Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by F(s)=int_0^infty e^{-st}f(t)dt where we assume s is a positive real number. For example, to find the Laplace transform of f(t) = e^{-t}, the following improper integral is evaluated using integration by parts: F(s)=int_0^infty e^{-st}e^{-t}dt=int_0^infty e^{-(s+1)t}dt=frac{1}{(s+1)} Verify the following Laplace transforms, where u is a real number. f(t)=1 rightarrow F(s)=frac{1}{s}

Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by F(s)=int_0^infty e^{-st}f(t)dt where we assume s is a positive real number. For example, to find the Laplace transform of f(t) = e^{-t}, the following improper integral is evaluated using integration by parts: F(s)=int_0^infty e^{-st}e^{-t}dt=int_0^infty e^{-(s+1)t}dt=frac{1}{(s+1)} Verify the following Laplace transforms, where u is a real number. f(t)=1 rightarrow F(s)=frac{1}{s}

Question
Laplace transform
asked 2020-10-18
Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by
\(F(s)=\int_0^\infty e^{-st}f(t)dt\)
where we assume s is a positive real number. For example, to find the Laplace transform of f(t) = e^{-t}, the following improper integral is evaluated using integration by parts:
\(F(s)=\int_0^\infty e^{-st}e^{-t}dt=\int_0^\infty e^{-(s+1)t}dt=\frac{1}{(s+1)}\)
Verify the following Laplace transforms, where u is a real number.
\(f(t)=1 \rightarrow F(s)=\frac{1}{s}\)

Answers (1)

2020-10-19
Step 1
\(f(t)=1 \rightarrow F(s)=\frac{1}{s}\)
Step 2
Given a function f(t), the Laplace transform is a new function F(s) is defined by
\(F(s)=\int_0^\infty e^{-st}f(t)dt\) where we assume s is a positive real number.
For f(t)=1, the Laplace transform is given by
\(F(s)=\int_0^\infty e^{-st}\cdot 1 dt,\)
\(\text{Let } u=-st\Rightarrow du=-s dt \Rightarrow dt=-\frac{du}{s}\)
\(\text{For } t=0, u=-s \cdot 0=0.\)
\(\text{For } t=\infty, u=-s \cdot \infty=-\infty\)
Therefore,
\(F(s)=-\int_0^{-\infty} \frac{(e^u)}{s} du\)
\(=\int_{-\infty}^0 \frac{(e^u)}{s} du [\because \int_a^bf(x)dx=-\int_b^af(x)dx]\)
\(=\frac{1}{s} \int_{-\infty}^0 (e^u) du\)
\(=\frac{1}{s}[e^u]_{-\infty}^0\)
\(=\frac{1}{s}[e^0-e^{-\infty}]\)
\(=\frac{1}{s}[1-0]\)
\(=\frac{1}{s}\)
Hence, verified.
0

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