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alkaholikd9

Answered 2021-12-10
Author has **4367** answers

Medicim6

Answered 2021-12-11
Author has **1769** answers

\(\displaystyle∵{L}{\left\lbrace{u}{\left({t}-{a}\right)}\right\rbrace}={\frac{{{1}}}{{{s}}}}{e}^{{-{a}{s}}}\)

\(\displaystyle\therefore{L}{\left\lbrace{u}{\left({t}\right)}-{u}{\left({t}-{1}\right)}\right\rbrace}={\frac{{{1}}}{{{s}}}}-{\frac{{{1}}}{{{s}}}}{e}^{{-{s}}}={\frac{{{1}-{e}^{{-{s}}}}}{{{s}}}}\)

\(\displaystyle\therefore{L}{\left\lbrace{u}{\left({t}\right)}-{u}{\left({t}-{1}\right)}\right\rbrace}={\frac{{{1}}}{{{s}}}}-{\frac{{{1}}}{{{s}}}}{e}^{{-{s}}}={\frac{{{1}-{e}^{{-{s}}}}}{{{s}}}}\)

asked 2021-12-11

Using the Laplace Transform Table in the textbook and Laplace Transform Properties,find the (unilateral) Laplace Transforms of the following functions:

\(\displaystyle{t}{\cos{{\left(\omega_{{0}}{t}\right)}}}{u}{\left({t}\right)}\)

\(\displaystyle{t}{\cos{{\left(\omega_{{0}}{t}\right)}}}{u}{\left({t}\right)}\)

asked 2021-09-06

Using only the Laplace transform table, obtain the Laplace transforms of the following function:

\(\displaystyle{\text{cosh}{{\left({4}{t}\right)}}}-{4}{t}\)

where ”cosh” stands for hyperbolic cosine and cosh \(\displaystyle{x}={\frac{{{e}^{{{x}}}-{e}^{{-{x}}}}}{{{2}}}}\)

\(\displaystyle{\text{cosh}{{\left({4}{t}\right)}}}-{4}{t}\)

where ”cosh” stands for hyperbolic cosine and cosh \(\displaystyle{x}={\frac{{{e}^{{{x}}}-{e}^{{-{x}}}}}{{{2}}}}\)

asked 2021-09-08

Use the transforms in the table below to find the inverse Laplace transform of the following function.

\(\displaystyle{F}{\left({s}\right)}={\frac{{{5}}}{{{s}^{{4}}}}}\)

\(\displaystyle{F}{\left({s}\right)}={\frac{{{5}}}{{{s}^{{4}}}}}\)

asked 2021-06-06

Use the table of Laplace transform and properties to obtain the Laplace transform of the following functions. Specify which transform pair or property is used and write in the simplest form.

a) \(x(t)=\cos(3t)\)

b)\(y(t)=t \cos(3t)\)

c) \(z(t)=e^{-2t}\left[t \cos (3t)\right]\)

d) \(x(t)=3 \cos(2t)+5 \sin(8t)\)

e) \(y(t)=t^3+3t^2\)

f) \(z(t)=t^4e^{-2t}\)

a) \(x(t)=\cos(3t)\)

b)\(y(t)=t \cos(3t)\)

c) \(z(t)=e^{-2t}\left[t \cos (3t)\right]\)

d) \(x(t)=3 \cos(2t)+5 \sin(8t)\)

e) \(y(t)=t^3+3t^2\)

f) \(z(t)=t^4e^{-2t}\)

asked 2021-09-08

Use the table of laplace transform to find the inverse Laplace transform of \(\displaystyle{F}{\left({s}\right)}={\frac{{{2}+{s}{\left({s}+{1}\right)}}}{{{s}{\left({s}^{{2}}-{s}-{6}\right)}}}}\)

asked 2021-09-15

\(L\left\{t^4+t^3-t^2-t+\sin \sqrt2t\right\}\)

asked 2021-12-08

Show that

\(\displaystyle{\int_{{0}}^{{\infty}}}{\frac{{{\sin{{\left({t}\right)}}}}}{{{t}}}}{\left.{d}{t}\right.}={\frac{{\pi}}{{{2}}}}\)

by using Laplace Transform method. I know that

\(\displaystyle{L}{\left\lbrace{\sin{{\left({t}\right)}}}\right\rbrace}={\int_{{0}}^{{\infty}}}{e}^{{-{s}{t}}}{\sin{{\left({t}\right)}}}{\left.{d}{t}\right.}={\frac{{{1}}}{{{s}^{{2}}+{1}}}}\)

\(\displaystyle{\int_{{0}}^{{\infty}}}{\frac{{{\sin{{\left({t}\right)}}}}}{{{t}}}}{\left.{d}{t}\right.}={\frac{{\pi}}{{{2}}}}\)

by using Laplace Transform method. I know that

\(\displaystyle{L}{\left\lbrace{\sin{{\left({t}\right)}}}\right\rbrace}={\int_{{0}}^{{\infty}}}{e}^{{-{s}{t}}}{\sin{{\left({t}\right)}}}{\left.{d}{t}\right.}={\frac{{{1}}}{{{s}^{{2}}+{1}}}}\)