:

1. I want to remove the transient part of the equation. So I can do that by using Expand[] and after that looking for each individual term that gives $$\displaystyle\lim_{t\to\infty}y_n(t)=0$$. But is there a way to let Mathematica do that?
2. After the transient part is removed, I want to find $$I_4(t)$$ in the standard form: $$I_4(t)=a\sin(\omega t+\varphi)\tag1$$ Where $$a$$, $$\omega$$ and $$\varphi$$ needs to be found using Mathematica.

Can someone help me with how I can do this in Mathematica? Thank you so much.

May be

Clear["Global*"]
expr = (58500000 (281 E^(-750000000000 t/281) π - 281 π Cos[10 π t]
+ 75000000000 Sin[10 π t]))/(5625000000000000000000 + 78961 π^2);

expr  = Expand[expr];
expr2 = Simplify[If[MatchQ[#,_.*Exp[__*t]],Limit[#,t->Infinity],#]&/@expr];

(Expand@Numerator[expr2]/.a_.*Cos[w_ t]+b_.*Sin[w_ t]
:>Sqrt[a^2+b^2]*Sin[w*t+ArcTan[b/a]])/Denominator[expr2] • Thank you very much. Jul 4 '21 at 11:04

Try this:

expr = (58500000 (281 E^(-750000000000 t/281) \[Pi] -
281 \[Pi] Cos[10 \[Pi] t] +
75000000000 Sin[10 \[Pi] t]))/(5625000000000000000000 +
78961 \[Pi]^2);

Then

expr1 = expr /. Exp[Rational[a_, b_]*t] -> 0 /.
a_*Cos[10 \[Pi] t] + b_*Sin[10*\[Pi]*t] ->
Sqrt[a^2 + b^2] Sin[10*\[Pi]*t + ArcTan[a, b]]

(*  -((58500000 Sin[10 \[Pi] t - ArcTan[75000000000/(281 \[Pi])]])/Sqrt[
5625000000000000000000 + 78961 \[Pi]^2])  *)

To see it better below I show the result as an image: Have fun!

• The problem with Exp[Rational[a_, b_]*t] -> 0 is that it also makes Exp[3/2*t] goes to zero. Which is not correct, as this term does not go to zero in the limit Jul 4 '21 at 11:34
• @Nasser You are right, but OP only asked about the expression containing a single specific exponential function. In such cases, for the sake of being fast in calculations, I always prefer to make a specific replacement, rather than a general one. In general case, of course you are right, and one easily corrects this as follows: Exp[Rational[a_, b_]*t] /; a < 0 && b > 0 -> 0 or likewise. Jul 4 '21 at 11:46