MATLAB: Ilaplace and laplace answer

functionsinverse laplace transformlaplace transformpartfrac

I am just learning how to use the Laplace/Inverse functions in Matlab and was hoping to get some understanding of the final answer below. When working out 1st and second order functions I can ascertain the answer as it does not contain symsum or rootof, or r4 etc. In any case I am unsure of what symsum and Rootof translates to for a final answer? Thanks
>> syms t s
>> F=5*(s+3)/((s^5+4*s^4-3*s^2+7*s+10))
F =
(5*s + 15)/(s^5 + 4*s^4 - 3*s^2 + 7*s + 10)
>> pretty(ans)
          5 s + 15
---------------------------
 5      4      2
s  + 4 s  - 3 s  + 7 s + 10
>> ilaplace(F)
ans =
15*symsum(exp(r4*t)/(5*r4^4 + 16*r4^3 - 6*r4 + 7), r4 in RootOf(s4^5 + 4*s4^4 - 3*s4^2 + 7*s4 + 10, s4)) + 5*symsum((r4*exp(r4*t))/(5*r4^4 + 16*r4^3 - 6*r4 + 7), r4 in RootOf(s4^5 + 4*s4^4 - 3*s4^2 + 7*s4 + 10, s4))
>>

Best Answer

If you have R2015a, you can invert it the old-fashion way with partfrac. Do a pratial fraction expansion and then invert it:
syms t s
Fsv=5*(s+3)/((s^5+4*s^4-3*s^2+7*s+10));
Fpf = partfrac(Fsv, s, 'FactorMode','real');
Ft = ilaplace(Fpf, s, t);
Ftv = vpa(Ft,5)
producing:
Ftv =
- exp(t*(0.97907 + 0.92185i))*(0.29419 + 0.30855i) - exp(t*(0.97907 - 0.92185i))*(0.29419 - 0.30855i) + exp(t*(- 1.1315 + 0.46501i))*(0.30547 - 0.6614i) + exp(t*(- 1.1315 - 0.46501i))*(0.30547 + 0.6614i) - 0.022557*exp(-3.6952*t)
I used vpa to make it easy to read. MATLAB maintains full precision.
Added —
This can be simplified further with rewrite and simplify:
Ftt = rewrite(Ftv, 'sincos');
Ftt = simplify(Ftt, 'Steps',10, 'IgnoreAnalyticConstraints',1);
Ftt = vpa(Ftt,5)
producing a much neater result:
Ftt =
1.3228*exp(-1.1315*t)*sin(0.46501*t) - 0.022557*exp(-3.6952*t) + 0.61094*exp(-1.1315*t)*cos(0.46501*t)
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