If L\left\{f(t)\right\}=F(s) , then L{f(4t)} (a)frac(1)(4) F(s) (b)F(frac(s)(4)) (c)4F(s) (d)frac(1)(4) F(frac(s)(4))

Ernstfalld 2021-09-23 Answered

If \(\displaystyle{L{f{{\left({t}\right)}}={F}{\left({s}\right)},\ \text{ then }\ {L}{f{{\left({4}{t}\right)}}}}}\)
a) \(\displaystyle{\frac{{{1}}}{{{4}}}}{F}{\left({s}\right)}\)
b) \(\displaystyle{F}{\left({\frac{{{s}}}{{{4}}}}\right)}\)
c) \(\displaystyle{4}{F}{\left({s}\right)}\)
d) \(\displaystyle{\frac{{{1}}}{{{4}}}}{F}{\left({\frac{{{s}}}{{{4}}}}\right)}\)

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Expert Answer

opsadnojD
Answered 2021-09-24 Author has 9358 answers
Step 1
Here we solve this from the definition of Laplace transformation.
step 2
\(\displaystyle{L}{\left({f{{\left({t}\right)}}}\right)}={F}{\left({s}\right)}={\int_{{0}}^{\infty}}{f{{\left({t}\right)}}}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}\)
\(\displaystyle\rightarrow{L}{\left({f{{\left({4}{t}\right)}}}\right)}={\int_{{0}}^{\infty}}{f{{\left({4}{t}\right)}}}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}\)
Step 3
Let us take transformation \(\displaystyle{z}={4}{t}\rightarrow{\left.{d}{t}\right.}={\frac{{{\left.{d}{z}\right.}}}{{{4}}}}\)
Then the integration becomes
\(\displaystyle{L}{\left({f{{\left({4}{t}\right)}}}\right)}={\int_{{0}}^{\infty}}{f{{\left({4}{t}\right)}}}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}={\int_{{0}}^{\infty}}{f{{\left({z}\right)}}}{e}^{{-{\frac{{{s}{z}}}{{{4}}}}}}{\frac{{{\left.{d}{z}\right.}}}{{{4}}}}\)
\(\displaystyle={\frac{{{1}}}{{{4}}}}{\int_{{0}}^{\infty}}{f{{\left({z}\right)}}}{e}^{{{\frac{{{s}}}{{{4}}}}{z}}}{\left.{d}{z}\right.}={\frac{{{1}}}{{{4}}}}{F}{\left({\frac{{{s}}}{{{4}}}}\right)}\)
Step 4
The final answer is \(\displaystyle{L}{\left({f{{\left({4}{t}\right)}}}\right)}={\frac{{{1}}}{{{4}}}}{F}{\left({\frac{{{s}}}{{{4}}}}\right)}\) so 4th option is correct.
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