(x(t) depends on the initial conditions)

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(x(t) depends on the initial conditions) Linear Differential Equations with Constant Coefficients: (Characteristic polynomial) If Different two roots (No oscillation) Repeated roots Complex roots (Oscillatory response) The form of the system response is determined by the eigenvalues of the system which are determined by m, c and k. Critical damping coefficient (Damping ratio) f(t) k c m x(t) Assume Homogeneous Form (x(t) depends on the initial conditions) 

Example: f(t): Input u(t): Output (response) 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)