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H(s)

H(s). Final value theorem teoremi:. x(t). y(t). Δ(s)=1. 1. u(t):Step function. 0. Impulse function. Laplace Transform:. Y(s)=X(s) H(s). Impulse Response:. : Inverse Laplace Transform of H(s). Eigenvalues:. a=[1,4,14,20];roots(a). -1±3i, -2. Partial fraction expansion:.

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H(s)

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  1. H(s) Final value theorem teoremi: x(t) y(t) Δ(s)=1 1 u(t):Step function 0 Impulse function Laplace Transform: Y(s)=X(s) H(s)

  2. 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), Δt=0.099, t∞=6.283 for the sysem

  3. ξ=0.3162 (s=-1±3i), Cevap için Δt=0.099, t∞=6.283 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)

  4. Eigenvalues : -1±3i , -2 ve s=0 Final value theorem: yss=0.15 Step Response: y(t) : Inverse Laplace Transformu of Y(s) 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

  5. 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)

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