example 6.22    Transfer function: 1st order   p.136

para1 = {a1 ->  4, c1 -> 40}
gs1 = c1/(s + a1) /. para1

impulse

gt11 = InverseLaplaceTransform[gs1, s, t]
zu11 = 
      Plot[gt11, {t, 0, 2.5}, PlotRange ->  {-20, 80}, 
        PlotStyle ->  Hue[0.9], 
        GridLines ->  Automatic]

step

gt12 = 
        InverseLaplaceTransform[gs1/s, s, t];
Expand[%]
zu12 = 
      Plot[gt12, {t, 0, 4}, PlotRange ->  {0, 25}, 
        PlotStyle ->  Hue[0.9], 
        GridLines ->  Automatic]

example 6.23   Transfer function: 2ndt order    p.137

para2 = {a1 ->  5, a2 ->  2, c1 ->  20, 
        c2 -> 40}
gs2 = (c2 s + c1)/(s^2 + a2 s + a1) /. para2

implse

gt21 = 
        InverseLaplaceTransform[gs2, s, t];
Expand[%]
FullSimplify[ExpToTrig[%]]
zu21 = 
      Plot[gt21, {t, 0, 2.5}, PlotRange ->  {-20, 80}, 
        PlotStyle ->  Hue[0.7], 
        GridLines ->  Automatic]

step

gt22 = 
        InverseLaplaceTransform[gs2/s, s, t];
Expand[%]
FullSimplify[ExpToTrig[%]]
zu22 = 
      Plot[gt22, {t, 0, 4}, PlotRange ->  {0, 25}, 
        PlotStyle ->  Hue[0.7], 
        GridLines ->  Automatic]

example 6.24      p.138

mode separation of control system

aa = {{0, 1, 0}, {0, 0, 1}, {-20, -13, -6}}; 
    MatrixForm[aa]
bb = {0, 0, 1}; MatrixForm[bb]
cc = {280, 260, 80}; MatrixForm[cc]
e3 = {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}

gs3 = Together[cc . Inverse[e3 s - aa] . bb]

Impluse

gt31 = InverseLaplaceTransform[gs3, s, t]
FullSimplify[ExpToTrig[%]]
zu31 = 
      Plot[gt31, {t, 0, 2.5}, PlotRange ->  {-20, 80}, 
        PlotStyle ->  Hue[0.5], 
        GridLines ->  Automatic]

Show[zu31, zu21, zu11]

step response

gt32 = 
        InverseLaplaceTransform[gs3/s, s, t];
Expand[%]
TrigReduce[FullSimplify[ExpToTrig[%]]]
zu32 = 
      Plot[gt32, {t, 0, 4}, PlotRange ->  {0, 25}, 
        PlotStyle ->  Hue[0.5], 
        GridLines ->  Automatic]

Show[zu32, zu22, zu12]