Find the Laplace transforms of the following time functions. Solve problem 1(a) and 1 (b) using the Laplace transform definition i.e. integration. For problem 1(c) and 1(d) you can use the Laplace Transform Tables. a)f(t)=1+2t b)f(t) =sin omega t text{Hint: Use Euler’s relationship, } sinomega t = frac{e^(jomega t)-e^(-jomega t)}{2j} c)f(t)=sin(2t)+2cos(2t)+e^{-t}sin(2t)

Question
Laplace transform
asked 2021-02-21
Find the Laplace transforms of the following time functions.
Solve problem 1(a) and 1 (b) using the Laplace transform definition i.e. integration. For problem 1(c) and 1(d) you can use the Laplace Transform Tables.
a)\(f(t)=1+2t\) b)\(f(t) =\sin \omega t \text{Hint: Use Euler’s relationship, } \sin\omega t = \frac{e^(j\omega t)-e^(-j\omega t)}{2j}\)
c)\(f(t)=\sin(2t)+2\cos(2t)+e^{-t}\sin(2t)\)

Answers (1)

2021-02-22
Step 1 To find the Laplace transform of the following functions.
Laplace transform of a function f(t) can be defined as \(L(f(t))=\int_0^\infty f(t)e^{-st}dt\)
Step 2
a) To find the Laplace transform of \(f(t)=1+2t\)
\(L(f(t))=L(1+2t)\)
\(=\int_0^\infty (1+2t)e^{-st}dt\)
\(=\int_0^\infty e^{-st}dt+\int_0^\infty(2t)e^{-st}dt\)
\(=\left[\frac{e^{-st}}{-s}\right]_0^\infty+2\left[(t)\cdot\frac{e^{-st}}{-s}\right]-\left[1\cdot \frac{e^{-st}}{s^2}\right]_0^\infty\)
\(=\left[\frac{0-1}{-s}\right]+2\left[(0-0)-(0-\frac{1}{s^2}\right]\)
\(=\frac{1}{s}+\frac{2}{s^2}\)
\(=\frac{s+2}{s^2}\) Thus, \(L(f(t))=\frac{s+2}{s^2}\)
Step 3
b) To find the Laplace transform of \(f(t)=\sin \omega t = \frac{e^{j\omega t}-e^{-j\omega t}}{2j}\)
\(L(f(t))=L(\sin\omega t)\)
\(=\int_0^\infty\sin\omega t \cdot e^{-st}dt\)
\(=\int_0^\infty\frac{e^{j\omega t}-e^{-j\omega t}}{2j} \cdot e^{-st}dt\) \(=\frac{1}{2j}\int_0^\infty e^{(j\omega-s)t}-e^{-(j\omega+s)t}dt\)
\(=\frac{1}{2j}\left[\left(\frac{e^{(j\omega-s)t)}}{j\omega-s}\right)-\left(\frac{e^{-(j\omega+s)t}}{-(j\omega+s)}\right)\right]_0^\infty\)
\(=\frac{1}{2j}\left[\left(\frac{e^{-(s-j\omega)t}}{-(s-j\omega)}\right)-\left(\frac{e^{-(j\omega+s)t}}{-(j\omega+s)}\right)\right]_0^\infty\)
\(=\frac{1}{2j}\left[\frac{0-1}{-(s-j\omega)}-\frac{0-1}{-(j\omega+s)}\right]\)
\(=\frac{1}{2j}\left[\frac{1}{(s-j\omega)}-\frac{1}{(j\omega+s)}\right]\)
\(=\frac{1}{2j}\left[\frac{1}{(s-j\omega)}-\frac{1}{(s+j\omega)}\right]\) \(=\frac{1}{2j}\left[\frac{(s+j\omega)-(s-j\omega)}{(s-j\omega)(s+j\omega)}\right]\)
\(=\frac{\omega}{s^2-(j\omega)^2}\)
\(=\frac{\omega}{s^2+\omega^2}\) Since \(\omega^2=-1\)
\(L(f(t))=\frac{\omega}{(s^2+omega^2)}\)
Step 4
c) To find the Laplace transform of \(f(t)=\sin(2t)+2\cos(2t)+e^{-t}\sin(2t)\)
We have from the Laplace transform table,
\(L(\sin at)=\frac{a}{(s^2+a^2)}\)
\(L(\cos at)=\frac{s}{(s^2+a^2)}\)
\(\text{If } L(f(t))=F(t) \text{ then } L(f(t))=F(s-a)\)
Consider,
\(L(f(t))=L(\sin(2t)+2\cos(2t)+e^{-t}\sin(2t))\)
\(=\frac{2}{(s^2+4)}+2\cdot \frac{s}{(s^2+4)}+\frac{2}{(s+1)^2+4}\)
\(=\frac{2}{(s^2+4)}+\frac{2s}{s^2+4}+\frac{2}{(s^2+2s+5)}\)
Thus, Laplace transform of \(f(t)=\sin(2t)+2\cos(2t)+e^{-t}\sin(2t)\) is \(\frac{2}{s^2+4}+\frac{2s}{s^2+4}+\frac{2}{(s^2+2s+5)}\)
0

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