8. Solution of Linear Differential Equations: f(t): Input,

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8. Solution of Linear Differential Equations: f(t): Input, u(t): Output Example 8.1 If f(t)=est then u(t)= H(s)est Transfer Function: where H(s) is the transfer function which transforms the input to the output s3H(s)est+4s2H(s)est+14sH(s)est+20H(s)est=3est+sest a=[1,4,14,20];roots(a) (s3+4s2+14s+20)H(s)=3+s Eigenvalues: -1±3i, -2 Exponential/Harmonic Input: s=-0.2+2.7i; hs=(s+3)/(s^3+4*s^2+14*s+20); abs(hs), angle(hs) RESONANCE

H(s) Laplace Transform: x(t) y(t) Y(s)=X(s) H(s) Δ(s)=1 1 Final value theorem: Impulse function Δ(s)=1 u(t): Step function 1

Inverse Laplace transform of H(s) Impulse Response: Inverse Laplace transform of H(s) Eigenvalues: a=[1,4,14,20];roots(a) -1±3i, -2 (Partial fraction expansion): p1=[1,3]; p2=[1,4,14,20]; [r,p,k]=residue(p1,p2) r(1)=-0.05-0.1833i, r(2)=-0.05+0.1833i, r(3)=0.1 z=-0.05+0.1833i 2*abs(z) phase(z) ξ=0.3162 (s=-1±3i), suitable values of Δt=0.099 and t∞=6.283

ξ=0. 3162 (s=-1±3i), use Δt=0. 099 and t∞=6 ξ=0.3162 (s=-1±3i), use Δt=0.099 and t∞=6.283 to plot the response h(t) clc;clear; t=0:0.099:6.283; yt=0.3801*exp(-t).*cos(3*t-1.837)+0.1*exp(-2*t) plot(t,yt)

y(t)  Inverse Laplace Transform of Y(s) Step Input Response: y(t)  Inverse Laplace Transform of Y(s) Singular Points (Eigenvalues) are -1±3i , -2 and s=0 Partial fraction expansion: p1=[1,3]; p2=[1,4,14,20,0]; [r,p,k]=residue(p1,p2) r(1)=-0.05+0.0333i, r(2)=-0.05-.0033i, r(3)=-0.05, r(4)=0.15 Final value theorem: yss=0.15

clc;clear; t=0:0.099:6.283; yt=0.1202*exp(-t).*cos(3*t+2.5536)-0.05*exp(-2*t)+0.15; plot(t,yt)

Solution due to the Initial Conditions (Homogeneous Solution): f=0 t=0 da p1=[-1.2,-2.3,-9.9]; p2=[1,4,14,20]; [r,p,k]=residue(p1,p2) r(1)=-0.095-0.0483i, r(2)=-0.095+0.0483i, r(3)=-1.01

a) Find the output θ(t) for yA(t)=0.2e-3tcos(17t-1.8) Problem: For the mechanical sysem given in the figure, yA is input and θ is output. Take m1=250 kg, m2=350 kg, k=37000 N/m, c=1500 Ns/m and L1=1.2 m. The governing differential equation of the system is given as m1, L1 θ yA m2 k c a) Find the output θ(t) for yA(t)=0.2e-3tcos(17t-1.8) Solution: yA(t)=est and θ(t)= H(s)est s=-3+17i ; h=(1800*s+44400)/(624*s^2+2160*s+53280) abs(h) angle(h)