Example 4 (Periodic Function)

Next consider a simple damped system y''+3y'+2y=f(t), y(0)=y'(0)=0 and its response to a periodic square wave 1 from 0 to Pi and -1 from Pi to 2Pi repeating with period 2Pi.

We cannot worry about infinite series, so instead we look at the solution over some interval, say 0 to 4Pi, so the input is a FINITE sum of square waves.

In[81]:=

  Clear[f]
  f[t_]:=(UnitStep[t]-UnitStep[t-Pi])-(UnitStep[t-Pi]-UnitStep[t-2 Pi])+
  (UnitStep[t-2 Pi]-UnitStep[t-3 Pi])-(UnitStep[t-3 Pi]-UnitStep[t-4 Pi])
  Plot[Evaluate[f[t]],{t,0,4 Pi}];

The function is written in that way for your benefit. If you had a longer interval, you would need to be clever about it.

In[82]:=

  Clear[g]
  g[t_]:=Sum[(-1)^k (UnitStep[t-k Pi]-UnitStep[t-(k+1) Pi]),{k,0,7}]
  Plot[Evaluate[g[t]],{t,0,8 Pi}];

In[83]:=

  Clear[de,tde,newtde,Y,y]

Here's the differential equation:

In[84]:=

  de=y''[t]+3 y'[t]+2 y[t]==f[t]

Out[84]=

  2 y[t] + 3 y'[t] + y''[t] == 
   
    UnitStep[t] + UnitStep[-4 Pi + t] - 2 UnitStep[-3 Pi + t] + 
   
     2 UnitStep[-2 Pi + t] - 2 UnitStep[-Pi + t]

Here is the transformed differential equation:

In[85]:=

  tde=LaplaceTransform[de,t,s]

Out[85]=

                                    2
  2 LaplaceTransform[y[t], t, s] + s  LaplaceTransform[y[t], t, s] + 
   
     3 (s LaplaceTransform[y[t], t, s] - y[0]) - s y[0] - y'[0] == 
   
    1       1           2           2          2
    - + --------- - --------- + --------- - -------
    s    4 Pi s      3 Pi s      2 Pi s      Pi s
        E       s   E       s   E       s   E     s

Let's make use of the initial values:

In[86]:=

  newtde=tde/.{y[0]->0,y'[0]->0}

Out[86]=

  2 LaplaceTransform[y[t], t, s] + 
   
     3 s LaplaceTransform[y[t], t, s] + 
   
      2
     s  LaplaceTransform[y[t], t, s] == 
   
    1       1           2           2          2
    - + --------- - --------- + --------- - -------
    s    4 Pi s      3 Pi s      2 Pi s      Pi s
        E       s   E       s   E       s   E     s

Now solve this equation for Y[s].

In[87]:=

  Solve[newtde,LaplaceTransform[y[t],t,s]];
  Y[s_]:=Evaluate[LaplaceTransform[y[t],t,s]/.Flatten[%]]
  Y[s]

Out[87]=

            Pi s      2 Pi s      3 Pi s    4 Pi s
    -1 + 2 E     - 2 E       + 2 E       - E
  -(----------------------------------------------)
                4 Pi s               2
               E       s (2 + 3 s + s )

Expect to wait for this one. Maybe go out for lunch.

In[88]:=

  y[t_]:=InverseLaplaceTransform[Y[s],s,t]
  y[t]

Out[88]=

                           8 Pi - 2 t
  1     1       -t    1   E              4 Pi - t
  - + ------ - E   + (- + ----------- - E        ) 
  2      2 t          2        2
      2 E
   
                                  6 Pi - 2 t      3 Pi - t
     UnitStep[-4 Pi + t] + (-1 - E           + 2 E        ) 
   
                                 4 Pi - 2 t      2 Pi - t
     UnitStep[-3 Pi + t] + (1 + E           - 2 E        ) 
   
                                  2 Pi - 2 t      Pi - t
     UnitStep[-2 Pi + t] + (-1 - E           + 2 E      ) 
   
     UnitStep[-Pi + t]

In[89]:=

  Plot[Evaluate[y[t]],{t,0,4 Pi}];

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