Question #233142

L-1{(e-2s)/[s2(s-1)]}


1
Expert's answer
2021-09-07T02:09:39-0400
1s2(s1)=As+Bs2+Cs1\dfrac{1}{s^2(s-1)}=\dfrac{A}{s}+\dfrac{B}{s^2}+\dfrac{C}{s-1}

=As(s1)+B(s1)+Cs2s2(s1)=\dfrac{As(s-1)+B(s-1)+Cs^2}{s^2(s-1)}

=As2As+BsB+Cs2s2(s1)=\dfrac{As^2-As+Bs-B+Cs^2}{s^2(s-1)}

s2:A+C=0s^2:A+C=0

s1:A+B=0s^1:-A+B=0

s0:B=1s^0:-B=1

A=1,B=1,C=1A=-1, B=-1, C=1

L1(e2ss2(s1))=H(t2)L1(1s1s2+1s1)(t2)L^{-1}(\dfrac{e^{-2s}}{s^2(s-1)})=H(t-2)L^{-1}(-\dfrac{1}{s}-\dfrac{1}{s^2}+\dfrac{1}{s-1})(t-2)

=H(t2)(H(t2)(t2)+et2)=H(t-2)(-H(t-2)-(t-2)+e^{t-2})



L1(e2ss2(s1))=H(t2)(H(t2)(t2)+et2)L^{-1}(\dfrac{e^{-2s}}{s^2(s-1)})=H(t-2)(-H(t-2)-(t-2)+e^{t-2})



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