Question 24.7: Given that when f(t) = {1 |t| ≤ 1 0 |t| > 1 F(ω) = 2sin ω......
Given that when f(t)={\left\{\begin{array}{l l}{1}&{|t|\leqslant1}\\ {0}&{|t|\gt 1}\end{array}\right.},F(\omega)={\frac{2\sin\omega}{\omega}}, apply the second shift theorem to find the Fourier transform of
g(t)={\left\{\begin{array}{l l}{1}&{1\leqslant t\leqslant3}\\ {0}&{{\mathrm{otherwise}}}\end{array}\right.}
Verify your result directly.
The function g(t) is depicted in Figure 24.3. Clearly g(t) is the function f (t) translated 2 units to the right, that is g(t) = f (t−2). Now F(\omega)={\frac{2\sin\omega}{\omega}} is the Fourier transform of f (t). Therefore, by the second shift theorem
{\mathcal{F}}\{g(t)\}={\mathcal{F}}\{f(t-2)\}=\mathrm{e}^{-2{\mathfrak{j}}\omega}F(\omega)={\frac{2\mathrm{e}^{-2{\mathfrak{j}}\omega}\sin\omega}{\omega}}
To verify this result directly we must evaluate
{\mathcal{F}}\{g(t)\}=\int_{-\infty}^{\infty}g(t)\,\mathrm{e}^{-\mathrm{j}\omega t}\,\mathrm{d}t=\int_{1}^{3}\mathrm{e}^{-\mathrm{j}\omega t}\,\mathrm{d}t=\left[{\frac{\mathrm{e}^{-\mathrm{j}\omega t}}{-\mathrm{j}\omega}}\right]_{1}^{3}
={\frac{\mathrm{e}^{-3\omega}-\mathrm{e}^{-\mathrm{j}\omega}}{-\mathrm{j}\omega}}=\mathrm{e}^{-2\mathrm{j}\omega}{\bigg(}{\frac{\mathrm{e}^{-\mathrm{j}\omega}-\mathrm{e}^{\mathrm{j}\omega}}{-\mathrm{j}\omega}}{\bigg)}
={\frac{2\mathrm{e}^{-2\mathrm{j}\omega}\sin\omega}{\omega}}
as required.
