Question

Derive Laplace Transform of the following:a)f(t)=e^(-at) (b) f(t)=t (c) f(t)=t^2 (d) cos h t

Laplace transform
ANSWERED
asked 2021-09-19

Derive Laplace Transform of the following:
a)\(\displaystyle{f{{\left({t}\right)}}}={e}^{{-{a}{t}}}\)
b) \(f(t)=t \)c) \(\displaystyle{f{{\left({t}\right)}}}={t}^{{2}}\)
d) \(\displaystyle{\cosh}{t}\)

Expert Answers (1)

2021-09-20

Step 1
we have given f(t).
We have to derive the following laplace transform
Step 2
1) Derive Laplace transform of the following
\(\displaystyle{f{{\left({t}\right)}}}={e}^{{-{a}{t}}}\)
We know that
\(L\left\{f(t)\right\}=\int_0^{\infty}e^{-st}f(t)dt\)
Now, \(L\left\{e^{-dt}\right\}=\int_0^{\infty}e^{-st} \cdot e^{-2t}dt=\int_0^{\infty} e^{-(s+a)t}dt=\left[\frac{e^{-(s+a)t}}{-(s+2)}_0^{\infty}\right]_0^{\infty}\)
Hence, \(L\left\{e^{-dt}\right\}=\left[0-\frac{e^0}{-(s+2)}\right]=\frac{1}{(s+2)}\)
b) f(t)=t
Now, \(L\left\{f(t)\right\}=L\left\{t\right\}=\int_0^{\infty} e^{-st} t dt\)
\(\displaystyle={{\left[{t}{\frac{{{e}^{{-{s}{t}}}}}{{-{s}}}}\right]}_{{0}}^{{\infty}}}-{\int_{{0}}^{{\infty}}}{\frac{{{\left.{d}{t}\right.}}}{{{\left.{d}{t}\right.}}}}{\frac{{{e}^{{-{s}{t}}}}}{{-{s}}}}{\left.{d}{t}\right.}\)
\(\displaystyle={\left({0}+{0}\right)}+{\frac{{{1}}}{{{s}}}}{{\left[{\frac{{{e}^{{-{s}{t}}}}}{{{\left(-{s}\right)}}}}\right]}_{{0}}^{{\infty}}}={\frac{{{1}}}{{{s}^{{2}}}}}{\left({0}+{1}\right)}={\frac{{{1}}}{{{s}^{{2}}}}}\)
Hence, \(L\left\{f(t)\right\}=L\left\{t\right\}=\frac{1}{s^2}\)
c)\(\displaystyle{f{{\left({t}\right)}}}={t}^{{2}}\)
Now,\(L\left\{f(t)\right\}=L\left\{t^2\right\}=\int_0^{\infty} e^{-st} t^2 dt\)
\(\displaystyle={{\left[{t}^{{2}}{\frac{{{e}^{{-{s}{t}}}}}{{{\left(-{s}\right)}}}}\right]}_{{0}}^{{\infty}}}-{\int_{{0}}^{{\infty}}}{2}{t}{\frac{{{e}^{{-{s}{t}}}}}{{{\left(-{s}\right)}}}}{\left.{d}{t}\right.}\)
\(\displaystyle={\left({0}-{0}\right)}+{\frac{{{2}}}{{{s}}}}{\left[{{\left[{t}{\frac{{{e}^{{-{s}{t}}}}}{{{\left(-{s}\right)}}}}\right]}_{{0}}^{{\infty}}}-{\int_{{0}}^{{\infty}}}{\frac{{{\left.{d}{t}\right.}}}{{{\left.{d}{t}\right.}}}}{\frac{{{e}^{{-{s}{t}}}}}{{{\left(-{s}\right)}}}}{\left.{d}{t}\right.}\right]}\)
\(\displaystyle={\frac{{{2}}}{{{s}}}}{\left({\left({0}-{0}\right)}\right)}+{\frac{{{2}}}{{{s}}}}{\frac{{{1}}}{{{s}}}}{{\left[{\frac{{{e}^{{-{s}{t}}}}}{{-{s}}}}\right]}_{{0}}^{{\infty}}}\)
Hence, \(L\left\{f(t)\right\}=L\left\{t^2\right\}=\frac{2}{s}(0)+\frac{2}{s^2}(0+\frac{1}{s})=\frac{2}{s^3}\)
d) \(\displaystyle{f{{\left({t}\right)}}}={\cos{{h}}}{t}\)
\(L\left\{f(t)\right\}=L\left\{\cos h t\right\}=\int_0^{\infty} e^{-st}\cos h t dt\)
\(\displaystyle={\int_{{0}}^{{\infty}}}{e}^{{-{s}{t}}}{\left({\frac{{{e}^{{t}}+{e}^{{-{t}}}}}{{{2}}}}\right)}{\left.{d}{t}\right.}\)
\(\displaystyle={\int_{{0}}^{{\infty}}}{\frac{{{e}^{{-{s}{t}}}{e}^{{t}}{\left.{d}{t}\right.}}}{{{2}}}}+{\int_{{0}}^{{\infty}}}{\frac{{{e}^{{-{s}{t}}}{e}^{{-{t}}}{\left.{d}{t}\right.}}}{{{2}}}}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{\left({\frac{{{1}}}{{{s}-{1}}}}\right)}+{\frac{{{1}}}{{{2}}}}{\left({\frac{{{1}}}{{{s}+{1}}}}\right)}={\frac{{{1}}}{{{2}}}}{\left({\frac{{{s}+{1}+{s}-{1}}}{{{\left({s}-{1}\right)}{\left({s}+{1}\right)}}}}\right)}\)
Hence, \(L\left\{\cos h t \right\}=\frac{s}{s^2-1}\)

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