the boxes are just graphics. There is no easy way in MapleSim to do what you want.

try this:

right-click the installer, select properties->compatibility

select run in compatibility mode for Windows 7

install

When using worksheet mode, you could press F3 or F4 to join or split Execution Groups.

you have imaginary numbers in your expression, so you cant plot h(x,t) directly. Do you need abs(h(x,t)) to be plotted?

You probably missed the multiplication sign between T and (Qin-Qp1)

Thanks Kitonum, meanwhile I had another idea. I used the Find/Replace dialogue from the edit menu for that.

okay, the example was too stripped down from the original problem. Check this:

eq1 := (q4+(-2*mu[2]^2+2*mu[2])*R[S]*B[n]*lambda[0]+(2-2*mu[2])*R[S]*B[n]*lambda[0]*mu[3]+(2*mu[2]-2)*R[S]*B[n]*lambda[0]*mu[3]*n^2*N[p]^2+(-2*mu[2]+2*mu[2]^2)*R[S]*B[n]*lambda[0]*n^2*N[p]^2+d^2*(g^2*(q5-q4+(2*mu[2]+2*mu[2]^2)*R[S]*B[n]*lambda[0]+(-2-2*mu[2])*R[S]*B[n]*lambda[0]*mu[3]+(2*mu[2]+2)*R[S]*B[n]*lambda[0]*mu[3]*n^2*N[p]^2+(-2*mu[2]-2*mu[2]^2)*R[S]*B[n]*lambda[0]*n^2*N[p]^2)+(-2*mu[2]-2*mu[2]^2)*R[S]*B[n]*lambda[0]+(-2-2*mu[2])*R[S]*B[n]*lambda[0]*mu[3]+(2*mu[2]+2)*R[S]*B[n]*lambda[0]*mu[3]*n^2*N[p]^2)+g^2*(q7+q6+q5-q4+(-2*mu[2]+2*mu[2]^2)*R[S]*B[n]*lambda[0]+(2-2*mu[2])*R[S]*B[n]*lambda[0]*mu[3]+(2*mu[2]-2)*R[S]*B[n]*lambda[0]*mu[3]*n^2*N[p]^2+(-2*mu[2]^2+2*mu[2])*R[S]*B[n]*lambda[0]*n^2*N[p]^2))/(mu[1]*(n^2*N[p]^2-1));

eq2 := subs((-2*mu[2]^2+2*mu[2])*R[S]*B[n]*lambda[0]+(2-2*mu[2])*R[S]*B[n]*lambda[0]*mu[3]+(2*mu[2]-2)*R[S]*B[n]*lambda[0]*mu[3]*n^2*N[p]^2+(-2*mu[2]+2*mu[2]^2)*R[S]*B[n]*lambda[0]*n^2*N[p]^2 = q11, eq1);

plots[polyhedraplot]([0, 0, 0], polytype = tetrahedron, scaling = constrained);

``(2*x+1);

%*3;

expand(%);

Suggestion: Place position sensors on all relevant mass points and connect them to a custom component. That component should just take the inputs of all sensors and calculate for each axis the center of mass according to

x_center=1/M*(x1*m1+x2*m2+...)

with M=sum of all masses.

yes, every bit helps. Going from Laplace to discrete after fitting makes sense.

But I still see the problem in the plot of discrete points from the inverse transformation. For large n, it diverges obviously, while the initial function exp(-n/5)*sin(n) converges against zero.

yes, good.

But still, I don't get the initial transfer function back.

invztrans(eq,z,n);seq((Re(%),n=1..10));
plots[listplot]([%],style=point);

This doesn't look like the damped sin  from initial list.

Thanks Georgios,your help is appreciated (rep+1).
I tried to check the result by doing an inverse transformation on eq.

invztrans(eq,z,n);seq(%,n=1..10);

I was hoping to get a sequence of points from exp(-n/5)*sin(n).
I tried to remove the small imaginary parts with

map(simplify,[%],zero);

Unfortunately, it doesn't remove all imaginaries.

restart:Q:= x + y + z - 1 = 0;
f:=unapply(lhs(Q),x,y,z);
f(a + 1, b + 2, c + 3);

restart:with(PolynomialTools):
Z1 := 1/(1/(s*L[1])+s*C[1])+s*C[2];
Z2:=normal(Z1);Z3:=Split(numer(Z2),s)/Split(denom(Z2),s):
convert(Z3,radical);