Find the inverse Laplace transform f(t) of the following transforms F(s). Match each of the following to the value of f(1.234) to two decimal places.

Name: Find the inverse Laplace transform f(t) of the following transforms F(s). Match each of the following to the value of f(1.234) to two decimal places.
Uploaded: 2022-02-23T17:22:57+00:00
Description: Find the inverse Laplace transform f(t) of the following transforms F(s). Match each of the following to the value of f(1.234) to two decimal places.

Question

nghodlokl · Accepted Answer

Inverse Laplace Lt{9s+13s2+2s+10}  Lt{9s+9+4s2+2s+10}  =Lt{a(s+1)+4(s+1)2+9}  =9Lt{s+1(s+1)2+9}+4Lt{1(s+1)2+9}  =9e−tLt{ss2+9}+4e−tLt{1s2+9}  Using Lt{F(s−a)}=eatt(t) (formula)  =9e−tcos⁡3t+43sin⁡3t  Hence Lt{9s+13s2+25+10}=9e−tcos⁡3t+43e−tsin⁡3t=f(t) (1)  Now f(1.234) (2)  From (1) and (2)  f(1.234)=9e−(1.234)cos⁡(3(1.234))+43e−(1.234)sin⁡(3(1.234))  =−2.42

Mason Hall · Accepted Answer

a) F(s)=9s+352+28+10  ⇒L−1{9s+3s2+28+10}=2  9s+3s2+28+10=9(s+1)(s+1)2+9=61(s+1)2+9  ⇒L−1{9s+3s2+28+10}=L−1{9(s+1)(s+1)2+9−6(s+1)2+9}  =9L−1{s+1(s+1)2+9}−6L−1{1(s+1)2+9}  =9e−tcos⁡(3t)−6e−t×13sin⁡3t  =9e−tcos⁡(3t)−2e−tsin⁡3t  F(s)=52−28+154−453+1052−12+5  Take partial fraction of  52−28+454−453+1052−125+5=34(5−1)2+14(52−28+5)  ⇒L−1{52−25+454−453+1052−125+5}=Lt{34(5−1)2+14(52−28+

karton · Accepted Answer

Lt{9s+13s2+2s+10}Lt{9s+9+4s2+2s+10}=Lt{a(s+1)+4(s+1)2+9}=9Lt{s+1(s+1)2+9}+4Lt{1(s+1)2+9}=9e−tLt{ss2+9}+4e−tLt{1s2+9}Lt{F(s−a)}=eatt(t) (formula)=9e−tcos⁡3t+43sin⁡3tLt{9s+13s2+25+10}=9e−tcos⁡3t+43e−tsin⁡3t=f(t) (1)f(1.234) (2)f(1.234)=9e−(1.234)cos⁡(3(1.234))+43e−(1.234)sin⁡(3(1.234))=−2.42