Question 24.14: Find, if possible, (a) the Laplace transform (b) the Fourier......
Find, if possible,
(a) the Laplace transform
(b) the Fourier transform
of f(t)=u(t)\,\mathrm{e}^{3t}. Comment upon the result.
(a) Either by integration, or from Table 21.1, we find
{\mathcal{L}}\{u(t)\,\mathrm{{e}}^{3t}\}={\frac{1}{s-3}}\qquad{\mathrm{provided}}\,s\gt 3
(b) {\mathcal{F}}\{u(t)\,{\mathrm{e}}^{3t}\}=\int_{0}^{\infty}\!\mathrm{e}^{3t}\,{\mathrm{e}}^{-{\mathrm{j}}\omega t}\,\mathrm{d}t=\int_{0}^{\infty}\!\mathrm{e}^{(3-\mathrm{j}\omega)t}\,\mathrm{d}t=\left[{\frac{\mathrm{e}^{(3-\mathrm{j}\omega)t}}{3-\mathrm{j}\omega}}\right]_{0}^{\infty}.
Now, as t~\rightarrow~\infty,\,\mathrm{e}^{3t}~\rightarrow~\infty, so that the integral fails to exist. Clearly, u(t)\,{\mathrm{e}}^{3t} has a Laplace transform but no Fourier transform.
| Table 21.1 The Laplace transforms of some common functions. |
|||
| Function, f (t ) | Laplace transform, F(s) | Function, f (t ) | Laplace transform, F(s) |
| 1 | \frac{1}{s} | \mathrm{e}^{-a t}\cos b t | \frac{s+a}{(s+a)^{2}+b^{2}} |
| t | \frac{1}{s^2} | \sinh\,b t | \frac{b}{s^{2}-b^{2}} |
| {\boldsymbol{t}}^{2} | \frac{2}{s^3} | \cosh b t | \frac{s}{s^{2}-b^{2}} |
| {\boldsymbol{t}}^{n} | \frac{n!}{s^{n+1}} | \mathrm{e}^{-a t}\sinh\ b t | \frac{b}{(s+a)^{2}-b^{2}} |
| \mathrm{e}^{at} | \frac{1}{s−a} | \mathrm{e}^{-a t}\cosh b t | \frac{s+a}{(s+a)^{2}-b^{2}} |
| \mathrm{e}^{−at} | \frac{1}{s+a} | t\sin b t | \frac{2b s}{(s^{2}+b^{2})^{2}} |
| t^{n}{\mathbf{e}}^{-a t} | \frac{n!}{(s\ +a)^{n+1}} | t\cos b t | \frac{s^{2}-b^{2}}{(s^{2}+b^{2})^{2}} |
| \sin b t | \frac{b}{s^{2}+b^{2}} | u(t ) unit step | \frac{ 1}{s} |
| \cos b t | \frac{s}{s^{2}+b^{2}} | u(t − d) | \frac{ \mathrm{e}^{-s d}}{s} |
| \mathrm{e}^{-a t}\sin b t | \frac{b}{(s+a)^{2}+ b^{2}} | \delta(t)
\delta(t-d) |
1
\mathrm{e}^{-s d} |