ME 422 – Lecture 2 Talk about homework Ask for questions Ask what’s funny about brake pedal of problem 1? Laplace transforms Much of the burden of doing Laplace transforms and inverse transforms is taken away from you by Matlab/Simulink. In fact you are unaware of all the math going on the background to do this for you. Need to spend a little time, just so you know that Laplace is there in the background and what it is doing for you. Definition of Laplace transform on page 2‐2, but usually transforms are gotten from tables (Tables 2.1 and 2.2). Solve a 1st‐order ODE via Laplace. 16
Solution is 2 0,
. Take Laplace transform of both sides. 16 ∙
∙
0 16
2 2
16
So we’ve found X(s), but we want x(t). So need L‐1{X(s)} = x(t). Getting the inverse Laplace transform is the problem. When I went through college in the 1970s, this was all done by hand. It took forever to work these problems manually. We didn’t work too many problems. If we look at the Table 2.1, the 5th entry is what we want. We pull the 2 out (L is linear). So 2
16
2∙
∙
We can plot this in Excel or in Matlab. Lecture 2 ‐ 1 To solve this system via Matlab, it’s a little complicated. You have to make a file with the system in it. function dx = F(t,x)
dx(1) = -16*x(1); % Same as xdot + 16*x = 0
Save this as dx.m. With this, can invoke the ODE solver ode45().
>> [t,x] = ode45( ‘dx’, [0,0.4], 2 );
This gives the vectors t and x. You now have x(t) and can plot it. Lecture 2 ‐ 2 >> plot( t,x )
Note that this 1st‐order system had no input, no exciting function. It was excited by an initial displacement, x0 = 2. You can also have 1st orders that are excited, say, by a step function. These have no 0 on the right‐hand side of the ODE. Usually you excite a system with 0 i.c.s and an external forcing function, or you excite it with non‐0 i.c.s and no forcing function. Theoretically you could have both—non‐0 i.c.s AND a forcing function, but this is not usually the case. If you have a forcing function (such as a step function), you have to use the Laplace tables to convert the forcing function into the s‐domain. Lecture 2 ‐ 3 Second‐order systems are a little more complicated than 1st‐order systems. 16
2∙
,
0,
0 Take L{} and continue. Lecture 2 ‐ 4