Question 24.4: (a) Show that the Fourier transform of f(t) = { 3 −2 ≤ t ≤ 2......
(a) Show that the Fourier transform of
f(t)={\left\{\begin{array}{l l}{3}&{-2\leqslant t\leqslant 2}\\ {0}&{{\mathrm{otherwise}}}\end{array}\right.}
is given by F(\omega)=\frac{6\sin2\omega}{\omega}.
(b) Use the first shift theorem to find the Fourier transform of \mathrm{e}^{-jt}f(t).
(c) Verify the first shift theorem by obtaining the Fourier transform of
\mathrm{e}^{-jt}f(t) directly.
(a) F(\omega)=3\int_{-2}^{2}\mathrm{e}^{-\mathrm{j}\omega t}\,\mathrm{d}t=3\biggl[{\frac{\mathrm{e}^{-\mathrm{j}\omega t}}{-\mathrm{j}\omega}}\biggr]_{-2}^{2}
=3\left({\frac{\mathrm{e}^{-2j \omega}-\mathrm{e}^{2j \omega}}{-\mathrm{j}\omega}}\right)
=6\left({\frac{\mathrm{e}^{2 j \omega}-\mathrm{e}^{-2j \omega}}{2\mathrm{j}\omega}}\right)
={\frac{6}{\omega}}\sin2\omega
(b) We have {\mathcal{F}}\{f(t)\}=F(\omega)={\frac{6\sin2\omega}{\omega}}. Using the first shift theorem with a = −1 we have
{\mathcal{F}}\{\mathrm{e}^{-j t}f(t)\}=F(\omega+1)={\frac{6}{\omega+1}}\sin2(\omega+1)
(c) \mathrm{e}^{-\mathrm{j}t}f(t)=\left\{{\begin{array}{c l c r}{3\mathrm{e}^{-\mathrm{j}t}}&{-2\leqslant t \leqslant 2}\\ {0}&{{\mathrm{otherwise}}}\end{array}}\right.
So to evaluate its Fourier transform directly we must find
{\mathcal{F}}\{\mathrm{e}^{-\mathrm{j}{t}}f(t)\}=3\int_{-2}^{2}\mathrm{e}^{-\mathrm{j}\omega t}\,\mathrm{d}t
=3\int_{-2}^{2}\mathrm{e}^{-(1+\omega)j t}\,\mathrm{d}t
=3\biggl[\frac{\mathrm{e}^{-(1+\omega)j t}}{-\mathrm{j}(1+\omega)}\biggr]_{-2}^{2}
={\frac{6}{1+\omega}}\biggl({\frac{\mathbf{e}^{2(1+\omega)j}-\mathbf{e}^{-2(1+\omega)j}}{2\bf j}}\biggr)
={\frac{6}{1+\omega}}\sin2(1+\omega)
as required.