## Customizing a Second Order Dynamical System

$\frac{{d}^{2}x}{d{t}^{2}}+2\zeta \omega \frac{dx}{dt}+{\omega }^{2}x=0$

The above classical textbook equation describes a 2nd order dynamical system. In this formulation, x(t) represents the displacement, ω represents the natural frequency, and ζ represents the damping ratio. To simulate this system with IVSOLVE, we transform it to two first order equations:

$\frac{dx}{dt}=v$ $\frac{dv}{dt}=-2\zeta \omega v-{\omega }^{2}x$

Using the named variables and initial values, we define the input to IVSOLVE as shown in Table 1.

 Variables with initial values Parameters Equations A B C D 1 2 t omega 1 3 x 1 zeta 0.25 4 v 0 5 6 dx/dt =v 7 dv/dt =-2*zeta*omega*v-omega^2*x

Next, we evaluate the array formula =IVSOLVE(B6:B7, (t,x,v), {0,12}) in array I1:K41 and obtain the solution shown Table 2 and plotted in Figure 1.

 I J K 1 t x v 2 0 1 0 3 0.4 0.926057 -0.35295 4 0.8 0.733004 -0.59143 5 1.2 0.470053 -0.70204 6 1.6 0.187507 -0.69215 7 2 -0.07064 -0.585 8 2.4 -0.2719 -0.41357 9 2.8 -0.39777 -0.21403 10 3.2 -0.4439 -0.02004 11 3.6 -0.41815 0.14165 12 4 -0.33723 0.253773 13 4.4 -0.22273 0.309249 14 4.8 -0.09711 0.31042 15 5.2 0.019632 0.26696 16 5.6 0.112406 0.193178 17 6 0.172275 0.10513 18 6.4 0.19664 0.018001 19 6.8 0.188456 -0.05592 20 7.2 0.154784 -0.10843 21 7.6 0.105069 -0.13591 22 8 0.049333 -0.13896 23 8.4 -0.00336 -0.12157 24 8.8 -0.04602 -0.08994 25 9.2 -0.07437 -0.05117 26 9.6 -0.08693 -0.01211 27 10 -0.08478 0.021603 28 10.4 -0.07088 0.046117 29 10.8 -0.04936 0.059586 30 11.2 -0.02468 0.062088 31 11.6 -0.00094 0.055252 32 12 0.018627 0.041749

The solution shows that the system is underdamped with an absolute overshoot over 0.4 at approximately a peak time of 3.3. In this exercise, we will customize the system response by optimizing ζ and ω to achieve an absolute overshoot of exactly 0.2 at t = 2.0.

### Solution

We define two constraints on the solution obtained in Table 2 to demand that x(2) = -0.2 and v(2) = 0, then use NLSOLVE to solve for ω and ζ that staisfy these constraints. The two constraint formulas are shown in Table 3.

 C 1 =ARRAYVAL(J7)-(-0.2) 2 =ARRAYVAL(K7)

J7 and K7 refere to x(2) and v(2) values from the solution array shown in Table 2, but what is ARRAYVAL? ARRAYVAL is one of the criterion functions that allow us to define dynamic constraints on the solution array. In this case we are simply picking a value from the solution but in general, a criterion function can be used to specify more complex constraint on the solution.

We evaluate the array formula =NLSOLVE(C1:C2, (w, zeta)) in array D1:E5 and obtain the solution shown in Table 4 and plotted in Figure 2. The new solution has an absolute overshoot of 0.2 at t=2 as desired.

 D E 1 omega 1.76493157 2 zeta 0.455950294 3 SSERROR 7.61331E-12 4 ITRN 11 5 TIME(s) 0.283

#### Exercise

What if we wanted our peak time to be at 2.5 instead of 2. Our solution array in Table 1 does not list the output point t=2.5. We have two options:

1. We can format the output to display the solution at t=2.5 (See Output Format section in IVSOLVE help page).

2. We define the constraints using another criterion function ODEVAL which allows us to interpolate the solution values at any point.

Using ODEVAL, we modify the constraints formulas as shown in Table 5:

 C 1 =ODEVAL(I2:I32,x,"INTERP",2.5)-(-0.2) 2 =ODEVAL(I2:I32,v,"INTERP",2.5)

The first argument to ODEVAL, is the time vector from the solution array in Table 2. The 2nd argument is the name of the variable we want to operate on. The 3rd argument is the operation (here interpolation), and the 4th argument is the time value we want to interpolate at.

Using the same formula but with updated constraint formulas, NLSOLVE computes the new solution shown in table 6 Below and plotted in Figure 3. Note that the peak time has shifted to 2.5.

 D E 1 omega 1.411995975 2 zeta 0.455958508 3 SSERROR 1.55938E-25 4 ITRN 8 5 TIME(s) 0.152
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