Linear Differential Equations with Constant Coefficients:

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Linear Differential Equations with Constant Coefficients: f(t): Input u(t): Output (response) Example: Homogeneous solution f(t)=0. u(t)=est With Matlab: Characteristic Equation: s3est + 4s2est + 14sest + 20est= 0 a=[1,4,14,20];roots(a) s3 + 4s2 + 14s + 20 = 0 Eigenvalues: -13i, - 2 uh(t) = C1e(-1+3i)t + C2 e(-1-3i)t + A2e-2t uh(t) = A1e-tcos(3t-φ)+A2e-2t

uh(t) = A1e-tcos(3t-φ)+A2e-2t Initial conditions: at t=0 -1.2 = A1 cosφ + A2 2.5 = -A1 cosφ +3A1 sinφ -2A2 -3.1= -8A1 cosφ - 6A1 sinφ + 4A2 A1, A2 and φ can be found by Newton-Raphson method.

Laplace Transform:

Laplace Transform of the Derivatives :

(shift in time or delay):

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

Partial fraction expansion: With Matlab; 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

Homogeneous solution : uh(t) = A1e-tcos(3t-φ)+A2e-2t With Matlab; z=-0.095+0.0483i A1=2*abs(z) fi=angle(z)

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

[r,p,k]=residue(num,den) EXAMPLES: 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. 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

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)