Step 1

29 (a) The pole diagram of F(s) tells us about long term behavior of f(t) .it tell us nothing about the near -term behavior .suppose we have just one pole at s=1. among the functions with the pole diagram we have:

\(F(s)=\frac{c}{s-1} , G(s)=\frac{(ce^{-as})}{(s-1)} ,H(s)=\frac{c}{(s-1)}+b\frac{1-e^{-as}}{(s-1)}\)

where c not zero .To be laplace transforms of real functions we must also assume them all to be real and a non negative then these are the laplace transform of

\(f(s)=ce^t , g(t) =\left\{\frac{ce^{t-a} \text{ for } t>a}{0 \text{ for } ta}{ce^t+b \text{ for } t

Step 2

\(i)F(s)=\frac{1}{s^2+1}\)

\(ii)F(s)=\frac{c(s+1)}{(s+1)^2+16}\)

\(iii)F(s) =\frac{1}{s-1}\)

laplace transformation of these functions are respectively:

\(i) f(t)=c\sin t\)

\(ii) f(t)=ce^{-t}\cos t\)

\(iii)f(t)=e^t\)

29 (a) The pole diagram of F(s) tells us about long term behavior of f(t) .it tell us nothing about the near -term behavior .suppose we have just one pole at s=1. among the functions with the pole diagram we have:

\(F(s)=\frac{c}{s-1} , G(s)=\frac{(ce^{-as})}{(s-1)} ,H(s)=\frac{c}{(s-1)}+b\frac{1-e^{-as}}{(s-1)}\)

where c not zero .To be laplace transforms of real functions we must also assume them all to be real and a non negative then these are the laplace transform of

\(f(s)=ce^t , g(t) =\left\{\frac{ce^{t-a} \text{ for } t>a}{0 \text{ for } ta}{ce^t+b \text{ for } t

Step 2

\(i)F(s)=\frac{1}{s^2+1}\)

\(ii)F(s)=\frac{c(s+1)}{(s+1)^2+16}\)

\(iii)F(s) =\frac{1}{s-1}\)

laplace transformation of these functions are respectively:

\(i) f(t)=c\sin t\)

\(ii) f(t)=ce^{-t}\cos t\)

\(iii)f(t)=e^t\)