Answer to Question #223816 in Chemical Engineering for Lokika

Question #223816

L^{-1} e^{-s}/√s }


1
Expert's answer
2021-08-10T05:54:45-0400

the time shifting property of Laplace transform:


f(ta)u(ta)=L1{easF(s)},a>0f(t-a)u(t-a)=L^{-1}\{e^{-as}F(s)\}, a>0

where u(t) is the Heaviside step function and


F(s)=L{f(t)}F(s)=L\{f(t)\}

hence


L1{es/s}=[a=1>0]=u(t1)f(t1)L^{-1}\{e^{-s}/\sqrt{s}\}=[a=1>0]=u(t-1)f(t-1)

where


F(s)=1s=[for p=12]=Γ(p+1)Γ(p+1)sp+1=Γ(12)Γ(12)sF(s)=\frac{1}{\sqrt{s}}=\Big[for \space p=-\frac 1 2 \Big]=\frac{\Gamma(p+1)}{\Gamma(p+1) s^{p+1}}=\frac{\Gamma(\frac 1 2)}{\Gamma(\frac 1 2) \sqrt{s}}

The Gamma function is an extension of the normal factorial function and


Γ(12)=π\Gamma \Big( \frac 1 2 \Big)=\sqrt {\pi}

for pth power of Laplace transform:


L1{Γ(p+1)sp+1}=[Re(p)>1]=tpL^{-1} \Bigg \{ \frac{\Gamma(p+1)}{s^{p+1}} \Bigg \}=[Re(p)>-1]=t^p

So


f(t)=L1{Γ(1/2)Γ(1/2)s}=1πL1{Γ(1/2)s}=1πtf(t)=L^{-1} \Bigg \{ \frac{\Gamma(1/2)}{\Gamma(1/2) \sqrt{s}} \Bigg \}=\frac{1}{\sqrt {\pi}}L^{-1} \Bigg \{ \frac{\Gamma(1/2)}{ \sqrt{s}} \Bigg \}=\frac{1}{\sqrt {\pi t}}L1{es/s}=u(t1)f(t1)=u(t1)π(t1)L^{-1}\{e^{-s}/\sqrt{s}\}=u(t-1)f(t-1)=\frac{u(t-1)}{\sqrt {\pi (t-1)}}

Answer: L1{es/s}=u(t1)π(t1)L^{-1}\{e^{-s}/\sqrt{s}\}=\frac{u(t-1)}{\sqrt {\pi (t-1)}}


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