1. Linear Differential Equations with Constant Coefficients: Example: f(t): Input u(t): Output (response) Characteristic Equation: Homogenous solution f(t)=0. u(t)=e st s 3 e st + 4s 2 e st + 14se st + 20e st = 0 s 3 + 4s 2 + 14s + 20 = 0 a=[1,4,14,20];roots(a) Eigenvalues:  1  3i,  2 u h (t) = C 1 e (-1+3i)t + C 2 e (-1-3i)t + A 2 e -2t u h (t) = A 1 e -t cos(3t-φ)+A 2 e -2t With Matlab:

2. u h (t) = A 1 e -t cos(3t-φ)+A 2 e -2t Initial conditions: at t=0 -1.2 = A 1 cosφ + A 2 2.5 = -A 1 cosφ +3A 1 sinφ -2A 2 -3.1= -8A 1 cosφ - 6A 1 sinφ + 4A 2 A 1, A 2 and φ can be found by Newton-Raphson method.

3. Laplace Transform:

4. Laplace Transform of the Derivative :

5. (shift in time or delay):

6. Laplace transform of the solution due to the initial conditions: Initial conditions: at t=0

7. Partial fraction expansion: num=[-1.2,-2.3,-9.9]; den=[1,4,14,20]; [r,p,k]=residue(num,den) r(1)=-0.095-0.0483i, r(2)=-0.095+0.0483i, r(3)=-1.01 With Matlab;

8. u h (t) = A 1 e -t cos(3t-φ)+A 2 e -2t z=-0.095+0.0483i A1=2*abs(z) fi=angle(z) With Matlab; Homogeneous solution :

9. EXAMPLES: m g θ Joint friction, B L The equation of the motion for the unforced motion of a simple pendulum is given as: m=2 kg B=4 Nms/rad L=2 m At t=0 are given. Find θ(t). Applying the Laplace transform,

10. EXAMPLES: clc;clear num=[4 10]; den=[8 4 39.24]; [r,p,k]=residue(num,den) r(2) A=2*abs(r(2)) Fi=angle(r(2)) Re 0.25 0.2556 Img Laplace transform of the homogenous solution (due to the initial conditions) Eigenvalues The system is stable because the real parts of all the roots are negative.

11. EXAMPLES: clc;clear dt=0.1418; ts=25.149; t=0:dt:ts; tetat=0.7151*exp(-0.25*t).*cos(2.2006*t-0.7965); plot(t,tetat)