Find the inverse Laplace transform of
The Correct Answer and Explanation is:
We are asked to find the inverse Laplace transform of the expression:L−1{slog(s−1s+1)}\mathcal{L}^{-1}\left\{s \log\left(\frac{s – 1}{s + 1}\right)\right\}L−1{slog(s+1s−1)}
Solution:
LetF(s)=slog(s−1s+1)F(s) = s \log\left(\frac{s – 1}{s + 1}\right)F(s)=slog(s+1s−1)
We aim to find f(t)=L−1{F(s)}f(t) = \mathcal{L}^{-1}\{F(s)\}f(t)=L−1{F(s)}
Step-by-step Explanation:
Let us denote:log(s−1s+1)=log(s−1)−log(s+1)\log\left(\frac{s – 1}{s + 1}\right) = \log(s – 1) – \log(s + 1)log(s+1s−1)=log(s−1)−log(s+1)
Thus,F(s)=slog(s−1)−slog(s+1)F(s) = s \log(s – 1) – s \log(s + 1)F(s)=slog(s−1)−slog(s+1)
Now, the Laplace transform of 1t(e−at−e−bt)\frac{1}{t} \left(e^{-at} – e^{-bt}\right)t1(e−at−e−bt) is:L{1t(e−at−e−bt)}=log(s+bs+a)\mathcal{L}\left\{ \frac{1}{t}(e^{-at} – e^{-bt}) \right\} = \log\left(\frac{s + b}{s + a}\right)L{t1(e−at−e−bt)}=log(s+as+b)
So by shifting variables and differentiating in the sss-domain, we can derive the inverse Laplace transform of slog(s−1s+1)s \log\left(\frac{s – 1}{s + 1}\right)slog(s+1s−1) using known results.
Let’s use known inverse Laplace transform results directly:
There is a known Laplace transform identity:L−1{slog(s−as+a)}=−2a2cosh(at)−1t\mathcal{L}^{-1}\left\{ s \log\left(\frac{s – a}{s + a} \right) \right\} = -2a^2 \frac{\cosh(at) – 1}{t}L−1{slog(s+as−a)}=−2a2tcosh(at)−1
Set a=1a = 1a=1, so:L−1{slog(s−1s+1)}=−2⋅cosh(t)−1t\mathcal{L}^{-1}\left\{ s \log\left(\frac{s – 1}{s + 1} \right) \right\} = -2 \cdot \frac{\cosh(t) – 1}{t}L−1{slog(s+1s−1)}=−2⋅tcosh(t)−1
Final Answer:
f(t)=−2⋅cosh(t)−1t\boxed{f(t) = -2 \cdot \frac{\cosh(t) – 1}{t}}f(t)=−2⋅tcosh(t)−1
Explanation:
This result is derived using the known Laplace transform pair involving logarithmic expressions. The hyperbolic cosine function cosh(t)=et+e−t2\cosh(t) = \frac{e^t + e^{-t}}{2}cosh(t)=2et+e−t arises naturally from the inverse transform of expressions involving log(s−as+a)\log\left(\frac{s – a}{s + a}\right)log(s+as−a). The factor 1t\frac{1}{t}t1 appears due to the s-multiplication in the Laplace domain, which corresponds to a differentiation operation or time division in the time domain.
