Question

Use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the convolution integral before transforming.(Write your answer as a function of s.) {L}{leftlbrace{e}^{{-{t}}}cdot{e}^{t} cos{{left({t}right)}}rightrbrace}

Laplace transform
ANSWERED
asked 2021-01-05
Use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the convolution integral before transforming.(Write your answer as a function of s.)
\({L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}\)

Answers (1)

2021-01-06
Step 1
Consider the provided question,
We have to find \({L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}\) Step 2
Now, the given Laplace transform is find as,
\({L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}={L}{\left\lbrace{e}^{{-{t}}}\right\rbrace}+{L}{\left\lbrace{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}\)
\(={\left(\frac{1}{{{s}+{1}}}\right)}\cdot\frac{{{s}-{1}}}{{{\left({s}-{1}\right)}^{2}+{1}^{2}}}{\left[{U}{s}{e},{L}{\left({e}^{{{a}{t}}}\right)}=\frac{1}{{{s}-{a}}}{\quad\text{and}\quad}{L}{\left({e}^{{{a}{t}}} \cos{{b}}{t}\right)}=\frac{{{s}-{a}}}{{{\left({s}-{a}\right)}^{2}+{b}^{2}}}\right]}\)
\(=\frac{{{s}-{1}}}{{{\left({s}+{1}\right)}{\left[{\left({s}-{1}\right)}^{2}+{1}\right]}}}\)
Thus , \({L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}=\frac{{{s}-{1}}}{{{\left({s}+{1}\right)}{\left[{\left({s}-{1}\right)}^{2}+{1}\right]}}}\)
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